Skip to main content
EnglandCombined ScienceSyllabus dot point

How do forces change the shape of objects, and how is the extension of a spring calculated?

Elastic and inelastic deformation, Hooke's law and the spring constant, the equation F = ke, the limit of proportionality, and the energy stored in a stretched spring.

A focused answer to Edexcel GCSE Combined Science Topic 15 (CP15), covering elastic and inelastic deformation, Hooke's law and the spring constant, the equation F = ke, the limit of proportionality, and the energy stored in a stretched spring.

Generated by Claude Opus 4.88 min answer

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

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. Elastic and inelastic deformation
  3. Hooke's law
  4. The limit of proportionality
  5. Energy stored in a spring
  6. Try this

What this dot point is asking

Edexcel wants you to distinguish elastic and inelastic deformation, state Hooke's law and use the equation F=keF = ke, explain the limit of proportionality, and describe the energy stored in a stretched spring.

Elastic and inelastic deformation

Hooke's law

The limit of proportionality

A graph of force against extension for a spring is a straight line through the origin while Hooke's law holds. The limit of proportionality is the point beyond which the line begins to curve, because the extension is no longer proportional to the force. Beyond this, stretching the spring further may also cause inelastic deformation, so it does not return to its original length.

Energy stored in a spring

When a force does work stretching a spring, energy is transferred to the elastic potential energy store of the spring. For a spring obeying Hooke's law, this stored energy equals the area under the force-extension graph. The energy is released when the spring returns to its original shape, which is how catapults, trampolines and clockwork devices work.

The core practical for this topic investigates how the extension of a spring depends on the force. You hang the spring from a clamp stand, measure its original length, then add masses one at a time. Each mass adds a known weight (force = mass times gg), and you measure the new length to find the extension. Plotting force against extension gives a straight line through the origin while the spring obeys Hooke's law, and the gradient of this line is the spring constant. The line begins to curve at the limit of proportionality.

The same ideas apply to compressing a spring, not just stretching it: a spring squashed by a force also stores elastic potential energy and obeys Hooke's law up to its limit. This is why springs are used in car suspensions and weighing scales, where a known force produces a predictable, reversible change in length. If a spring is stretched beyond its limit it deforms inelastically and no longer returns to its original length, so it can no longer be used to measure force reliably.

Try this

Q1. State Hooke's law. [1 mark]

  • Cue. The extension of a spring is directly proportional to the force applied (up to the limit of proportionality).

Q2. A spring with a spring constant of 50 N/m50\,N/m is stretched by a force of 10 N10\,N. Calculate the extension. [2 marks]

  • Cue. e=F/k=10/50=0.2 me = F / k = 10 / 50 = 0.2\,m.

Exam-style practice questions

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

Edexcel 20203 marksA spring has a spring constant of 25 N/m25\,\text{N/m}. A force of 5 N5\,\text{N} is applied to it. Calculate the extension of the spring, stating the equation you use.
Show worked answer β†’

A 3-mark calculation using Hooke's law.

Use F=keF = ke, rearranged to e=Fke = \dfrac{F}{k} (1 mark). Substitute: e=525e = \dfrac{5}{25} (1 mark). So e=0.2 me = 0.2\,\text{m} (1 mark).

Markers reward stating and rearranging F=keF = ke, the substitution, and the correct extension with the unit metres. A common error is multiplying instead of dividing.

Edexcel 20214 marksDescribe the core practical to investigate the relationship between the force applied to a spring and its extension, and explain what is meant by the limit of proportionality.
Show worked answer β†’

A 4-mark required-practical question.

Hang the spring from a clamp and measure its original length. Add masses one at a time (each adding a known weight, force = mass times g), and measure the new length each time to find the extension (2 marks). Plot a graph of force against extension; it is a straight line through the origin while the spring obeys Hooke's law (1 mark). The limit of proportionality is the point beyond which the line curves, because extension is no longer proportional to the force (1 mark).

Markers reward the method (adding known forces and measuring extension), the straight-line graph, and the limit of proportionality as where the graph stops being straight.

Related dot points

Sources & how we know this