Changes in energy: (assuming the limit of proportionality has not been exceeded)

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Q: Is this Changes in energy: (assuming the limit of proportionality has not been exceeded) guide aligned to the AQA specification?

Yes. This guide is linked to the Changes in energy: (assuming the limit of proportionality has not been exceeded) topic in the AQA GCSE specification.

This condition means the equation Eₑ = ½ke² is valid only while the spring obeys Hooke’s law, where force is directly proportional to extension.

Study notes

From the checked teaching notes for this specification point.

The idea

What earns the marks

State that the limit of proportionality is the point beyond which force is no longer directly proportional to extension.
Explain that Eₑ = ½ke² assumes a linear force–extension relationship, so it applies only up to the limit of proportionality.
Recognise that beyond the limit, the force–extension graph curves and the equation Eₑ = ½ke² does not apply.
Describe how to identify the limit from a force–extension graph as the point where the line stops being straight.

Common mistakes and how to put them right

Assuming the equation always applies no matter how far the spring is stretched.
Put it right by checking the force–extension graph is linear.
Confusing the limit of proportionality with the elastic limit.
Put it right by stating that the limit of proportionality is where proportionality ends, while the elastic limit is where permanent deformation begins.
Using the equation after the graph has curved.
Put it right by recognising the equation is invalid beyond the limit of proportionality.

Exam tips

If a question mentions 'assuming the limit of proportionality has not been exceeded', use Eₑ = ½ke² directly.
When interpreting a force–extension graph, mark the limit of proportionality where the straight line ends.

Try the lesson

The start of the Changes in energy: (assuming the limit of proportionality has not been exceeded) lesson, made from checked teaching notes for this specification point.

Have a go first

A student calculates the elastic potential energy stored in a stretched spring using the equation E_e = \frac{1}{2} k e^2. What condition must be met for this equation to be valid?

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Listen

This condition means the equation Eₑ = ½ke² is valid only while the spring obeys Hooke’s law, where force is directly proportional to extension. The limit of proportionality is the point beyond which force is no longer proportional to extension; the spring may still stretch but the graph curves. If the limit is exceeded, the simple equation no longer gives the stored elastic potential energy accurately and must not be used. For example, a spring with k = 200 N/m stretched by 0.10 m stores 1.0 J, but if stretched beyond its limit, the linear relationship breaks down. In calculations, check that the extension is within the linear region before using Eₑ = ½ke². A strong answer earns marks for these points. State that the limit of proportionality is the point beyond which force is no longer directly proportional to extension. Explain that Eₑ = ½ke² assumes a linear force–extension relationship, so it applies only up to the limit of proportionality. Recognise that beyond the limit, the force–extension graph curves and the equation Eₑ = ½ke² does not apply. Describe how to identify the limit from a force–extension graph as the point where the line stops being straight. A common slip: Assuming the equation always applies no matter how far the spring is stretched. Instead, checking the force–extension graph is linear. Exam tip: If a question mentions 'assuming the limit of proportionality has not been exceeded', use Eₑ = ½ke² directly.

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