Physics I: Classical Mechanics

In 1999 the Mars Climate Orbiter fired its engine to settle into orbit and was never heard from again. The spacecraft was healthy; the bookkeeping was not. One team supplied thruster figures in pound-force seconds while the navigation software expected newton-seconds, and the silent factor of about 4.45 between them steered the probe too deep into the Martian atmosphere. Nobody there was bad at physics. They simply failed to carry the units, and this lesson is about the habit that prevents exactly that.

Physics is a quantitative science, so every measurement needs a unit. A number without a unit is meaningless: "5" could be 5 meters, 5 seconds, or 5 kilograms, and each describes a completely different physical situation. The international standard is the SI system (Systeme International), built on a small set of base units. In mechanics we use just three: the meter (m) for length, the kilogram (kg) for mass, and the second (s) for time.

Every other mechanical unit is a combination of these, called a derived unit. Speed is meters per second (m/s), and force, as you will see, is measured in newtons, where one newton equals one kg times m/s squared. Because the whole of mechanics is assembled from just these three building blocks, keeping careful track of them is the single most reliable habit a physicist can have.

Key idea: Every mechanical quantity is built from the meter, the kilogram, and the second, so a unit is never decoration - it is part of the answer.

The SI base units and prefixes

Real problems span an enormous range of sizes, from the diameter of an atom to the distance to a galaxy. Rather than write out long strings of zeros, SI uses prefixes that multiply a unit by a power of ten. A kilometer (km) is 10^3 meters; a millimeter (mm) is 10^-3 meters; a microsecond is 10^-6 seconds. Learning the common prefixes - kilo, centi, milli, micro, mega, giga - lets you move fluidly between scales. The trick is that a prefix is just a shorthand for a numerical factor, so "3.0 km" and "3000 m" are literally the same quantity written two ways.

One trap deserves flagging now, because it costs more marks than any other unit error: when a prefixed unit is raised to a power, the prefix is raised too. A square meter is not 100 square centimeters. Since 1 m = 100 cm, squaring gives 1 m^2 = (100 cm)^2 = 10000 cm^2, and cubing gives 1 m^3 = 10^6 cm^3. Convert first, raise to the power second.

Key idea: A prefix is a multiplying factor, and when the unit is squared or cubed the factor is squared or cubed with it.

Dimensional analysis

Each quantity has a dimension: length [L], mass [M], or time [T]. The dimension is more fundamental than the unit - length is length whether you measure it in meters, feet, or light-years. A correct equation must have matching dimensions on both sides, because you cannot add a length to a time any more than you can add apples to hours. Checking dimensions is a fast way to catch mistakes before you ever plug in a number. Consider the kinematics equation x = v t.

On the right, velocity [L/T] times time [T] gives [L], which matches the position on the left. The equation is dimensionally consistent. Now test a wrong guess, x = v t squared: the right side has dimension [L/T][T squared] = [L T], which is not a length, so the equation cannot be correct no matter what numbers you use. Dimensional analysis will not catch a missing factor of one half, but it reliably flags any error that changes the kind of quantity.

Two further rules sharpen the technique. Only quantities of the same dimension may be added or subtracted, so "length plus time" is not merely wrong but meaningless. And the argument of a sine, cosine, exponential, or logarithm must be dimensionless: the pendulum formula reads theta = theta_max cos(omega t), never cos(t), because omega t has dimension [1/T][T] = 1. A bare time inside a trigonometric function always means a constant has gone missing.

Key idea: Dimensional analysis can prove an equation wrong but never proves it right, because it is blind to dimensionless factors such as 1/2, 2, or pi.

Significant figures

A measured value carries only as much precision as the instrument allows. The significant figures are the digits that are known reliably plus one estimated digit. A ruler marked in millimeters lets you report a length to the nearest tenth of a millimeter by estimating between the marks, but no further. When you multiply or divide, the answer keeps the same number of significant figures as the least precise input, because the result can be no more certain than its shakiest ingredient.

When you add or subtract, you instead line up the decimal places and keep the answer only to the least precise decimal position. If a car travels 100.0 m (4 significant figures) in 8.2 s (2 significant figures), the speed is 100.0 / 8.2 = 12.195..., which we report as 12 m/s - two significant figures, matching the time. Reporting "12.195 m/s" would be dishonest: it claims a precision the data does not support.

Accuracy, precision, and exact numbers

Two words that everyday speech treats as synonyms differ here. Accuracy is how close a measurement lies to the true value; precision is how finely it is resolved. A miscalibrated scale reading 5.2731 kg every time for a 4 kg mass is exquisitely precise and badly inaccurate. Significant figures speak only to precision, never to whether the instrument was calibrated.

Not every number limits precision. Exact numbers - counted objects, and defined factors such as the 1000 in "1 km = 1000 m" - carry unlimited significant figures. Converting 12.4 km gives 12400 m with three significant figures, not five, because 1000 is a definition, not a measurement. Finally, carry extra digits through intermediate steps and round only at the end.

Key idea: Significant figures track precision, not accuracy, and exact defined or counted numbers never limit how many you may keep.

Worked example: unit conversion

Given: a speed of 90 km/h. Find: the speed in m/s.
Solution: Multiply by conversion factors chosen so the unwanted units cancel. A conversion factor is just a ratio equal to one, such as (1000 m / 1 km), so multiplying by it never changes the physical quantity, only its units.

There are 1000 m in a km and 3600 s in an hour.
90 km/h x (1000 m / 1 km) x (1 h / 3600 s) = 90 x 1000 / 3600 = 25 m/s.
Notice how "km" cancels against "km" and "h" cancels against "h", leaving m/s. So 90 km/h equals 25 m/s. A useful rule of thumb falls out of this: to convert km/h to m/s, divide by 3.6.

Sanity check: 25 m/s is a plausible motorway speed - a person walks near 1.4 m/s and a sprinter tops out near 10 m/s, so a car belongs in the tens, not the hundreds. The direction of the change is right too: the answer had to come out smaller than 90. Multiplying by 3.6 instead would have given 324 m/s, faster than sound, which the estimate rejects at a glance.

Worked example: repairing a formula with dimensions

Given: a student proposes that air drag on a car is F = (1/2) C rho A v, with C a dimensionless drag coefficient, rho air density in kg/m^3, A frontal area in m^2, and v speed in m/s.
Find: (a) whether the formula can be correct, and (b) if not, the power of v that repairs it.

Solution (a): write out each factor's dimensions. Force is [M L T^-2], density [M L^-3], area [L^2], speed [L T^-1]; the dimensionless 1/2 and C contribute nothing. Multiplying the right side:
[M L^-3] x [L^2] x [L T^-1] = [M L^(-3 + 2 + 1) T^-1] = [M L^0 T^-1] = [M T^-1].
That is a mass per unit time, not a force, so the proposal is dimensionally impossible and no choice of numbers can rescue it.

Solution (b): replace v with v^n and solve for n.
[M L^-3] x [L^2] x [L^n T^-n] = [M L^(n - 1) T^-n].
Matching this to [M L T^-2] gives two independent conditions: the time exponent requires -n = -2, so n = 2, and the length exponent requires n - 1 = 1, so n = 2 again. Both conditions agree, so the drag force must go as the square of speed: F = (1/2) C rho A v^2.

Sanity check: substitute realistic numbers for a car at motorway speed - C = 0.30, rho = 1.2 kg/m^3, A = 2.2 m^2, v = 30 m/s:
0.5 x 0.30 = 0.15; 0.15 x 1.2 = 0.18; 0.18 x 2.2 = 0.396; 0.396 x 900 = 356.4 N.
The units confirm it: (kg/m^3)(m^2)(m^2/s^2) = kg m/s^2 = N. Is 356 N the right order of magnitude? The power needed to overcome it is P = F v = 356.4 x 30 = 10692 W, about 10.7 kW or 14 horsepower - close to what a real car burns at motorway speed on level ground. The direction is right too: drag opposes motion, so it points backward.

Worked example: a multi-step conversion with significant figures

Given: a cyclist covers 12.4 km in 27.5 min. Both figures are measured to three significant figures.
Find: the average speed in m/s and in km/h, reported to the correct precision.

Solution: convert into base SI units first, using exact defined factors that do not limit precision.
d = 12.4 km x (1000 m / 1 km) = 12400 m (still 3 significant figures)
t = 27.5 min x (60 s / 1 min) = 1650 s (still 3 significant figures)
v = d / t = 12400 / 1650 = 7.51515... m/s, reported as 7.52 m/s.
For km/h, convert the time to hours instead:
t = 27.5 / 60 = 0.458333... h, so v = 12.4 / 0.458333 = 27.0545... km/h, reported as 27.1 km/h.

Sanity check: the two answers must agree through the rule of thumb above, and they do: 7.51515 x 3.6 = 27.0545 km/h. The magnitude is believable - 27 km/h is a brisk but ordinary cycling pace, several times walking speed and well under a car's.

Common wrong turns

• Squaring a length but not its prefix. 1 m^2 = 100 cm^2 is wrong by a factor of 100; convert first, then apply the exponent.
• Using the multiplication rule for addition. Products match the fewest significant figures; sums match the fewest decimal places. So 3.6 + 12.42 = 16.0, not 16.02.
• Letting an exact number limit the answer. The 1000 in "1 km = 1000 m" is a definition with unlimited precision. Only measured values set significant figures.
• Believing a dimensional check proves an equation right. x = 2 v t and x = v t are both dimensionally sound and only one is correct.
• Inverting a conversion factor. Write it as a fraction and confirm the unwanted unit cancels before touching the calculator.

Why it matters

Getting units and precision right is not busywork. The Mars Climate Orbiter is only the most expensive example; the same error yearly produces mis-dosed medications and mis-cut steel. When a result surprises you, check the dimensions before anything else - that is usually where the fault lies.

Recap

• Mechanics is built on three SI base units - meter, kilogram, second - and every other unit derives from them.
• Prefixes are multiplying powers of ten, and a unit raised to a power raises its prefix too.
• Dimensional analysis can disprove a formula but never confirm one, since it is blind to factors like 1/2 and pi.
• Only same-dimension quantities may be added, and any trigonometric, exponential, or logarithmic argument must be dimensionless.
• Significant figures report precision, not accuracy; exact counted and defined numbers never limit them.

Sources

1. OpenStax. (2016). 1.2 Units and standards. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 1.3 Unit conversion. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 1.4 Dimensional analysis. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2016). 1.6 Significant figures. In University Physics Volume 1. Rice University. openstax.org
5. OpenStax. (2022). 1.3 Accuracy, precision, and significant figures. In College Physics 2e. Rice University. openstax.org
6. National Institute of Standards and Technology. (n.d.). The NIST reference on constants, units, and uncertainty: International System of Units (SI). physics.nist.gov
7. Nave, R. (n.d.). Physical units. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu

Two tugboats pull on a stranded barge, each with a winch rated at 40 kN. Depending on nothing but the angle between the tow lines, the barge feels anywhere from 80 kN down to zero. Nothing about the boats changed - only the geometry. That is the whole reason physics needs vectors, and why the arithmetic you already know is not enough on its own.

Some quantities need only a size to be fully described. Mass, time, temperature, and speed are scalars - a single number with a unit. Other quantities also need a direction. Displacement, velocity, acceleration, and force are vectors: they have both a magnitude (size) and a direction. Walking 3 km is a distance (scalar); walking 3 km east is a displacement (vector). This distinction runs through all of mechanics, because the direction of a force or a velocity changes everything about what happens next. Two forces of equal size can add up, cancel out, or produce something in between, entirely depending on their directions.

Representing vectors

A vector is drawn as an arrow: its length shows the magnitude and the way it points shows the direction. We write a vector's magnitude with plain symbols, so a force of magnitude F might point at some angle. Two vectors are equal if they have the same length and direction, regardless of where they are drawn on the page - sliding an arrow around without rotating or resizing it does not change the vector it represents. This freedom is what lets us line vectors up head to tail when we add them.

Components

The most reliable way to work with vectors is to break each one into perpendicular pieces called components along the x and y axes. This turns a hard geometry problem into simple arithmetic. A vector of magnitude V pointing at an angle theta above the x-axis has components Vx = V cos(theta) and Vy = V sin(theta).

These come straight from the definitions of sine and cosine in the right triangle the vector forms with the axes. To reverse the process, the magnitude is V = sqrt(Vx squared + Vy squared) by the Pythagorean theorem, and the reference angle is arctan(|Vy| / |Vx|). Components can be positive or negative: a vector pointing down and to the left has both Vx and Vy negative, which automatically encodes its direction.

The quadrant trap

Here the notation hides a genuine hazard. A calculator's arctan only ever returns an angle between -90 and +90 degrees, so it can never point into the second or third quadrant. Feed it arctan(-80 / -60) and the two minus signs cancel inside the division: it returns +53.1 degrees, up and to the right, when the true vector points down and to the left at 233.1 degrees - the exact opposite direction.

The fix is to handle signs separately from magnitude. Take arctan of the absolute values to get a reference angle in the first quadrant, then place it in the correct quadrant by inspecting the signs of Vx and Vy: both positive means the first quadrant (use the reference angle as is), Vx negative with Vy positive means the second (180 - ref), both negative means the third (180 + ref), and Vx positive with Vy negative means the fourth (360 - ref). Always sketch the vector roughly before trusting a number.

Key idea: Components carry the direction in their signs, and a calculator's arctan discards that information - recover the quadrant from the signs, never from the arctan alone.

Adding vectors

To add two vectors, add their x-components to get the total x, and add their y-components to get the total y. The two sums are the components of the resultant, the single vector that has the same effect as the originals combined. This component method works for any number of vectors and any directions, which is why physicists prefer it over drawing and measuring.

To subtract a vector, add its negative (the same arrow reversed), which flips the signs of both components. The graphical picture that matches this arithmetic is the head-to-tail rule: place the tail of the second arrow at the head of the first, and the resultant runs from the very first tail to the very last head.

Two limits are worth memorizing as a sanity check on any sum. The resultant of two vectors can never be larger than the sum of their magnitudes (achieved when they point the same way) and never smaller than the difference of their magnitudes (achieved when they oppose). A 40 kN and a 40 kN tow line must therefore produce between 0 and 80 kN, exactly as the barge in the opening showed. If your components ever hand back a resultant outside that window, the arithmetic is wrong.

Key idea: Vectors add component by component, and the resultant magnitude always lies between the difference and the sum of the individual magnitudes.

Worked example: two displacements

Given: you walk 6.0 m east, then 8.0 m north. Find: your total displacement (magnitude and direction).
Solution: East is the +x direction and north is +y. The components of the total are x = 6.0 m and y = 8.0 m. The magnitude is sqrt(6.0 squared + 8.0 squared) = sqrt(36 + 64) = sqrt(100) = 10.0 m.

The direction is arctan(8.0 / 6.0) = arctan(1.33) = 53 degrees north of east. So your displacement is 10.0 m at 53 degrees north of east. Notice this straight-line displacement (10 m) is shorter than the total distance walked (14 m), because displacement measures only the net change in position, not the path length.

Sanity check: both components are positive, so the vector sits in the first quadrant and the raw arctan needs no correction here. The magnitude must lie between 8.0 - 6.0 = 2.0 m and 8.0 + 6.0 = 14.0 m, and 10.0 m does. Since the northward leg is the longer one, the answer should tilt more than halfway toward north, and 53 degrees is indeed past 45.

Worked example: adding two forces

Given: two forces act on a ring: F1 = 30 N pointing east, and F2 = 40 N pointing at 90 degrees to it, north. Find: the resultant force.
Solution: The components add directly: total Fx = 30 N, total Fy = 40 N. The magnitude is sqrt(30 squared + 40 squared) = sqrt(900 + 1600) = sqrt(2500) = 50 N, at arctan(40/30) = 53 degrees north of east. The single resultant of 50 N would have exactly the same effect as the two forces together, which is why we can always replace a set of forces with their vector sum.

Worked example: a resultant that lands in the third quadrant

Given: two forces act on a bracket, measured counterclockwise from the +x axis: F1 = 50 N at 210 degrees and F2 = 30 N at 300 degrees.
Find: the magnitude and direction of the resultant.

Solution: resolve each force, keeping every sign.
F1x = 50 cos(210) = 50 x (-0.8660) = -43.30 N
F1y = 50 sin(210) = 50 x (-0.5000) = -25.00 N
F2x = 30 cos(300) = 30 x (0.5000) = +15.00 N
F2y = 30 sin(300) = 30 x (-0.8660) = -25.98 N
Sum the columns:
Rx = -43.30 + 15.00 = -28.30 N
Ry = -25.00 - 25.98 = -50.98 N
R = sqrt(28.30^2 + 50.98^2) = sqrt(800.9 + 2599.1) = sqrt(3400.0) = 58.3 N.

Now the direction, handled in two stages. The reference angle from the absolute values is arctan(50.98 / 28.30) = arctan(1.8014) = 60.96 degrees. Both components are negative, so the resultant lies in the third quadrant, and the true bearing is 180 + 60.96 = 241.0 degrees. The answer is 58.3 N at 241 degrees, pointing down and to the left.

Sanity check: rebuild the components from the answer. 58.3 cos(241.0) = 58.3 x (-0.4853) = -28.3 N and 58.3 sin(241.0) = 58.3 x (-0.8743) = -51.0 N, reproducing the sums. The magnitude also sits inside the required window, between 50 - 30 = 20 N and 50 + 30 = 80 N. Watch what a careless calculator does here: keying arctan(-50.98 / -28.30) gives +60.96 degrees, up and to the right, which is 180 degrees wrong. The minus signs cancelled inside the division and the quadrant was lost.

Worked example: the force that balances two others

Given: two cables pull on a ring, T1 = 120 N at 40 degrees and T2 = 90 N at 160 degrees.
Find: the single third force (the equilibrant) that would hold the ring at rest.

Solution: first add the two known pulls.
T1x = 120 cos(40) = 120 x 0.7660 = 91.93 N, T1y = 120 sin(40) = 120 x 0.6428 = 77.13 N
T2x = 90 cos(160) = 90 x (-0.9397) = -84.57 N, T2y = 90 sin(160) = 90 x 0.3420 = 30.78 N
Rx = 91.93 - 84.57 = 7.36 N, Ry = 77.13 + 30.78 = 107.91 N
R = sqrt(7.36^2 + 107.91^2) = sqrt(54.2 + 11644.6) = sqrt(11698.8) = 108.2 N
Both components are positive, so the resultant is in the first quadrant at arctan(107.91 / 7.36) = arctan(14.66) = 86.1 degrees.

The equilibrant is the exact opposite of this resultant, so it has the same magnitude and points 180 degrees away: 108.2 N at 266.1 degrees, that is, very nearly straight down.

Sanity check: the two cables pull up and to the right and up and to the left respectively, and their horizontal parts almost cancel (91.9 N right against 84.6 N left), so the combined pull should be nearly vertical - and 86.1 degrees is only 3.9 degrees off straight up. The equilibrant must therefore hang nearly straight down, which it does. The magnitude 108 N also falls between the difference (30 N) and the sum (210 N), as any vector sum must. Notice that the equilibrant is not 210 N: adding magnitudes instead of components would overstate the needed force by nearly a factor of two.

Common wrong turns

• Adding magnitudes instead of components. A 6-unit and an 8-unit vector at right angles give 10, not 14. Only parallel vectors add arithmetically.
• Trusting the raw arctan. It returns only -90 to +90 degrees. Take arctan of the absolute values, then place the quadrant from the signs of the components.
• Swapping sine and cosine. With theta measured from the x-axis, the x-component uses cosine and the y-component uses sine. If the angle is measured from the y-axis instead, the two swap - so always state which axis you measured from.
• Confusing distance with displacement. A closed loop has a large distance and zero displacement; only the vector one can be added component by component.
• Dropping a sign when a component points along a negative axis. The sign is not decoration; it is the only record of the direction once you switch to components.

Recap

• Scalars need only a magnitude; vectors such as displacement, velocity, and force need a direction as well.
• A vector resolves into Vx = V cos(theta) and Vy = V sin(theta) when theta is measured from the x-axis, and rebuilds as V = sqrt(Vx^2 + Vy^2).
• Vectors add component by component, and the resultant is the single vector with the same effect as the whole set.
• The resultant magnitude always lies between the difference and the sum of the individual magnitudes.
• Recover direction from the reference angle plus the signs of the components, never from arctan alone.
• The equilibrant of a set of forces is equal in magnitude to their resultant and opposite in direction.

