AQA AS Level Physics Paper 2, June 2022: Question 27

1 mark · Medium difficulty · Multiple Choice

Calculate the rate of energy transfer to the brake and rope for a climber descending at a steady speed.

Practise this question

Question

A multiple-choice question featuring a diagram of a climber descending a rope using a brake attached to a harness. The climber descends at a steady speed of 0.50 m s^-1, and the combined mass is 60 kg. Below the diagram, four options A (15 W), B (29 W), C (150 W), and D (290 W) are provided alongside selection boxes.
Question text

27 A climber wears a harness attached to a rope. The rope passes through a brake. There is

friction between the rope and the brake.

The climber uses the brake to descend at a steady speed of 0.50 m s−1.

The combined mass of the climber, the harness and the brake is 60 kg.

What is the rate of energy transfer to the brake and rope?

[1 mark]

A 15 W

B 29 W

C 150 W

D 290 W

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer is option D (290 W).

27 D (AO2) 290 W

How to answer it

Climber Descent & Rate of Energy Transfer

What this question tests

This question assesses your ability to link gravitational potential energy, power (rate of energy transfer), and steady-state motion at constant velocity. You are required to apply physics equations to a practical, real-world context involving friction and mechanical work.

Question 27 (Multiple Choice)

Determining the Rate of Energy Transfer to the Brake and Rope

✅ Correct Answer

D (290 W)

Marks available: 1 mark (AO2)

💡 Key Knowledge

  • Power is defined as the rate of transfer of energy: P = ΔE / Δt .
  • When an object descends at a steady speed, the loss in gravitational potential energy per second is equal to the thermal energy generated by friction in the brake.
  • Gravitational potential energy change: ΔE = m g Δh .
  • Combining these gives power: P = m g (Δh / Δt) , where Δh / Δt is the vertical velocity ( v ). Therefore, P = m g v .

🧠 Exam Technique

Don't get bogged down looking for a formula for "friction power". Recognise that energy conservation applies: gravitational potential energy is being converted into thermal energy via the brake at a constant rate because speed is constant.

❌ Common Errors

  • Gravity value trap: Forgetting to multiply by standard gravitational field strength ( g = 9.81 m s⁻² ), using g = 9.8 , or incorrectly using 10 m s⁻² which yields 300 W (tempting students to pick a wrong distractor).
  • Kinetic energy confusion: Adding a kinetic energy term ( 0.5 m v² ). Because speed is steady (constant), there is zero change in kinetic energy!

📐 Step-by-Step Calculation

Break the calculation down logically to ensure full accuracy and safeguard against arithmetic slips:

  1. Identify given values: Mass m = 60 kg , Velocity v = 0.50 m s⁻¹ , Gravity g = 9.81 m s⁻² .
  2. State the working equation: P = m × g × v
  3. Substitute values: P = 60 × 9.81 × 0.50
  4. Calculate intermediate value: 60 × 0.50 = 30
  5. Final multiplication: 30 × 9.81 = 294.3 W
  6. Apply significant figures: Round to 2 significant figures to match input data ( 60 kg and 0.50 m s⁻¹ are given to 2 sf), yielding 290 W .

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.