OxfordAQA International AS and A-level Physics: Work, power, efficiency and conservation of energy (9630) – 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 are aligned to the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, sections 3.2.7 (work, energy and power) and 3.2.8 (conservation of energy). Both are shared International AS and A-level content. Take g = 9.81 N kg⁻¹.

Read the study guide and revision notes first. See also the course page, the checklist and the free diagnostics.

Questions

1. A porter carries a 12 kg bag at constant speed for 20 m along a level corridor. (a) State the work done by the porter’s upward force on the bag over the 20 m, and explain your answer. [2] (b) The porter then lifts the bag at constant speed from the floor onto a shelf 0.90 m high. Calculate the work done against gravity. [2]

2. A boy pulls a loaded trolley 15 m along level ground with a rope. The tension is 40 N and the rope makes 28° with the ground. (a) Calculate the work done by the tension. [2] (b) The pull takes 20 s. Calculate the average power developed by the tension. [2]

3. A winch takes in energy at a rate of 2.4 kW and does useful work at a rate of 1.5 kW. (a) Calculate its efficiency as a percentage. [1] (b) Calculate the energy dissipated in one minute. [2]

4. A 1.5 kg trolley starts from rest on a smooth horizontal track. A horizontal force acts along the track. It rises uniformly from 0 to 18 N over the first 0.40 m, stays at 18 N from 0.40 m to 0.80 m, then falls uniformly to 6.0 N at 1.00 m. (a) Calculate the work done on the trolley over the 1.00 m. [3] (b) Calculate the speed of the trolley at 1.00 m. [2]

5. A car has a maximum power output of 55 kW. The resistive force is a constant 1400 N. Its mass is 1100 kg. (a) Calculate its maximum speed on a level road. [2] (b) Calculate its acceleration at 20 m s⁻¹ with full power. [3] (c) Explain why the acceleration falls to zero as the car approaches its maximum speed. [1]

6. A pendulum bob hangs from a light string of length 0.80 m. It is pulled aside until the string makes 60° with the vertical and released from rest. Ignore air resistance. (a) Calculate the speed of the bob at the lowest point. [3] (b) State the effect on this speed of doubling the mass of the bob. [1]

7. A ball is dropped onto a hard floor and rebounds to a lower height than it was released from. Describe the energy transfers until it reaches the top of the rebound, and explain why the rebound is lower. [4]

8. A catapult band behaves like a spring with spring constant 300 N m⁻¹. It is stretched 0.20 m and used to fire a 0.025 kg stone horizontally. (a) Calculate the elastic energy stored. [2] (b) Calculate the stone’s speed if all the energy becomes its kinetic energy. [2] (c) In practice 15% of the stored energy is dissipated. Calculate the actual launch speed. [2]

9. A skater of mass 62 kg passes the top of a slope at 5.0 m s⁻¹. The top is 3.0 m above the bottom, and the slope is 14 m long. She reaches the bottom at 8.0 m s⁻¹. (a) Calculate the work done against resistive forces. [4] (b) Calculate the mean resistive force along the slope. [1]

10. A motorised conveyor raises 4.0 kg of gravel every second through a vertical height of 5.5 m. Its electrical power input is 350 W. The gravel gains no kinetic energy. (a) Calculate the useful power output. [2] (b) Calculate the efficiency of the conveyor. [1] (c) Suggest one reason why the efficiency is less than 100%. [1]

11. A lift cabin and passengers have a total mass of 900 kg. They are raised at a steady 2.0 m s⁻¹, with friction in the guide rails giving a constant resistive force of 1500 N. (a) Calculate the force the cable must exert. [2] (b) Calculate the useful power delivered to the cabin. [2] (c) The motor is 70% efficient. Calculate its input power. [2]

12. A cyclist and bicycle have a combined mass of 85 kg. They freewheel 120 m down a straight road inclined at 4.0° to the horizontal, without pedalling. Their speed rises from 3.0 m s⁻¹ to 11 m s⁻¹. (a) Calculate the loss in gravitational potential energy. [2] (b) Calculate the gain in kinetic energy. [2] (c) Calculate the work done against resistive forces and hence the mean resistive force. [2] (d) Explain why the resistive force at the bottom is likely to be greater than this mean value. [1]

Answers

1. (a) The work done is zero [1] because the upward force is at 90° to the horizontal displacement, so cos 90° = 0 [1]. [2] (b) W = mgΔh = 12 × 9.81 × 0.90 [1] = 106 J [1]. [2] Examiner insight: Explain using the angle between force and displacement. “Nothing moves” is wrong here, because the bag does move.

2. (a) W = Fs cos θ = 40 × 15 × cos 28° [1] = 5.3 × 10² J (530 J) [1]. [2] (b) P = W ÷ t = 530 ÷ 20 [1] = 26.5 W [1]. [2] Examiner insight: Using F × s without cos θ gives 600 J, which is a different quantity. Carry the unrounded work into part (b).

3. (a) Efficiency = 1.5 ÷ 2.4 = 0.625, so 62.5% (or 63%) [1]. [1] (b) Dissipated power = 2.4 − 1.5 = 0.9 kW [1]. Energy = 900 × 60 = 5.4 × 10⁴ J (54 kJ) [1]. [2] Examiner insight: Convert kW to W before multiplying by the time in seconds.

