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How do we calculate the energy in moving, raised and stretched objects?

Kinetic, gravitational potential and elastic potential energy: calculating each store and using conservation of energy to link them.

A focused answer to AQA GCSE Physics 4.1.1, covering the kinetic energy, gravitational potential energy and elastic potential energy equations and how conservation of energy links them in a transfer.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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

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What this dot point is asking
Kinetic energy
Gravitational potential energy
Elastic potential energy
Linking the stores with conservation
Try this

What this dot point is asking

AQA wants you to calculate the energy in the kinetic, gravitational potential and elastic potential stores, and to use conservation of energy to follow energy as it moves between these stores during a change. This is part of topic 4.1.1 of the AQA GCSE Physics (8463) specification, and the kinetic and gravitational potential equations are given on the equation sheet while the elastic potential equation must be recalled.

Kinetic energy

Anything that is moving has energy in its kinetic store. The amount depends on the mass and, much more strongly, on the speed, because speed is squared in the equation.

The squared dependence on speed is one of the most exam-relevant ideas in this topic. It explains why stopping distances grow so rapidly with speed (a car at twice the speed has four times the kinetic energy and needs roughly four times the braking distance to remove it) and why high-speed collisions are so much more destructive than low-speed ones.

Gravitational potential energy

The gravitational potential store changes whenever an object is raised or lowered in a gravitational field. Lifting an object higher gives it more gravitational potential energy.

Only the change in height matters for the change in this store, so you measure $h$ from the starting position to the finishing position, not from the centre of the Earth. AQA usually supplies the value of $g$ , but on Earth it is about $9.8\,N/kg$ .

Elastic potential energy

The elastic potential store is the energy stored in a stretched or compressed spring (or other elastic object). It is recovered when the spring returns to its original shape.

Like kinetic energy, the elastic store depends on the square of a quantity (here the extension), so doubling the extension stores four times the energy, provided the spring is still behaving elastically.

These three equations come up constantly, and the most reliable way to handle them is to write down the equation, list the known values in the correct units, substitute and then evaluate. Watch for the units in particular: mass must be in kilograms, lengths and heights in metres, and speed in metres per second. A frequent slip is to leave a mass in grams or a height in centimetres, which makes the answer wrong by a factor of a thousand or a hundred. The kinetic and gravitational potential energy equations are provided on the AQA equation sheet, but you save time by knowing them, and the elastic potential energy equation must be recalled from memory.

Linking the stores with conservation

Worked example: a falling ball

A ball is dropped from a height of $h = 5\,\text{m}$ . Ignoring air resistance, find its speed just before it hits the ground. Take $g = 9.8\,\text{N/kg}$ .

Step 1: Apply conservation of energy to link the two stores

When air resistance is ignored, all the energy in the gravitational potential store transfers to the kinetic store. Setting the two expressions equal gives the central equation:

mgh = \tfrac{1}{2}mv^2

Step 2: Cancel the mass and rearrange for speed

The mass $m$ appears on both sides, so it cancels. This is a useful result: the speed at the bottom does not depend on how heavy the ball is. Rearranging for $v$ :

gh = \tfrac{1}{2}v^2 \implies v^2 = 2gh \implies v = \sqrt{2gh}

Step 3: Substitute the values and evaluate

Replacing $g$ and $h$ with their values:

v = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} \approx 9.9\,\text{m/s}

Final answer: $v \approx 9.9\,\text{m/s}$ .

Try this

Q1. Calculate the kinetic energy of a $2\,kg$ ball moving at $3\,m/s$ . [2 marks]

Cue. $E_k = \frac{1}{2} \times 2 \times 3^2 = 9\,J$ .

Q2. A $0.5\,kg$ book is lifted $1.2\,m$ . Calculate the gain in gravitational potential energy ( $g = 9.8\,N/kg$ ). [2 marks]

Cue. $E_p = 0.5 \times 9.8 \times 1.2 = 5.88\,J$ .

Exam-style practice questions

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

AQA 20194 marksA car of mass

1200\,\text{kg}

is travelling at

15\,\text{m/s}

. Calculate the kinetic energy of the car, then calculate the new kinetic energy if the car's speed doubles to

30\,\text{m/s}

Show worked answer →

Use $E_k = \frac{1}{2}mv^2$ . At $15\,\text{m/s}$ : $E_k = \frac{1}{2} \times 1200 \times 15^2 = \frac{1}{2} \times 1200 \times 225 = 135{,}000\,\text{J}$ , which is $1.35 \times 10^{5}\,\text{J}$ (2 marks). At $30\,\text{m/s}$ : $E_k = \frac{1}{2} \times 1200 \times 30^2 = \frac{1}{2} \times 1200 \times 900 = 540{,}000\,\text{J}$ , which is $5.4 \times 10^{5}\,\text{J}$ (2 marks). Markers reward squaring the speed before multiplying and the recognition that doubling the speed gives four times the kinetic energy. The commonest error is to forget to square $v$ .

AQA 20215 marksA ball of mass

0.20\,\text{kg}

is dropped from a height of

1.8\,\text{m}

. Assuming no energy is transferred to the surroundings, calculate the speed of the ball just before it hits the ground. Take

g = 9.8\,\text{N/kg}

Show worked answer →

By conservation of energy, all the gravitational potential energy is transferred to the kinetic store, so $mgh = \frac{1}{2}mv^2$ (1 mark). First find the gravitational potential energy: $E_p = mgh = 0.20 \times 9.8 \times 1.8 = 3.528\,\text{J}$ (1 mark). Setting this equal to the kinetic energy, $\frac{1}{2}mv^2 = 3.528$ , so $v^2 = \frac{2 \times 3.528}{0.20} = 35.28$ (1 mark) and $v = \sqrt{35.28} = 5.9\,\text{m/s}$ (1 mark). The mass cancels in the algebra, which is worth noting (1 mark). Markers reward equating the stores, correct rearrangement, and the final square root. Many candidates forget to take the square root at the end.

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Sources & how we know this

AQA GCSE Physics (8463) specification — AQA (2016)