OCR GCSE Physics A: Energy stores, transfers and energy calculations (J249)

This guide is aligned to the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), for first assessment in 2018. It covers topic P7.1 Work done, statements P7.1a to P7.1e. Every statement in this unit is assessed in both the Foundation and Higher Tier papers, so nothing here is labelled Higher Tier only.

Find the course page at the OCR GCSE Physics hub and tick off topics on the printable checklist. Use the revision notes for quick recall and the practice questions to test yourself. Related pages: OCR GCSE Physics: Forces and OCR GCSE Physics: Matter. To find weak spots first, try the free diagnostics.

What this unit covers

Statement	What you must be able to do
P7.1a	Describe that, in an energy transfer in a closed system, there is no net change to the total energy (qualitative only)
P7.1b	Describe how the stores holding energy change when a system alters, in everyday situations
P7.1c	Describe the energy changes when a system is changed by heating, by work done by forces and by work done when a current flows
P7.1d	Calculate energy changes for mechanical, electrical and thermal processes, and use kW h for domestic energy use
P7.1e	Calculate the energy stored in a moving body, a stretched spring and an object raised above the ground

Energy is stored, not used up (P7.1a)

Energy is not a fuel that gets used up. Think of it as a quantity that sits in stores and moves between them. A system is the object or group of objects you choose to look at.

The law of conservation of energy says that energy is never created or destroyed; it only moves from one store to another. In a closed system there is no net change to the total energy. Energy leaving one store appears in another.

Do not say that energy is “lost”. When a ball stops bouncing, its energy has gone to the thermal store of the ball, the floor and the air. That energy is still there. It is just spread out and no longer useful. Resting objects also have energy: a book on a shelf has gravitational potential energy, and everything has a thermal store.

Energy stores and the three ways to change them (P7.1b, P7.1c)

The stores you need are kinetic, gravitational potential, elastic potential and thermal. A store changes in one of three ways:

1. Heating: energy moves because of a temperature difference.
2. Work done by forces: a force moves something through a distance.
3. Work done when a current flows: charge moves through a potential difference.

Always name the store that goes down and the store that goes up. A sentence such as “the kinetic store decreases and the thermal store of the brakes increases” gets the idea across. A vague “energy is transferred” does not.

The situations the specification lists

Situation	Store that decreases	Store that increases
Object projected upwards	Kinetic	Gravitational potential
Object moving up a slope	Kinetic	Gravitational potential, plus thermal if there is friction
Moving object hits an obstacle	Kinetic	Thermal (and sound), plus elastic if things deform
Object accelerated by a constant force	Chemical or electrical store of whatever supplies the force	Kinetic
Vehicle slowing down	Kinetic	Thermal in the brakes, tyres and road
Kettle bringing water to the boil	Electrical (work done by the current)	Thermal store of the water, then the kettle and surroundings

For a ball thrown upwards, the kinetic store falls to zero at the top as the gravitational potential store rises. If you ignore air resistance, the two changes are equal.

The energy equations

Energy	Equation	Symbols
Kinetic	E = ½mv²	m in kg, v in m/s
Gravitational potential	E = mgh	g in N/kg, h in m
Elastic potential	E = ½kx²	k in N/m, x extension in m
Work done by a force	W = Fs	F in N, s in m along the line of the force
Thermal	ΔE = mcΔθ	c in J/kg °C, Δθ in °C
Electrical	E = QV, E = Pt, P = VI, Q = It	Q in C, V in V, P in W, t in s

Energy is measured in joules (J). Work done and energy transferred are measured on the same scale, which lets you compare mechanical, electrical and thermal changes on one common scale.

Kinetic energy and gravitational potential energy (P7.1e)

Kinetic energy depends on speed squared. Doubling the speed makes the kinetic energy four times as large.

Worked example 1. A stone of mass 0.30 kg is thrown straight up at 12 m/s. Take g = 9.8 N/kg and ignore air resistance. Find its kinetic energy at launch and the greatest height it reaches.

E = ½mv² = ½ × 0.30 × 12² = 21.6 J
At the top all of this is gravitational potential energy:
mgh = 21.6
h = 21.6 / (0.30 × 9.8) = 7.35 m = 7.3 m (2 s.f.)

Notice that the mass cancels if you write it out: h = v²/2g. A heavier stone thrown at the same speed would go just as high.

Worked example 2. A 2.5 kg weight is lifted 0.80 m by a small motor. A joulemeter shows 30 J supplied. Find the gain in gravitational potential energy and the energy that goes elsewhere.

E = mgh = 2.5 × 9.8 × 0.80 = 19.6 J
Energy to other stores = 30 − 19.6 = 10.4 J

The 10.4 J goes to the thermal store of the motor and surroundings, and some leaves as sound. Total energy is still 30 J.

