A Level Physics 5.2 GPE and Kinetic Energy MCQs – 30 Practice Questions with Answers (Cambridge 9702)

Free practice MCQs for Cambridge International AS Level Physics (9702), Section 5.2 – gravitational potential energy and kinetic energy. Every question is original, with a fully worked answer; take g = 9.81 m s−2 throughout.

Formula recap: ΔEp = mgΔh (Δh is the VERTICAL height change) and Ek = ½mv2. See all the equations together on the A Level Physics Formula Sheet.

Gravitational Potential Energy

Q1. A mass of 2.0 kg is lifted vertically through a height of 3.0 m at constant speed. What is the increase in its gravitational potential energy?

  • (A) 5.9 J
  • (B) 59 J
  • (C) 29 J
  • (D) 118 J
Show Answer

Answer: B – ΔEp = mgΔh = 2.0 × 9.81 × 3.0 = 58.9 J ≈ 59 J.

Q2. A trolley of mass 4.0 kg is pulled up a ramp of length 5.0 m. The top of the ramp is 2.0 m vertically above the bottom. What is the gain in gravitational potential energy of the trolley?

  • (A) 196 J
  • (B) 39 J
  • (C) 78 J
  • (D) 8.0 J
Show Answer

Answer: C – Only the vertical rise matters: ΔEp = mgΔh = 4.0 × 9.81 × 2.0 = 78.5 J ≈ 78 J. Using the ramp length 5.0 m (option A) is the classic trap.

Q3. A student of mass 70 kg climbs a flight of stairs with a vertical rise of 4.5 m. What is the increase in the student’s gravitational potential energy?

  • (A) 3.1 kJ
  • (B) 1.5 kJ
  • (C) 315 J
  • (D) 6.2 kJ
Show Answer

Answer: A – ΔEp = 70 × 9.81 × 4.5 = 3090 J ≈ 3.1 kJ.

Q4. A book of mass 1.5 kg rests on a shelf 1.20 m above the floor. A table top is 0.40 m above the floor. What is the gravitational potential energy of the book measured relative to the table top?

  • (A) 17.7 J
  • (B) 5.9 J
  • (C) 23.5 J
  • (D) 11.8 J
Show Answer

Answer: D – GPE is measured from a chosen zero level. Height above the table = 1.20 − 0.40 = 0.80 m, so Ep = 1.5 × 9.81 × 0.80 = 11.8 J ≈ 12 J. (Relative to the floor it would be 17.7 J.)

Q5. A crane lifts a load of mass 500 kg through a vertical height of 12 m at constant speed. What is the minimum work done by the crane against gravity?

  • (A) 5.9 kJ
  • (B) 59 kJ
  • (C) 118 kJ
  • (D) 6.0 kJ
Show Answer

Answer: B – At constant speed the work done against gravity equals the GPE gain: W = mgΔh = 500 × 9.81 × 12 = 58 860 J ≈ 59 kJ.

Q6. In a hydroelectric dam, 1000 kg of stored water falls through a vertical drop of 50 m. What is the gravitational potential energy lost by the water?

  • (A) 4.9 × 104 J
  • (B) 5.0 × 104 J
  • (C) 4.9 × 105 J
  • (D) 9.8 × 105 J
Show Answer

Answer: C – ΔEp = mgΔh = 1000 × 9.81 × 50 = 490 500 J ≈ 4.9 × 105 J. This is the energy available to the turbines (before efficiency losses).

Q7. A crate of mass 3.0 kg is lowered vertically by 2.5 m at constant speed. What is the change in its gravitational potential energy?

  • (A) −74 J
  • (B) +74 J
  • (C) −7.5 J
  • (D) −147 J
Show Answer

Answer: A – Δh = −2.5 m (height decreases), so ΔEp = 3.0 × 9.81 × (−2.5) = −73.6 J ≈ −74 J. The negative sign shows GPE is lost.

Q8. A sledge of mass 12 kg is pulled up a snowy slope to a point 8.0 m vertically above the start. What is the gain in gravitational potential energy?

  • (A) 96 J
  • (B) 1.9 kJ
  • (C) 0.47 kJ
  • (D) 0.94 kJ
Show Answer

Answer: D – ΔEp = 12 × 9.81 × 8.0 = 942 J ≈ 0.94 kJ. The length and angle of the slope are irrelevant.

Q9. A pump lifts 200 kg of water every minute from a well 15 m deep. What is the minimum useful power of the pump?

  • (A) 29 kW
  • (B) 0.49 kW
  • (C) 4.9 kW
  • (D) 0.20 kW
Show Answer

Answer: B – Energy per minute = mgΔh = 200 × 9.81 × 15 = 29 430 J. Power = 29 430 / 60 = 490 W ≈ 0.49 kW. (Forgetting to divide by 60 s gives the trap answer A.)

