OCR GCSE Physics A: Work done, stored energy and power from forces (J249) – Revision Notes

These notes condense statements P2.2m, P2.2n and P2.2o of the OCR GCSE (9-1) Gateway Science Physics A specification (J249), Version 5.0, July 2026, first assessment in 2018. All three statements are assessed in both tiers. For full explanations and longer worked examples, read the study guide.

Useful links: OCR GCSE Physics hub, printable checklist, practice questions, OCR GCSE Physics forces revision notes and the free 10-minute diagnostics.

Key definitions

• Work done: energy transferred when a force moves an object along the line of action of the force.
• Joule (J): the work done when a force of 1 N moves an object 1 m along the line of action of the force. So 1 N m = 1 J.
• Power: the rate of energy transfer, which is the same as the rate of doing work.
• Watt (W): one joule per second.

Equations and units

Quantity	Equation	Units
Work done	W = F × s	J = N × m
Power	P = W ÷ t	W = J ÷ s
Gravitational potential energy	E = m × g × h	J = kg × N/kg × m
Kinetic energy	E = ½ × m × v²	J = kg × (m/s)²

Rearranged forms:

To find	Use
Force	F = W ÷ s
Distance	s = W ÷ F
Work done	W = P × t
Time	t = W ÷ P

Prefix reminders: 1 kJ = 1,000 J, 1 kW = 1,000 W, 1 MJ = 1,000,000 J.

Method in steps: work done and power

1. Write down what you are given with units. Convert kJ, km, minutes and so on into J, m and s.
2. Check the distance is along the line of action of the force.
3. Write the equation in words, then in symbols.
4. Rearrange if needed before substituting.
5. Substitute, calculate, and give the unit. Round sensibly (usually to the same significant figures as the least precise data, or as the question asks).
6. Ask whether the size is sensible.

Work done: points to remember

• No movement means no work done by that force.
• Movement at right angles to a force means that force does no work. A force that holds a load up while it moves sideways does no work on it.
• Work done by a force lifting at steady speed equals the gain in gravitational potential energy.
• Work done against friction at steady speed goes to thermal energy stores. The object and the surface warm up.
• Work done by a braking force equals the kinetic energy removed, so braking force × braking distance = ½ × m × v² when the vehicle stops.

Energy transfers at a glance

Force acts to…	Main store that increases
Lift an object at steady speed	Gravitational potential
Speed up an object	Kinetic
Overcome friction at steady speed	Thermal (object and surroundings)
Stop a moving object using friction	Thermal (kinetic store decreases)

Worked reminders

Work done. A 240 N force moves a cart 0.75 m along its own line of action. W = 240 × 0.75 = 180 J.

Distance from work. An 810 J push with a 45 N force: s = 810 ÷ 45 = 18 m.

Lifting. A 4.0 kg block is raised 2.5 m. With g = 10 N/kg, GPE gain = 4.0 × 10 × 2.5 = 100 J. The lifting force at steady speed is 40 N.

Kinetic energy. A 0.50 kg trolley at 6.0 m/s: E = 0.5 × 0.50 × 6.0² = 0.5 × 0.50 × 36 = 9.0 J.

Power. 600 J transferred in 12 s: P = 600 ÷ 12 = 50 W.

Energy from power. A 2.5 kW heater runs for 40 s: W = P × t = 2,500 × 40 = 100,000 J.

Two-step reminders

Braking. An 800 kg van travels at 10 m/s and stops with a constant braking force of 400 N.

KE = 0.5 × 800 × 10² = 40,000 J
s = W ÷ F = 40,000 ÷ 400 = 100 m

Lifting and power. A hoist raises a 36 kg load by 30 m in 45 s (g = 10 N/kg).

W = m × g × h = 36 × 10 × 30 = 10,800 J
P = W ÷ t = 10,800 ÷ 45 = 240 W

Command words and what to write

Command word	What the answer needs
State / recall	The equation or definition, with units
Calculate	Working shown, rearranged equation, answer with unit
Describe the energy transfer	Name the store that gains energy and the store that loses it
Explain	A reason linking the cause to the effect, such as “distance along the line of action is zero, so no work is done”

Must-know distinctions

Idea	Not the same as	Difference
Work done	Force	Work is energy transferred (J). Force is a push or pull (N)
Power	Energy	Power is the rate of transfer (W). Energy is the amount (J)
Work done on an object	Effort felt by a person	Holding a load still does no work on it, though muscles still tire
Distance along the line of action	Total distance travelled	Only the movement along the force counts
N m as work	N m as a moment	Here N m means joules of work done
Energy “used up”	Energy transferred	Energy is transferred to other stores, not destroyed

Explaining power with examples

• Two people climb the same stairs. The faster person does the same work in less time, so has the greater power.
• Two motors lift the same load through the same height. The motor that does it in less time has the greater power.
• A device with a higher power rating transfers more energy each second.

Keep the wording tied to the situation, for example “the work done is the same but the time is shorter, so power = work ÷ time is larger”.

Quick self-test

1. State the equation for work done and name the unit of each quantity.
2. A force of 60 N moves a box 4.5 m along the line of the force. How much work is done?
3. A force does 2.4 kJ of work in moving a trolley. The force is 150 N. How far does the trolley move?
4. A 4.0 kg object is lifted 2.5 m at steady speed. How much energy is stored (g = 10 N/kg)?
5. How much kinetic energy has a 0.50 kg trolley moving at 6.0 m/s?
6. 600 J of work is done in 12 s. Calculate the power.
7. A 2.5 kW heater runs for 40 s. How much energy does it transfer?
8. A motor does 1,800 J of work at a power of 450 W. How long does this take?
9. Machine A does 900 J in 3.0 s. Machine B does 1,500 J in 6.0 s. Which has the greater power, and by how much?
10. Why does a person carrying a bag at constant height along a level road do no work on the bag?
11. Convert 18 N m to joules.

Answers

1. Work done (J) = force (N) × distance moved along the line of action of the force (m).
2. 60 × 4.5 = 270 J.
3. 2.4 kJ = 2,400 J. s = 2,400 ÷ 150 = 16 m.
4. 4.0 × 10 × 2.5 = 100 J.
5. 0.5 × 0.50 × 36 = 9.0 J.
6. 600 ÷ 12 = 50 W.
7. 2,500 × 40 = 100,000 J (100 kJ).
8. t = 1,800 ÷ 450 = 4.0 s.
9. A: 900 ÷ 3.0 = 300 W. B: 1,500 ÷ 6.0 = 250 W. A has more power, by 50 W.
10. The upward force and the horizontal movement are at right angles. Nothing moves along the force, so its work is zero.
11. 1 N m = 1 J, so 18 J.

Where marks are usually lost

• Not converting kJ, minutes or centimetres before substituting.
• Writing the answer with no unit, or with N instead of J.
• Using the mass instead of the weight as the force. Weight = m × g, in newtons.
• Calculating GPE with the wrong height, such as the length of a slope rather than the vertical height.
• Saying energy is “lost” or “used up” rather than transferred to thermal stores of the surroundings.
• Describing power as “how strong” a machine is, or as “the energy”, instead of the rate of transfer.
• Forgetting that distance must be along the line of action.
• Writing too few significant figures too early and carrying a rounded value into a later step.

Official syllabus

OCR, GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0, July 2026, first assessment in 2018. Statements P2.2m, P2.2n and P2.2o.