AQA AS Level Physics Paper 2, June 2024: Question 27

1 mark · Medium difficulty · Multiple Choice

Calculate the efficiency of a lift motor raising a load of 750 N through a vertical distance of 3.0 m in 1.5 s, given a motor current of 12 A and potential difference of 200 V.

Practise this question

Question

Multiple-choice question 27 asking for the efficiency of a lift motor that raises a 750 N load through 3.0 m in 1.5 s, with a motor current of 12 A and potential difference of 200 V. Four options are provided: A 16%, B 63%, C 88%, and D 94%.
Question text

27 The electric motor of a lift raises a load of 750 N at a constant speed. The load moves

through a vertical distance of 3.0 m in 1.5 s. As the load is being raised, the current in the

motor is 12 A and the potential difference across the motor is 200 V.

What is the efficiency of the lift?

[1 mark]

A 16%

B 63%

C 88%

D 94%

Mark scheme

Show the mark scheme Mark scheme indicating question 27 has the correct answer B, corresponding to 63%.

27 B 63%

How to answer it

Calculating the Efficiency of an Electric Motor

AQA AS Level Physics • Mechanics & Electricity

What this question tests

This question tests your ability to link fundamental mechanics concepts (work done, gravitational potential energy, power) with electrical power equations (P = IV) to calculate the efficiency of an energy transfer system. You are required to distinguish between total input electrical energy and useful output mechanical energy.

Question 27: Efficiency Multiple Choice

Exam breakdown and step-by-step solution

✅ Correct Answer

Option B: 63%

Awarded 1 mark for selecting the correct option.

💡 Key Knowledge

  • Efficiency formula: Efficiency = (Useful output energy / Total input energy) × 100% (or using power: Useful output power / Total input power).
  • Electrical Power: P = I × V (Current × Potential difference).
  • Mechanical Work Done: Work = Force × Distance ( F × Δh ), which represents the useful output energy gained by the load.

🧠 Exam Technique

When dealing with rates (like lifting a load over a time interval), it is often easiest to calculate either total energy in time t or power values. Working with power values avoids calculating energy for a specific time and directly compares rates of energy transfer.

❌ Common Errors

  • Inverting the ratio: Dividing input power by output power, resulting in a value greater than 100% (or picking distractor values like option D).
  • Forgetting time or distance mismatch: Incorrectly multiplying or dividing by the given time 1.5 s in inconsistent parts of the calculation.

📐 Step-by-Step Calculation

  1. Calculate the total input electrical power ( P_in ):
    P_in = I × V = 12 A × 200 V = 2400 W
  2. Calculate the useful output mechanical power ( P_out ):
    Useful work done = Force × distance = 750 N × 3.0 m = 2250 J
    Output power = Work / time = 2250 J / 1.5 s = 1500 W
    Alternatively, calculate velocity v = 3.0 / 1.5 = 2.0 m s⁻¹ , then P_out = F × v = 750 × 2.0 = 1500 W .
  3. Calculate the efficiency:
    Efficiency = (P_out / P_in) × 100%
    Efficiency = (1500 / 2400) × 100% = 0.625 × 100% = 62.5%
  4. Round to appropriate significant figures:
    Rounding 62.5% gives 63%, matching Option B.

Topics

Physics · 3.4 Mechanics and materials · 3.5 Electricity

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.