Q10. The same object is raised from the ground to a shelf 2.0 m above it, once by lifting it straight up and once by pushing it up a long ramp. Friction is negligible in both cases. How do the GPE gains compare?

  • (A) Larger for the straight lift because the distance is shorter
  • (B) Larger for the ramp because more distance is covered
  • (C) Equal, because only the vertical height change matters
  • (D) Equal, because the object’s mass has not changed
Show Answer

Answer: C – ΔEp = mgΔh depends only on the vertical height change, not on the path taken.

Kinetic Energy

Q11. A car of mass 1000 kg travels at 20 m s−1. What is its kinetic energy?

  • (A) 200 kJ
  • (B) 20 kJ
  • (C) 400 kJ
  • (D) 10 kJ
Show Answer

Answer: A – Ek = ½mv2 = 0.5 × 1000 × 202 = 200 000 J = 200 kJ.

Q12. A ball of mass 0.50 kg speeds up from 4.0 m s−1 to 8.0 m s−1. By what factor does its kinetic energy increase?

  • (A) 2
  • (B) 8
  • (C) 16
  • (D) 4
Show Answer

Answer: D – Ek ∝ v2. Doubling the speed multiplies KE by 22 = 4 (here from 4.0 J to 16 J).

Q13. Two trolleys travel at the same speed. Trolley X has mass 2.0 kg and trolley Y has mass 4.0 kg. How does the kinetic energy of Y compare with that of X?

  • (A) Four times as large
  • (B) Twice as large
  • (C) The same
  • (D) Half as large
Show Answer

Answer: B – At fixed speed, Ek ∝ m, so doubling the mass doubles the KE.

Q14. An object of mass 5.0 kg has momentum 20 kg m s−1. What is its kinetic energy? (Use Ek = p2 / 2m.)

  • (A) 100 J
  • (B) 200 J
  • (C) 40 J
  • (D) 10 J
Show Answer

Answer: C – Ek = p2 / 2m = 202 / (2 × 5.0) = 400 / 10 = 40 J.

Q15. A car of mass 1200 kg travelling at 25 m s−1 brakes to rest over a distance of 40 m. Assuming a constant braking force, what is its magnitude?

  • (A) 9.4 kN
  • (B) 4.7 kN
  • (C) 750 N
  • (D) 15 kN
Show Answer

Answer: A – Work done by the braking force = KE lost. Ek = 0.5 × 1200 × 252 = 375 000 J. F = W / d = 375 000 / 40 = 9375 N ≈ 9.4 kN.

Q16. A constant force of 500 N pushes an object of mass 10 kg over 20 m on a frictionless surface, starting from rest. What is the final speed of the object?

  • (A) 20 m s−1
  • (B) 100 m s−1
  • (C) 32 m s−1
  • (D) 45 m s−1
Show Answer

Answer: D – Work done = Fd = 500 × 20 = 10 000 J = KE gained. v = √(2Ek/m) = √(2 × 10 000 / 10) = √2000 = 44.7 ≈ 45 m s−1.

Q17. An object of mass 2.0 kg accelerates from 3.0 m s−1 to 7.0 m s−1. What is the change in its kinetic energy?

  • (A) 16 J
  • (B) 40 J
  • (C) 49 J
  • (D) 80 J
Show Answer

Answer: B – ΔEk = ½m(v2 − u2) = 0.5 × 2.0 × (49 − 9) = 40 J.

Q18. A ball of mass 0.20 kg has kinetic energy 45 J. What is its speed?

  • (A) 15 m s−1
  • (B) 450 m s−1
  • (C) 21 m s−1
  • (D) 6.7 m s−1
Show Answer

Answer: C – v = √(2Ek/m) = √(2 × 45 / 0.20) = √450 = 21.2 ≈ 21 m s−1.

Q19. Which object has the greatest kinetic energy?

  • (A) Mass 2.0 kg moving at 10 m s−1
  • (B) Mass 4.0 kg moving at 6.0 m s−1
  • (C) Mass 1.0 kg moving at 13 m s−1
  • (D) Mass 8.0 kg moving at 4.0 m s−1
Show Answer

Answer: A – A: 0.5 × 2.0 × 100 = 100 J. B: 0.5 × 4.0 × 36 = 72 J. C: 0.5 × 1.0 × 169 = 84.5 J. D: 0.5 × 8.0 × 16 = 64 J. Speed matters more than mass because KE depends on v2.

Q20. Two objects have the same momentum. Object P has mass 1.0 kg and object Q has mass 4.0 kg. How do their kinetic energies compare? (Use Ek = p2 / 2m.)

  • (A) Q has four times the KE of P
  • (B) They are equal
  • (C) P has twice the KE of Q
  • (D) P has four times the KE of Q
Show Answer

Answer: D – With p fixed, Ek = p2/2m ∝ 1/m. P has one quarter of the mass of Q, so P has four times the kinetic energy.


