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How do forces do work and stretch springs?

Work done and elasticity: work done by a force, the link to energy, Hooke's law, the spring constant and elastic potential energy, and the required practical.

A focused answer to AQA GCSE Physics 4.5.2 and 4.5.3, covering work done by a force and its link to energy transfer, Hooke's law and the spring constant, elastic potential energy, and the required practical investigating force and extension.

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 dot point is asking
Work done
Hooke's law
Elastic potential energy
Required practical: force and extension
Try this

What this dot point is asking

AQA wants you to calculate the work done by a force, link work to energy transfer, apply Hooke's law and the spring constant, calculate elastic potential energy, and describe the required practical on force and extension. This covers topics 4.5.2 and 4.5.3 of the AQA GCSE Physics (8463) specification.

Work done

Work and energy are two sides of the same idea: the work done by a force is exactly equal to the energy transferred by that force. When you lift a box you do work against gravity and transfer energy to its gravitational potential store; when you push a box across a rough floor you do work against friction and transfer energy to the thermal store. The unit makes this clear, since one joule is one newton metre. A subtle point AQA tests is that only the component of the force along the direction of motion does work, which is why $s$ in the equation is the distance moved along the line of the force.

Hooke's law

An important distinction is between elastic and inelastic deformation. A deformation is elastic if the object returns to its original shape once the force is removed (as a spring does within its limit), and inelastic if it stays permanently changed. Hooke's law and the elastic potential energy equation only apply while the deformation is elastic and the limit of proportionality has not been passed. Beyond that point the force-extension graph curves away from the straight line and the spring may not return to its original length.

Elastic potential energy

Required practical: force and extension

You hang a spring vertically, add masses one at a time, and record the extension for each force (the weight added). Plotting force against extension gives a straight line through the origin while Hooke's law holds; the line curves once the limit of proportionality is passed. The gradient relates to the spring constant.

Worked example: finding the spring constant

A force of $6\,N$ stretches a spring by $0.04\,m$ . We want to find the spring constant $k$ , which measures how stiff the spring is.

Step 1: Identify the known quantities

Hooke's law states $F = ke$ , where $F$ is the applied force and $e$ is the extension. Both are given, so we rearrange to find $k$ :

F = 6\,N, \quad e = 0.04\,m

Step 2: Rearrange Hooke's law for $k$

Dividing both sides of $F = ke$ by the extension $e$ :

k = \dfrac{F}{e} = \dfrac{6}{0.04}

Step 3: Calculate the spring constant

k = 150\,N/m

Final answer: the spring constant is $150\,N/m$ , meaning each newton of force stretches this spring by $\frac{1}{150}\,m$ .

Try this

Q1. State Hooke's law. [2 marks]

Cue. Extension is directly proportional to the force, up to the limit of proportionality.

Q2. A spring with spring constant $200\,N/m$ is stretched by $0.05\,m$ . Calculate the force. [2 marks]

Cue. $F = ke = 200 \times 0.05 = 10\,N$ .

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 20205 marksA spring has a spring constant of

250\,\text{N/m}

. A force is applied that stretches it by

0.080\,\text{m}

, without exceeding the limit of proportionality. Calculate the force applied and the elastic potential energy stored in the spring.

Show worked answer →

First use Hooke's law $F = ke = 250 \times 0.080 = 20\,\text{N}$ (2 marks). Then use the elastic potential energy equation $E_e = \frac{1}{2}ke^2 = \frac{1}{2} \times 250 \times 0.080^2 = \frac{1}{2} \times 250 \times 0.0064 = 0.80\,\text{J}$ (3 marks). It is essential to square the extension before multiplying: $0.080^2 = 0.0064$ , not $0.080$ . Markers reward the correct force from Hooke's law, squaring the extension, and including the factor of one half. A common error is to forget the squaring or the half, or to use the total length rather than the extension.

AQA 20184 marksDescribe the required practical to investigate the relationship between the force applied to a spring and its extension, and explain how the results show whether the spring obeys Hooke's law.

Show worked answer →

Hang the spring vertically from a clamp and measure its original length, then add masses one at a time and record the new length each time, calculating the extension as the new length minus the original length (1 mark for the method, 1 mark for using extension not total length). The force is the weight of the masses added, found from $W = mg$ (1 mark). Plot a graph of force (vertical axis) against extension (horizontal axis): if the spring obeys Hooke's law the graph is a straight line through the origin, showing extension is directly proportional to force; the line curves once the limit of proportionality is passed, showing Hooke's law no longer holds (1 mark). Markers reward a clear method, measuring extension, and interpreting the straight-line graph.

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Sources & how we know this

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