Sources

1. OpenStax. (2016). 2.1 Scalars and vectors. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 2.2 Coordinate systems and components of a vector. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 2.3 Algebra of vectors. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2022). 3.3 Vector addition and subtraction: Analytical methods. In College Physics 2e. Rice University. openstax.org
5. Nave, R. (n.d.). Basic vector operations. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu
6. Feynman, R. P., Leighton, R. B., & Sands, M. (2013). Chapter 11: Vectors. In The Feynman Lectures on Physics, Volume I (New Millennium ed., online). California Institute of Technology. feynmanlectures.caltech.edu
7. PhET Interactive Simulations. (n.d.). Vector addition [Simulation]. University of Colorado Boulder. phet.colorado.edu

A driver travelling at 25 m/s sees a hazard and stops as fast as the car allows. Almost everyone guesses the stopping distance far too short, because they picture only the braking and forget the fifth of a second - or the whole second - before the pedal even moves. By the end of this lesson you will be able to compute that distance exactly, and the answer is a little alarming. The tools are four equations and a habit of writing down signs.

Kinematics describes how things move without yet asking why - that "why" waits for Newton's laws in Module 3. In one dimension, position is measured along a line, with one direction chosen as positive. Displacement is the change in position, delta x = x_final - x_initial, and it can be negative if the object ends up behind where it started. Divide displacement by the time taken and you get average velocity, v = delta x / delta t.

Velocity differs from speed in that it has a sign showing direction: a velocity of -5 m/s means moving in the negative direction at 5 m/s. Acceleration is the rate of change of velocity, a = delta v / delta t, measured in m/s squared. Crucially, acceleration is about changing velocity, so an object can be accelerating even while slowing down - deceleration is simply acceleration opposite to the motion.

Key idea: Displacement, velocity, and acceleration are vectors; in one dimension their direction is carried entirely by a plus or minus sign.

Sign conventions: a decision you make once

More one-dimensional problems are lost to sign errors than to anything else, and the cure is a single discipline: choose a positive direction before you write any equation, then obey it without exception. If you pick "up is positive," a downward acceleration is -9.8 m/s^2 whether the object is rising, falling, or momentarily at rest. If you pick "east is positive," a westward velocity is negative even when the object is speeding up.

The confusing case is a negative acceleration. It does not mean "slowing down." It means "pointing in the negative direction." A car moving in the -x direction with a negative acceleration is speeding up, because velocity and acceleration point the same way. The reliable test is the product: when v and a have the same sign the object speeds up, and when they have opposite signs it slows down. Nothing about the individual signs alone tells you which.

Key idea: An object speeds up when velocity and acceleration share a sign and slows down when their signs differ, regardless of which direction you called positive.

Average versus instantaneous

Average velocity smooths over a whole time interval, but often we want the velocity at a single moment - the reading on a speedometer. This is the instantaneous velocity, the value the average approaches as the time interval shrinks toward zero. On a position-versus-time graph, average velocity is the slope of the line connecting two points, while instantaneous velocity is the slope of the tangent at one point. The same distinction applies to acceleration. For the constant-acceleration problems in this lesson, the average and instantaneous accelerations are equal, which is exactly what makes the equations below so clean.

Average speed and average velocity are also different animals, and the difference is not pedantry. Average velocity is displacement over time and can be zero for a long, exhausting journey that ends where it began. Average speed is total path length over time and is never negative. The two agree only when the motion never reverses direction. A further trap: average speed is not the arithmetic mean of the speeds of each leg unless those legs took equal times, because the slower leg occupies more of the clock and therefore counts for more.

Reading motion graphs

Graphs encode the same information and often make it obvious. On a position-time graph the slope is the velocity, so a straight line means constant velocity, a curve means acceleration, and a horizontal stretch means the object is at rest. On a velocity-time graph the slope is the acceleration and, just as usefully, the area between the curve and the time axis is the displacement - with area below the axis counting as negative. That area rule is why the equation x = (1/2)(v0 + v) t works: for constant acceleration the velocity graph is a straight line, and the area under it is a trapezoid whose average height is (v0 + v)/2.

Key idea: Slope on a position-time graph is velocity, slope on a velocity-time graph is acceleration, and area under a velocity-time graph is displacement.

The kinematic equations

When acceleration is constant, four equations connect the five quantities of motion: initial velocity v0, final velocity v, acceleration a, displacement x, and time t. Each equation leaves out exactly one variable, so the art of solving a problem is to identify what you know and what you want, then pick the equation that omits the quantity you neither know nor need.

Equation	Leaves out
v = v0 + a t	x
x = v0 t + (1/2) a t squared	v
v squared = v0 squared + 2 a x	t
x = (1/2)(v0 + v) t	a

A reliable recipe: (1) write down every known quantity with its sign, (2) identify the unknown, (3) choose the equation that contains your knowns and unknown but not the irrelevant variable, and (4) solve algebraically before substituting numbers. Keeping the algebra symbolic until the last step reduces arithmetic mistakes and makes the physics visible.

Worked example: an accelerating car

Given: a car starts from rest (v0 = 0) and reaches v = 27 m/s in t = 5.0 s. Find: its acceleration and the distance it covers.
Solution: For acceleration, use v = v0 + a t. Solving, a = (v - v0) / t = (27 - 0) / 5.0 = 5.4 m/s squared. For distance, use x = (1/2)(v0 + v) t = (1/2)(0 + 27)(5.0) = 67.5 m. So the car accelerates at 5.4 m/s squared and travels 67.5 m. As a check, the average velocity over the interval is (0 + 27)/2 = 13.5 m/s, and 13.5 x 5.0 = 67.5 m, which agrees.

Worked example: stopping distance

Given: a car moving at v0 = 20 m/s brakes with acceleration a = -5.0 m/s squared until it stops (v = 0). Find: the stopping distance.
Solution: Time is not asked for, so use v squared = v0 squared + 2 a x. Then 0 = 20 squared + 2(-5.0) x, so 0 = 400 - 10 x, giving x = 40 m.

The car needs 40 m to stop. Notice the acceleration is negative because it opposes the motion. An important consequence hides in the v0 squared term: stopping distance grows with the square of speed, so a car at 40 m/s needs four times the distance of one at 20 m/s. This is the physics behind highway speed limits and following-distance rules.

Sanity check: cross-check with a different route. The stopping time is t = (0 - 20)/(-5.0) = 4.0 s, and the average velocity over that interval is (20 + 0)/2 = 10 m/s, so x = 10 x 4.0 = 40 m. The two methods agree. The sign story is consistent too: velocity is positive and acceleration negative, so the car slows, exactly as intended - had both been negative the car would have been speeding up backwards.

Worked example: the real stopping distance, reaction time included

Given: a car travels at v0 = 25 m/s (90 km/h). The driver's reaction time is 0.70 s, during which the car coasts at constant speed. The brakes then produce a = -6.0 m/s^2 until the car stops.
Find: (a) the distance covered while reacting, (b) the braking distance, (c) the total stopping distance, and (d) the total time.

Solution (a): during the reaction the acceleration is zero, so the motion is uniform.
x1 = v0 t = 25 x 0.70 = 17.5 m.

Solution (b): time is not wanted yet, so use the timeless equation with v = 0.
0 = 25^2 + 2(-6.0) x2
12 x2 = 625, so x2 = 52.083 = 52.1 m.

Solution (c): x_total = 17.5 + 52.1 = 69.6 m.
Solution (d): the braking time is t2 = (0 - 25)/(-6.0) = 4.17 s, so t_total = 0.70 + 4.17 = 4.87 s.

Sanity check: verify the braking leg a second way. Its average velocity is (25 + 0)/2 = 12.5 m/s, and 12.5 x 4.167 = 52.08 m, matching part (b). Is roughly 70 m plausible? That is about fourteen car lengths, and it agrees with the stopping distances published in national driving handbooks for 90 km/h. The direction is right as well: every distance came out positive, as it must for a car that never reverses. The striking lesson is that the reaction phase alone contributes 17.5 m - a quarter of the total - even though the driver did nothing wrong.

Worked example: average speed is not average velocity

Given: a runner covers 400 m east in 80 s, then turns and jogs 250 m west in 70 s.
Find: the average velocity and the average speed for the whole trip.

Solution: take east as the positive direction.
displacement = +400 - 250 = +150 m
total time = 80 + 70 = 150 s
average velocity = 150 / 150 = +1.0 m/s, that is 1.0 m/s east.
For speed, use path length rather than displacement:
total distance = 400 + 250 = 650 m
average speed = 650 / 150 = 4.33 m/s.

Sanity check: the average speed must exceed the magnitude of the average velocity whenever the motion reverses, and 4.33 m/s comfortably exceeds 1.0 m/s. The sign is right too - the runner finished east of the start, so the average velocity points east. One more trap worth noticing: the two leg speeds are 400/80 = 5.00 m/s and 250/70 = 3.57 m/s, whose plain average is 4.29 m/s. That is not the answer, because the legs took different times; only the total-distance-over-total-time calculation is correct.

Common wrong turns

• Reading a negative acceleration as "slowing down." It means "directed along the negative axis." Compare the signs of v and a: matching signs speed the object up.
• Using a kinematic equation when the acceleration is not constant. All four equations assume constant a. A trip with a reaction phase and a braking phase needs two separate calculations, as above.
• Averaging the speeds of two legs. That is valid only when the legs take equal times. Otherwise divide total distance by total time.
• Forgetting that final velocity is zero when something "stops." The word "stops" supplies a free equation, v = 0, that many students leave unused.
• Mixing sign conventions midway. If up is positive at the start of a problem it stays positive to the end, even after the object turns around.

Recap

• Displacement is the change in position and can be negative; distance is path length and cannot.
• Choose one positive direction at the outset and keep it; signs then carry all the directional information.
• An object speeds up when velocity and acceleration share a sign, and slows down when they differ.
• Four kinematic equations apply when acceleration is constant, each omitting one variable; pick the one that omits what you neither know nor need.
• Position-time slope is velocity, velocity-time slope is acceleration, and velocity-time area is displacement.
• Stopping distance grows with the square of speed, and reaction distance grows only in proportion to it.

Sources

1. OpenStax. (2016). 3.1 Position, displacement, and average velocity. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 3.2 Instantaneous velocity and speed. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 3.3 Average and instantaneous acceleration. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2016). 3.4 Motion with constant acceleration. In University Physics Volume 1. Rice University. openstax.org
5. OpenStax. (2022). 2.5 Motion equations for constant acceleration in one dimension. In College Physics 2e. Rice University. openstax.org
6. OpenStax. (2022). 2.8 Graphical analysis of one-dimensional motion. In College Physics 2e. Rice University. openstax.org
7. Nave, R. (n.d.). Constant acceleration motion. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu

Ask most people whether a bowling ball or a tennis ball hits the ground first and they will hesitate, then guess the bowling ball. The intuition is ancient, it is nearly universal, and it is wrong. Air resistance, not gravity, is what makes a leaf drift while a stone plummets, and once you strip the air away the two fall in perfect step. This lesson is where that becomes a calculation rather than a claim.

Near Earth's surface, every object in free fall - moving under gravity alone, with air resistance neglected - accelerates downward at the same rate. This acceleration due to gravity is written g and equals about 9.8 m/s squared.

A feather and a hammer dropped in a vacuum land together, a fact famously demonstrated by astronaut David Scott on the Moon in 1971. Mass does not affect the rate of fall; it only affects the gravitational force, as Module 4 explains. This is one of the most counterintuitive results in all of physics, because everyday experience - a leaf drifting while a stone plummets - seems to say the opposite. The difference is entirely air resistance, not gravity.

Setting up vertical motion

Free fall is just one-dimensional kinematics with a specific acceleration. Choose up as positive; then the acceleration is a = -g, pointing down, throughout the motion - on the way up, at the top, and on the way down alike. All four kinematic equations still apply, with the vertical position usually written as y. The key insight is that the same equations govern an object dropped, thrown down, or thrown up; only the initial velocity v0 changes. This means you do not learn new physics for free fall, only a new value for the acceleration.

One point of vocabulary trips people up constantly. "Free fall" does not mean "falling downward." It means gravity is the only force acting. A ball still climbing after you throw it is in free fall; so is a ball at the very top of its arc, momentarily motionless; so is a satellite curving around the planet at 7.7 km/s. In every one of those cases the acceleration is g downward and nothing else. Whether the object happens to be moving up, down, or sideways is irrelevant to the acceleration.

Key idea: Free fall means gravity is the only force acting, so the acceleration is g downward throughout - rising, at the peak, and falling alike.

Mass and weight are not the same thing

Because free-fall problems mix the two constantly, it is worth being blunt. Mass is a scalar measured in kilograms; it is a property of the object itself, the measure of its inertia, and it is identical on Earth, on the Moon, and in deep space. Weight is a vector force measured in newtons, equal to W = m g, and it changes with location because g does. A 70 kg astronaut has a mass of 70 kg everywhere, a weight of about 70 x 9.8 = 686 N on Earth, and a weight of about 70 x 1.6 = 112 N on the Moon.

Notice also what g is and is not. It is an acceleration, 9.8 m/s^2 - not a force. Writing "the force of gravity is 9.8" is a units error as well as a concept error: the force is m g in newtons, the acceleration is g in meters per second squared.

Key idea: Mass in kilograms measures inertia and never changes; weight in newtons is the force m g and varies with location.

Why mass does not matter

It is worth pausing on why a heavy object and a light one fall together. A heavier object is pulled harder by gravity, but it also has more inertia and is correspondingly harder to accelerate. These two effects cancel exactly, leaving the same acceleration for every mass. Galileo argued this centuries before Newton by a thought experiment: if heavy things fell faster, then tying a light object to a heavy one should slow the heavy one down (the light one lagging), yet the combination is heavier still and should fall faster - a contradiction that can only be resolved if all objects fall at the same rate.

Worked example: a dropped object

Given: a ball is dropped from rest (v0 = 0) from a height of 44.1 m. Take g = 9.8 m/s squared. Find: the time to reach the ground and the speed at impact.
Solution: Using y = (1/2) g t squared for the magnitude of the fall, 44.1 = (1/2)(9.8) t squared = 4.9 t squared, so t squared = 9.0 and t = 3.0 s.

The impact speed is v = g t = 9.8 x 3.0 = 29.4 m/s. So the ball falls for 3.0 s and hits at 29.4 m/s. Note that the fall distance grows as time squared, so in the first second the ball falls only 4.9 m, but in the third second it covers over 24 m - it speeds up as it goes.

Worked example: thrown straight up

Given: a ball is thrown straight up at v0 = 19.6 m/s. Find: the time to reach the highest point and the maximum height.
Solution: At the highest point the velocity is momentarily zero. Using v = v0 - g t with v = 0: 0 = 19.6 - 9.8 t, so t = 2.0 s.

For the height, v squared = v0 squared - 2 g y gives 0 = 19.6 squared - 2(9.8) y, so y = 384.16 / 19.6 = 19.6 m. The ball rises for 2.0 s to a height of 19.6 m. By symmetry it takes another 2.0 s to fall back, returning at the same 19.6 m/s it left with, but now directed downward.

The symmetry of flight and real air

The up-and-down symmetry - equal rise and fall times, equal launch and return speeds - is exact only when air resistance is ignored. A real thrown ball feels a drag force that grows with speed and always opposes motion, so it actually spends slightly longer falling than rising and returns a little slower than it left.

When drag balances gravity entirely, an object stops accelerating and settles at a constant terminal velocity; this is why a skydiver reaches a steady speed and why raindrops do not arrive at lethal velocity despite falling from great heights. For the idealized problems in this course we neglect air, but it is good to know exactly which assumption we are making.

Worked example: thrown up from a rooftop

Given: a ball is thrown straight up at v0 = 12.0 m/s from a rooftop 25.0 m above the ground. Take g = 9.8 m/s^2 and let up be positive, so a = -9.8 m/s^2 and the ground sits at y = 0 with the release point at y0 = 25.0 m.
Find: (a) the maximum height above the ground, (b) the total time until it lands, and (c) the impact speed.

Solution (a): at the peak the velocity is momentarily zero. Using the timeless equation for the rise above the roof:
0 = 12.0^2 + 2(-9.8) h, so 19.6 h = 144 and h = 7.35 m.
Above the ground that is 25.0 + 7.35 = 32.3 m.

Solution (b): use the position equation and set y = 0.
0 = 25.0 + 12.0 t - 4.9 t^2, or 4.9 t^2 - 12.0 t - 25.0 = 0.
t = [12.0 +/- sqrt(144 + 4(4.9)(25.0))] / (2 x 4.9) = [12.0 +/- sqrt(634)] / 9.8 = [12.0 +/- 25.18] / 9.8.
The negative root, -1.35 s, describes a time before the throw and is discarded. The physical answer is t = 37.18 / 9.8 = 3.79 s.

Solution (c): v = v0 + a t = 12.0 - 9.8(3.79) = -25.2 m/s, so the ball lands at a speed of 25.2 m/s, the minus sign showing it is moving downward.

Sanity check: get the impact speed a second way, without the time. v^2 = v0^2 + 2a(y - y0) = 144 + 2(-9.8)(0 - 25.0) = 144 + 490 = 634, so v = 25.2 m/s - the same. Break the timing apart as a third check: the rise takes 12.0/9.8 = 1.22 s, and falling 32.3 m from rest takes sqrt(2 x 32.3 / 9.8) = sqrt(6.59) = 2.57 s, and 1.22 + 2.57 = 3.79 s, matching part (b). Are the magnitudes sensible? The ball must land faster than the 12.0 m/s it left with, because it finishes 25 m lower than it started, and 25.2 m/s duly exceeds it. Note that the up-and-down symmetry that holds for level ground fails here: the ball spends 1.22 s going up and 2.57 s coming down.

Worked example: a sandbag dropped from a rising balloon

Given: a hot-air balloon is climbing steadily at 4.0 m/s when a sandbag is released at an altitude of 80.0 m. Up is positive, g = 9.8 m/s^2.
Find: (a) the bag's highest point, (b) the time it takes to reach the ground, and (c) its impact speed.

Solution: the crucial step is the initial velocity. The bag was travelling with the balloon at the moment of release, so v0 = +4.0 m/s upward, not zero. Releasing something is not the same as throwing it down.

(a) 0 = 4.0^2 - 2(9.8) h, so h = 16 / 19.6 = 0.82 m above the release point, giving a peak of 80.8 m.
(b) 0 = 80.0 + 4.0 t - 4.9 t^2, or 4.9 t^2 - 4.0 t - 80.0 = 0.
t = [4.0 + sqrt(16 + 4(4.9)(80.0))] / 9.8 = [4.0 + sqrt(1584)] / 9.8 = [4.0 + 39.80] / 9.8 = 4.47 s.
(c) v^2 = 4.0^2 + 2(-9.8)(0 - 80.0) = 16 + 1568 = 1584, so v = 39.8 m/s downward.

Sanity check: compare with a bag simply dropped from rest at the same 80.0 m, which takes sqrt(2 x 80.0/9.8) = 4.04 s and lands at sqrt(2 x 9.8 x 80.0) = 39.6 m/s. The upward release buys a noticeably longer flight, 4.47 s against 4.04 s, but almost no extra speed, 39.8 against 39.6 m/s - because the extra time comes from the 0.82 m detour up and back, while the extra speed comes only from that same 0.82 m of additional drop. The orders of magnitude are right: an 80 m fall takes a few seconds and ends near 40 m/s, about 140 km/h.

Common wrong turns

• Setting v0 = 0 for anything "released." An object let go from a moving carrier keeps the carrier's velocity. The sandbag above starts at +4.0 m/s.
• Claiming acceleration is zero at the top of the flight. The velocity is zero for an instant; the acceleration is still 9.8 m/s^2 downward, which is precisely what turns the motion around.
• Assuming the rise time always equals the fall time. That symmetry holds only when launch and landing heights match. From a rooftop, the fall takes longer.
• Confusing mass with weight. Mass in kilograms never changes; weight in newtons is m g and depends on where you are.
• Making g positive on the way up and negative on the way down. Choose one sign convention and keep it; the acceleration never flips.
• Keeping the negative root of the quadratic. It usually describes a time before the motion began and must be discarded on physical grounds.

Recap

• Free fall means gravity is the only force acting, so the acceleration is g downward whether the object rises, hovers at its peak, or falls.
• All objects fall at the same rate in the absence of air, because greater gravitational pull is exactly offset by greater inertia.
• Free-fall problems are ordinary one-dimensional kinematics with a = -g once up is taken as positive.
• Mass is a scalar in kilograms that never changes; weight is a force in newtons equal to m g.
• An object released from a moving carrier inherits the carrier's velocity as its v0.
• Rise and fall are symmetric only for equal launch and landing heights, and only when air resistance is neglected.

Sources

1. OpenStax. (2016). 3.5 Free fall. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 5.4 Mass and weight. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 6.4 Drag force and terminal speed. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2022). 2.7 Falling objects. In College Physics 2e. Rice University. openstax.org
5. Nave, R. (n.d.). Mass, weight, density. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu
6. National Institute of Standards and Technology. (n.d.). CODATA value: standard acceleration of gravity. The NIST Reference on Constants, Units, and Uncertainty. physics.nist.gov
7. O'Connor, J. J., & Robertson, E. F. (n.d.). Galileo Galilei. MacTutor History of Mathematics Archive, University of St Andrews. mathshistory.st-andrews.ac.uk

Fire a bullet horizontally from a rifle and, at the same instant, drop an identical bullet from the same height. They hit the ground together. The fired bullet travels perhaps half a kilometer downrange in that time, and none of it buys a single extra millisecond in the air. That result offends intuition and it is the whole content of this lesson: the two directions do not talk to each other.

