Changes in energy KE, GPE and EPE — AQA GCSE Physics

SPEC 4.1.1.2 • ENERGY — CHANGES IN ENERGY

KE, GPE and EPE

Three equations, three traps. The calculations behind kinetic, gravitational potential, and elastic potential energy — and the recurring errors that lose marks every series.

Three Equations, Three Traps

This lesson is calculation-heavy. Every Paper 1 series has at least one KE, GPE, or EPE calculation — often a multi-step chain at grade 7+. The marks come from showing your working: each line is a separate marking point.

3

Equations to Master

2

Recurring Traps

⭐

High-Priority Topic

By the end of this lesson

You'll calculate the energy in any of the three "calculate" stores, choose the right equation by inspecting the question, chain two equations together for a Grade 7+ problem, and avoid both the squaring trap (KE) and the unit-conversion trap (EPE).

Which Equation Do I Use?

Three equations, three triggers. Read the question, spot the keyword, pick the equation.

Is it MOVING?

Has a speed or velocity

Eₖ = ½ m v²

Kinetic store

Is it RAISED?

At a height in a gravity field

Eₚ = m g h

Gravitational potential store

Is it STRETCHED?

Or compressed (a spring)

Eₑ = ½ k e²

Elastic potential store

Equation Sheet

All three equations have been printed on the AQA equation sheet provided in the exam since June 2022. You still need to recognise them, choose between them, and apply them correctly. The equation sheet doesn't help if you don't know which one to pick.

Kinetic Energy

Kinetic energy is the energy stored in a moving object. The bigger the mass, the more kinetic energy. The bigger the speed — the kinetic energy goes up squared.

Equation

Eₖ = ½ m v²

Eₖ = kinetic energy in joules (J)

m = mass in kilograms (kg)

v = speed in metres per second (m/s)

The squared matters

Doubling the speed of a car doesn't double its kinetic energy — it quadruples it. Tripling the speed gives 9× the energy. This is the physics behind why braking distance grows so much faster than speed.

Worked Example — Kinetic Energy

A car of mass 1500 kg is travelling at 20 m/s. Calculate its kinetic energy. 3 marks

EEquation:Eₖ = ½ m v²

SSubstitute:Eₖ = 0.5 × 1500 × 20² = 0.5 × 1500 × 400

UUnit: joules (J)

AAnswer:Eₖ = 300 000 J = 3.0 × 10⁵ J

Mark allocation

1 mark for the equation, 1 mark for the substitution, 1 mark for the answer with the right unit. Show every line — examiners can't award follow-on marks if there's no working.

The Squared Trap

The single most-cited error on KE calculations is forgetting to square the speed. AQA examiner reports flag this every year.

✗ Loses 2 of 3 marks

Eₖ = 0.5 × m × v

= 0.5 × 1500 × 20

= 15 000 J

The number is 20× too small. The mark scheme can't credit this.

✓ Full marks

Eₖ = 0.5 × m × v²

= 0.5 × 1500 × 20²

= 0.5 × 1500 × 400

= 300 000 J

Squaring is the difference between right and wrong.

Habit to build

When you write the substitution line, write the number with a small² immediately. "v² = 400" as a separate working step makes the squaring impossible to miss — and gives you a clean intermediate value to substitute.

Gravitational Potential Energy

Gravitational potential energy is the energy stored when an object is raised in a gravity field. Lift it higher, give it more mass, or move to a stronger gravity field — all increase the GPE.

Equation

Eₚ = m g h

Eₚ = gravitational potential energy in joules (J)

m = mass in kilograms (kg)

g = gravitational field strength in newtons per kilogram (N/kg)

h = height in metres (m)

Value of g

On Earth, g = 9.8 N/kg. The question will give you the value (sometimes 9.8, sometimes 10 N/kg for round-number questions). Use the value the question gives — never invent your own.

Worked Example — Gravitational PE

A 50 kg student climbs a flight of stairs 4.5 m high. Calculate the gain in gravitational potential energy. Take g = 9.8 N/kg. 3 marks

EEquation:Eₚ = m g h

SSubstitute:Eₚ = 50 × 9.8 × 4.5

UUnit: joules (J)

AAnswer:Eₚ = 2205 J ≈ 2200 J (to 2 s.f.)

