AP Physics 1 · Lesson 1

Work and kinetic energy

Know Know it

Work is how a force transfers energy as an object moves. For a constant force F acting through a displacement d, W=Fdcos⁡θW = Fd\cos\theta, where θ is the angle between the force and the displacement. Work is a scalar, measured in joules (1 J = 1 N·m).

Only the part of a force along the motion does work. A force at right angles to the motion, like the tension in a string that whirls a ball in a horizontal circle, does no work.

Kinetic energy is K=12mv2K = \tfrac{1}{2}mv^2. The work-energy theorem says the net work on an object equals its change in kinetic energy: Wnet=ΔKW_{\text{net}} = \Delta K. When a force changes as the object moves, its work is the area under its force-position graph.

The sign of work
Angle between force and displacementWorkExample
0° (along the motion)Positive: W=FdW = FdA tow rope pulling a car forward
90° (perpendicular)ZeroThe normal force on a cart rolling across a level floor
180° (against the motion)Negative: W=−FdW = -FdKinetic friction on a sliding puck

Calculator On the exam

On the exam you may use a four-function, scientific or graphing calculator on both sections, and the testing app also has built-in scientific and graphing calculators. You get an equation sheet with reference tables too, so spend your effort on choosing the right relationship, not memorizing it. In these lessons, use g = 9.8 m/s² unless a problem gives another value. Set the calculator to degrees before you take the cosine of an angle in W=Fdcos⁡θW = Fd\cos\theta.

Worked example Easy

A 0.060 kg tennis ball leaves a racket at 40 m/s. What is its kinetic energy?

  1. 1.2 J
  2. 2.4 J
  3. 48 JAnswer
  4. 96 J

How to solve it

  1. K=12mv2K = \tfrac{1}{2}mv^2.
  2. (40)2=1600(40)^2 = 1600, and 12(0.060)(1600)=48\tfrac{1}{2}(0.060)(1600) = 48.
  3. K = 48 J.

Why each choice is right or wrong

  • A. Incorrect. 1.2 J computes 12mv\tfrac{1}{2}mv, leaving out the square on the speed.
  • B. Incorrect. 2.4 J is mvmv, which is the ball’s momentum in kg·m/s, not its kinetic energy.
  • C. Correct. K=12mv2=12(0.060)(40)2=48K = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(0.060)(40)^2 = 48 J.
  • D. Incorrect. 96 J is mv2mv^2, which leaves out the ½.

Trap Work needs motion along the force

Holding a heavy bag still, or carrying it across a level room at a steady speed, does no work on the bag, even though it is tiring. In the first case there is no displacement; in the second, your upward force is perpendicular to the motion. Work needs a force component along the displacement.

Explain Explain it

Task verbs: calculate, justify

To use the work-energy theorem, list every force on the object and find the work each one does, with its sign. Add them to get the net work, then set that equal to 12mvf2−12mvi2\tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2.

To justify the sign of a work, compare directions: a force with a component along the motion does positive work, transferring energy to the object, and a force with a component against the motion does negative work. The net work of all the forces decides whether the kinetic energy rises or falls.

Worked example Medium

A dockworker pulls a 40 kg cart from rest across a level floor by a handle that makes a 35° angle above the horizontal. The pull along the handle is a steady 90 N, and friction is small enough to ignore. How fast is the cart moving after it has gone 6.0 m?

  1. 3.3 m/s
  2. 3.9 m/s
  3. 4.7 m/sAnswer
  4. 5.2 m/s

How to solve it

  1. Only the horizontal part of the pull does work: W=Fdcos⁡θ=(90)(6.0)(0.819)=442W = Fd\cos\theta = (90)(6.0)(0.819) = 442 J.
  2. Work-energy theorem from rest: 442=12(40)v2442 = \tfrac{1}{2}(40)v^2.
  3. v=884/40=22.1≈4.7v = \sqrt{884/40} = \sqrt{22.1} \approx 4.7 m/s.

Why each choice is right or wrong

  • A. Incorrect. 3.3 m/s leaves out the ½ in 12mv2\tfrac{1}{2}mv^2, setting the work equal to mv2mv^2.
  • B. Incorrect. 3.9 m/s uses sin⁡35∘\sin 35^\circ, the vertical part of the pull, which is perpendicular to the motion and does no work.
  • C. Correct. The work is W=(90)(6.0)cos⁡35∘=442W = (90)(6.0)\cos 35^\circ = 442 J, and all of it becomes kinetic energy: v=2(442)/40≈4.7v = \sqrt{2(442)/40} \approx 4.7 m/s.
  • D. Incorrect. 5.2 m/s treats the whole 90 N pull as horizontal; only 90cos⁡35∘90\cos 35^\circ acts along the floor.

Justify Your turn

A 2.0 kg cart rolls along a level track at 3.0 m/s, and a constant braking force brings it to rest. Calculate the work done by the braking force and justify its sign. Then say how the braking work would change for a 4.0 kg cart that starts at the same speed.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Gives the work done by the braking force as -9.0 J
  • Justifies the negative sign by saying the braking force points opposite to the cart’s motion, or that it removes the cart’s kinetic energy
  • Says the 4.0 kg cart needs twice as much braking work, -18 J

Model answer

By the work-energy theorem the braking work equals the change in kinetic energy: W=0−12(2.0)(3.0)2=−9.0W = 0 - \tfrac{1}{2}(2.0)(3.0)^2 = -9.0 J. It is negative because the braking force points opposite to the cart’s displacement, so it takes kinetic energy away. A 4.0 kg cart at 3.0 m/s has twice the kinetic energy, 18 J, so the braking force must do twice as much work, -18 J, to stop it.

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