When an object moves in two dimensions under gravity, the secret is a powerful idea called the independence of motion: the horizontal and vertical parts of the motion do not affect each other. Horizontally, gravity exerts no force, so velocity is constant. Vertically, the object is in free fall with acceleration g.

This means a two-dimensional problem is really two one-dimensional problems running in parallel, linked only by the shared clock. A ball rolled off a table and a ball dropped from the same height hit the floor at the same instant, because their vertical motions are identical - the horizontal motion of the first ball changes where it lands but not when.

The strategy: split, solve, recombine

Every projectile problem follows the same three steps. First, resolve the initial velocity into horizontal and vertical components. Second, treat each direction with its own kinematics: horizontal uses x = vx t with no acceleration, while vertical uses the free-fall equations with a = -g. Third, use the time t as the bridge between the two, since both motions happen over the same interval. The moment you internalize that time is the only quantity the two directions share, projectile motion stops feeling like a special topic and becomes routine.

Key idea: Horizontal and vertical motion are independent and share only the clock; solve each as a separate one-dimensional problem and let t link them.

Velocity partway through the flight

A question that trips people up is "how fast is it going at some moment in the middle?" The answer is a vector reconstruction, not a single formula. At any instant, vx = v0 cos(theta) is unchanged, while vy = v0 sin(theta) - g t has shrunk, passed through zero at the apex, and gone negative. The speed is then v = sqrt(vx^2 + vy^2), and the direction of travel is the angle whose tangent is vy / vx - taken below the horizontal once vy is negative.

Two consequences follow. The projectile is slowest at the top of its arc, where only vx survives, and it is never at rest. And on level ground the impact speed equals the launch speed, with the impact angle mirroring the launch angle below the horizontal - but from a raised launch point neither symmetry holds.

Key idea: The horizontal velocity component never changes, so the projectile is slowest at its apex, where the vertical component is momentarily zero.

Horizontal launch

For an object launched horizontally at speed vx from height h, the initial vertical velocity is zero, so the vertical motion determines the fall time from h = (1/2) g t squared. The horizontal distance (the range) is then simply x = vx t. The path traced out is a parabola, curving downward more and more steeply as the growing vertical speed combines with the steady horizontal speed.

Worked example: off a cliff

Given: a ball leaves a 20 m high cliff horizontally at vx = 15 m/s. Take g = 9.8. Find: the time in the air and how far from the base it lands.
Solution: Fall time from 20 = (1/2)(9.8) t squared = 4.9 t squared gives t squared = 4.08, so t = 2.02 s.

Horizontal range is x = 15 x 2.02 = 30.3 m. The ball is airborne for 2.02 s and lands 30.3 m out. The horizontal launch speed had no effect on the fall time - the ball would hit the ground at the same moment whether it left the cliff at 1 m/s or 100 m/s.

Angled launch

Launch at speed v0 and angle theta above the horizontal. Split the initial velocity: v0x = v0 cos(theta) stays constant, while v0y = v0 sin(theta) behaves exactly like a ball thrown straight up. Three useful results follow, valid when launch and landing heights are equal:

• Time of flight: t = 2 v0 sin(theta) / g
• Maximum height: H = (v0 sin(theta)) squared / (2 g)
• Range: R = v0 squared sin(2 theta) / g

Because of the sin(2 theta) term, the range is greatest at a launch angle of 45 degrees, where sin(2 theta) = sin 90 = 1. A neat consequence is that angles equally above and below 45 - such as 30 and 60, or 40 and 50 - give the same range, though the higher angle produces a taller, slower-arriving arc. This is why a punt in football and a low line-drive can travel the same distance by very different routes.

Worked example: a kicked ball

Given: a ball is kicked at v0 = 20 m/s at theta = 30 degrees. Take g = 9.8. Find: the maximum height and the range.
Solution: The vertical launch speed is v0 sin 30 = 20 x 0.5 = 10 m/s. Maximum height is H = 10 squared / (2 x 9.8) = 100 / 19.6 = 5.1 m.

For range, sin(2 x 30) = sin 60 = 0.866, so R = 20 squared x 0.866 / 9.8 = 400 x 0.866 / 9.8 = 35.3 m. The ball rises 5.1 m and lands 35.3 m away. To confirm with the split method: the horizontal speed is 20 cos 30 = 17.3 m/s, time of flight is 2 x 10 / 9.8 = 2.04 s, and 17.3 x 2.04 = 35.3 m, which matches.

When the range formula does not apply

The three boxed results above carry a condition that is easy to forget: the launch and landing heights must be equal. Throw from a cliff, kick onto a raised green, or shoot at a target uphill, and R = v0^2 sin(2 theta)/g is simply wrong. So is the claim that 45 degrees maximizes the range. Launch from above the landing point and the best angle drops below 45 degrees, because extra hang time is already supplied by the drop and it pays to spend more of the launch speed going forward.

When the heights differ there is no shortcut: return to split, solve, recombine. Write the vertical position equation including the height difference, solve the quadratic for the flight time, and multiply by the constant horizontal speed. That procedure never fails, which is why it deserves more trust than any memorized formula.

Key idea: The range, apex, and time-of-flight formulas and the 45-degree rule all assume equal launch and landing heights; otherwise solve the vertical quadratic directly.

Worked example: a kick from a cliff top

Given: a ball is launched at v0 = 18.0 m/s at theta = 35 degrees above the horizontal from a cliff 12.0 m above level ground. Take g = 9.8 m/s^2, up positive.
Find: (a) the time of flight, (b) the horizontal distance from the cliff base, and (c) the impact speed and angle.

Solution: resolve the launch velocity first.
v0x = 18.0 cos(35) = 18.0 x 0.8192 = 14.74 m/s (constant for the whole flight)
v0y = 18.0 sin(35) = 18.0 x 0.5736 = 10.32 m/s

(a) The heights differ, so solve the vertical quadratic with y = 0 at the ground and y0 = 12.0 m:
0 = 12.0 + 10.32 t - 4.9 t^2, or 4.9 t^2 - 10.32 t - 12.0 = 0.
t = [10.32 + sqrt(106.6 + 235.2)] / 9.8 = [10.32 + sqrt(341.8)] / 9.8 = [10.32 + 18.49] / 9.8 = 2.94 s.

(b) x = v0x t = 14.74 x 2.94 = 43.3 m.

(c) At impact, vx = 14.74 m/s still, while vy = 10.32 - 9.8(2.94) = -18.49 m/s.
v = sqrt(14.74^2 + 18.49^2) = sqrt(217.3 + 341.9) = sqrt(559.2) = 23.6 m/s,
at arctan(18.49 / 14.74) = arctan(1.254) = 51.4 degrees below the horizontal.

Sanity check: get the impact speed independently, without ever using the time. Only the vertical component changes, so v^2 = v0^2 + 2 g (drop) = 18.0^2 + 2(9.8)(12.0) = 324 + 235.2 = 559.2, giving v = 23.6 m/s - identical. Are the answers plausible? The ball must land faster than the 18.0 m/s it left with, since it finishes 12 m lower, and it does. It must also land steeper than the 35 degrees it left at, because the vertical speed grew while the horizontal stayed put, and 51.4 degrees is duly steeper. Finally, compare the 43.3 m range with what the level-ground formula would have predicted, 18.0^2 sin(70)/9.8 = 31.1 m: the cliff adds about 12 m, which is why using that formula here would have been a serious error.

Worked example: does it clear the wall?

Given: a ball is struck at v0 = 18.0 m/s at theta = 30 degrees from ground level. A wall 2.5 m high stands 25.0 m away.
Find: (a) whether the ball clears the wall, and (b) the minimum launch speed at the same angle that just would.

Solution (a): the question asks for a height at a specified horizontal position, so find the time to reach the wall and then the height at that time.
v0x = 18.0 cos(30) = 15.59 m/s, v0y = 18.0 sin(30) = 9.00 m/s
t = 25.0 / 15.59 = 1.604 s
y = 9.00(1.604) - 4.9(1.604)^2 = 14.43 - 4.9(2.572) = 14.43 - 12.60 = 1.83 m.
The ball arrives at only 1.83 m, so it strikes the wall about 0.67 m below the top and does not clear it.

Solution (b): use the trajectory equation, which gives height directly as a function of horizontal distance:
y = x tan(theta) - g x^2 / (2 v0^2 cos^2(theta)).
Setting y = 2.5, x = 25.0, theta = 30:
2.5 = 25.0(0.5774) - 9.8(625) / (2 v0^2 x 0.75)
2.5 = 14.43 - 4083 / v0^2, so 4083 / v0^2 = 11.93 and v0^2 = 342.2, giving v0 = 18.5 m/s.

Sanity check: feed 18.5 m/s back through part (a). v0x = 16.02 m/s, v0y = 9.25 m/s, t = 25.0/16.02 = 1.560 s, and y = 9.25(1.560) - 4.9(1.560)^2 = 14.43 - 11.93 = 2.50 m - exactly the wall height, so the ball just grazes it. The direction of the correction is right: needing more height means needing more speed, not less. And the magnitude is instructive - a mere 2.8 percent increase in launch speed converts a clear failure into a clear success, because range depends on v0^2. Note too that the ball is already descending at the wall, since its apex sits at x = 14.3 m, well short of 25 m.

Common wrong turns

• Using g in the horizontal equation. There is no horizontal force, so vx is constant and x = vx t has no acceleration term.
• Applying the range formula when the heights differ. R = v0^2 sin(2 theta)/g and the 45-degree optimum both assume level ground. From a height, solve the vertical quadratic instead.
• Swapping sine and cosine. With the angle measured from the horizontal, vx takes cosine and vy takes sine. Check a limiting case: at 90 degrees the launch must be purely vertical.
• Thinking acceleration is zero at the apex. Only vy vanishes there; the acceleration stays at g downward throughout.
• Answering a "how high at that point" question with the maximum height. If the question names a horizontal distance, find the time to reach it first.

Recap

• Horizontal and vertical motions are independent and share only the elapsed time.
• Horizontally, vx is constant and x = vx t; vertically, the motion is free fall with a = -g.
• Speed at any instant is sqrt(vx^2 + vy^2), so the projectile is slowest, never stationary, at its apex.
• On level ground the range is v0^2 sin(2 theta)/g, maximized at 45 degrees, and complementary angles give equal ranges.
• When launch and landing heights differ, abandon the range formula and solve the vertical quadratic for the flight time.
• Launching from a height makes the projectile land both faster and at a steeper angle than it left.

Sources

1. OpenStax. (2016). 4.1 Displacement and velocity vectors. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 4.3 Projectile motion. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 4.5 Relative motion in one and two dimensions. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2022). 3.4 Projectile motion. In College Physics 2e. Rice University. openstax.org
5. Nave, R. (n.d.). Trajectories. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu
6. PhET Interactive Simulations. (n.d.). Projectile motion [Simulation]. University of Colorado Boulder. phet.colorado.edu
7. Dourmashkin, P., Fisher, P., & Tomasik, M. (2016). 8.01SC Classical mechanics [Course materials]. MIT OpenCourseWare. ocw.mit.edu

For two thousand years the best minds in the world believed that a moving object needs a continuous push to keep moving. It is an entirely reasonable conclusion from everyday life, where carts stop and thrown stones fall. It took until 1687 for that intuition to be overturned by three sentences that turned out to describe the motion of everything from a rolling ball to a planet. This lesson is those three sentences and what they cost you to believe.

Kinematics tells us how objects move; dynamics tells us why. The foundation is three laws set out by Isaac Newton in his 1687 work the Principia, which together explain the motion of everything from a rolling ball to the planets. A force is a push or pull, a vector measured in newtons (N).

What matters for motion is never a single force in isolation but the net force, the vector sum of all forces acting on an object. Getting comfortable with the idea of net force is the whole game: once you can total up the forces on an object, Newton's second law tells you exactly what it will do.

The first law: inertia

An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted on by a net force. This tendency to resist changes in motion is called inertia, and mass is its quantitative measure - a loaded truck has far more inertia than a bicycle. The first law overturns the old Aristotelian intuition that a force is needed to keep something moving.

In reality, moving objects slow down only because forces like friction and air resistance act on them; remove those forces, as on a frictionless surface or in the vacuum of space, and no force at all is needed to maintain constant velocity forever. A spacecraft coasting between planets with its engines off is the first law in its purest form.

Key idea: Constant velocity requires zero net force, not a steady push; only a change in velocity demands a force.

Inertial frames: where the laws are true

The first law does more than describe inertia - it quietly defines the arena in which all three laws apply. A reference frame in which a force-free object really does move at constant velocity is called an inertial frame. Newton's laws hold in inertial frames and fail in others.

Slam the brakes in a car and a bag on the seat slides forward. From the roadside, an inertial frame, this is unremarkable: the bag continues at its previous velocity while the car slows beneath it. From inside the decelerating car, though, the bag appears to accelerate forward with nothing pushing it, flatly contradicting the second law. The car is a non-inertial frame, and to keep the equations working from inside it you must invent a fictitious force - the same trick that produces the centrifugal and Coriolis forces in rotating frames. The practical rule is to set up every problem in a frame attached to the ground, and no fictitious force is ever needed.

Key idea: Newton's laws hold only in inertial frames, so measure accelerations relative to the ground, never relative to an accelerating vehicle.

The second law: F = ma

When a net force does act, it produces an acceleration in the same direction, given by the most important equation in mechanics: F_net = m a. The acceleration is directly proportional to the net force and inversely proportional to the mass. Double the force and you double the acceleration; double the mass and you halve it.

This single equation is the bridge between the force concepts of this module and the kinematics of Module 2: find the net force, divide by mass to get acceleration, and then the kinematic equations tell you the motion. It is worth noting that the first law is really just the special case of the second when the net force is zero, giving zero acceleration and hence constant velocity.

Three details are easy to skate past. The equation is a vector statement, so it holds separately in each direction and the acceleration always points along the net force, never along the velocity. The m is mass in kilograms, not weight in newtons; dividing a net force by a weight is a units error that produces answers off by a factor of 9.8. And the F is the net force, the vector sum of everything acting, which is why the free-body diagram of the next lesson comes before the algebra.

Key idea: F_net = m a is a vector equation in which the acceleration points along the net force, and m is mass in kilograms, never weight in newtons.

Worked example: pushing a car

Given: a 1200 kg car experiences a net forward force of 3600 N. Find: its acceleration.
Solution: From F_net = m a, solve a = F_net / m = 3600 / 1200 = 3.0 m/s squared. The car accelerates at 3.0 m/s squared. If you then wanted the speed after 5 seconds starting from rest, you would hand this acceleration to kinematics: v = a t = 3.0 x 5 = 15 m/s. This handoff between dynamics and kinematics is the pattern behind most mechanics problems.

The third law: action and reaction

For every action there is an equal and opposite reaction. If object A pushes on object B with some force, then B pushes back on A with a force equal in magnitude and opposite in direction. The single most important thing to understand about the third law is that these two forces act on different objects, which is precisely why they do not cancel each other out - cancellation only happens between forces on the same object.

When you walk, your foot pushes backward on the ground and the ground pushes forward on you; that forward reaction force is what propels you, and it acts on you, not on the ground. A rocket works the same way, pushing exhaust gas downward so the gas pushes the rocket upward, which is why rockets function even in the vacuum of space where there is nothing to "push against" but their own exhaust.

A related distinction separates students who can draw a free-body diagram from those who cannot. Forces that balance and forces that form a third-law pair are different things. A book on a table feels its weight down and the normal force up; these are equal and opposite, but they act on the same body and are not a third-law pair - the proof is that they stop being equal the moment the table sits in an accelerating lift. The genuine partner of the table's upward push on the book is the book's downward push on the table. Third-law partners always act on different bodies and are equal in every circumstance whatsoever.

Key idea: Third-law partners act on different bodies and are always equal; two balanced forces act on the same body and need not be.

Worked example: two blocks and the string between them

Given: blocks of mass m1 = 3.0 kg and m2 = 5.0 kg rest on a frictionless horizontal surface, joined by a light string. A horizontal force F = 24 N is applied to m2, pulling both to the right.
Find: (a) the acceleration of the pair and (b) the tension in the connecting string.

Solution (a): the string force is internal to the two-block system, so treat both blocks as one body of mass m1 + m2 = 8.0 kg. Only the applied 24 N is external in the horizontal direction.
a = F / (m1 + m2) = 24 / 8.0 = 3.0 m/s^2, to the right.

Solution (b): to expose the tension, isolate a single block. Take the trailing block m1: the only horizontal force on it is the string tension T, pulling it forward.
T = m1 a = 3.0 x 3.0 = 9.0 N.

Sanity check: verify with the other block, whose free-body diagram carries the applied force forward and the tension backward (by the third law, the string pulls back on m2 exactly as hard as it pulls m1 forward).
F - T = m2 a gives 24 - 9.0 = 15.0 N, and m2 a = 5.0 x 3.0 = 15.0 N. The two routes agree. Is 9.0 N sensible? The tension must be less than the 24 N applied force, because the string is only responsible for accelerating the 3.0 kg block, not the whole 8.0 kg - and the ratio works out to m1/(m1 + m2) = 3/8 of the applied force, which is exactly 0.375 x 24 = 9.0 N. Note the asymmetry: had the same 24 N been applied to m1 instead, the string would have had to accelerate the heavier 5.0 kg block and the tension would have been (5/8) x 24 = 15 N. Which end you pull from matters.

Worked example: two forces at an angle

Given: a 6.0 kg crate sits on a frictionless floor. Two horizontal forces act: F1 = 30 N due east and F2 = 40 N at 60 degrees north of east.
Find: the magnitude and direction of the crate's acceleration.

Solution: the second law is a vector equation, so find the net force by components first.
F1x = 30 N, F1y = 0
F2x = 40 cos(60) = 40 x 0.500 = 20.0 N, F2y = 40 sin(60) = 40 x 0.8660 = 34.64 N
Fx = 30 + 20.0 = 50.0 N, Fy = 0 + 34.64 = 34.64 N
F_net = sqrt(50.0^2 + 34.64^2) = sqrt(2500 + 1200) = sqrt(3700) = 60.8 N
a = F_net / m = 60.8 / 6.0 = 10.1 m/s^2
Direction: both components are positive, so the first quadrant applies directly, and arctan(34.64 / 50.0) = arctan(0.693) = 34.7 degrees north of east.

Sanity check: the net force must lie between the difference and the sum of the two magnitudes, that is between 10 N and 70 N, and 60.8 N does. Its direction must lie between the two force directions, 0 and 60 degrees, and 34.7 degrees does - tilted past the halfway mark of 30 degrees toward the larger force, exactly as expected. Because the surface is frictionless the acceleration points along the net force rather than along any single push, which is the vector content of the second law. Finally, 10.1 m/s^2 is a large but believable acceleration for a modest crate under 61 N of push; had we divided by the crate's weight of 6.0 x 9.8 = 58.8 N instead of its mass, we would have got a meaningless 1.03 with no correct units at all.

Common wrong turns

• Believing a moving object needs a forward force. Constant velocity means zero net force. Only a change of velocity requires one.
• Dividing by weight instead of mass. a = F/m takes mass in kilograms. Using the weight in newtons scales every answer wrongly by g.
• Treating a third-law pair as cancelling. The two forces act on different bodies, so they can never cancel on a single free-body diagram.
• Working in an accelerating frame. Measure accelerations relative to the ground, or fictitious forces become necessary.
• Assuming the acceleration points along the velocity. It points along the net force, which is why a projectile accelerates downward while moving sideways.

Why it matters

These three laws are not abstract - they underlie seatbelts (your inertia keeps you moving forward when the car stops suddenly), the recoil of a fired gun, the way a swimmer pushes water backward to move forward, and the design of every vehicle and structure. Master them and you have the conceptual core of all classical mechanics.

Recap

• The first law says an object keeps its velocity unless a net force acts, and it defines the inertial frames in which the laws apply.
• The second law, F_net = m a, is a vector equation; the acceleration points along the net force and m is mass, not weight.
• The third law pairs every force with an equal and opposite force on a different body, which is why the pair never cancels.
• Balanced forces on one body and third-law partners on two bodies are different ideas and must not be confused.
• For a connected system, external forces set the shared acceleration and isolating one body exposes the internal forces such as tension.
• The tension in a string joining two blocks is always less than the applied force, and depends on which end you pull from.

