Find elastic energy from a force-extension graph
Interpret the area under a force-extension graph as work done and calculate elastic potential energy within the proportional region.
AS Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 6. The wire investigation is an original study guide and data-analysis practice, not a supervised laboratory activity or proof of practical competence. Full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use force, area, length, extension and graph gradient with consistent SI units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does the area under a force-extension graph represent?
What you will learn
- Relate area under a force-extension graph to work done
- Calculate elastic energy from graph area within the proportional limit
- Use Eₚ = ½Fx = ½kx² in the linear region
Graph area is the work done in stretching
The work done as a load extends a material is the area under its force-extension graph. For a straight line from the origin, this area is a triangle, so W = ½ × extension × final force.
Within the proportional region, F = kx. Substituting into the triangular area gives elastic potential energy Eₚ = ½Fx = ½kx². This formula applies to the linear elastic range described by Hooke's law.
Use the correct graph shape and boundary
If the graph is not a straight line, the energy is still represented by the area under the force-extension curve, but it may need to be estimated from sections or supplied areas rather than by the simple triangle formula.
The syllabus formula Eₚ = ½Fx = ½kx² is for deformation within the limit of proportionality. Beyond that region, do not assume one constant spring value or a triangular area from the origin.
Pause and explain
When is Eₚ = ½kx² valid in this syllabus model?
Worked example
A spring with k = 200 N m⁻¹ is extended by 0.050 m within its proportional limit. Find the elastic potential energy stored.
Show the worked solution
- Use Eₚ = ½kx².
- Eₚ = 0.5 × 200 × 0.050².
- Alternatively, F = kx = 10 N and Eₚ = ½Fx.
Answer Eₚ = 0.25 J.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A spring extends 0.10 m and the force increases linearly from zero to 20 N. Find the work done and elastic energy stored.
Give me a hint
Use the triangular area under the straight force-extension graph.
Compare my reasoning
- Area = ½ × base × height.
- Eₚ = 0.5 × 0.10 × 20.
- The graph area is work done in stretching.
1.0 J.
Look for these in your work
- Triangle area chosen
- Force and extension used
- Energy interpretation stated
A spring has k = 500 N m⁻¹ and is extended by 0.040 m within its proportional limit. Find its elastic potential energy.
Give me a hint
Use Eₚ = ½kx² and square the extension.
Compare my reasoning
- x² = 0.040² = 0.0016 m².
- Eₚ = 0.5 × 500 × 0.0016.
- Report joules.
0.40 J.
Look for these in your work
- Linear-range formula used
- Extension squared
- Energy unit stated
A force-extension graph is a straight line through the origin with gradient 300 N m⁻¹ up to 0.020 m. Find the stored energy at that extension and state whether the same formula is justified at 0.030 m if proportionality ended at 0.020 m.
Give me a hint
Calculate the in-range energy, then check whether the supplied model still applies beyond that extension.
Compare my reasoning
- At 0.020 m, Eₚ = ½kx² = 0.5 × 300 × 0.020² = 0.060 J.
- The proportional region ends at 0.020 m.
- At 0.030 m the same constant-k formula is not justified without a new force-extension model.
0.060 J at 0.020 m. The ½kx² equation cannot be assumed at 0.030 m because that extension lies beyond the stated proportional limit.
Look for these in your work
- In-range energy calculated
- Boundary checked
- Extrapolation limit stated
Common mistakes
- Using the full rectangular area Fx instead of the triangular area for a straight line from the origin
- Applying ½kx² beyond the proportional limit
- Reading area from a graph with mismatched force or extension units
What can you explain now?
What does the area under a force-extension graph represent?
Compare with the explanation
Work done in extending the material
Force multiplied by a small extension gives an energy transfer.
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Make it stick
How does the triangle area under a linear force-extension graph produce the elastic-energy formula?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.