Power and efficiency

Energy

GCSE Physics (8463)

ORIGINAL

Input energy from power × time, efficiency as useful ÷ total, and the useful power output — with a check that efficiency is below 1.

Board and spec code confirmed against AQA 8463 · registry checked 2026-07-17How this checking works

Question

An electric motor has an input power of

400\,\text{W}

. In

6.0\,\text{s}

it transfers

1800\,\text{J}

of energy usefully to lift a load. Calculate (a) the total energy supplied to the motor, (b) the efficiency of the motor, (c) the useful power output of the motor.

Every step worked, with the reasoning.

1
$E = Pt = 400 \times 6.0 = 2400\,\text{J}$ supplied
Total input energy = input power × time.
2
$\text{efficiency} = \dfrac{\text{useful output energy transfer}}{\text{total input energy transfer}} = \dfrac{1800}{2400}$
Useful output goes on top — an efficiency can never be greater than 1.
3
$\text{efficiency} = 0.75$ , or $75\%$
Multiply the decimal by 100 to give a percentage.
4
$P_{\text{useful}} = \dfrac{E}{t} = \dfrac{1800}{6.0} = 300\,\text{W}$
Power is the rate of energy transfer: useful energy ÷ time.
5
Check: $\dfrac{300}{400} = 0.75$ ; the other $2400 - 1800 = 600\,\text{J}$ is wasted, dissipated to the surroundings.
Efficiency can also be found from powers: useful power output ÷ total power input. Both routes must agree.

Answer: (a) $2400\,\text{J}$ ; (b) $0.75$ ( $75\%$ ); (c) $300\,\text{W}$

Question

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