Work, energy and power practice problems AP Physics 1, Unit 3
Work, kinetic and potential energy, the work-energy theorem, 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.
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Key equations for work, energy and power
| K = ½mv2 | kinetic energy |
|---|---|
| W = Fd cosθ | work done by a constant force |
| ΔUg = mgΔy | gravitational potential energy near the Earth |
| Us = ½kΔx2 | elastic potential energy of a spring |
| P = ΔE / Δt | average power |
All questions use g = 9.8 m/s², as on the AP Physics 1 equation sheet. A calculator is allowed, as in the exam. The full sheet is in our AP Physics 1 equation sheet.
Warm-up questions
Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.
Hint
K = ½mv2. Square the speed first.
Worked solution
- K = ½ × 0.15 × 302 = 0.075 × 900.
- K = 67.5 J.
Hint
Only the component of the force along the motion does work.
Worked solution
- W = Fd cosθ, where θ is the angle between the force and the displacement.
- W = 50 × 8.0 × cos 60° = 400 × 0.5.
- W = 200 J.
Hint
Work depends on the angle between the force and the direction of motion.
Worked solution
- In a circular orbit, gravity points to the centre and the velocity is along the tangent.
- The angle is 90°, and cos 90° = 0, so gravity does no work at any instant.
- This matches the constant speed: no work means no change in kinetic energy. The answer is A.
Hint
At steady speed, all the motor's work goes into gravitational potential energy.
Worked solution
- Energy gained: mgΔy = 200 × 9.8 × 15 = 29,400 J.
- P = ΔE / Δt = 29,400 / 20.
- P = 1470 W, about 1.5 kW.
Exam-style questions
Questions at the level of the AP Physics 1 exam. The ones marked Challenge combine more than one idea.
Hint
Only the drop in height matters: energy lost as potential energy becomes kinetic energy.
Worked solution
- Conservation of energy: mgΔh = ½mv2, and the mass cancels.
- The car has dropped Δh = 25 − 10 = 15 m.
- v = √(2 × 9.8 × 15) = √294.
- v = 17.1 m/s.
Hint
The spring's elastic energy becomes the ball's kinetic energy.
Worked solution
- Elastic energy: ½kΔx2 = ½ × 400 × 0.102 = 2.0 J.
- Set 2.0 = ½ × 0.050 × v2, so v2 = 80.
- v = 8.94 m/s.
Hint
The work done by friction removes all the box's kinetic energy. Does the mass matter?
Worked solution
- Work-energy theorem: μmg d = ½mv2, so the mass cancels.
- μ = v2 / (2gd) = 25 / (2 × 9.8 × 3.0) = 25 / 58.8.
- μ = 0.425, about 0.43 (no units).
Hint
Compare their energies at the top of the cliff and at the ground.
Worked solution
- Both start with the same kinetic energy (same speed) and the same potential energy (same height).
- Both lose the same potential energy falling to the ground.
- So both arrive with the same kinetic energy, and the same speed. The answer is C. Only their times differ.
Hint
Compare the potential energy at the top with the kinetic energy at the bottom.
Worked solution
- Potential energy lost: mgh = 60 × 9.8 × 40 = 23,520 J.
- Kinetic energy gained: ½ × 60 × 202 = 12,000 J.
- Energy transferred by friction and air resistance: 23,520 − 12,000 = 11,520 J, about 11.5 kJ.
Hint
Kinetic energy depends on the square of the speed.
Worked solution
- K = ½mv2, so K is proportional to v2.
- Tripling v multiplies K by 32 = 9.
- The answer is C. This is why stopping distances grow so quickly with speed.
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.
Qualitative/quantitative translation style · 6 points · about 22 minutes
A block of mass m is released from rest at height h on a frictionless curved ramp. At the bottom it slides onto a rough horizontal surface with coefficient of kinetic friction μ, and comes to rest after sliding a distance d along it.
- (a) Derive an expression for d in terms of h and μ.
- (b) A student claims that a heavier block released from the same height would slide further, because it has more energy at the bottom. Explain whether the student is correct.
- (c) The block is released from height 2h instead. By what factor does d change? Justify your answer using part (a).
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)
The potential energy mgh is all removed by the work done by friction, μmg d. So mgh = μmg d, which gives d = h / μ.
Part (b)
The student is wrong. A heavier block does have more kinetic energy at the bottom, but the friction force μmg is larger by the same factor. The mass cancels, so d = h/μ is the same for any mass.
Part (c)
d = h/μ is directly proportional to h, so doubling the height doubles the stopping distance.
Your mark: 0 of 6
Common mistakes in work, energy and power
- Using the total height instead of the change in height in mgΔh.
- Forgetting cosθ when the force is not along the motion.
- Assuming mechanical energy is conserved when friction or air resistance acts.
- Mixing up energy (J) with power (W = J/s).
Keep going
Revise the ideas behind these questions:
Then try real exam questions: AP Physics 1 past papers, with official scoring guidelines. · All AP Physics 1 units