Energy Conversions

Q21. A stone is dropped from rest from a height of 20 m. Ignoring air resistance, what is its speed just before it hits the ground?

  • (A) 14 m s−1
  • (B) 20 m s−1
  • (C) 28 m s−1
  • (D) 40 m s−1
Show Answer

Answer: B – GPE lost = KE gained: mgh = ½mv2, so v = √(2gh) = √(2 × 9.81 × 20) = √392 = 19.8 ≈ 20 m s−1.

Q22. A ball of mass 0.50 kg falls freely through a vertical distance of 5.0 m. What is its gain in kinetic energy (ignoring air resistance)?

  • (A) 2.5 J
  • (B) 49 J
  • (C) 25 J
  • (D) 12 J
Show Answer

Answer: C – KE gained = GPE lost = mgΔh = 0.50 × 9.81 × 5.0 = 24.5 J ≈ 25 J.

Q23. A pendulum bob is released from rest at a point 0.45 m vertically above its lowest position. Ignoring friction, what is its speed at the lowest point?

  • (A) 3.0 m s−1
  • (B) 2.1 m s−1
  • (C) 4.4 m s−1
  • (D) 1.5 m s−1
Show Answer

Answer: A – v = √(2gΔh) = √(2 × 9.81 × 0.45) = √8.83 = 2.97 ≈ 3.0 m s−1.

Q24. A roller-coaster car starts from rest at a height of 30 m and descends to a point 12 m above the ground. Assuming no friction, what is its speed at the lower point?

  • (A) 24 m s−1
  • (B) 13 m s−1
  • (C) 15 m s−1
  • (D) 19 m s−1
Show Answer

Answer: D – The height FALLEN is 30 − 12 = 18 m. v = √(2 × 9.81 × 18) = √353 = 18.8 ≈ 19 m s−1.

Q25. A ball is thrown vertically upwards with a speed of 14 m s−1. Ignoring air resistance, to what maximum height does it rise?

  • (A) 7.0 m
  • (B) 20 m
  • (C) 10 m
  • (D) 14 m
Show Answer

Answer: C – KE at launch = GPE at the top: ½mv2 = mgh, so h = v2/2g = 196 / 19.62 = 9.99 m ≈ 10 m.

Q26. A block slides from rest down a frictionless slope, descending a vertical height of 6.0 m. What is its speed at the bottom?

  • (A) 7.7 m s−1
  • (B) 11 m s−1
  • (C) 15 m s−1
  • (D) It depends on the mass of the block
Show Answer

Answer: B – v = √(2gΔh) = √(2 × 9.81 × 6.0) = √118 = 10.8 ≈ 11 m s−1. The mass cancels, and the slope angle does not matter – only the vertical drop.

Q27. A block of mass 5.0 kg slides down a rough slope, descending a vertical height of 4.0 m. Friction does 60 J of work against the block. What is the kinetic energy of the block at the bottom (it started from rest)?

  • (A) 136 J
  • (B) 196 J
  • (C) 256 J
  • (D) 60 J
Show Answer

Answer: A – KE gained = GPE lost − work done against friction = (5.0 × 9.81 × 4.0) − 60 = 196 − 60 = 136 J. The missing 60 J becomes internal (thermal) energy.

Q28. A spring of spring constant 200 N m−1 is compressed by 0.10 m. What is the elastic potential energy stored in it? (E = ½kx2.)

  • (A) 2.0 J
  • (B) 20 J
  • (C) 0.50 J
  • (D) 1.0 J
Show Answer

Answer: D – E = ½kx2 = 0.5 × 200 × 0.102 = 0.5 × 200 × 0.010 = 1.0 J. (Elastic PE is covered fully in Section 6, Deformation of Solids.)

Q29. In a small hydroelectric scheme, 500 kg of water falls through 20 m and the generator produces 78 kJ of electrical energy. What is the efficiency of the conversion?

  • (A) 64%
  • (B) 98%
  • (C) 80%
  • (D) 39%
Show Answer

Answer: C – Input = mgΔh = 500 × 9.81 × 20 = 98 100 J ≈ 98 kJ. Efficiency = 78 / 98 = 0.80 = 80%.

Q30. A skier moving at 6.0 m s−1 descends a frictionless slope through a vertical drop of 3.0 m. What is the skier’s speed at the bottom?

  • (A) 7.7 m s−1
  • (B) 9.7 m s−1
  • (C) 12 m s−1
  • (D) 6.8 m s−1
Show Answer

Answer: B – Energies add: ½mv2 = ½mu2 + mgΔh, so v2 = u2 + 2gΔh = 36 + 2 × 9.81 × 3.0 = 36 + 58.9 = 94.9, giving v = 9.74 ≈ 9.7 m s−1. Note the speeds do NOT simply add.

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