Sources

1. OpenStax. (2016). 5.1 Forces. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 5.2 Newton's first law. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 5.3 Newton's second law. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2016). 5.5 Newton's third law. In University Physics Volume 1. Rice University. openstax.org
5. OpenStax. (2022). 4.4 Newton's third law of motion: Symmetry in forces. In College Physics 2e. Rice University. openstax.org
6. Nave, R. (n.d.). Newton's laws. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu
7. O'Connor, J. J., & Robertson, E. F. (n.d.). Isaac Newton. MacTutor History of Mathematics Archive, University of St Andrews. mathshistory.st-andrews.ac.uk

To apply Newton's second law, you first need a clear picture of every force acting on your object. A free-body diagram is that picture: the object is isolated and drawn as a dot or box, and every force acting on it is drawn as an arrow pointing in its direction of action. Forces the object exerts on other things do not appear - only forces acting on the chosen body. Getting this diagram right is more than half the battle in any dynamics problem; most mistakes trace back to a missing force, an extra force that should not be there, or a wrong direction.

The common forces

A handful of forces appear again and again in mechanics. Weight (W = m g) always points straight down toward Earth's center and acts on every object with mass. The normal force (N) is the support a surface pushes back with, always perpendicular to that surface. Tension (T) is the pull transmitted along a rope, string, or cable, directed along its length away from the object.

Friction (covered next lesson) acts along a surface, opposing relative sliding. And there may be any applied push or pull from a hand, engine, or other agent. Learning to recognize which of these act in a given situation is a skill that comes quickly with practice.

The solution recipe

Once the diagram is drawn, the procedure is always the same: choose x and y axes, write Newton's second law separately for each direction (sum of Fx = m ax and sum of Fy = m ay), and solve.

On a flat surface with no vertical acceleration and nothing else pushing vertically, the vertical equation gives N - W = 0, so the normal force balances the weight and N = m g. If the object is in equilibrium - at rest or moving at constant velocity - then every acceleration is zero, so the forces in each direction must sum to zero. This equilibrium condition is the basis of statics, the engineering of bridges and buildings that must not move.

The procedure, in order

Every problem in this module yields to the same five steps, and following them in order is what separates a two-line solution from a page of confusion.

1. Choose the body. Decide precisely which object you are isolating. Draw a boundary around it and cut everything that touches it.
2. Draw every force on that body. Weight always. Then one force for each thing touching it: a normal force from each surface, a tension along each rope, an applied push. Nothing else. If you cannot name what exerts a force, it does not exist.
3. Choose axes. Align one axis with the acceleration if you know its direction - along the slope on a ramp, toward the center in circular motion.
4. Write the second law once per axis. sum Fx = m ax and sum Fy = m ay, with every force entering with its correct sign.
5. Solve symbolically, then substitute. Keeping letters until the last step exposes cancellations and makes errors visible.

Key idea: A free-body diagram shows only forces acting on the chosen body, one for each thing touching it plus gravity - and if you cannot name the agent, the force is not real.

The normal force is almost never just mg

The single most common error in this module is writing N = m g by reflex. That equation is the result of a vertical equation in one special case: a horizontal surface, no vertical acceleration, and no other force with a vertical component. Break any of those three conditions and it fails.

On an incline of angle theta the surface is tilted, so N = m g cos(theta), which is always less than the weight. In an elevator accelerating upward, N = m(g + a), which is more. If you push a crate at a downward angle, the push adds to what the floor must support and N exceeds m g; pull at an upward angle and N falls below it. Always derive the normal force from the perpendicular equation rather than assuming it. This matters enormously in the next lesson, because friction is proportional to N, so every error in the normal force propagates straight into the friction force.

Key idea: Find the normal force from the equation perpendicular to the surface every time; N = m g is a special case, not a definition.

Worked example: a box on a frictionless incline

Given: a box slides down a frictionless ramp tilted at theta = 30 degrees. Take g = 9.8. Find: its acceleration down the ramp.
Solution: The clever move is to tilt the axes so x points down the slope and y points away from the surface, matching the geometry. Then only the weight needs resolving. The component of gravity along the slope is W sin(theta) = m g sin(theta), and the component pressing into the surface is W cos(theta) = m g cos(theta).

Newton's second law along the slope gives m g sin(theta) = m a, so the mass cancels and a = g sin(theta) = 9.8 x sin 30 = 9.8 x 0.5 = 4.9 m/s squared. The box accelerates at 4.9 m/s squared down the ramp, regardless of its mass. This is why the steepness, not the weight, sets how fast a frictionless slide speeds up. Perpendicular to the slope there is no acceleration, so N = m g cos(theta), which is less than the full weight - a result that matters as soon as friction enters.

Worked example: an elevator

Given: a 70 kg person stands in an elevator that accelerates upward at 2.0 m/s squared. Take g = 9.8. Find: the normal force from the floor (the "apparent weight").
Solution: Two forces act on the person: the normal force N up and the weight W = mg = 686 N down. Newton's second law upward gives N - mg = m a, so N = m(g + a) = 70 x (9.8 + 2.0) = 70 x 11.8 = 826 N.

The person feels 826 N pressing up, heavier than their true weight of 686 N - the familiar sensation of being pushed into the floor as an elevator starts up. Were the elevator accelerating downward, the normal force would be less than the weight, and in free fall (a = -g) it would be zero, which is the origin of apparent weightlessness.

Sanity check: test the formula at its limits. With a = 0 it returns N = m g = 686 N, the resting value. With a = -g it returns N = 0, matching free fall. Both limits behave, so the algebra is trustworthy in between. The size is also believable: a 2.0 m/s^2 start adds about 20 percent to the apparent weight, which is roughly what a brisk lift feels like. Note that the person's mass is still 70 kg throughout - it is the normal force, not the mass or the true weight, that changes.

Worked example: a block on a table pulling a hanging mass

Given: a block of mass m1 = 4.0 kg sits on a frictionless horizontal table. A light string runs from it, over a frictionless massless pulley at the table edge, down to a hanging block of mass m2 = 6.0 kg. Take g = 9.8 m/s^2.
Find: the acceleration of the system and the tension in the string.

Free-body diagrams in words. For m1: weight down, normal force up (these cancel, since it has no vertical acceleration), and tension T horizontally toward the pulley. For m2: its weight m2 g down and the same tension T up. Because the string is inextensible, both blocks share one acceleration magnitude a; because the pulley is massless and frictionless, the tension is the same on both sides.

Solution: take the direction of motion as positive for each block - right for m1, down for m2.
m1: T = m1 a
m2: m2 g - T = m2 a
Add the two equations to eliminate T:
m2 g = (m1 + m2) a, so a = m2 g / (m1 + m2) = (6.0 x 9.8) / 10.0 = 58.8 / 10.0 = 5.88 m/s^2.
Then T = m1 a = 4.0 x 5.88 = 23.5 N.

Sanity check: substitute back into the second equation. m2 g - T = 58.8 - 23.5 = 35.3 N, and m2 a = 6.0 x 5.88 = 35.3 N. They match. Now test the physics rather than the arithmetic. The tension must be less than the hanging block's 58.8 N weight, or that block could not accelerate downward at all. The acceleration must be less than g, since the table block is dragged along for the ride. Two limiting cases confirm the formula: with m1 = 0 it gives a = g and T = 0, a plain free fall; with m1 enormous, a approaches zero and T approaches the full 58.8 N of a stationary weight. A tempting wrong answer is to assume the tension equals m2 g, which holds only when nothing accelerates.

Worked example: a push angled downward changes the normal force

Given: a 30 kg crate on a frictionless floor is pushed with a force of 120 N directed 25 degrees below the horizontal. Take g = 9.8 m/s^2.
Find: (a) the normal force from the floor and (b) the crate's acceleration.

Solution: resolve the push into components. The horizontal part drives the crate forward; the vertical part presses it into the floor.
Fx = 120 cos(25) = 120 x 0.9063 = 108.8 N
Fy = -120 sin(25) = -120 x 0.4226 = -50.7 N (downward)

(a) There is no vertical acceleration, so the vertical forces sum to zero. The floor must now support the weight and the downward component of the push:
N - m g - 50.7 = 0
N = (30 x 9.8) + 50.7 = 294 + 50.7 = 344.7 = 345 N.

(b) a = Fx / m = 108.8 / 30 = 3.63 m/s^2, horizontally forward.

Sanity check: the normal force came out larger than the 294 N weight, exactly as pushing downward should make it - the floor has more to hold up. Applying the same 120 N at 25 degrees above the horizontal would instead have lifted rather than pressed, dropping N to 294 - 50.7 = 243 N. Notice a subtlety that becomes crucial next lesson: on this frictionless floor the acceleration is 108.8/30 either way. Add friction and the two cases diverge sharply, because friction is proportional to N - which is precisely why movers pull rather than push. Writing N = m g = 294 N by reflex would have been wrong by 17 percent.

Common wrong turns

• Assuming N = m g. True only on a horizontal surface, with no vertical acceleration and no other vertical force component. Derive it from the perpendicular equation instead.
• Drawing forces the object exerts on other things. A free-body diagram carries only the forces acting on the chosen body.
• Inventing a "force of motion." A moving object does not carry a forward force with it. If nothing is touching it in that direction, no such arrow belongs on the diagram.
• Assuming tension equals the hanging weight. That holds only in equilibrium. Once the system accelerates, tension is less than m g for a falling mass and more for a rising one.
• Keeping horizontal and vertical axes on an incline. Tilting the axes to align with the slope reduces the problem to one dimension and lets the mass cancel.
• Confusing mass with weight in the second law. F = m a uses kilograms; W = m g converts that mass into a force in newtons.

Recap

• A free-body diagram isolates one object and shows every force acting on it, one per contact plus gravity.
• Weight acts downward, normal force perpendicular to the surface, tension along the rope away from the body.
• Write the second law once per axis, choosing axes aligned with the acceleration wherever possible.
• The normal force must be derived from the perpendicular equation; m g, m g cos(theta), and m(g + a) are all special cases.
• Connected bodies joined by an inextensible string share one acceleration, and an ideal pulley gives the same tension on both sides.
• Apparent weight is the normal force, which rises when accelerating upward and falls to zero in free fall.

Sources

1. OpenStax. (2016). 5.6 Common forces. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 5.7 Drawing free-body diagrams. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 6.1 Solving problems with Newton's laws. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2022). 4.5 Normal, tension, and other examples of forces. In College Physics 2e. Rice University. openstax.org
5. OpenStax. (2022). 4.6 Problem-solving strategies. In College Physics 2e. Rice University. openstax.org
6. Nave, R. (n.d.). Free-body diagram. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu
7. PhET Interactive Simulations. (n.d.). Forces and motion: Basics [Simulation]. University of Colorado Boulder. phet.colorado.edu

Push a heavy filing cabinet across a floor and you will feel the whole of this lesson in your arms. Nothing happens, nothing happens, nothing happens - and then it lurches forward and suddenly feels lighter. That sequence is not your imagination or your technique. It is two distinct physical regimes with two different coefficients, and once you can put numbers to them you can predict exactly when the lurch will come.

Friction is the force that resists sliding between two surfaces in contact. It arises from microscopic roughness and molecular adhesion where the surfaces touch, and it always acts along the surface in the direction that opposes relative motion or the tendency toward motion. Friction is often treated as a nuisance, but it is also indispensable: without it you could not walk, drive, hold a pencil, or tie a knot. Learning to handle friction turns the idealized frictionless problems of the last lesson into realistic ones.

Two kinds of friction

There are two distinct regimes. Static friction acts when the surfaces are not yet sliding. It is a responsive force that adjusts itself to exactly match whatever is trying to cause motion, up to a maximum value - push a heavy crate gently and static friction pushes back just as hard, so nothing moves; push harder and it pushes back harder, until you exceed its limit and the crate breaks free.

Kinetic friction acts once the surfaces are actually sliding, and it is generally a bit weaker than the maximum static friction and roughly constant regardless of sliding speed. This is why it takes a big initial shove to get something moving but less effort to keep it sliding.

The friction model

Both types depend on how hard the surfaces are pressed together, that is, on the normal force N. The kinetic friction force is f_k = mu_k N, where mu_k is the coefficient of kinetic friction, a dimensionless number set by the pair of materials in contact. The maximum static friction is f_s(max) = mu_s N, with its own coefficient mu_s. Below that maximum, the actual static friction is only as large as it needs to be to prevent sliding.

These coefficients are typically between about 0.1 for slippery pairs (ice on steel) and 1.0 or more for grippy ones (rubber on dry concrete). A surprising feature of this model is that friction does not depend on the contact area - a brick slides with the same friction whether laid on its broad face or its narrow edge, because spreading the same weight over more area reduces the pressure proportionally.

Static friction is an inequality, not an equation

This is the point where more marks are lost than anywhere else in the module, so it deserves to be stated as starkly as possible. The static friction law is not f_s = mu_s N. It is

f_s <= mu_s N.

Static friction is a responsive force: it takes exactly the value needed to prevent sliding, and no more. Push a stationary crate with 10 N and static friction is 10 N; push with 60 N and, if it still does not move, static friction is 60 N. The quantity mu_s N is only the ceiling. Writing f_s = mu_s N for a box sitting untouched on a table predicts a large friction force pushing on nothing at all, which is absurd.

The routine is therefore always two steps: compute the ceiling mu_s N, then compare it with the force trying to cause sliding. If the driving force is smaller, the object stays put and friction equals that driving force. If larger, the object breaks free and friction drops to the fixed kinetic value mu_k N - hence the lurch.

Key idea: Static friction equals whatever is needed to prevent sliding, up to a maximum of mu_s N; only kinetic friction is genuinely equal to mu_k N.

Friction on a slope and the angle of repose

On an incline the normal force is no longer the weight, so the friction available changes too. With the axes tilted along the slope, the perpendicular equation gives N = m g cos(theta) and the driving force along the slope is m g sin(theta). An object therefore begins to slide when

m g sin(theta) > mu_s m g cos(theta), that is, when tan(theta) > mu_s.

The mass cancels entirely, so whether something slips depends only on the slope and the materials, never on how heavy it is. The critical angle theta = arctan(mu_s) is the angle of repose, which is why a pile of dry sand always settles into the same cone however much you pour. It also gives a neat measurement method: tilt a plank until the object just slides and you have measured mu_s.

Key idea: An object slips on a slope when tan(theta) exceeds mu_s, a condition independent of mass.

Worked example: pushing a box on the floor

Given: a 10 kg box on a level floor is pushed horizontally with 40 N. The coefficient of kinetic friction is mu_k = 0.30. Take g = 9.8. Find: the friction force and the acceleration.
Solution: On a level floor with no vertical acceleration, the normal force equals the weight: N = m g = 10 x 9.8 = 98 N. The kinetic friction force is f_k = mu_k N = 0.30 x 98 = 29.4 N, opposing the push.

The net horizontal force is 40 - 29.4 = 10.6 N, so the acceleration is a = 10.6 / 10 = 1.06 m/s squared. The box speeds up at about 1.06 m/s squared. Note carefully what this calculation assumes: the box is already sliding, which is the only condition under which f_k = mu_k N applies. Whether a 40 N push could have started the box moving in the first place is a separate question, answered by comparing 40 N against the static ceiling mu_s N - and mu_s is a different number from mu_k, always the larger of the two. Never use the kinetic coefficient to decide whether something breaks loose.

Worked example: will it slide?

Given: a 15 kg crate sits on a floor with mu_s = 0.40. You push horizontally with 50 N. Take g = 9.8. Find: whether the crate moves.
Solution: First find the maximum static friction: N = mg = 147 N, so f_s(max) = mu_s N = 0.40 x 147 = 58.8 N.

Your push of 50 N is less than this 58.8 N limit, so static friction rises to match your 50 N and the crate does not move. You would need to push beyond 58.8 N to break it loose. This kind of threshold check - compare the applied force to mu_s N - is how you decide in any problem whether an object is about to slide.

Why static friction matters

Static friction is what lets you walk without slipping and lets car tires grip the road. A car accelerates because static friction from the road pushes the tires forward; the tire surface in contact with the road is momentarily at rest relative to it, so it is static, not kinetic, friction at work.

If you demand more force than the maximum (mu_s N), the tires spin and you skid. Because mu_s is usually larger than mu_k, a rolling, gripping tire has more traction than a skidding one, which is exactly the physics behind anti-lock braking systems that prevent the wheels from locking up so they keep gripping rather than sliding.

Worked example: sliding down a rough incline

Given: a 12 kg crate slides down a ramp inclined at theta = 30 degrees. The coefficients are mu_s = 0.35 and mu_k = 0.25. Take g = 9.8 m/s^2.
Find: (a) whether the crate slides from rest, and (b) if so, its acceleration down the slope.

Solution (a): the slip condition compares tan(theta) with mu_s.
tan(30) = 0.577 and mu_s = 0.35. Since 0.577 > 0.35, gravity's pull along the slope beats the static ceiling and the crate does slide. Equivalently, the angle of repose here is arctan(0.35) = 19.3 degrees, and 30 degrees is well past it.

Solution (b): tilt the axes so x runs down the slope and y is perpendicular to it.
Perpendicular (no acceleration): N = m g cos(30) = 12 x 9.8 x 0.8660 = 117.6 x 0.8660 = 101.8 N.
Kinetic friction, acting up the slope: f_k = mu_k N = 0.25 x 101.8 = 25.5 N.
Along the slope: m g sin(30) - f_k = m a
117.6 x 0.5000 - 25.5 = 58.8 - 25.5 = 33.3 N
a = 33.3 / 12 = 2.78 m/s^2, directed down the slope.

Sanity check: the mass should cancel, and it does - factoring it out gives a = g(sin(theta) - mu_k cos(theta)) = 9.8(0.5000 - 0.2165) = 9.8 x 0.2835 = 2.78 m/s^2, the same answer with no mass in sight. The value must also be smaller than the frictionless result g sin(30) = 4.90 m/s^2, and it is, by a bit over 40 percent. Note that using the full weight for the normal force, 117.6 N instead of 101.8 N, would have overstated friction by 15 percent and given a wrongly small acceleration.

Worked example: below the threshold and above it

Given: a 45 kg crate rests on a level floor with mu_s = 0.50 and mu_k = 0.35. Take g = 9.8 m/s^2.
Find: the friction force and the acceleration when you push horizontally with (a) 180 N and then (b) 250 N.

Solution: first establish the static ceiling, which both cases are measured against.
N = m g = 45 x 9.8 = 441 N
f_s(max) = mu_s N = 0.50 x 441 = 220.5 N

(a) The 180 N push is below the 220.5 N ceiling, so the crate does not move. Static friction rises to match the push exactly: f_s = 180 N, and the acceleration is 0. It is not 220.5 N - if it were, the crate would accelerate backwards toward you.

(b) The 250 N push exceeds the ceiling, so the crate breaks loose. The moment it slides, friction switches to the kinetic value:
f_k = mu_k N = 0.35 x 441 = 154.4 N
F_net = 250 - 154.4 = 95.6 N
a = 95.6 / 45 = 2.13 m/s^2.

Sanity check: in case (a) the crate is in equilibrium, so the forces must sum to zero - and 180 N of push against 180 N of friction does. In case (b), notice what happened at the instant of breaking free: the resisting force dropped abruptly from 220.5 N to 154.4 N, a fall of 66 N with no change in the push. That is exactly the lurch you feel when a heavy object finally gives, and it is why mu_s > mu_k is not a technicality. Note also that 250 N is only 13 percent above the 220.5 N threshold, yet it produces a real 2.13 m/s^2 - because friction fell as soon as motion began.

Common wrong turns

• Writing f_s = mu_s N. That is only the maximum. Below the threshold, static friction equals whatever force is trying to cause sliding.
• Using mu_k to decide whether something starts moving. The starting question is always answered with mu_s; mu_k applies only once sliding is under way.
• Using the weight as the normal force on an incline. On a slope N = m g cos(theta), so friction is smaller than a flat-surface calculation would suggest.
• Thinking friction depends on contact area. In this model it depends only on N and the coefficient. A brick on edge and a brick on its face slide alike.
• Assuming friction always opposes motion. It opposes relative sliding at the contact. The static friction that drives a car forward points in the direction the car is travelling.
• Forgetting that friction stops when the driving force does. Friction never pushes a resting object into motion on its own.

Recap

• Friction acts along the contact surface and opposes relative sliding or the tendency toward it.
• Static friction obeys f_s <= mu_s N and adjusts itself; kinetic friction is fixed at f_k = mu_k N.
• Because mu_s exceeds mu_k, the resisting force drops the instant an object breaks free, producing the familiar lurch.
• Both forms scale with the normal force, not with the contact area.
• On a slope an object slips when tan(theta) > mu_s, a mass-independent condition defining the angle of repose.
• Static friction is what lets tires grip and feet push off, which is why anti-lock brakes work to prevent sliding.