Sig figs

Round the final answer to the same number of significant figures as the data given. 50, 9.8, 4.5 are all 2 s.f. (50 is borderline) — so 2 s.f. on the answer is safe. Show the unrounded value first, then round at the end.

The cm → m Trap

If the question gives the height in centimetres, you must convert to metres before substituting. Substituting a height in cm gives a number 100× too big.

✗ Wrong substitution

"A 0.5 kg ball is lifted 80 cm. Calculate GPE."

Eₚ = 0.5 × 9.8 × 80

= 392 J

The 80 should have been 0.80 m. Answer is 100× too big.

✓ Convert at substitution

"A 0.5 kg ball is lifted 80 cm. Calculate GPE."

h = 80 cm = 0.80 m

Eₚ = 0.5 × 9.8 × 0.80

= 3.92 J

Conversion done first, on its own line.

Habit to build — convert ALL units at substitution

Whenever a question gives length in cm, mm, or km, convert to metres on a separate line before substituting.

The same applies to mass: if the question gives mass in grams, convert to kg before substituting (divide by 1000). A 25 g ball is 0.025 kg — not 25 kg, and not 25 g.

Elastic Potential Energy

Elastic potential energy is the energy stored when an elastic object — like a spring — is stretched or compressed. The stiffer the spring, or the more you stretch it, the more energy stored.

Equation (on equation sheet)

Eₑ = ½ k e²

Eₑ = elastic potential energy in joules (J)

k = spring constant in newtons per metre (N/m)

e = extension in metres (m)

Two things to notice

The extension is squared — doubling the stretch quadruples the stored energy. Same trap as KE.

Extension is in metres, but questions often give it in cm or mm. Convert first.

The equation works identically for a compressed spring — e is the change in length, whether stretched or compressed (think spring-loaded toys, dart launchers, jack-in-the-box).

Worked Example — Elastic PE

A spring with spring constant 200 N/m is stretched by 5.0 cm. Calculate the elastic potential energy stored. 3 marks

EEquation:Eₑ = ½ k e²

SSubstitute:Convert e = 5.0 cm = 0.050 m

Eₑ = 0.5 × 200 × (0.050)² = 0.5 × 200 × 0.0025

UUnit: joules (J)

AAnswer:Eₑ = 0.25 J

Notice the two-trap defence

The substitution line did both the unit conversion (cm → m) and showed the squaring (0.050²). Two traps neutralised in one line. This is the habit that earns full marks every time.

When the Equation Stops Working

The Eₑ = ½ k e² equation is built on Hooke's law (force is proportional to extension). Beyond a certain point — the limit of proportionality — Hooke's law breaks down and the equation no longer applies.

Limit of Proportionality

The extension beyond which a spring no longer obeys Hooke's law — force is no longer proportional to extension. Beyond this point, the spring deforms and may not return to its original shape.

Mark-scheme caveat

The spec specifically states the EPE equation applies "assuming the limit of proportionality has not been exceeded". If a question shows a force-extension graph that curves at the top, the EPE equation only works for extensions before the curve starts.

Reading the graph

In the linear (Hooke's law) region, the area under the line is a triangle: ½ × base × height = ½ × e × (ke) = ½ k e². That's where the equation comes from. Beyond the limit of proportionality, the graph curves, and the area no longer matches ½ k e².

Chained Calc — GPE → KE

The Grade 7+ pattern: use one equation to find an energy, then use a second equation to find a different quantity. Mark schemes step-mark these — every substitution line earns its own mark.

Question

A 2.0 kg mass is dropped from a height of 5.0 m. Assuming no air resistance, calculate its speed just before it hits the ground. Take g = 9.8 N/kg. 5 marks

E1Equation 1 (GPE at top):Eₚ = m g h

S1Substitute:Eₚ = 2.0 × 9.8 × 5.0 = 98 J

→Conservation: with no air resistance, all GPE at top → KE at bottom, so Eₖ = 98 J.

E2Equation 2 (KE at bottom):Eₖ = ½ m v² → rearrange: v² = 2 Eₖ / m

S2Substitute:v² = (2 × 98) / 2.0 = 98 → v = √98

AAnswer:v ≈ 9.9 m/s

Step-marking rules

Even if you make an arithmetic error in step 1, your step 2 substitution can still earn marks — provided your working is shown. This is called error-carried-forward (ECF). Show every line.