Elastic potential energy (P7.1e)

A stretched or compressed spring stores energy in its elastic store, provided it is not stretched past its limit. The energy depends on the extension squared.

Worked example 3. A spring of constant 250 N/m is compressed by 0.080 m and then used to launch a ball of mass 0.050 kg along a smooth horizontal surface. Find the speed of the ball as it leaves the spring.

E = ½kx² = ½ × 250 × 0.080² = 0.80 J
All of it becomes kinetic energy:
½mv² = 0.80
v² = 2 × 0.80 / 0.050 = 32
v = 5.66 = 5.7 m/s (2 s.f.)

Keep extension in metres. A compression of 8.0 cm must be written as 0.080 m before you square it.

Work done by forces (P7.1c, P7.1d)

Work done is the energy transferred by a force moving an object. Use W = Fs, where s is the distance the object moves in the direction the force acts.

Worked example 4. A 25 kg sledge on smooth ice is pushed from rest by a constant 40 N force over 6.0 m. Find the final speed.

W = Fs = 40 × 6.0 = 240 J
This becomes kinetic energy: ½ × 25 × v² = 240
v² = 2 × 240 / 25 = 19.2
v = 4.38 = 4.4 m/s

Worked example 5. A car of mass 1100 kg travelling at 15 m/s brakes to rest. The braking force is a constant 3500 N. Find the energy transferred to the thermal store and the braking distance.

Kinetic energy = ½ × 1100 × 15² = 123 750 J = 1.2 × 10⁵ J (2 s.f.)
Work done by brakes: Fs = 123 750
s = 123 750 / 3500 = 35.4 = 35 m (2 s.f.)

The kinetic store has decreased by 123 750 J and the thermal store of the brakes, tyres and road has increased by the same amount. Because kinetic energy depends on v², a car at twice the speed needs four times the braking distance for the same force.

Heating (P7.1c, P7.1d)

The energy needed to change the temperature of a material depends on its mass, its specific heat capacity and the temperature change. Revisit specific heat capacity in the matter page if you need the definition.

Worked example 6. Find the energy needed to heat 0.50 kg of aluminium (c = 900 J/kg °C) from 20 °C to 60 °C.

ΔE = mcΔθ = 0.50 × 900 × (60 − 20) = 18 000 J

Worked example 7. A kettle contains 1.5 kg of water at 20 °C (c = 4200 J/kg °C). Its power is 2.4 kW. Find the energy needed to reach 100 °C and the time taken if all the electrical energy heats the water.

ΔE = 1.5 × 4200 × 80 = 504 000 J
P = 2400 W, so t = E / P = 504 000 / 2400 = 210 s

A real kettle takes longer, because some energy heats the kettle itself and the room. The extra energy still ends up in thermal stores. It is not lost from the system.

Work done when a current flows (P7.1c, P7.1d)

When a current flows, the cell or supply does work moving charge through the circuit. Energy transferred = charge × potential difference.

Worked example 8. A charge of 450 C passes through a lamp across 12 V. A heater uses 230 V and 4.0 A for 5.0 minutes. Find both energies.

Lamp: E = QV = 450 × 12 = 5400 J
Heater: P = VI = 230 × 4.0 = 920 W
t = 5.0 × 60 = 300 s
E = Pt = 920 × 300 = 276 000 J

Kilowatt-hours (P7.1d)

At home, energy use is measured in kW h. One kW h is the energy transferred by a 1 kW appliance running for 1 hour. Use E = Pt with power in kW and time in hours.

Worked example 9. A 2.4 kW heater runs for 3.5 hours. Electricity costs 28 p per kW h. Find the energy in kW h, the cost, and the energy in joules.

E = 2.4 × 3.5 = 8.4 kW h
Cost = 8.4 × 28 = 235.2 p = £2.35
1 kW h = 1000 W × 3600 s = 3 600 000 J
E = 8.4 × 3 600 000 = 30 240 000 J = 3.0 × 10⁷ J (2 s.f.)

Common errors

• Writing “energy is used up” or “energy is lost”. Say where it was transferred to.
• Forgetting to square the speed or the extension. Doubling v multiplies kinetic energy by 4, not 2.
• Leaving a spring extension in cm or a mass in g.
• Mixing kW with seconds, or W with hours. Keep kW with hours, or W with seconds.
• Using the length of a slope instead of the vertical rise when finding gravitational potential energy.
• Naming only one store. Energy changes always involve at least two.

Official syllabus

This page is based on the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), for first assessment in 2018, published by OCR (Oxford Cambridge and RSA). Topic P7.1 Work done.