Sources

1. OpenStax. (2016). 6.2 Friction. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 6.1 Solving problems with Newton's laws. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 5.6 Common forces. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2022). 5.1 Friction. In College Physics 2e. Rice University. openstax.org
5. OpenStax. (2022). 5.2 Drag forces. In College Physics 2e. Rice University. openstax.org
6. Nave, R. (n.d.). Friction. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu
7. PhET Interactive Simulations. (n.d.). Friction [Simulation]. University of Colorado Boulder. phet.colorado.edu

Swing a bucket of water in a vertical circle fast enough and the water stays in, even at the top when the bucket is upside down. Nothing holds it there. The water is falling the whole time - it simply cannot fall faster than the bucket is moving out of its way. Understanding why is the same thing as understanding orbits, banked roads, and centrifuges.

An object moving in a circle at constant speed is in uniform circular motion. The word "uniform" refers only to the speed; the velocity is emphatically not constant, because velocity is a vector and its direction is continually changing as the object rounds the circle. Since any change in velocity - even a change in direction alone - is an acceleration, a circling object is always accelerating, and that acceleration points toward the center of the circle. This center-pointing acceleration is called centripetal acceleration, from the Latin for "center-seeking." Grasping that constant speed can coexist with nonzero acceleration is the conceptual key to this entire topic.

The centripetal equations

For an object moving at speed v in a circle of radius r, the centripetal acceleration is a_c = v squared / r, directed toward the center. Two things about this formula are worth noticing: the acceleration grows with the square of the speed, so doubling the speed quadruples the acceleration, and it shrinks with a larger radius, so a gentle wide curve requires less acceleration than a tight one at the same speed. By Newton's second law, producing this acceleration requires a net inward force, the centripetal force, of magnitude F_c = m a_c = m v squared / r.

Centripetal force is a role, not a new force

It is vital to understand that centripetal force is not a new kind of force to add to your list alongside gravity, tension, and friction. It is the name for the role played by whichever real force happens to point toward the center in a given situation. For a car turning on a flat road, the centripetal force is the static friction between tires and road. For a ball whirled on a string, it is the tension. For the Moon circling Earth, it is gravity.

For a rider on a loop-the-loop, it can be the normal force from the track. Identifying which real force is doing the job is the crucial first step in any circular-motion problem. Remove that force - cut the string, hit an icy patch - and the object does not fly outward; it flies off along a straight line tangent to the circle, exactly as Newton's first law predicts, because with no inward force there is nothing to bend its path.

The "centrifugal" feeling

When a car turns sharply you feel flung toward the outside of the curve, which tempts you to believe an outward force is acting. There is no such force on you. What you feel is your own inertia: your body tends to continue in a straight line, so the car door must push inward on you to force you around the curve, and you experience that inward push from the door as if you were being pressed outward against it. The outward "centrifugal force" is an illusion of being in an accelerating (rotating) reference frame, not a real force in the fixed frame we use for these problems.

Worked example: a car rounding a curve

Given: a car takes a curve of radius r = 45 m at v = 15 m/s. Find: the centripetal acceleration.
Solution: a_c = v squared / r = 15 squared / 45 = 225 / 45 = 5.0 m/s squared, directed toward the center of the curve. The static friction between the tires and road must supply the force that produces this acceleration. If the required friction ever exceeds what the tires can provide (mu_s N), the car slides outward off the curve - which is why you must slow down for a tight or icy bend.

Worked example: a puck on a string

Given: a 0.50 kg puck slides on a frictionless horizontal table, tethered by a string to a pin at the center so that it travels a circle of radius 1.5 m at v = 6.0 m/s. Find: the tension in the string.
Solution: The table's normal force supports the puck's weight, so those two vertical forces cancel and the string tension is purely horizontal - the only force with a component toward the center. The centripetal acceleration is a_c = 6.0 squared / 1.5 = 36 / 1.5 = 24 m/s squared.

The tension provides the centripetal force: F_c = m a_c = 0.50 x 24 = 12 N. The string must pull with 12 N toward the center to keep the puck circling. If you sped the puck up, the required tension would climb with the square of the speed, which is why a string eventually snaps if you spin something too quickly.

The horizontal table matters more than it looks. Whirl a ball on a string in the air and the string cannot be horizontal, because then nothing would hold the ball up against gravity. The string must slope upward to the hand, so its tension has a vertical component supporting the weight and only its horizontal component supplies the centripetal force. That arrangement is a conical pendulum, and there the tension satisfies T sin(theta) = m v^2 / r and T cos(theta) = m g together, making T larger than the simple m v^2 / r above. Whenever a problem says "horizontal circle," check whether something other than the string is carrying the weight.

Circular motion in terms of period

Speed is often the hardest quantity to measure on a rotating object; the period T, the time for one revolution, is usually the easiest. Because the object covers one circumference per period, v = 2 pi r / T, and substituting into a_c = v^2 / r gives

a_c = 4 pi^2 r / T^2.

This form is the workhorse for orbits, centrifuges, and rotating machinery. It also carries a warning about intuition: for a fixed period, the centripetal acceleration grows in direct proportion to the radius, so an outer point on a spinning disc experiences more acceleration than an inner one even though both take the same time to go round. For a fixed speed, by contrast, a larger radius means less acceleration. The two statements sound contradictory and are both correct - which is exactly why you must be clear about what is being held constant.

Key idea: Centripetal acceleration is v^2/r at fixed speed and 4 pi^2 r / T^2 at fixed period, and the radius enters those two with opposite effect.

Worked example: the fastest safe speed around a flat curve

Given: a car rounds an unbanked curve of radius r = 60 m. The static friction coefficient between tires and dry road is mu_s = 0.70. Take g = 9.8 m/s^2.
Find: (a) the maximum speed at which the car can hold the curve, and (b) the same speed on a wet road where mu_s falls to 0.35.

Solution: the only horizontal force available is static friction from the road, and it must supply the entire centripetal force. The car is on the verge of skidding when friction reaches its ceiling.
mu_s m g = m v^2 / r
The mass cancels from both sides, leaving v = sqrt(mu_s g r).

(a) v = sqrt(0.70 x 9.8 x 60) = sqrt(411.6) = 20.3 m/s, about 73 km/h.
(b) v = sqrt(0.35 x 9.8 x 60) = sqrt(205.8) = 14.3 m/s, about 52 km/h.

Sanity check: halving mu_s halves v^2, so the safe speed should fall by a factor of sqrt(2) = 1.414. Indeed 20.3 / 1.414 = 14.4 m/s, matching part (b). Note what cancelled: the mass. A loaded truck and a small car skid off the same curve at the same speed, provided their tires have the same mu_s - heavier vehicles get more friction but need proportionally more centripetal force, and the two effects cancel exactly. The magnitudes are believable too: 73 km/h on a fairly tight 60 m bend is about what a road sign would advise, and the wet-weather figure of 52 km/h explains why those signs are set conservatively. Note also that the friction here is static, because the tires roll rather than slide; once the car does skid, mu_k takes over and it is lower still, which is why a skid tends to worsen.

Worked example: a ball at the top of a vertical circle

Given: a 0.30 kg ball is whirled in a vertical circle of radius r = 1.2 m on a light string. Take g = 9.8 m/s^2.
Find: (a) the minimum speed at the top for the string to stay taut, and (b) the string tension at the top when the ball passes through at 5.0 m/s.

Solution: at the very top, both the tension and the weight point downward, which is toward the center of the circle. So both contribute to the centripetal force:
T + m g = m v^2 / r.

(a) A string can pull but never push, so the limiting case is T = 0, where gravity alone supplies the whole centripetal force:
m g = m v^2 / r, so v_min = sqrt(g r) = sqrt(9.8 x 1.2) = sqrt(11.76) = 3.43 m/s.
Below this the string goes slack and the ball leaves its circular path.

(b) At 5.0 m/s, rearrange for the tension:
T = m v^2 / r - m g = (0.30 x 25) / 1.2 - (0.30 x 9.8) = 6.25 - 2.94 = 3.31 N.

Sanity check: the answer must be positive, since a taut string cannot pull the wrong way, and 3.31 N is. It is also consistent with part (a): 5.0 m/s comfortably exceeds the 3.43 m/s minimum, so a positive tension is exactly what we should expect, and had we been asked about 3.0 m/s the formula would have returned a negative tension - the signal that the situation is impossible rather than an answer. Compare the bottom of the circle, where the weight points away from the center and the equation flips to T - m g = m v^2 / r, giving T = 6.25 + 2.94 = 9.19 N at the same speed. The string works nearly three times harder at the bottom, which is why ropes and swings break at the low point of a swing, not the high point.

Common wrong turns

• Adding a centrifugal force to the free-body diagram. In a ground-based frame there is no outward force. Only real, inward-pointing forces belong on the diagram.
• Treating centripetal force as a separate force to add. It is a role filled by tension, gravity, friction, or a normal force. Adding it alongside those double-counts.
• Forgetting to square the speed. a_c = v^2/r, so doubling the speed quadruples the requirement. Writing v/r understates it badly.
• Assuming the string in a horizontal circle is horizontal. Unless a surface carries the weight, the string must slope and only its horizontal component is centripetal.
• Using mu_k for a rolling tire. A tire that grips is in static contact, so the ceiling is mu_s N.
• Thinking an object flies radially outward when the force is removed. It flies off along the tangent, in the direction it was already moving.

Recap

• Uniform circular motion has constant speed but continuously changing velocity, so the object is always accelerating.
• The centripetal acceleration is v^2/r, or equivalently 4 pi^2 r / T^2, always directed toward the center.
• Centripetal force is a role, not a new force; identify which real force is pointing inward before writing any equation.
• On a flat curve the limiting speed is sqrt(mu_s g r), independent of the vehicle's mass.
• At the top of a vertical circle, weight and tension both point inward, so the minimum speed for a taut string is sqrt(g r).
• The outward "centrifugal" sensation is inertia observed from a rotating frame, not a force acting on the body.

Sources

1. OpenStax. (2016). 4.4 Uniform and nonuniform circular motion. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 6.3 Centripetal force. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2022). 6.2 Centripetal acceleration. In College Physics 2e. Rice University. openstax.org
4. OpenStax. (2022). 6.3 Centripetal force. In College Physics 2e. Rice University. openstax.org
5. OpenStax. (2022). 6.4 Fictitious forces and non-inertial frames: The Coriolis force. In College Physics 2e. Rice University. openstax.org
6. Nave, R. (n.d.). Centripetal force. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu
7. Nave, R. (n.d.). Circular motion and rotation. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu

Before 1687, the heavens and the Earth were understood to run on different rules: things fell down here, and things circled up there, and no one had a reason to connect the two. The single most consequential idea in the history of physics is that they are the same phenomenon. The apple and the Moon obey one equation, and that equation is short enough to fit on a line.

Newton's great insight was that the same force pulling an apple to the ground also holds the Moon in orbit - that terrestrial and celestial motion obey one law. His law of universal gravitation states that every pair of masses attracts each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers:

F = G m1 m2 / r squared

Here m1 and m2 are the two masses, r is the distance between their centers, and G is the gravitational constant, 6.674 x 10^-11 in SI units (newton meters squared per kilogram squared). The force is always attractive and acts along the line joining the two masses, with each mass pulling on the other equally in accordance with Newton's third law.

Because G is such a tiny number, gravity between everyday objects is utterly negligible - two people standing side by side attract each other with a force far too small to feel. Gravity becomes significant only when at least one of the masses is astronomically large, like a planet or star.

The inverse-square law

The 1 / r squared dependence is the heart of the law and deserves careful attention. Double the distance and the force drops to a quarter; triple it and the force drops to a ninth; increase it tenfold and the force falls to a hundredth. This rapid falloff is why gravity weakens quickly with altitude yet, because 1/r squared never actually reaches zero, extends its reach across the entire universe, binding galaxies together over unimaginable distances. The inverse-square pattern is not arbitrary; it reflects the way influence spreads out over the surface of an expanding sphere, whose area grows as r squared.

Key idea: Gravitational force is proportional to the product of the masses and inversely proportional to the square of the separation, and it is always attractive.

Why r is measured from the center

A detail that quietly does a great deal of work: the law as written applies to point masses, yet we happily use it for planets that are thousands of kilometers across. The justification is Newton's shell theorem, which he had to invent calculus to prove. It states that a uniform spherical shell attracts an outside object exactly as though all its mass were concentrated at its center. A solid sphere is just a stack of such shells, so Earth pulls on you as if its entire 5.97 x 10^24 kg sat at a single point 6371 km beneath your feet.

This is why r is measured center to center, never surface to surface. Standing on the ground, you are 6371 km from Earth's "center of gravity," not zero meters from it, and forgetting this is one of the most common errors in gravitation problems. A satellite orbiting at 400 km altitude has r = 6371 + 400 = 6771 km, not 400 km - a distinction that changes the answer by a factor of nearly 300.

The theorem carries a second, more surprising half: inside a uniform shell the gravitational field is exactly zero everywhere, not merely at the center. All the pulls cancel. That is why gravity weakens as you descend into a mine rather than intensifying, and why an object at Earth's exact center would feel no gravitational force at all.

Key idea: By the shell theorem a spherical body pulls as if all its mass sat at its center, so r is always measured center to center.

Weight is gravitation

An object's weight is simply the gravitational force the Earth exerts on it - the two are the same phenomenon, not merely related. Setting m g = G M_Earth m / R_Earth squared and canceling the object's mass m shows that g = G M_Earth / R_Earth squared.

Plugging in Earth's mass (5.97 x 10^24 kg) and radius (6.37 x 10^6 m) yields about 9.8 m/s squared, exactly the free-fall value used throughout Module 2. This is a beautiful moment of unification: the constant g you have been using since the second module is a direct consequence of Newton's gravitation applied to the specific mass and size of Earth. On the Moon, with far less mass, the same formula gives a surface gravity of only about 1.6 m/s squared, which is why astronauts bounded so easily across it.

Gravity as the centripetal force for orbits

Combining this lesson with the last one explains orbits completely. A satellite or moon travels in a near-circular path because gravity supplies exactly the centripetal force needed to bend its motion into a circle. Setting the gravitational force equal to the centripetal force, G M m / r squared = m v squared / r, and canceling the orbiting mass, gives the orbital speed v = sqrt(G M / r).

Everything in orbit is, in truth, continuously falling toward Earth - it simply moves sideways fast enough that it keeps missing, curving around the planet instead of hitting it. Newton himself pictured this with a cannon on a mountaintop firing ever faster until the ball's fall matched Earth's curvature and it circled the globe.

Worked example: comparing gravity to weight

Given: a satellite orbits at an altitude equal to one Earth radius, so its distance from Earth's center is 2 R_Earth. Find: how its gravitational acceleration compares to surface gravity.
Solution: Because gravity follows 1 / r squared, doubling the distance divides the acceleration by 2 squared = 4.

So the acceleration is 9.8 / 4 = 2.45 m/s squared, about one quarter of the surface value. The satellite still very much feels gravity - that inward pull is precisely the centripetal force keeping it in orbit. Astronauts aboard such a satellite float not because gravity is absent but because they and their spacecraft fall together, an idea explored in the assignment.

Worked example: surface gravity on the Moon

Given: the Moon has mass M = 7.35 x 10^22 kg and radius R = 1.74 x 10^6 m. Use G = 6.674 x 10^-11 N m^2/kg^2.
Find: (a) the surface gravity on the Moon, and (b) the weight there of an astronaut whose mass, including the suit, is 120 kg.

Solution (a): set the gravitational force equal to m g and cancel the small mass:
g = G M / R^2
Numerator: 6.674 x 10^-11 x 7.35 x 10^22 = 4.905 x 10^12
Denominator: (1.74 x 10^6)^2 = 3.028 x 10^12
g = 4.905 x 10^12 / 3.028 x 10^12 = 1.62 m/s^2.

Solution (b): W = m g = 120 x 1.62 = 194 N, against 120 x 9.81 = 1177 N on Earth.

Sanity check: the ratio 9.81 / 1.62 = 6.05 reproduces the familiar statement that lunar gravity is about one sixth of Earth's, which is a strong independent confirmation. The exponent bookkeeping also checks out: 10^-11 x 10^22 = 10^11, divided by 10^12, gives 10^-1, and 4.905 x 10^11 / 3.028 x 10^12 is indeed of order 1, not 100 or 0.01. Physically the answer makes sense: the Moon has about 1.2 percent of Earth's mass but only 27 percent of its radius, and since g scales as M/R^2 the small radius partly compensates for the small mass. The astronaut's mass is still 120 kg on the Moon - only the weight changed - which is why moving a heavy object there is still hard to start and hard to stop, even though lifting it is easy.

Worked example: the speed and period of the space station

Given: the International Space Station orbits at an altitude of about 420 km. Earth's mass is M = 5.97 x 10^24 kg and its radius 6371 km.
Find: (a) the orbital radius, (b) the orbital speed, (c) the orbital period, and (d) the local value of g at that altitude.

Solution (a): radii are measured from Earth's center, so
r = 6371 + 420 = 6791 km = 6.791 x 10^6 m.

Solution (b): gravity supplies the centripetal force, G M m / r^2 = m v^2 / r. The station's own mass cancels, leaving v = sqrt(G M / r).
G M = 6.674 x 10^-11 x 5.97 x 10^24 = 3.984 x 10^14
G M / r = 3.984 x 10^14 / 6.791 x 10^6 = 5.867 x 10^7
v = sqrt(5.867 x 10^7) = 7.66 x 10^3 m/s, about 7.66 km/s.

Solution (c): T = 2 pi r / v = (6.2832 x 6.791 x 10^6) / 7660 = 4.267 x 10^7 / 7660 = 5571 s, which is 5571 / 60 = 92.8 minutes.

Solution (d): g = G M / r^2 = 3.984 x 10^14 / (6.791 x 10^6)^2 = 3.984 x 10^14 / 4.612 x 10^13 = 8.64 m/s^2.

Sanity check: the published orbital period of the ISS is about 93 minutes, so the calculation lands almost exactly on the real value - the strongest possible confirmation. Part (d) settles a persistent misconception: at station altitude, gravity is 8.64 / 9.81 = 88 percent of its surface strength, nowhere near zero. Astronauts float because they and the station are in free fall together, not because gravity has switched off. The direction makes sense too: gravity points inward, which is exactly the direction a centripetal force must point. As a final check on magnitude, 7.66 km/s is about 27,600 km/h, roughly twenty-two times the speed of sound, which is why reaching orbit is so much harder than reaching altitude.

Common wrong turns

• Using altitude instead of orbital radius. r is measured from the planet's center. For a 420 km orbit, r is 6791 km, not 420 km.
• Forgetting to square the distance. Tripling the separation divides the force by nine, not by three.
• Believing gravity vanishes in orbit. At station altitude it is still 88 percent of its surface value; weightlessness comes from free fall, not from absent gravity.
• Concluding that a heavier satellite needs a different orbital speed. The orbiting mass cancels, so v = sqrt(G M / r) depends only on the central body and the radius.
• Confusing G with g. G is a universal constant of 6.674 x 10^-11; g is a local acceleration of about 9.8 m/s^2 that depends entirely on which planet you stand on.
• Thinking gravity increases as you descend into a mine. By the shell theorem the mass above you contributes nothing, so g actually falls toward zero at the center.

Recap

• Every pair of masses attracts with F = G m1 m2 / r^2, always along the line joining them and always attractive.
• G = 6.674 x 10^-11 N m^2/kg^2 is tiny, so gravity matters only when at least one mass is astronomical.
• The shell theorem lets a spherical body be treated as a point at its center, so r is measured center to center.
• Surface gravity is g = G M / R^2, which is why the same free-fall acceleration applies to every object on a given planet.
• For a circular orbit, gravity is the centripetal force, giving v = sqrt(G M / r) independent of the orbiting mass.
• Orbiting objects are in continuous free fall, which is the real reason astronauts appear weightless.

Sources

1. OpenStax. (2016). 13.1 Newton's law of universal gravitation. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 13.2 Gravitation near Earth's surface. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 13.4 Satellite orbits and energy. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2022). 6.5 Newton's universal law of gravitation. In College Physics 2e. Rice University. openstax.org
5. National Institute of Standards and Technology. (n.d.). CODATA value: Newtonian constant of gravitation. The NIST Reference on Constants, Units, and Uncertainty. physics.nist.gov
6. Nave, R. (n.d.). Gravity concepts and applications. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu
7. PhET Interactive Simulations. (n.d.). Gravity and orbits [Simulation]. University of Colorado Boulder. phet.colorado.edu

Hold a heavy suitcase at arm's length until your shoulder burns. You have done, in the physicist's sense, exactly zero work on it. Now carry that same suitcase a kilometer along a level platform: still zero. The definition about to be introduced is deliberately narrow, and its narrowness is what makes it powerful - it isolates the one thing that actually changes an object's motion.

In everyday speech, work means effort. In physics, work has a precise and sometimes surprising meaning: it is done only when a force moves an object through a distance. Specifically, W = F d cos(theta), where F is the magnitude of the force, d is the magnitude of the displacement, and theta is the angle between the force and the displacement. Work is a scalar measured in joules (J), where one joule is one newton-meter. This definition captures the idea that a force accomplishes something mechanical only to the extent that it acts along the direction of motion.