Other common rearrangements

The same approach works in reverse. Given energy and asked for height: h = Eₚ / (m g). Given energy and asked for extension: e = √(2 Eₑ / k). Given energy and asked for spring constant: k = 2 Eₑ / e².

Examiner Alert

The two recurring errors on this topic

1. Forgetting to square the speed in KE. Documented in AQA examiner reports across 2023 and 2024 — students lose 2 of 3 marks on a calc that's otherwise correct.

2. Substituting cm into the EPE equation without converting. Recurring across every year (2022–2024). Mark schemes (e.g. Nov 2021 P2 Q02.6) explicitly cap the marks at 1 of 3 if two unconverted values are used.

Two errors documented every series: forgetting to square (KE) and not converting cm to m (EPE). Square the speed. Convert the extension. — Pattern across AQA P1 examiner reports

The defence

Always write the conversion on its own line: "e = 5.0 cm = 0.050 m". Always write the squared value as an intermediate: "v² = 400" or "e² = 0.0025". Two simple habits, two traps neutralised.

Common Errors That Cost Marks

Students often writeCorrect version

Eₖ = 0.5 × m × vEₖ = 0.5 × m × v² (square the speed)

Substituting e in cm into Eₑ = ½ k e²Convert e to m on a separate line: e = 5.0 cm = 0.050 m

Eₚ = m × h (forgetting g)Eₚ = m × g × h, with g = 9.8 N/kg as given in the question

Using g = 10 N/kg when the question said 9.8Use exactly the value the question gives

Final answer without unit ("Eₖ = 300 000")Always include the unit — "Eₖ = 300 000 J"

Final answer to too many sig figsRound to the s.f. of the input data — usually 2 or 3 s.f.

No working shown — just the final numberShow the equation, the substitution, then the answer — earn the step-marks

Practice Question

A small ball of mass 25 g is dropped from a window 12 m above the ground.

Calculate the speed of the ball just before it hits the ground. Assume air resistance is negligible. Take g = 9.8 N/kg. 4 marks

E1

GPE at top: Eₚ = m g h

S1

Convert mass: m = 25 g = 0.025 kg. Eₚ = 0.025 × 9.8 × 12 = 2.94 J

→

Conservation: all GPE at top → KE at bottom (no air resistance), so Eₖ = 2.94 J

E2

KE at bottom: Eₖ = ½ m v² → rearrange: v² = 2 Eₖ / m

S2

v² = (2 × 2.94) / 0.025 = 235.2 → v = √235.2

A

v ≈ 15 m/s (to 2 s.f.)

Checkpoint

✅ You can now

Recall and apply Eₖ = ½ m v² for kinetic energy.

Recall and apply Eₚ = m g h for gravitational potential energy.

Recognise and apply Eₑ = ½ k e² for elastic potential energy (on the equation sheet).

Pick the right equation by inspecting the question — moving / raised / stretched.

Square the speed in KE and the extension in EPE.

Convert cm → m before substituting (and g → kg if mass is in grams).

Chain GPE → KE for a Grade 7+ multi-step calculation.

Acknowledge the limit of proportionality for the EPE equation.

Coming Next

Lesson 1.03 — Specific heat capacity and the first required practical (RP1). The thermal-energy equation: ΔE = m c Δθ. One of the most-examined RPs in the qualification.

Summary

Three Equations to Master

Eₖ = ½ m v² — kinetic. Squared speed. Most common error: dropping the squared.

Eₚ = m g h — gravitational potential. Always include g (= 9.8 N/kg on Earth, given in the question).

Eₑ = ½ k e² — elastic potential. On the equation sheet. Squared extension. Convert cm → m first. Only valid below limit of proportionality.

Habits that earn marks

Show the equation in symbols

Convert units on a separate line

Show the squared value: "v² = 400"

Include the unit on the final answer

Errors that lose marks

Forgetting to square (KE / EPE)

Substituting cm into the equation

Forgetting g in GPE

Final answer without unit

Exam Tip

For chained calcs (GPE → KE, EPE → KE), show every line. The mark scheme awards each step separately, and an arithmetic slip in step 1 doesn't kill step 2 if your working is visible (error-carried-forward).