OCR A Level Physics A: Work, energy, power and efficiency (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.3 (Work, energy and power) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. Use g = 9.81 N kg⁻¹. Start with the study guide or the revision notes.

Questions

1. Define the joule. [2]

2. A child pulls a toy cart 7.5 m along a level path using a rope. The tension is 22 N and the rope is at 40° to the path. Calculate the work done by the tension. [2]

3. A porter carries a box horizontally across a room at constant velocity.

(a) Explain why the upward force the porter exerts on the box does no work. [2] (b) The box is then pushed along the floor and slows to rest. State the sign of the work done on it by friction and name the energy transfer. [1]

4. Derive Ep = mgh for a mass m raised through a height h in a uniform gravitational field. [3]

5. A dart of mass 0.020 kg travels at 14 m s⁻¹.

(a) Find its kinetic energy. [2] (b) Calculate the speed at which its kinetic energy would be 5.0 J. [2]

6. A block of mass 0.80 kg is released from rest at a height of 1.2 m on a smooth curved track. At the bottom it slides onto a rough horizontal surface and stops after 3.0 m.

(a) Calculate its speed at the bottom of the track. [2] (b) Calculate the average frictional force on the rough surface. [3]

7. A crane raises a load of mass 640 kg through 12 m at constant speed in 20 s. The electrical power supplied to its motor is 5.2 kW.

(a) Calculate the useful output power. [2] (b) Calculate the efficiency of the crane. [2]

8. (a) Derive P = Fv for a constant force F moving an object at constant speed v. [2] (b) A cyclist rides at a steady 8.0 m s⁻¹ and the driving force is 45 N. Calculate the power developed at the wheels. [2]

9. A pump raises 900 kg of water every minute through a height of 8.0 m. Its electrical input power is 1.5 kW.

(a) Calculate the useful power output. [2] (b) Calculate the efficiency. [1] (c) State where most of the wasted energy goes. [1]

10. A car of mass 1250 kg travelling at 22 m s⁻¹ is braked to rest over 46 m by a constant braking force.

(a) Calculate the initial kinetic energy. [2] (b) Calculate the braking force. [2] (c) The car now travels at twice the speed and the same braking force is applied. Calculate the stopping distance. [2]

11. A cyclist and bicycle have a combined mass of 82 kg. Without pedalling, they descend a road 600 m long that falls 35 m vertically. Their speed increases from 3.0 m s⁻¹ to 11.0 m s⁻¹.

(a) Calculate the loss of gravitational potential energy and the gain in kinetic energy. [3] (b) Calculate the energy dissipated and the average resistive force. [3] (c) On a level road the cyclist rides at a constant 11.0 m s⁻¹ against a resistive force of 48 N. Calculate the power needed at the wheels. [1] (d) The cyclist’s body is 24% efficient at supplying this power. Calculate the total power the body must produce. [1]

12. A crate of mass 9.0 kg is pulled 4.0 m up a slope inclined at 25° to the horizontal by a force of 85 N acting 20° above the slope surface. Friction is a constant 18 N and the crate starts from rest.

(a) Calculate the work done by the 85 N force. [2] (b) Calculate the gain in gravitational potential energy. [2] (c) Calculate the speed of the crate after 4.0 m. [3]

Answers

1. One joule is the work done (energy transferred) when a force of one newton [1] moves its point of application one metre in the direction of the force [1]. [2] Examiner insight: “Force times distance” is an incomplete definition. Include “in the direction of the force”, since that phrase carries the physics.

2. W = Fx cos θ = 22 × 7.5 × cos 40° [1] = 126 J [1]. [2] Examiner insight: Show the substitution and the angle used. Writing the equation and the numbers makes your method visible, even if a slip occurs later.

3. (a) The force acts upwards, at 90° to the horizontal displacement [1]. Work depends on the force component along the displacement, and cos 90° = 0, so no energy is transferred [1]. (b) Friction acts against the motion, so its work is negative [1]: kinetic energy is transferred to thermal energy. [3] Examiner insight: Refer to the direction of the force relative to the displacement, not to “no movement”. The box does move, so “nothing moves” is wrong.

4. Lift at constant speed, so the lifting force equals the weight, F = mg [1]. Work done = mg × h, the force multiplied by the distance moved along its own direction [1]. This energy is stored as gravitational potential energy, so Ep = mgh [1]. [3] Examiner insight: A derivation needs the condition (constant speed, so no change in kinetic energy) and the link between work done and energy gained. Quoting the final formula alone shows no reasoning.