When work is zero, positive, or negative

The cos(theta) factor makes work behave in ways that clash with intuition. If you push in the direction of motion, theta = 0 and cos 0 = 1, so W = F d is maximal and positive. A force perpendicular to the motion (theta = 90 degrees) does zero work, because cos 90 = 0 - which is why carrying a heavy box across a level floor does no work on the box against gravity, since the upward carrying force is perpendicular to the horizontal motion.

And a force opposing the motion (theta = 180 degrees, cos 180 = -1) does negative work, meaning it removes energy; friction is the classic example. Positive work speeds things up or stores energy; negative work slows them down or releases it.

Note carefully what kind of quantity this is. Work is a scalar, so it has no direction and work done by different forces simply adds arithmetically - but it does carry a sign, and the sign is not optional. A crate dragged across a floor might have +900 J done on it by the rope and -480 J by friction, for a net of +420 J. There is no "direction" to that 420 J, yet the minus sign on the friction term is the whole story of why the crate does not accelerate as fast as the rope alone would suggest.

Key idea: Work is a signed scalar, W = F d cos(theta), and only the component of force along the displacement contributes.

Work done by a varying force

The formula W = F d cos(theta) assumes the force is constant over the whole displacement. Many important forces are not. The classic case is a spring, which obeys Hooke's law, F = k x: the further you stretch or compress it, the harder it resists.

For a varying force the work is the area under the force-versus-position graph. For a spring that graph is a straight line rising from zero to k x, so the area is a triangle of base x and height k x, giving

W = (1/2) k x^2.

Notice this is not k x times x. Using the maximum force over the whole distance would double the answer, because the force only reaches its maximum at the very end. Averaging correctly - the mean of 0 and k x is k x / 2 - gives the factor of one half that the graph makes obvious. The same area-under-the-graph reasoning handles any varying force you meet later.

Key idea: When a force varies, work is the area under the force-position graph, which for a spring gives (1/2) k x^2.

Worked example: pushing a crate

Given: you push a crate with F = 50 N over d = 8.0 m in the direction of the push. Find: the work done.
Solution: W = F d cos(0) = 50 x 8.0 x 1 = 400 J. If instead you pushed at 60 degrees to the motion, W = 50 x 8.0 x cos 60 = 50 x 8.0 x 0.5 = 200 J - only half as much, because only the component of force along the motion, F cos(theta), does work. The perpendicular component merely presses the crate into or out of the floor without contributing to its forward progress.

Kinetic energy

Kinetic energy is the energy an object has by virtue of its motion, given by KE = (1/2) m v squared, also measured in joules. Two features stand out. First, kinetic energy is always positive, since it depends on speed squared and mass, both positive. Second, and more importantly, it grows with the square of speed, so doubling an object's speed quadruples its kinetic energy and tripling it multiplies the energy ninefold. This quadratic dependence is why a small increase in highway speed sharply raises both stopping distance and collision severity, and why a bullet, though light, carries devastating energy at high speed.

The work-energy theorem

Work and kinetic energy are tied together by one of the most useful results in mechanics, the work-energy theorem: the net work done on an object equals its change in kinetic energy, W_net = KE_final - KE_initial. This holds regardless of the details of the forces or the path.

Its great value is that it lets you relate forces and speeds directly, bypassing acceleration and time entirely - if you know the net work done, you know the change in speed, and vice versa. Whenever a problem gives you forces and distances and asks about speeds (or the reverse), the work-energy theorem is often the fastest route to the answer.

Worked example: speeding up a cart

Given: a 1500 kg cart starts from rest and a net work of 30000 J is done on it. Find: its final speed.
Solution: By the work-energy theorem, W_net = (1/2) m v squared - 0, so 30000 = (1/2)(1500) v squared = 750 v squared. Then v squared = 40 and v = 6.3 m/s. The cart ends up moving at about 6.3 m/s. Notice we never needed the force, the acceleration, or the time - only the total work, which is the whole point of the energy method.

Power: the rate of doing work

Two engines can do the same total work but differ greatly in how fast they do it. Power is the rate at which work is done, P = W / t, measured in watts (W), where one watt is one joule per second. A powerful motor does a given amount of work in less time. Power also equals force times velocity (P = F v) for a force acting along the motion, which is why a car needs far more engine power to sustain high speed against air resistance than to cruise slowly. This is why engines and appliances are rated in watts or kilowatts.

Worked example: every force's work on a dragged crate

Given: a 25 kg crate is dragged 12 m across a level floor by a rope pulling with 90 N at 30 degrees above the horizontal. The kinetic friction coefficient is mu_k = 0.20, and the crate starts from rest. Take g = 9.8 m/s^2.
Find: the work done by each force, the net work, and the crate's final speed.

Solution: start with the normal force, because friction depends on it and the rope has an upward component.
Fy = 90 sin(30) = 45.0 N upward, so N = m g - Fy = 245 - 45.0 = 200 N.
f_k = mu_k N = 0.20 x 200 = 40.0 N, opposing the motion.

Now take each force in turn.
W_rope = F d cos(30) = 90 x 12 x 0.8660 = +935 J
W_friction = f_k d cos(180) = 40.0 x 12 x (-1) = -480 J
W_gravity = 0 and W_normal = 0, because both are perpendicular to the horizontal displacement.
W_net = 935 - 480 + 0 + 0 = +455 J.

By the work-energy theorem, starting from rest:
455 = (1/2)(25) v^2 = 12.5 v^2, so v^2 = 36.4 and v = 6.04 m/s.

Sanity check: get the same answer through forces and kinematics. The net horizontal force is 90 cos(30) - 40.0 = 77.94 - 40.0 = 37.9 N, so a = 37.9/25 = 1.52 m/s^2, and v^2 = 2 a d = 2 x 1.52 x 12 = 36.4, giving v = 6.04 m/s. The two routes agree exactly, as they must. Now check the trap: had we written N = m g = 245 N by reflex, friction would have come out at 49.0 N instead of 40.0 N, the net work at 347 J, and the speed at 5.27 m/s - about 13 percent too slow. The upward tilt of the rope genuinely lightens the crate on the floor, which is precisely why you pull a heavy load rather than push it. Note finally that the normal force and gravity did zero work despite both being large; being perpendicular to the motion, they change the crate's direction of contact but not its speed.

Worked example: a spring launcher and its power

Given: a spring of stiffness k = 320 N/m is compressed x = 0.15 m and then released, pushing a 0.40 kg block along a frictionless horizontal track. The spring returns to its natural length in 0.060 s.
Find: (a) the work the spring does on the block, (b) the block's launch speed, and (c) the average power delivered.

Solution (a): the spring force varies from k x = 320 x 0.15 = 48 N down to zero, so use the area under the graph:
W = (1/2) k x^2 = 0.5 x 320 x (0.15)^2 = 0.5 x 320 x 0.0225 = 3.6 J.

Solution (b): the track is frictionless and horizontal, so this is the entire net work.
3.6 = (1/2)(0.40) v^2 = 0.20 v^2, so v^2 = 18.0 and v = 4.24 m/s.

Solution (c): P = W / t = 3.6 / 0.060 = 60 W.

Sanity check: test the magnitude of the work against something familiar. 3.6 J would lift the same 0.40 kg block a height of 3.6 / (0.40 x 9.8) = 0.92 m, which is a plausible amount of stored energy for a hand-compressed spring - not the 3600 J a car engine might deliver, nor a millijoule. The launch speed of 4.24 m/s is a brisk walking-to-jogging pace, again believable for a toy launcher. Check the factor of one half by the wrong route: using the peak force over the full compression would give 48 x 0.15 = 7.2 J, exactly double, because the spring only reaches 48 N at the instant of release. The 60 W figure is the power of a bright light bulb, delivered for just six hundredths of a second - which is the whole point of the distinction between work and power.

Common wrong turns

• Believing effort implies work. Holding a weight still, or carrying it horizontally, does zero work on it however tiring it feels.
• Dropping the sign on friction's work. Friction removes energy, so its work is negative and must be subtracted, not added.
• Using k x times x for spring work. The force rises from zero, so the work is (1/2) k x^2 - half what the peak force would suggest.
• Applying the work-energy theorem with only one force's work. The theorem uses net work, the sum over every force acting.
• Forgetting that kinetic energy scales with v^2. Tripling the speed multiplies the energy ninefold, not threefold.
• Using N = m g when a rope pulls at an angle. The vertical component changes the normal force and therefore the friction and its work.

Recap

• Work is F d cos(theta), a signed scalar measured in joules; only the force component along the displacement counts.
• A perpendicular force does no work, which is why normal forces and circular-motion tensions never change speed.
• Kinetic energy is (1/2) m v^2, always positive and quadratic in speed.
• The work-energy theorem says net work equals the change in kinetic energy, bypassing time and acceleration entirely.
• For a varying force, work is the area under the force-position graph, giving (1/2) k x^2 for a spring.
• Power is the rate of doing work, P = W/t or P = F v, measured in watts.

Sources

1. OpenStax. (2016). 7.1 Work. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 7.2 Kinetic energy. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 7.3 Work-energy theorem. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2016). 7.4 Power. In University Physics Volume 1. Rice University. openstax.org
5. OpenStax. (2022). 7.1 Work: The scientific definition. In College Physics 2e. Rice University. openstax.org
6. OpenStax. (2022). 7.2 Kinetic energy and the work-energy theorem. In College Physics 2e. Rice University. openstax.org
7. Nave, R. (n.d.). Work, energy and power. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu

A roller-coaster car released from the top of a 40 m hill arrives at the bottom at a very specific speed, and you can compute it in one line without knowing anything about the shape of the track, the mass of the car, or how long the journey took. Force analysis would need every one of those. This is the single biggest labor-saving idea in mechanics, and it comes with one condition that you must learn to check.

Potential energy is stored energy that depends on an object's position or configuration rather than its motion. It represents work that has been done to place an object somewhere and that can be recovered later as motion. The most familiar form is gravitational potential energy, the energy an object has because of its height: PE = m g h, where h is the height above a chosen reference level.

Lift an object and you do work against gravity that is stored as potential energy; release it and that stored energy is converted back into motion as the object falls. A second common form is elastic potential energy stored in a stretched or compressed spring: PE = (1/2) k x squared, where k is the spring's stiffness (its "spring constant") and x is the distance it is stretched or compressed from its natural length.

The reference level is your choice

An important subtlety: only changes in gravitational potential energy have physical meaning, so you are free to place the h = 0 reference wherever is convenient - the floor, a tabletop, the ground below a cliff. A book on a shelf has a different potential energy depending on whether you measure height from the desk or the floor, but the change in potential energy as it falls a given distance is the same either way. Choosing a sensible reference level, usually the lowest point in the problem, keeps the numbers simple.

Conservation of mechanical energy

The mechanical energy of a system is the sum of its kinetic and potential energy: E = KE + PE. Here is the powerful result: when only conservative forces like gravity and ideal spring forces act, and no energy is lost to friction, this total remains constant. This is the conservation of mechanical energy, written KE_initial + PE_initial = KE_final + PE_final. Energy is never created or destroyed; it merely transforms back and forth between kinetic and potential forms.

A conservative force is one for which the work done depends only on the start and end points, not on the path taken - lift a book straight up or along a winding ramp to the same shelf and gravity does the same work either way. Friction is not conservative, because the energy it removes as heat depends on the path length and cannot be recovered.

When mechanical energy is conserved, and when it is not

Because KE_i + PE_i = KE_f + PE_f is so convenient, it gets applied where it does not belong more often than any other equation in this course. The condition is precise: mechanical energy is conserved only when no non-conservative force does work on the system. In practice that means no friction, no air resistance, no collision that deforms anything, and no external agent pushing or pulling along the motion.

So it does apply to a ball in free fall or on a frictionless track, a pendulum swinging on a light string over a short time, a block launched by an ideal spring on a smooth surface, and a satellite in orbit. It does not apply to a crate sliding across a rough floor, a parachutist falling at terminal velocity, a car braking, or any real collision in which objects crumple or stick.

Crucially, failure of mechanical-energy conservation does not mean energy vanished. Total energy is conserved absolutely; the mechanical part simply leaked into thermal energy, sound, or permanent deformation. When friction is present, the bookkeeping becomes

KE_i + PE_i + W_nc = KE_f + PE_f,

where W_nc is the work done by non-conservative forces - negative for friction, since it removes mechanical energy. For sliding friction over a path length d, that term is simply -f_k d, and note that it depends on the path length, which is exactly why no potential-energy function can be written for friction.

Key idea: Mechanical energy is conserved only when no non-conservative force does work; otherwise add W_nc = -f_k d to the ledger, and total energy still balances.

Worked example: a falling object

Given: a ball is dropped from rest at height h = 10 m. Take g = 9.8. Find: its speed just before hitting the ground.
Solution: At the top, all the energy is potential: PE = m g h, and KE = 0 since it starts from rest. At the bottom, all of that energy has become kinetic: KE = (1/2) m v squared, and PE = 0 at the reference level.

Setting initial energy equal to final energy, m g h = (1/2) m v squared. The mass cancels, giving v squared = 2 g h = 2 x 9.8 x 10 = 196, so v = 14 m/s. The ball lands at 14 m/s - the same answer kinematics gives, but reached with no time calculation at all.

Worked example: a roller coaster

Given: a coaster car starts from rest at the top of a 20 m frictionless hill. Take g = 9.8. Find: its speed at the bottom.
Solution: Again m g h = (1/2) m v squared, so v squared = 2 g h = 2 x 9.8 x 20 = 392 and v = 19.8 m/s.

Notice two things: the mass never mattered, and neither did the shape of the track - only the total height dropped. A car descending a straight ramp, a gentle curve, or a wild corkscrew all arrive at the bottom with the same speed, provided friction is negligible. That is the elegance and the power of the energy method: it cuts straight through geometric complexity that would defeat a force-by-force analysis.

Where does the energy go with friction?

In real systems friction and air resistance drain mechanical energy, so a real roller coaster arrives a bit slower than the ideal calculation predicts. But energy is still conserved overall - the "lost" mechanical energy has simply been converted into thermal energy, warming the track, the wheels, and the air. This broader conservation of energy, counting heat and every other form, is one of the deepest and most universal laws in all of physics: the total energy of an isolated system is absolutely constant, no matter what transformations occur within it. Mechanical energy conservation is the special case that holds when no energy leaks into thermal or other forms.

Worked example: down a smooth ramp, across a rough floor

Given: a 2.0 kg block is released from rest at a height of 1.8 m on a curved frictionless ramp. At the bottom it slides onto a horizontal floor where mu_k = 0.30. Take g = 9.8 m/s^2.
Find: (a) its speed at the bottom of the ramp, and (b) how far it travels along the floor before stopping.

Solution (a): the ramp is frictionless, so mechanical energy is conserved on that stretch. Take the floor as the reference level.
m g h = (1/2) m v^2
The mass cancels: v = sqrt(2 g h) = sqrt(2 x 9.8 x 1.8) = sqrt(35.28) = 5.94 m/s.

Solution (b): on the floor, friction is non-conservative, so mechanical energy is not conserved. Use the extended ledger instead. All the kinetic energy at the bottom is removed by friction over the sliding distance d.
Kinetic energy at the bottom: KE = m g h = 2.0 x 9.8 x 1.8 = 35.28 J
Friction force on the level floor: f_k = mu_k m g = 0.30 x 2.0 x 9.8 = 5.88 N
35.28 = 5.88 d, so d = 6.0 m.

Sanity check: notice that the mass cancels from the whole calculation. Setting m g h = mu_k m g d gives d = h / mu_k = 1.8 / 0.30 = 6.0 m, confirming the arithmetic and revealing that a boulder and a pebble released from the same height slide the same distance on the same surface. Is 6.0 m sensible? A drop of 1.8 m on a rough-ish floor carrying an object six meters feels about right - and the formula behaves correctly at its limits, since a frictionless floor (mu_k = 0) would give an infinite distance and a very grippy floor a very short one. The direction of the energy flow is right too: 35.28 J of gravitational potential energy became 35.28 J of kinetic energy and then 35.28 J of thermal energy warming the floor and the block. Nothing was destroyed; it merely stopped being mechanical.

Worked example: a spring launcher up an incline

Given: a spring with k = 500 N/m is compressed 0.20 m and released, launching a 1.5 kg block horizontally and then up a 25 degree incline. Take g = 9.8 m/s^2.
Find: (a) the distance the block travels up a frictionless incline, and (b) the same distance if the incline instead has mu_k = 0.15.

Solution (a): the stored spring energy becomes gravitational potential energy at the highest point, where the block is momentarily at rest.
PE_spring = (1/2) k x^2 = 0.5 x 500 x (0.20)^2 = 0.5 x 500 x 0.04 = 10.0 J
Travelling a distance d along the slope raises the block by h = d sin(25), so
10.0 = m g d sin(25) = 1.5 x 9.8 x d x 0.4226 = 6.213 d
d = 10.0 / 6.213 = 1.61 m along the incline, a vertical rise of 1.61 x 0.4226 = 0.680 m.

Solution (b): with friction, add the non-conservative term. On the incline the normal force is m g cos(25), so f_k = mu_k m g cos(25) and the energy lost over distance d is f_k d.
10.0 = m g d sin(25) + mu_k m g cos(25) d = m g d [sin(25) + mu_k cos(25)]
10.0 = 14.7 d [0.4226 + 0.15 x 0.9063] = 14.7 d [0.4226 + 0.1359] = 14.7 d (0.5586) = 8.211 d
d = 10.0 / 8.211 = 1.22 m.

Sanity check: the rough incline must give a shorter distance than the smooth one, and 1.22 m is duly less than 1.61 m, by about 24 percent. Verify part (a) independently: 10.0 J should lift 1.5 kg through 10.0 / (1.5 x 9.8) = 0.680 m, exactly the vertical rise computed. Notice what falls out of the frictionless case - the height reached does not depend on the incline angle at all, only the distance along the slope does. That stops being true once friction enters, because a shallower ramp also means a longer path over which friction acts, so the block ends up lower as well as farther.

Common wrong turns

• Using energy conservation when friction acts. Check for non-conservative forces first; if any do work, add W_nc to the ledger.
• Saying friction "destroys" energy. It converts mechanical energy into thermal energy. Total energy is always conserved.
• Forgetting that gravitational potential energy depends on the reference level. Only changes matter, so pick one zero level and use it consistently for every term.
• Writing spring energy as k x or k x^2. It is (1/2) k x^2, and x is measured from the spring's natural length.
• Assuming a heavier object arrives faster. The mass cancels from v = sqrt(2 g h), and it also cancels from the sliding-distance result d = h / mu_k.
• Using the vertical height where the slope distance belongs, or the reverse. Gravitational potential energy uses the vertical rise; friction on a ramp uses the distance along the ramp.

Recap

• Potential energy is stored by position or configuration: m g h for gravity, (1/2) k x^2 for a spring.
• Only changes in gravitational potential energy are physical, so the h = 0 reference is yours to choose.
• Mechanical energy KE + PE is conserved when only conservative forces do work.
• When friction or drag acts, use KE_i + PE_i + W_nc = KE_f + PE_f with W_nc = -f_k d.
• Conservative forces are path-independent, which is exactly what allows a potential energy function to exist for them.
• Energy methods bypass time, acceleration, and track shape, which is why they defeat problems that force analysis cannot reach.

Sources

1. OpenStax. (2016). 8.1 Potential energy of a system. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 8.2 Conservative and non-conservative forces. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 8.3 Conservation of energy. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2022). 7.3 Gravitational potential energy. In College Physics 2e. Rice University. openstax.org
5. OpenStax. (2022). 7.5 Nonconservative forces. In College Physics 2e. Rice University. openstax.org
6. Nave, R. (n.d.). Elastic potential energy. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu
7. PhET Interactive Simulations. (n.d.). Energy skate park [Simulation]. University of Colorado Boulder. phet.colorado.edu

An egg dropped onto concrete breaks; the same egg dropped from the same height onto a pillow does not. Gravity gave both eggs exactly the same momentum to lose. The difference is entirely how long the pillow took to remove it. That single sentence is the design principle behind airbags, crumple zones, helmets, crash mats, and the reason a boxer rolls with a punch.

Linear momentum measures how much motion an object carries, combining its mass and velocity into a single vector: p = m v. It points in the direction of the velocity and is measured in kilogram meters per second. Momentum captures something that neither mass nor velocity alone conveys: a slow-moving loaded truck and a fast-flying bullet can carry comparable momentum, because momentum weighs mass and speed together. As you will see in the next lesson, momentum is conserved in interactions even when kinetic energy is not, which makes it in some ways an even more fundamental quantity for analyzing collisions.

Momentum is a vector

Because momentum is a vector, direction matters and signs count. Two objects of equal mass moving toward each other at the same speed have momenta that are equal in size but opposite in sign, so their total momentum is zero even though each is moving. This vector nature is exactly what makes momentum so useful for collisions, where objects approach from different directions and the bookkeeping of pluses and minuses does the work of tracking who ends up going which way.

