From energy to power (two steps)

Energy

GCSE Physics (8463)

ORIGINAL

Step 1 finds the energy transferred, step 2 divides by the time — carrying the unrounded energy through.

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Question

A lift motor raises a lift of mass

540\,\text{kg}

at a steady speed through a vertical height of

24\,\text{m}

18\,\text{s}

. Gravitational field strength

= 9.8\,\text{N/kg}

. Calculate the useful power output of the motor. Give your answer in kilowatts to 2 significant figures.

Every step worked, with the reasoning.

1
Step 1, the energy: $E_p = mgh$
Power needs an energy and a time. The useful energy here is the increase in the lift's gravitational potential energy store.
2
$E_p = 540 \times 9.8 \times 24 = 127\,008\,\text{J}$
Keep this unrounded value for the next step.
3
Step 2, the power: $P = \dfrac{E}{t} = \dfrac{127\,008}{18}$
Power is the energy transferred per second.
4
$P = 7056\,\text{W} = 7.056\,\text{kW}$
Divide by 1000 to convert watts to kilowatts.
5
$P = 7.1\,\text{kW}$ (2 s.f.)
Round once, at the end, to the significant figures asked for.

Answer: $7.1\,\text{kW}$