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Work, energy and power practice problems AP Physics C: Mechanics, Unit 3

Written by STEM-Tutor's tutors · Updated October 2026 · 8 questions and 1 free-response question

Work as an integral, the work-energy theorem, potential energy and force, conservation of energy, and power. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.

On this page
  1. Key equations
  2. Warm-up questions
  3. Exam-style questions
  4. Free-response practice
  5. Common mistakes
  6. Keep going

Key equations for work, energy and power

W = ∫ F dxwork by a variable force
Wnet = ΔKwork-energy theorem
Fx = −dU/dxforce from potential energy
Us = ½kx2spring potential energy
P = dW/dt = Fvpower

Use g = 9.8 m/s², as on the AP reference sheet. You can type answers like 4.9, 1/3 or 3.6e2. The full sheet is in our AP Physics C equation sheet.

Warm-up questions

Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.

Q1Warm-up
Find the kinetic energy of a 2.0 kg object moving at 3.0 m/s.
J

Hint

K = ½mv2.

Worked solution
  1. K = ½ × 2.0 × 9.0 = 9 J.
Q2Warm-up
A force F(x) = 3x2 N acts along the x-axis. Find the work it does from x = 0 to x = 2 m.
J

Hint

W = ∫ F dx.

Worked solution
  1. W = ∫02 3x2 dx = [x3]02.
  2. W = 8 J.
Q3Warm-up
A block slides along a horizontal floor. How much work does the normal force do on it?

Hint

Work depends on the angle between force and displacement.

Worked solution
  1. The normal force is perpendicular to the displacement, so cos 90° = 0.
  2. It does zero work: C.

Exam-style questions

Questions at the level of the AP Physics C: Mechanics exam. The ones marked Challenge combine more than one idea.

Q4Exam-style
A spring with k = 200 N/m is compressed 0.10 m and launches a 0.050 kg ball horizontally on a frictionless surface. Find the ball's launch speed.
m/s

Hint

Spring energy becomes kinetic energy.

Worked solution
  1. ½kx2 = ½mv2, so v = x√(k/m).
  2. v = 0.10 × √(200/0.050) = 0.10 × 63.2.
  3. v = 6.32 m/s.
Q5Exam-style
A particle has potential energy U(x) = 4x2 − x4 joules, with x in metres. Find the force on it at x = 1 m.
N

Hint

F = −dU/dx.

Worked solution
  1. dU/dx = 8x − 4x3, which is 4 at x = 1.
  2. F = −dU/dx = −4 N, in the negative x direction.
Q6Exam-style
A car travels at a constant 20 m/s against a total resistive force of 600 N. Find the power its engine delivers.
W

Hint

At constant speed the driving force equals the resistance.

Worked solution
  1. Driving force = 600 N.
  2. P = Fv = 600 × 20 = 12000 W (12 kW).
Q7Exam-style
On a graph of potential energy U against position x, a particle is in stable equilibrium where:

Hint

F = −dU/dx. What does the force do if the particle moves slightly away?

Worked solution
  1. At a minimum, dU/dx = 0, so F = 0.
  2. Moving either way makes F point back towards the minimum, so the equilibrium is stable: B.
Q8Challenge
A block slides from rest down a frictionless ramp from a height of 2.0 m, then across a rough horizontal floor with μk = 0.25. How far along the floor does it slide before stopping?
m

Hint

All the gravitational energy is removed by friction on the floor.

Worked solution
  1. mgh = μkmg d, and m and g cancel.
  2. d = h/μk = 2.0/0.25 = 8 m.

Free-response practice

One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.

Mathematical routines style · 6 points · about 20 minutes

A 0.50 kg block is attached to a nonlinear spring that exerts a force F(x) = −kx − βx3, with k = 50 N/m and β = 200 N/m3. The block moves on a frictionless horizontal surface.

  1. (a) Derive an expression for the potential energy U(x), taking U(0) = 0.
  2. (b) The block is released from rest at x = 0.20 m. Find its speed at x = 0.
  3. (c) Is x = 0 a position of stable equilibrium? Justify your answer.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

U = −∫0x F dx = ½kx2 + ¼βx4.

Part (b)

U(0.20) = 25(0.04) + 50(0.0016) = 1.08 J. ½(0.50)v2 = 1.08, so v = 2.08 m/s.

Part (c)

Yes: U has a minimum at x = 0, and for any small displacement F points back towards x = 0.

Your mark: 0 of 6

Common mistakes in work, energy and power

Keep going

Revise the ideas behind these questions:

Then try real exam questions: AP Physics C: Mechanics past papers, with official scoring guidelines. · All AP Physics C: Mechanics units