The vector nature has one consequence that catches almost everyone the first time. When an object reverses direction, the momentum change is not the difference of the speeds but their sum, because the initial and final momenta have opposite signs. A ball that arrives at 25 m/s and rebounds at 20 m/s has changed its velocity by 45 m/s, not by 5 m/s. This is why a bouncy collision delivers a far larger impulse than a dead-stop one, and why a hailstone that bounces off a roof does more damage than a raindrop of equal mass that splatters.

Key idea: Momentum is a vector, so a reversal makes the initial and final contributions add rather than subtract - a bounce delivers a bigger impulse than a stop.

Momentum and kinetic energy measure different things

Both quantities grow with mass and speed, so it is tempting to treat them as two versions of the same idea. They are not, and the distinction settles many otherwise confusing questions.

Momentum is linear in speed and is a vector; kinetic energy is quadratic in speed and is a scalar. Combining p = m v and KE = (1/2) m v^2 gives the useful relation KE = p^2 / (2m). Read that carefully: for a fixed momentum, kinetic energy is inversely proportional to mass. Two objects carrying identical momentum therefore carry wildly different energies if their masses differ - the lighter, faster one has far more. A 2000 kg vehicle at 6.0 m/s and a 60 kg motorcyclist at 200 m/s both carry 12000 kg m/s, but the rider carries about 33 times the kinetic energy.

The practical upshot arrives in the next lesson: momentum is conserved in every collision without exception, while kinetic energy is conserved only in the special elastic case. Knowing which quantity survives an interaction is what tells you which equation you are allowed to write.

Key idea: Since KE = p^2/(2m), two objects with equal momentum have very unequal kinetic energies unless their masses match.

Impulse

To change an object's momentum you must apply a force over a period of time. The product of the force and the time interval over which it acts is called impulse: J = F t. The impulse-momentum theorem states that the impulse delivered to an object equals its change in momentum: F t = delta p = m v_final - m v_initial.

This is really just Newton's second law rearranged, since F = m a = m (delta v / delta t) gives F delta t = m delta v. The theorem is the workhorse for any situation involving forces acting over short times, especially impacts, where the force is large and brief and hard to measure directly but the change in momentum is easy to compute.

Worked example: catching a baseball

Given: a 0.145 kg baseball arrives at 40 m/s and is brought to rest by a catcher in 0.020 s. Find: the average force on the ball.
Solution: The change in momentum is delta p = m v_final - m v_initial = 0.145 x 0 - 0.145 x 40 = -5.8 kg m/s.

The average force is F = delta p / t = -5.8 / 0.020 = -290 N. The magnitude is 290 N, and the negative sign shows the force opposes the ball's motion. That is a substantial force - about the weight of a 30 kg mass - delivered in a fraction of a second, which is why catching a fast ball stings.

Why time matters for safety

The impulse-momentum theorem explains an enormous amount of safety engineering, and understanding it changes how you see the world. Rearrange it to F = delta p / t: for a fixed change in momentum - say, bringing a moving body to rest - the force is inversely proportional to the stopping time. Stretch the collision out over a longer time and the peak force drops proportionally.

This single insight is the design principle behind airbags, which inflate to extend the time over which your head decelerates; automobile crumple zones, which crush progressively to lengthen the crash; padded dashboards and helmets; the netting under a trapeze artist; and the instinct to bend your knees when you land from a jump. Catching a raw egg by drawing your hands back as it arrives is the very same physics as a car's crumple zone, scaled down. In every case the change in momentum is fixed by the situation, and safety comes entirely from managing the time.

Worked example: comparing stopping times

Given: the same baseball as above, but a skilled catcher lets the glove move back so the ball stops in 0.10 s instead of 0.020 s. Find: the new average force.
Solution: The momentum change is unchanged at 5.8 kg m/s. Now F = 5.8 / 0.10 = 58 N, one-fifth the earlier 290 N. By extending the stopping time fivefold, the catcher cuts the force to a fifth - which is exactly why experienced catchers "give" with the ball rather than holding a rigid glove.

Worked example: a ball that bounces back off a wall

Given: a 0.150 kg ball strikes a wall head-on at 25 m/s and rebounds straight back at 20 m/s. The contact lasts 0.0050 s.
Find: (a) the impulse delivered to the ball and (b) the average force the wall exerts on it.

Solution: take the direction toward the wall as positive. This sign convention is the entire problem.
p_i = m v_i = 0.150 x (+25) = +3.75 kg m/s
p_f = m v_f = 0.150 x (-20) = -3.00 kg m/s (negative because the ball now travels away from the wall)
J = delta p = p_f - p_i = -3.00 - (+3.75) = -6.75 kg m/s.
The magnitude is 6.75 N s, directed away from the wall.

(b) F_avg = delta p / t = -6.75 / 0.0050 = -1350 N, so the wall pushes on the ball with 1350 N directed away from itself.

Sanity check: the sign work is where this problem is won or lost. Subtracting the speeds as if they were both positive gives 3.75 - 3.00 = 0.75 kg m/s and a force of only 150 N - nine times too small. The rebound reverses the momentum, so the two contributions add. A second useful comparison: had the ball simply stuck to the wall and stopped, the impulse would have been 3.75 N s and the force 750 N. Bouncing back nearly doubles the load, which is why a bouncing hailstone damages a roof more than an equal-mass raindrop that splatters. Is 1350 N plausible? It corresponds to an acceleration of 1350/0.150 = 9000 m/s^2, more than 900 times gravity, sustained for five thousandths of a second. Large accelerations over tiny times are exactly what impacts are, and the ball's own weight of 1.47 N is utterly negligible beside 1350 N - which is why we ignore gravity during the contact.

Worked example: a baseball struck back and upward

Given: a 0.145 kg baseball arrives horizontally at 38 m/s and leaves the bat at 45 m/s at 30 degrees above the horizontal, travelling back the way it came. Contact lasts 0.70 ms = 7.0 x 10^-4 s.
Find: the impulse delivered and the average force on the ball.

Solution: momentum is a vector, so work in components. Let +x be the ball's original direction of travel and +y be upward.
p_ix = 0.145 x 38 = +5.51 kg m/s, p_iy = 0
After the hit the velocity is (-45 cos(30), +45 sin(30)) = (-38.97, +22.50) m/s, so
p_fx = 0.145 x (-38.97) = -5.65 kg m/s, p_fy = 0.145 x (+22.50) = +3.26 kg m/s

Take the difference component by component:
delta p_x = -5.65 - 5.51 = -11.16 kg m/s
delta p_y = 3.26 - 0 = +3.26 kg m/s
|J| = sqrt(11.16^2 + 3.26^2) = sqrt(124.6 + 10.6) = sqrt(135.2) = 11.6 N s
F_avg = 11.6 / (7.0 x 10^-4) = 1.66 x 10^4 N, about 16600 N.

The direction: the reference angle is arctan(3.26/11.16) = 16.3 degrees, and with delta p_x negative and delta p_y positive the impulse lies in the second quadrant, at 180 - 16.3 = 163.7 degrees - mostly a reversal, tilted slightly upward.

Sanity check: the horizontal component of the impulse (11.16) dwarfs the vertical one (3.26), which matches the picture - the bat's main job is to reverse the ball, and the upward tilt is a secondary effect. Measured bat-ball forces really are of order 10^4 N, so the magnitude is right. Note again how thoroughly gravity can be ignored: over the 0.70 ms of contact, gravity supplies an impulse of only m g t = 0.145 x 9.8 x 7.0 x 10^-4 = 1.0 x 10^-3 N s, roughly ten thousand times smaller than the bat's. A common error here is to subtract the speeds (45 - 38 = 7) and treat the problem as one-dimensional; that discards both the reversal and the angle and understates the force by more than an order of magnitude.

Common wrong turns

• Subtracting speeds when an object reverses. Assign signs from a chosen positive direction; a reversal makes the contributions add.
• Treating momentum as a scalar. In two dimensions, resolve into components and subtract component by component.
• Believing an airbag reduces your change in momentum. It does not - you still go from your travelling speed to zero. It reduces the peak force by lengthening the time.
• Equating momentum with kinetic energy. Since KE = p^2/(2m), equal momentum means very unequal energy when the masses differ.
• Including gravity during a brief impact. Over a few milliseconds the weight's impulse is thousands of times smaller than the impact's and is safely neglected.
• Reporting the average force as if it were the peak. Real impact forces spike well above the average; the average is what delivers the same total impulse.

Recap

• Momentum p = m v is a vector in kilogram meters per second, pointing along the velocity.
• Impulse J = F t equals the change in momentum, which is Newton's second law rearranged.
• A reversal makes the initial and final momenta add, so bounces deliver larger impulses than dead stops.
• For a fixed momentum change, force is inversely proportional to the contact time - the whole basis of impact safety design.
• In two dimensions, compute the impulse component by component and recombine at the end.
• Momentum and kinetic energy are different quantities related by KE = p^2/(2m), and they are conserved under different conditions.

Sources

1. OpenStax. (2016). 9.1 Linear momentum. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 9.2 Impulse and collisions. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 9.7 Rocket propulsion. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2022). 8.1 Linear momentum and force. In College Physics 2e. Rice University. openstax.org
5. OpenStax. (2022). 8.2 Impulse. In College Physics 2e. Rice University. openstax.org
6. Nave, R. (n.d.). Impulse of force. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu
7. Nave, R. (n.d.). Conservation laws. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu

An accident investigator arrives at an intersection hours after a crash. The cars are gone; only the skid marks remain. From the direction of those marks and the masses of the vehicles, she can reconstruct how fast each was travelling before impact - without knowing anything about the forces, the deformation, or how long the collision lasted. The tool that makes this possible is the most robust conservation law in mechanics.

One of the deepest and most reliable principles in physics is the conservation of momentum: in the absence of a net external force, the total momentum of a system of objects stays constant, no matter how the objects interact within it. This follows directly from Newton's third law - when two objects push on each other, the forces are equal and opposite, so the momentum one gains the other loses, and the total is unchanged.

For any interaction, then, total momentum before = total momentum after. This one statement handles collisions, explosions, and recoil with equal ease, and because it involves only the states before and after, it sidesteps the complicated and often unknowable forces that act during the interaction itself.

What "isolated" means

Momentum conservation requires an isolated system, meaning no net external force acts on the objects you are tracking. In a collision the forces the objects exert on each other are internal and cancel, but forces from outside - a sustained push, or friction with the ground - are external and can change the total momentum. In practice, collisions happen so quickly that external forces like friction have negligible time to act, so momentum is very nearly conserved during the brief instant of impact even when it is not conserved over longer times. Choosing the right system and the right instants is part of the skill.

Key idea: Momentum is conserved whenever no net external force acts, and internal forces - however violent - always cancel in pairs by the third law.

Momentum conserves component by component

Momentum is a vector, so its conservation is not one equation but one per direction. In a two-dimensional collision you write

sum p_x(before) = sum p_x(after) and sum p_y(before) = sum p_y(after),

and the two are completely independent. This is what makes intersection collisions tractable: a car moving east contributes nothing to the north-south ledger, and a truck moving north contributes nothing to the east-west one, yet after they lock together the wreck's single velocity must satisfy both books at once. Its direction is then fixed entirely by the ratio of the two momentum components - which is precisely the calculation an accident investigator runs backwards from the skid marks.

A warning that follows directly: you may never add momentum magnitudes across different directions. A car with 30000 kg m/s east and a truck with 30000 kg m/s north do not produce 60000 kg m/s of anything. The vector sum is sqrt(30000^2 + 30000^2) = 42400 kg m/s at 45 degrees.

Key idea: In two dimensions, conserve momentum separately in each direction and recombine only at the end.

Two kinds of collision

Momentum is conserved in every collision, but kinetic energy is a different story. In an elastic collision, kinetic energy is also conserved: the objects bounce apart with no energy lost to heat, sound, or permanent deformation. Collisions between hard objects like billiard balls or steel bearings come close to elastic, and collisions between atoms and subatomic particles can be perfectly elastic.

In an inelastic collision, some kinetic energy is converted into other forms - heat, sound, and the energy of bending metal - so the total kinetic energy afterward is less than before, even though momentum is unchanged. The extreme case is a perfectly inelastic collision, in which the objects stick together and move off as a single combined mass; this loses the most kinetic energy consistent with conserving momentum. Real-world crashes are almost always inelastic, which is precisely why they are destructive: the "lost" kinetic energy is what crumples the cars.

Worked example: a perfectly inelastic collision

Given: a 2000 kg car moving at 20 m/s rear-ends a stationary 3000 kg truck, and they lock together. Find: their common velocity just after impact.
Solution: Conserve momentum. Before the collision, the total is p = 2000 x 20 + 3000 x 0 = 40000 kg m/s, all carried by the car. After, the two vehicles move together with combined mass 5000 kg at a common velocity v, so p = 5000 v.

Setting before equal to after, 40000 = 5000 v, giving v = 8.0 m/s. The wreck moves off at 8.0 m/s in the car's original direction. As a check on the "inelastic" label, the kinetic energy before is (1/2)(2000)(20 squared) = 400000 J, while after it is (1/2)(5000)(8 squared) = 160000 J - well over half the kinetic energy has been converted to deformation and heat, exactly the energy that damages the vehicles.

Worked example: recoil

Given: a 60 kg skater at rest on frictionless ice throws a 2.0 kg ball horizontally at 10 m/s. Find: the skater's recoil speed.
Solution: The total momentum starts at zero, since nothing is moving, so it must remain zero after the throw. Therefore 0 = m_ball v_ball + m_skater v_skater = 2.0 x 10 + 60 x v_skater. Solving, 60 v_skater = -20, so v_skater = -0.33 m/s.

The skater glides backward at about 0.33 m/s, in the direction opposite the ball. This is exactly the physics that propels a rocket (throwing exhaust one way to move the other), recoils a fired gun, and lets a swimmer push off a wall - all cases where a system with zero initial momentum splits into parts moving in opposite directions so the total stays zero.

Why momentum, not just energy

You might wonder why we need momentum at all when energy served so well in Module 5. The reason is that in an inelastic collision, kinetic energy is not conserved - an unknown amount vanishes into heat and deformation - so you cannot use energy conservation to find the final velocities.

Momentum, however, is always conserved regardless of how much energy is lost, so it gives you a reliable equation every time. For elastic collisions you can use both conservation laws together, momentum and kinetic energy, which provides enough equations to solve for two unknown final velocities. Knowing which conservation laws apply to which type of collision is the key to choosing your approach.

Worked example: an elastic head-on collision

Given: a 0.50 kg ball moving at 4.0 m/s strikes a stationary 0.30 kg ball head-on. The collision is perfectly elastic.
Find: the velocity of each ball afterward.

Solution: two conservation laws hold here, so two unknowns can be found. Solving them together yields the standard results for a moving mass striking a stationary one:
v1' = [(m1 - m2)/(m1 + m2)] v1 and v2' = [2 m1/(m1 + m2)] v1.
With m1 + m2 = 0.80 kg and m1 - m2 = 0.20 kg:
v1' = (0.20/0.80) x 4.0 = 0.25 x 4.0 = 1.0 m/s (still forward)
v2' = (2 x 0.50/0.80) x 4.0 = 1.25 x 4.0 = 5.0 m/s (forward).

Sanity check: test both conservation laws directly rather than trusting the formulas.
Momentum before: 0.50 x 4.0 = 2.00 kg m/s. After: 0.50 x 1.0 + 0.30 x 5.0 = 0.50 + 1.50 = 2.00 kg m/s. Conserved.
Kinetic energy before: 0.5 x 0.50 x 4.0^2 = 4.00 J. After: 0.5 x 0.50 x 1.0^2 + 0.5 x 0.30 x 5.0^2 = 0.25 + 3.75 = 4.00 J. Also conserved, confirming the collision really is elastic.
The physics is sensible too: the lighter target is knocked away faster (5.0 m/s) than the incoming ball was travelling (4.0 m/s), while the heavier incoming ball keeps moving forward but slowed. Check the formula at a familiar limit: if the masses were equal, m1 - m2 = 0 would give v1' = 0 and v2' = v1 - the incoming ball stops dead and the target leaves with the full original speed, exactly what a cue ball does in a square billiard strike.

Worked example: a two-dimensional crash at an intersection

Given: a 1500 kg car travelling east at 20 m/s collides with a 2500 kg truck travelling north at 12 m/s. They lock together.
Find: (a) the velocity of the wreckage immediately after impact and (b) the kinetic energy lost.

Solution (a): conserve momentum separately in each direction. Let +x be east and +y be north.
p_x = 1500 x 20 = 30000 kg m/s (all from the car; the truck contributes nothing eastward)
p_y = 2500 x 12 = 30000 kg m/s (all from the truck)
The combined mass is 1500 + 2500 = 4000 kg, so
vx = 30000 / 4000 = 7.5 m/s and vy = 30000 / 4000 = 7.5 m/s
v = sqrt(7.5^2 + 7.5^2) = sqrt(112.5) = 10.6 m/s at arctan(7.5/7.5) = 45 degrees north of east.

Solution (b):
KE before = 0.5(1500)(20^2) + 0.5(2500)(12^2) = 300000 + 180000 = 480000 J
KE after = 0.5(4000)(10.6^2) = 0.5 x 4000 x 112.5 = 225000 J
KE lost = 480000 - 225000 = 255000 J, about 53 percent of the original.

Sanity check: the two momentum components came out identical at 30000 kg m/s each, so by symmetry the wreck must move at exactly 45 degrees - and it does, which validates the whole calculation in one glance. The final speed of 10.6 m/s is less than either vehicle's original speed, as it must be when a much larger combined mass carries the same total momentum. The energy loss of 53 percent is the physical signature of a perfectly inelastic collision, and that quarter-megajoule is exactly what crumples the metal and generates the noise and heat. Notice what would have been wrong: adding the momentum magnitudes to get 60000 kg m/s would give a speed of 15 m/s, badly overstating the answer, because momenta in perpendicular directions must be combined as vectors. Note finally that momentum was conserved even though more than half the kinetic energy was not - which is precisely why momentum, not energy, is the investigator's tool.

Common wrong turns

• Assuming kinetic energy is conserved in every collision. It is conserved only in elastic collisions; real crashes lose a large fraction of it.
• Adding momenta in different directions as numbers. Momentum is a vector; conserve each component separately and recombine with Pythagoras.
• Dropping a sign for an object moving the other way. A head-on approach means one velocity is negative in your chosen convention.
• Applying conservation when a large external force acts over the interval. Momentum is conserved for the isolated system; a sustained external push or long-acting friction breaks it.
• Using energy conservation to find velocities after an inelastic collision. An unknown amount of energy is lost, so only momentum gives a usable equation.
• Forgetting that "stick together" means one final velocity. A perfectly inelastic collision has a single unknown, not two.

Recap

• Total momentum is conserved whenever no net external force acts, in every type of collision without exception.
• Conservation is a vector statement: apply it separately to each direction and recombine at the end.
• Elastic collisions also conserve kinetic energy; inelastic ones convert some of it into heat, sound, and deformation.
• A perfectly inelastic collision leaves one combined object with one velocity and loses the most kinetic energy consistent with conserving momentum.
• For an elastic strike on a stationary target, v1' = [(m1 - m2)/(m1 + m2)] v1 and v2' = [2 m1/(m1 + m2)] v1.
• Momentum conservation is what lets a rocket accelerate in vacuum, a gun recoil, and an investigator reconstruct a crash.

Sources

1. OpenStax. (2016). 9.3 Conservation of linear momentum. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 9.4 Types of collisions. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 9.5 Collisions in multiple dimensions. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2022). 8.3 Conservation of momentum. In College Physics 2e. Rice University. openstax.org
5. OpenStax. (2022). 8.4 Elastic collisions in one dimension. In College Physics 2e. Rice University. openstax.org
6. OpenStax. (2022). 8.6 Collisions of point masses in two dimensions. In College Physics 2e. Rice University. openstax.org
7. Nave, R. (n.d.). Elastic and inelastic collisions. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu

A bench grinder spinning at 1800 revolutions per minute puts every point on its rim under an acceleration of more than four hundred times gravity. That is why the wheels are rated by speed and why one that exceeds its rating does not merely wobble - it explodes. Everything you need to compute that number is a direct translation of the linear kinematics you already know, and this lesson is that dictionary.

So far objects have moved along paths, but a huge fraction of the motion around us is rotation: wheels, gears, turbines, planets, hard drives, and spinning tops. Rotational motion has its own set of quantities, and the good news is that they parallel the linear ones you already know so closely that learning rotation is largely a matter of translation.

Instead of position we track angular displacement theta, the angle an object has turned through, measured in radians. Instead of velocity we have angular velocity omega, the rate of spin in radians per second. Instead of acceleration we have angular acceleration alpha, the rate at which the spin speeds up or slows down, in radians per second squared.