5. (a) Ek = ½mv² = ½ × 0.020 × 14² [1] = 1.96 J [1]. (b) v = √(2Ek/m) = √(2 × 5.0 ÷ 0.020) [1] = 22 m s⁻¹ (22.4) [1]. [4] Examiner insight: Rearrange before substituting. A common slip is to forget to square the speed in part (a).

6. (a) Loss of Ep = gain of Ek, so v = √(2gh) [1] = √(2 × 9.81 × 1.2) = 4.85 m s⁻¹ [1]. (b) Energy dissipated on the rough surface = Ek at the bottom = mgh = 0.80 × 9.81 × 1.2 = 9.42 J [1]. Work done by friction = F × 3.0 [1], so F = 9.42 ÷ 3.0 = 3.1 N (3.14 N) [1]. [5] Examiner insight: The mass cancels in part (a), but you need it in part (b). Equating the work done by friction to the kinetic energy removed is the neat route.

7. (a) Useful power = mgh/t = 640 × 9.81 × 12 ÷ 20 [1] = 3.8 kW (3767 W) [1]. (b) Efficiency = 3767 ÷ 5200 × 100% [1] = 72% [1]. [4] Examiner insight: Put both powers in the same unit before dividing. Efficiency has no unit; show a percentage sign.

8. (a) Work done W = Fx, so P = W/t = Fx/t [1]. At constant speed x/t = v, so P = Fv [1]. (b) P = Fv = 45 × 8.0 [1] = 360 W [1]. [4] Examiner insight: State that x/t is the constant speed. This is the step that turns the derivation from an assertion into an argument.

9. (a) Useful power = mgh/t = 900 × 9.81 × 8.0 ÷ 60 [1] = 1.18 kW (1177 W) [1]. (b) Efficiency = 1177 ÷ 1500 × 100% = 78% [1]. (c) Thermal energy (in the motor and the pipework, with some sound) [1]. [4] Examiner insight: Convert “per minute” to seconds. Name the final form of the wasted energy rather than saying it is “lost”.

10. (a) Ek = ½ × 1250 × 22² [1] = 3.0 × 10⁵ J (302 500 J) [1]. (b) Work done by the brakes = Ek, so F × 46 = 302 500 [1], giving F = 6.6 kN (6576 N) [1]. (c) At twice the speed, Ek is four times as large [1]. With the same force, the distance is four times as large: 4 × 46 = 184 m [1]. [6] Examiner insight: In part (c), explain the effect of v² on Ek. Simply multiplying 46 by two is a common wrong answer.

11. (a) Loss of Ep = mgh = 82 × 9.81 × 35 = 2.82 × 10⁴ J (28 155 J) [1]. Gain in Ek = ½ × 82 × (11.0² − 3.0²) [1] = 4.59 × 10³ J [1]. (b) Energy dissipated = 28 155 − 4592 = 2.36 × 10⁴ J [1]. This equals the work done against resistive forces, F × 600 [1], so F = 23 563 ÷ 600 = 39 N (39.3 N) [1]. (c) P = Fv = 48 × 11.0 = 528 W [1] (d) Total power = 528 ÷ 0.24 = 2200 W [1] [8] Examiner insight: Use the vertical height of 35 m in mgh and the road length of 600 m in the work done against resistance. Mixing the two distances is the usual error.

12. (a) W = Fx cos θ = 85 × 4.0 × cos 20° [1] = 319 J [1]. (b) Height gained = 4.0 sin 25° = 1.69 m [1]. ΔEp = 9.0 × 9.81 × 1.69 = 149 J [1]. (c) Work against friction = 18 × 4.0 = 72 J [1]. Gain in Ek = 319.5 − 149.3 − 72 = 98.2 J [1]. v = √(2 × 98.2 ÷ 9.0) = 4.7 m s⁻¹ (4.67 m s⁻¹) [1]. [7] Examiner insight: Set out the energy balance as work done by the pull = ΔEp + ΔEk + friction work. A clear balance lets a slip in one term be traced.

Where marks are usually lost

• Using Fx instead of Fx cos θ, or measuring θ from the wrong line.
• Using the length of the slope where the vertical height belongs in mgh.
• Forgetting to square the speed in ½mv², or squaring only part of it.
• Leaving out the energy dissipated by friction in an energy-balance question.
• Mixing units: kW with W, minutes with seconds, kW h with J.
• Writing a derivation without the starting condition (from rest, or constant speed) or the link between work and energy.
• Describing energy as “used up” instead of transferred to thermal energy.
• Giving an efficiency above 100%, or leaving it without a percentage sign.
• Rounding intermediate values and drifting from the correct answer. Carry full values and round at the end.

Next steps

• Go over the revision notes and the study guide.
• Return to the OCR A Level Physics hub and tick items off the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Section 3.3 Work, energy and power.