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Elastic Potential Energy

FoundationHigherCombined & TripleAQAEdexcelOCR

Build confidence in Elastic Potential Energy for GCSE Physics with this free worksheet and full mark scheme — Foundation and Higher exam-style questions with worked answers for AQA, Edexcel and OCR. Stretching or compressing a spring stores elastic potential energy of ½ke², as long as the limit of proportionality is not exceeded.

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These worksheets and mark schemes are original, written for Virtus Academy and checked against the current AQA, Edexcel and OCR specifications. Every worksheet comes with a full mark scheme.

Topic overview

The elastic potential energy store of a stretched or compressed spring depends on the spring constant and the extension. The equation is elastic potential energy = 0.5 × spring constant × extension².

The extension is the increase in length, not the total length. A spring of natural length 10 cm stretched to 14 cm has an extension of 4 cm, and using 14 cm would give an answer far too large.

This equation assumes the limit of proportionality has not been exceeded. Beyond that limit the spring no longer obeys Hooke's law and the equation no longer applies. Extension must be in metres and the spring constant in newtons per metre.

Revision notes

The equation

Elastic potential energy = 0.5 × spring constant × extension², or Eₑ = ½ke².

Spring constant in newtons per metre, extension in metres, energy in joules. The equation assumes the limit of proportionality has not been exceeded.

Extension, not total length

The extension is the increase in length from the natural, unstretched length.

A spring of natural length 10 cm stretched to 14 cm has an extension of 4 cm, which is 0.04 m. Using the total length is the most frequent error in this topic.

The limit of proportionality

Up to the limit of proportionality, extension is directly proportional to the force applied.

Beyond it, the spring no longer obeys Hooke's law, the graph curves, and the equation for elastic potential energy no longer applies.

Key points

  • Elastic potential energy = 0.5 × spring constant × extension².
  • Extension is the increase in length.
  • Extension must be in metres.
  • Spring constant is in newtons per metre.
  • The equation assumes the limit of proportionality is not exceeded.
  • Beyond that limit Hooke's law no longer applies.

Worked examples

Example 1

A spring has a spring constant of 200 N/m and is extended by 0.1 m. Calculate the elastic potential energy stored. [3 marks]

Model answer

Elastic potential energy = 0.5 × spring constant × extension² = 0.5 × 200 × 0.1² (1 mark). 0.1² = 0.01, so 0.5 × 200 × 0.01 (1 mark) = 1 J (1 mark).

Example 2

A spring of natural length 12 cm is stretched to 20 cm. State the extension in metres. [2 marks]

Model answer

Extension = 20 − 12 = 8 cm (1 mark), which is 0.08 m (1 mark).

Example 3

Explain what is meant by the limit of proportionality. [2 marks]

Model answer

It is the point beyond which the extension is no longer directly proportional to the force applied (1 mark), so the spring no longer obeys Hooke's law and the force-extension graph is no longer a straight line (1 mark).

Common mistakes

  • Using the total length instead of the extension.

    Subtract the natural length to find the extension first.

  • Forgetting to convert centimetres to metres.

    Extension must be in metres, so divide centimetres by 100.

  • Not squaring the extension.

    The extension is squared in the equation.

  • Applying the equation beyond the limit of proportionality.

    It only holds while the spring obeys Hooke's law.

Exam tips

  • Work out the extension before doing anything else.
  • Convert centimetres to metres by dividing by 100.
  • Square the extension as a separate step.
  • State the assumption about the limit of proportionality if asked.

Key terms

Elastic potential energy
Energy stored in a stretched or compressed spring.
Spring constant
The stiffness of a spring, in newtons per metre.
Extension
The increase in length from the natural length.
Limit of proportionality
The point beyond which extension is no longer proportional to force.

Written and reviewed against the current AQA, Edexcel and OCR specifications. Spotted an error? Let us know.