Why radians

Angles could be measured in degrees, but physics uses radians because they connect rotation to distance so naturally. One radian is the angle that subtends an arc equal in length to the radius, and a full turn is 2 pi radians, about 6.28. The reason radians are the natural choice is the clean relationship they give: the arc length s traced by a point at radius r through an angle theta is simply s = r theta, with no awkward conversion factor. Every linear-angular link in this lesson depends on measuring angles in radians.

Linking linear and angular motion

A point on a rotating object at radius r from the axis moves along a circle, and its linear motion connects to the object's rotation through two key relations. Its linear (tangential) speed is v = r omega, and its linear tangential acceleration is a = r alpha.

The farther a point sits from the axis, the faster it moves for the same rotation - which is why the rim of a spinning wheel moves faster than a point near its hub, why the outer horses on a carousel travel faster than the inner ones, and why a longer propeller blade has faster-moving tips. All points share the same angular velocity omega, but their linear speeds differ in proportion to their distance from the axis.

Key idea: Every point on a rigid rotating body shares one theta, omega, and alpha, while its linear quantities scale with its distance r from the axis.

A rotating point has two accelerations

Here is a subtlety that separates a careful answer from a wrong one. A point on a body whose spin rate is changing has two perpendicular accelerations at once, and the total is their vector sum.

• Tangential acceleration, a_t = r alpha, points along the direction of travel and describes the point speeding up or slowing down.
• Centripetal acceleration, a_c = r omega^2, points inward toward the axis and describes the direction of travel changing. Note that this is just v^2/r rewritten, since v = r omega.

They are perpendicular, so the total magnitude is a = sqrt(a_t^2 + a_c^2). In uniform circular motion alpha = 0, the tangential part vanishes, and only the centripetal survives - which is the case Module 4 treated. But whenever a wheel is spinning up or braking, both are present.

The two behave very differently as the spin builds. With constant alpha, the tangential acceleration is constant, while the centripetal acceleration grows as the square of the angular velocity. Within a few seconds of spin-up the centripetal term utterly dominates, which is exactly why the failure mode of an over-speeding flywheel is being torn apart radially rather than twisted.

Key idea: A point on a spinning-up wheel has a constant tangential acceleration r alpha and a centripetal acceleration r omega^2 that grows with the square of the spin rate.

The rotational kinematic equations

Because the angular definitions parallel the linear ones exactly, the constant-angular-acceleration equations have precisely the same form as the linear kinematic equations, with each linear symbol replaced by its angular partner. Anything you learned to do in Module 2 you can now do for rotation by simple substitution.

Linear	Rotational
v = v0 + a t	omega = omega0 + alpha t
x = v0 t + (1/2) a t squared	theta = omega0 t + (1/2) alpha t squared
v squared = v0 squared + 2 a x	omega squared = omega0 squared + 2 alpha theta

Worked example: a spinning wheel

Given: a wheel starts from rest (omega0 = 0) and reaches omega = 12 rad/s in t = 4.0 s. Find: its angular acceleration and the number of revolutions it makes.
Solution: Using omega = omega0 + alpha t, solve alpha = (omega - omega0)/t = (12 - 0)/4.0 = 3.0 rad/s squared. The wheel's spin increases at 3.0 rad/s squared.

For the angle turned, use theta = omega0 t + (1/2) alpha t squared = 0 + (1/2)(3.0)(4.0 squared) = 24 radians. Dividing by 2 pi gives 24 / 6.28 = 3.8 revolutions. Every equation from linear kinematics carries straight over, so once you master linear motion, rotation costs you almost nothing extra.

Worked example: a point on the rim

Given: the same wheel has radius 0.30 m. Find: the linear speed of a point on its rim at the moment omega = 12 rad/s.
Solution: v = r omega = 0.30 x 12 = 3.6 m/s. A point on the rim is moving at 3.6 m/s, while a point halfway to the center, at r = 0.15 m, moves at only 0.15 x 12 = 1.8 m/s - half as fast, because it is half as far from the axis.

Rotational inertia: the analog of mass

The dictionary has one more entry, and it is the only one that is not a straight substitution. The rotational counterpart of mass is the moment of inertia I, and unlike mass it depends not only on how much matter an object has but on where that matter sits relative to the axis. Each bit of mass contributes in proportion to the square of its distance from the axis.

The consequence is dramatic. A hoop and a solid disk of identical mass and radius have I = m R^2 and I = (1/2) m R^2 respectively - the hoop is twice as hard to spin up, purely because its mass all sits at the rim. With I in hand the rest of the dictionary follows: Newton's second law becomes tau = I alpha, the subject of the next lesson, and kinetic energy becomes KE = (1/2) I omega^2. A rolling object therefore carries both translational and rotational kinetic energy, which is why a hoop rolls down a ramp more slowly than a disk of the same mass and radius.

Key idea: Moment of inertia is the rotational analog of mass, but it depends on how far the mass sits from the axis, not merely on how much there is.

Worked example: a flywheel coasting to a stop

Given: a flywheel spinning at 240 rev/min is braked uniformly and comes to rest in 15 s.
Find: (a) its angular acceleration and (b) the number of revolutions it turns while stopping.

Solution: convert to SI units first. Every rotational equation requires radians per second, never revolutions per minute.
omega0 = 240 rev/min x (2 pi rad / 1 rev) x (1 min / 60 s) = 240 x 6.2832 / 60 = 25.13 rad/s.

(a) alpha = (omega - omega0)/t = (0 - 25.13)/15 = -1.68 rad/s^2. The minus sign records that the angular acceleration opposes the rotation.

(b) theta = omega0 t + (1/2) alpha t^2 = 25.13(15) + 0.5(-1.68)(15^2) = 376.99 - 188.50 = 188.5 rad
revolutions = 188.5 / 6.2832 = 30.0 rev.

Sanity check: because the deceleration is uniform, the average angular velocity is simply the midpoint, (25.13 + 0)/2 = 12.57 rad/s, and 12.57 x 15 = 188.5 rad, matching part (b). Better still, the whole thing can be checked without ever leaving revolutions per minute: the average spin rate is 120 rev/min, sustained for 15 s = 0.25 min, giving 120 x 0.25 = 30 revolutions exactly. Two independent routes to 30.0 rev is strong confirmation. The sign is right too - a wheel slowing from positive rotation must have negative alpha, while the total angle turned stays positive because the wheel never reverses.

Worked example: both accelerations on a grinding wheel

Given: a grinding wheel of radius 0.12 m starts from rest and reaches 1800 rev/min in 8.0 s with constant angular acceleration.
Find: (a) the angular acceleration, (b) the rim speed at full spin, and (c) the tangential, centripetal, and total acceleration of a rim point at full spin.

Solution (a): omega = 1800 x 2 pi / 60 = 30 x 6.2832 = 188.5 rad/s, so
alpha = (188.5 - 0)/8.0 = 23.56 rad/s^2.

Solution (b): v = r omega = 0.12 x 188.5 = 22.6 m/s.

Solution (c):
a_t = r alpha = 0.12 x 23.56 = 2.83 m/s^2
a_c = r omega^2 = 0.12 x (188.5)^2 = 0.12 x 35531 = 4264 m/s^2
a = sqrt(2.83^2 + 4264^2) = sqrt(8.0 + 18181696) = 4264 m/s^2, to three figures indistinguishable from the centripetal part alone.

Sanity check: the centripetal acceleration is 4264 / 9.8 = 435 times gravity, which is exactly why grinding wheels carry maximum-speed ratings and why exceeding one makes a wheel disintegrate rather than merely wobble. The tangential contribution of 2.83 m/s^2 is smaller by a factor of about 1500 and is safely ignored at full speed - though it was not negligible early on. At t = 1.0 s the angular velocity is only 23.56 rad/s, giving a_c = 0.12 x 555 = 66.6 m/s^2 against the same constant a_t = 2.83 m/s^2, a ratio of just 24. That contrast confirms the general rule that a_t stays fixed while a_c grows as omega^2. Cross-check the rim speed independently: a_c = v^2/r = 22.6^2/0.12 = 510.8/0.12 = 4257 m/s^2, agreeing with the r omega^2 route to rounding. Finally, 22.6 m/s is about 81 km/h at the rim, a believable and rather sobering figure for a small bench tool.

Common wrong turns

• Leaving angles in degrees or revolutions. Every relation here - s = r theta, v = r omega, a_t = r alpha - requires radians. Convert first.
• Reporting theta in radians when revolutions were asked for. Divide by 2 pi; one revolution is 6.283 rad.
• Using only the centripetal acceleration on a wheel that is speeding up. There is a tangential component too, and the total is their vector sum.
• Assuming every point on a wheel moves at the same speed. They share omega, not v; the rim outruns the hub in proportion to r.
• Treating moment of inertia as just the mass. Where the mass sits matters, and each element counts in proportion to r^2.
• Forgetting that alpha is negative when a wheel slows. The sign convention works exactly as it does in linear kinematics.

Recap

• Rotational motion is described by angular displacement, velocity, and acceleration, measured in radians and their derivatives.
• Radians are the natural unit because they make s = r theta, v = r omega, and a_t = r alpha hold without conversion factors.
• The constant-angular-acceleration equations mirror the linear ones symbol for symbol.
• All points on a rigid body share omega and alpha, while linear quantities scale with distance from the axis.
• A point on a spinning-up body has a constant tangential acceleration r alpha and a centripetal acceleration r omega^2 that grows quadratically.
• Moment of inertia plays the role of mass and depends on how the mass is distributed about the axis.

Sources

1. OpenStax. (2016). 10.1 Rotational variables. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 10.2 Rotation with constant angular acceleration. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 10.3 Relating angular and translational quantities. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2016). 10.4 Moment of inertia and rotational kinetic energy. In University Physics Volume 1. Rice University. openstax.org
5. OpenStax. (2022). 10.1 Angular acceleration. In College Physics 2e. Rice University. openstax.org
6. OpenStax. (2022). 10.2 Kinematics of rotational motion. In College Physics 2e. Rice University. openstax.org
7. Nave, R. (n.d.). Rotational quantities. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu

Hold a 5 kg dumbbell in your hand with your forearm horizontal. Your biceps is pulling with roughly 460 newtons - about nine times the weight you are actually holding. The muscle attaches only four centimeters from your elbow while the dumbbell sits thirty-two centimeters away, and that geometric mismatch is the entire explanation. By the end of this lesson you will be able to compute that number, and a great many like it.

Force causes linear acceleration; its rotational counterpart, torque, causes angular acceleration. Torque measures the turning effect of a force, and the crucial insight is that it depends not just on how hard you push but on where and in what direction.

Its magnitude is tau = r F sin(theta), where r is the distance from the axis to the point where the force is applied, F is the force, and theta is the angle between the direction of r and the direction of the force. Torque is measured in newton-meters. The combination r sin(theta) is called the lever arm or moment arm - it is the perpendicular distance from the axis to the line along which the force acts.

Why placement matters

The dependence on r explains a great deal of everyday mechanics. A longer wrench loosens a stubborn bolt more easily because a larger r multiplies the same force into a larger torque. A door handle is placed far from the hinges, not near them, for the same reason: pushing far from the axis produces more turning effect. The angle matters too.

Pushing perpendicular to the door (theta = 90 degrees, sin 90 = 1) gives the maximum torque, while pushing straight toward the hinge (theta = 0, sin 0 = 0) produces no torque at all, no matter how hard you shove - the door simply will not swing. This is why you instinctively push a door near its outer edge and at right angles to its face.

Note which trigonometric function appears. Torque takes sine, whereas work took cosine, and the two are opposites in exactly the situation where students most often confuse them: a force pointing straight along r does maximum work and zero torque, while a force perpendicular to r does maximum torque. If you remember that pushing a door toward its hinge accomplishes nothing rotationally, you will never mix them up.

Key idea: Torque is r F sin(theta), so only the force component perpendicular to r turns anything, and a force aimed at the pivot produces no torque at all.

Torque has a sign

Torque is not a plain number to be added arithmetically. Before summing torques you must choose a positive sense of rotation - by universal convention, counterclockwise is positive - and then every torque enters the sum with the sign of the direction it would turn the body. A clockwise torque of 40 N m and a counterclockwise torque of 40 N m give a net of zero, not 80 N m.

This is what makes equilibrium a genuine equation rather than a slogan. The rotational equilibrium condition sum tau = 0 is only meaningful once the signs are in place, and forgetting them is the single most common source of a "balanced" answer that is off by a factor of two.

Key idea: Assign every torque a sign from a chosen rotational sense before summing; counterclockwise positive is the standard convention.

Choosing the pivot to make unknowns vanish

Full mechanical equilibrium requires two conditions at once - sum F = 0 and sum tau = 0. The second comes with a gift that is easy to miss: when a body is in equilibrium the net torque is zero about every point, not merely the real pivot, so you may take torques wherever you like.

The strategy that follows is worth more than any formula here. Take torques about the point where an unknown force acts. That force then has a zero lever arm and drops out entirely, leaving one equation in one unknown. Solve it, then use sum F = 0 to recover what you eliminated. Reworking the problem about a different point is also the best available check, since the two routes must agree.

Key idea: For a body in equilibrium the net torque is zero about any point, so take torques about an unknown force's line of action to eliminate it.

Worked example: a wrench

Given: you push with F = 20 N perpendicular to a wrench, r = 0.50 m from the bolt (theta = 90 degrees). Find: the torque.
Solution: tau = r F sin(90) = 0.50 x 20 x 1 = 10 N m. You apply 10 N m of torque.

If you pushed at only 30 degrees to the wrench instead, the torque would drop to 0.50 x 20 x sin 30 = 0.50 x 20 x 0.5 = 5 N m, half as effective. And if you gripped a wrench twice as long, r = 1.0 m, the same 20 N perpendicular push would give 1.0 x 20 x 1 = 20 N m, twice the torque - the mechanical advantage of a long handle.

Rotational equilibrium

An object is in rotational equilibrium when the total torque on it is zero, so it does not start spinning or change a spin it already has. Torques that tend to cause clockwise rotation must exactly balance those that tend to cause counterclockwise rotation. This is the principle of the lever and the seesaw: a lighter person can balance a heavier one by sitting farther from the pivot, trading distance for weight so that the two torques match.

Full mechanical equilibrium actually requires two conditions - the net force must be zero (so the object does not accelerate linearly) and the net torque must be zero (so it does not angularly accelerate) - and engineers apply both when designing any structure that must stay put.

Worked example: balancing a seesaw

Given: a child weighing 300 N sits 2.0 m to the left of the pivot. On the right, an adult weighs 600 N. Find: where the adult must sit to balance.
Solution: For balance, the counterclockwise torque from the child must equal the clockwise torque from the adult: 300 x 2.0 = 600 x d.

So 600 = 600 d, giving d = 1.0 m. The adult must sit 1.0 m from the pivot, exactly half the child's distance because the adult weighs twice as much - the extra weight is offset by the shorter lever arm. This torque balance is the foundation of levers, balance scales, cranes, and countless machines that trade force for distance.

Levers as force multipliers

The same balance principle lets a small force overcome a large one. On a lever, if you push down on the long end, a short lever arm on the other side can lift a heavy load, because equal torques on both sides mean the side with the longer arm needs less force.

A crowbar, a bottle opener, a wheelbarrow, and a pair of pliers all exploit this: the ratio of the two lever arms sets the mechanical advantage. Archimedes captured the idea when he reportedly said that given a long enough lever and a place to stand, he could move the Earth. Torque is the physics that makes that boast, in principle, true.

Worked example: a plank on two supports

Given: a uniform plank of mass 30 kg and length 4.0 m rests on a support at each end. A person of mass 70 kg stands 1.0 m from the left end. Take g = 9.8 m/s^2.
Find: the upward force from each support.

Set-up. Two unknowns, N_left and N_right, so we need two equations. The plank is uniform, so its whole weight acts at its center, 2.0 m from either end.
W_plank = 30 x 9.8 = 294 N at 2.0 m from the left
W_person = 70 x 9.8 = 686 N at 1.0 m from the left

Solution: take torques about the left support, so that N_left has zero lever arm and disappears. Counterclockwise positive; N_right lifts the far end counterclockwise, while both weights turn the plank clockwise about that point.
N_right (4.0) - 294 (2.0) - 686 (1.0) = 0
4.0 N_right = 588 + 686 = 1274
N_right = 318.5 N.
Now use the force condition:
N_left + N_right = 294 + 686 = 980 N, so N_left = 980 - 318.5 = 661.5 N.

Sanity check: redo the torque equation about the right support, which gives N_left directly and independently. The person is now 3.0 m away and the plank's center 2.0 m away:
4.0 N_left = 294 (2.0) + 686 (3.0) = 588 + 2058 = 2646, so N_left = 661.5 N - exactly the value found the other way. The physical picture agrees: the person stands nearer the left support, so that support carries the larger share, and the two forces sum to the full 980 N. As a final plausibility test, standing at the midpoint would split the load evenly at 490 N each, and standing directly over the left support would load it with 686 + 147 = 833 N - the computed 661.5 N sits sensibly between those extremes.

Worked example: the biceps as a losing lever

Given: a forearm holds a 5.0 kg dumbbell. The biceps tendon attaches 4.0 cm from the elbow joint and pulls vertically upward. The dumbbell is 32 cm from the elbow. The forearm and hand together have mass 1.8 kg, with their center of gravity 15 cm from the elbow. The forearm is held horizontal and still. Take g = 9.8 m/s^2.
Find: (a) the force the biceps must exert and (b) the force the elbow joint exerts on the forearm.

Solution (a): the elbow joint force is unknown in both magnitude and direction, so take torques about the elbow to eliminate it in one stroke. Convert every distance to meters.
W_dumbbell = 5.0 x 9.8 = 49.0 N at 0.32 m
W_forearm = 1.8 x 9.8 = 17.64 N at 0.15 m
Taking counterclockwise as positive, the biceps lifts while both weights pull down:
F_biceps (0.040) - 49.0 (0.32) - 17.64 (0.15) = 0
0.040 F_biceps = 15.68 + 2.646 = 18.326
F_biceps = 18.326 / 0.040 = 458 N.

Solution (b): now apply the force condition vertically. The biceps pulls up with 458 N while the two weights pull down with a total of 49.0 + 17.64 = 66.6 N. For the sum to vanish, the joint must pull down on the forearm:
F_joint = 458 - 66.6 = 391 N downward.

Sanity check: the biceps force is 458 / 49.0 = 9.3 times the weight held, and the lever arms predict exactly that - the load's 0.32 m arm is eight times the muscle's 0.040 m arm, with the forearm's own weight accounting for the rest. This is a lever of disadvantage: the body trades force for speed and range of motion, since a small contraction sweeps the hand through a large arc. The joint force of 391 N, roughly eight times the dumbbell's weight, explains why modest hand-held loads place heavy demands on the elbow. A common error is to omit the forearm's own weight, giving 15.68/0.040 = 392 N - about 14 percent low.

Common wrong turns

• Using cosine instead of sine. Torque is r F sin(theta). A force aimed straight at the pivot produces no torque however large it is.
• Adding torques without signs. Choose counterclockwise positive and give every torque its sign, or opposing torques will wrongly reinforce each other.
• Believing zero net force means equilibrium. Two equal and opposite forces offset from each other form a couple: no net force, but a very real net torque.
• Forgetting the object's own weight. A uniform beam's weight acts at its center and contributes a real torque about any other point.
• Assuming torques must be taken about the physical pivot. In equilibrium any point works, and choosing an unknown force's line of action removes it from the equation.

Recap

• Torque is tau = r F sin(theta), measured in newton-meters, where r sin(theta) is the perpendicular lever arm.
• Only the force component perpendicular to r turns anything; a force through the pivot produces no torque.
• Torques carry signs from a chosen rotational sense, conventionally counterclockwise positive.
• Full equilibrium needs both sum F = 0 and sum tau = 0; zero net force alone does not prevent rotation.
• For a body in equilibrium the net torque vanishes about every point, so pick the pivot that kills an unknown.
• Levers trade force for distance, which is why a long wrench multiplies your effort and why the biceps must far exceed the weight it lifts.

Sources

1. OpenStax. (2016). 10.6 Torque. In University Physics Volume 1. Rice University. openstax.org
2. OpenStax. (2016). 10.7 Newton's second law for rotation. In University Physics Volume 1. Rice University. openstax.org
3. OpenStax. (2016). 12.1 Conditions for static equilibrium. In University Physics Volume 1. Rice University. openstax.org
4. OpenStax. (2016). 12.2 Examples of static equilibrium. In University Physics Volume 1. Rice University. openstax.org
5. OpenStax. (2022). 9.2 The second condition for equilibrium. In College Physics 2e. Rice University. openstax.org
6. OpenStax. (2022). 9.6 Forces and torques in muscles and joints. In College Physics 2e. Rice University. openstax.org
7. Nave, R. (n.d.). Torque and equilibrium. In HyperPhysics. Georgia State University, Department of Physics and Astronomy. hyperphysics.phy-astr.gsu.edu