4. (a) Triangle: ½ × 0.40 × 18 = 3.6 J [1]. Rectangle: 0.40 × 18 = 7.2 J [1]. Trapezium: ½ × (18 + 6.0) × 0.20 = 2.4 J. Total = 13.2 J [1]. [3] (b) Smooth track, so the work becomes kinetic energy: ½ × 1.5 × v² = 13.2 [1]. v = √17.6 = 4.2 m s⁻¹ [1]. [2] Examiner insight: Show each shape’s area separately, so a slip in one can be spotted. The speed comes from equating the work to kinetic energy.

5. (a) At maximum speed driving force equals resistive force, 1400 N [1]. v = P ÷ F = 55 000 ÷ 1400 = 39 m s⁻¹ [1]. [2] (b) Driving force = 55 000 ÷ 20 = 2750 N [1]. Resultant = 2750 − 1400 = 1350 N [1]. a = 1350 ÷ 1100 = 1.2 m s⁻² [1]. [3] (c) As speed rises at fixed power the driving force P ÷ v falls until it equals the resistive force, so the resultant force is zero [1]. [1] Examiner insight: At 20 m s⁻¹ the driving force is P ÷ v, not the 1400 N resistive force. Quote F = ma only for the resultant.

6. (a) Height: Δh = 0.80 (1 − cos 60°) = 0.40 m [1]. mgΔh = ½mv² [1]. v = √(2 × 9.81 × 0.40) = 2.8 m s⁻¹ [1]. [3] (b) No effect (the mass cancels) [1]. [1] Examiner insight: Measure the height vertically from the lowest point, not along the arc. A small triangle diagram shows where 0.40 m comes from.

7. Gravitational potential energy transfers to kinetic energy as the ball falls [1]. On impact, kinetic energy goes to elastic energy as the ball and floor deform, then back to kinetic energy as it rebounds [1]. Some energy is dissipated as thermal energy and sound in the ball and floor [1]. Air resistance also does work, so less energy remains and the rebound height is lower [1]. [4] Examiner insight: Name each store and give the transfers in order. “Energy is lost” does not say where it has gone.

8. (a) Energy = ½kΔL² = ½ × 300 × 0.20² [1] = 6.0 J [1]. [2] (b) ½mv² = 6.0 so v = √(2 × 6.0 ÷ 0.025) [1] = 22 m s⁻¹ (21.9) [1]. [2] (c) Kinetic energy = 0.85 × 6.0 = 5.1 J [1]. v = √(2 × 5.1 ÷ 0.025) = 20 m s⁻¹ (20.2) [1]. [2] Examiner insight: Take the 15% off the energy, not off the speed.

9. (a) Initial kinetic energy = ½ × 62 × 5.0² = 775 J [1]. Potential energy lost = 62 × 9.81 × 3.0 = 1825 J [1]. Final kinetic energy = ½ × 62 × 8.0² = 1984 J [1]. Work against resistance = 775 + 1825 − 1984 = 616 J [1]. [4] (b) Force = 616 ÷ 14 = 44 N [1]. [1] Examiner insight: Use 3.0 m for the potential energy and 14 m for the work against the resistive force.

10. (a) Useful power = mgΔh per second = 4.0 × 9.81 × 5.5 [1] = 216 W [1]. [2] (b) Efficiency = 215.8 ÷ 350 = 0.617, so 62% [1]. [1] (c) For example, friction in the roller bearings, or thermal energy in the motor windings [1]. [1] Examiner insight: State that the useful output is the potential energy gained per second, since the gravel gains no kinetic energy.

11. (a) Constant speed, so the cable force equals weight plus resistance: 900 × 9.81 + 1500 [1] = 8829 + 1500 = 1.03 × 10⁴ N [1]. [2] (b) P = Fv = 10 329 × 2.0 [1] = 2.1 × 10⁴ W (20.7 kW) [1]. [2] (c) Input = 20 658 ÷ 0.70 [1] = 3.0 × 10⁴ W (29.5 kW) [1]. [2] Examiner insight: Say that the forces balance at constant speed. To find input from output, divide by the efficiency.

12. (a) Δh = 120 sin 4.0° = 8.37 m [1]. ΔEp = 85 × 9.81 × 8.37 = 6.98 × 10³ J [1]. [2] (b) ΔEk = ½ × 85 × (11² − 3.0²) [1] = ½ × 85 × 112 = 4.76 × 10³ J [1]. [2] (c) Work against resistive forces = 6980 − 4760 = 2.2 × 10³ J [1]. Mean force = 2220 ÷ 120 = 18.5 N [1]. [2] (d) Air resistance increases with speed, so it is larger at 11 m s⁻¹ than the average over the descent [1]. [1] Examiner insight: The kinetic energy change uses the difference of the squares of the speeds. Write the energy balance in words first.

Where marks are usually lost

• Leaving out cos θ when the force is not along the displacement, or using sin where cos belongs.
• Using slope length in mgΔh, or vertical height in work against friction.
• Using peak force times total distance instead of the area under the graph.
• Using the resistive force in P = Fv below maximum speed.
• Subtracting speeds before squaring in the change in kinetic energy.
• Writing “energy is lost” instead of naming where it goes.
• Mixing up input and output when using efficiency, which gives a value above 100%.
• Missing units, or mixing kJ and J without converting.

Next steps

• Review the revision notes and study guide.
• Open the course page and the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.2.7 and 3.2.8.