A car of mass 1000\,kg travels at 20\,m/s. Calculate its kinetic energy. [2 marks]

OCR OCR 2018-style practice: A person pushes a box with a force of 40\,N across a floor, moving it 6\,m in the direction of the force. Calculate the work done.

A P7 Calculate question on the recall equation W = Fd (work done equals force times distance moved in the direction of the force). Write the values: force F = 40\,N and distance d = 6\,m (1 mark for the equation). Substitute: W = Fd = 40 × 6 = 240\,J (2 marks for the calculation and the unit joules). Markers reward the correct equation, substitution and answer in joules. A common error is to forget the unit, or to use a distance that is not in the direction of the force.

OCR OCR 2021-style practice: A motor lifts a 50\,kg load through a height of 4\,m in 8\,s. Taking g = 10\,N/kg, calculate the gravitational potential energy gained by the load and the useful power output of the motor.

A P7 Calculate question using two equations. First the gravitational potential energy: E_p = mgh = 50 × 10 × 4 = 2000\,J (2 marks for the equation, substitution and answer with units). The useful **power** is the rate of transferring energy: P = Et = 20008 = 250\,W (2 marks for the power equation and the answer in watts). Markers reward the GPE of 2000\,J and the power of 250\,W. A common error is to divide the energy by the height instead of the time, or to forget that power is energy per second.

EnglandPhysicsSyllabus dot point

How do we calculate work done, kinetic energy, gravitational potential energy and power?

Work done as energy transferred by a force, the work done equation, the kinetic energy and gravitational potential energy equations, and power as the rate of doing work or transferring energy.

A focused answer to OCR Gateway GCSE Physics A topic P7 on work, energy and power, covering work done as energy transferred by a force, the work done equation, the kinetic energy and gravitational potential energy equations, and power as the rate of doing work or transferring energy.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Work done
Kinetic energy
Gravitational potential energy
Power
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What this topic is asking

OCR wants you to define work done as energy transferred by a force, use the work done equation, use the kinetic energy and gravitational potential energy equations, and define and calculate power. This is topic P7.1 (and part of P7.2) of the OCR Gateway Physics A (J249) specification.

Work done

If a force acts but the object does not move, or moves at right angles to the force, then no work is done by that force. Work done against friction is transferred to the thermal stores of the surfaces (so they warm up).

Kinetic energy

Gravitational potential energy

For a falling object with no air resistance, the gravitational potential energy lost equals the kinetic energy gained, so $mgh = \tfrac{1}{2}mv^2$ , which lets you find the speed after falling a given height.

Power

Finding the power of a runner climbing stairs

A student of mass $60\,\text{kg}$ runs up a flight of stairs of height $5\,\text{m}$ in $4\,\text{s}$ . Taking $g = 10\,\text{N/kg}$ , calculate the useful power output.

step 1: Find the work done (energy transferred)

The work done lifting the student equals the gravitational potential energy gained: $E_p = mgh = 60 \times 10 \times 5 = 3000\,\text{J}$ .

step 2: Choose the power equation

Power is the rate of transferring energy: $P = \dfrac{E}{t}$ .

step 3: Substitute and calculate

$P = \dfrac{3000}{4} = 750\,\text{W}$ .

step 4: State the answer

The useful power output is $750\,\text{W}$ . Running up the same stairs more slowly would transfer the same energy but in a longer time, so the power would be lower.

Try this

Q1. A car of mass $1000\,\text{kg}$ travels at $20\,\text{m/s}$ . Calculate its kinetic energy. [2 marks]

Cue. $E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2} \times 1000 \times 20^2 = 200\,000\,\text{J}$ ( $200\,\text{kJ}$ ).

Q2. State what is meant by a power of one watt. [1 mark]

Cue. An energy transfer (or work done) of one joule per second.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20183 marksA person pushes a box with a force of

40\,\text{N}

across a floor, moving it

6\,\text{m}

in the direction of the force. Calculate the work done.

Show worked answer →

A P7 Calculate question on the recall equation $W = Fd$ (work done equals force times distance moved in the direction of the force). Write the values: force $F = 40\,\text{N}$ and distance $d = 6\,\text{m}$ (1 mark for the equation). Substitute: $W = Fd = 40 \times 6 = 240\,\text{J}$ (2 marks for the calculation and the unit joules). Markers reward the correct equation, substitution and answer in joules. A common error is to forget the unit, or to use a distance that is not in the direction of the force.

OCR 20214 marksA motor lifts a

50\,\text{kg}

load through a height of

4\,\text{m}

8\,\text{s}

. Taking

g = 10\,\text{N/kg}

, calculate the gravitational potential energy gained by the load and the useful power output of the motor.

Show worked answer →

A P7 Calculate question using two equations. First the gravitational potential energy: $E_p = mgh = 50 \times 10 \times 4 = 2000\,\text{J}$ (2 marks for the equation, substitution and answer with units). The useful power is the rate of transferring energy: $P = \dfrac{E}{t} = \dfrac{2000}{8} = 250\,\text{W}$ (2 marks for the power equation and the answer in watts). Markers reward the GPE of $2000\,\text{J}$ and the power of $250\,\text{W}$ . A common error is to divide the energy by the height instead of the time, or to forget that power is energy per second.

Related dot points

Sources & how we know this

OCR GCSE (9-1) Physics A (Gateway Science) J249 specification — OCR (2016)