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Elastic Potential Energy (Force–Extension Graph)

Key idea: Determine work done and elastic potential energy from force–extension area, using one-half kx squared only in the Hooke's-law region.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Represent fields and relate work done by a field to potential-energy change.
  • Use force–extension graphs to determine elastic potential energy.

1. Definitions (Must Know)

A. Extension, x (m)

Extension, x, is the increase in length of an object: extension = final length − original length.

B. Force–extension graph

A force–extension graph plots force F (N) against extension x (m).

C. Elastic potential energy, Eₑ (J)

Elastic potential energy, Eₑ, is energy stored when an object is elastically deformed.

D. “Area under the graph”

The area under a graph is the area between the curve and the x-axis.

For an F–x graph, the unit is: N m = J

2. Key Ideas (What Earns Marks)

  • Elastic potential energy stored in a deformed material equals the area under the force–extension graph:
Eₑ = area under the F–x graph
  • If Hooke’s law applies (F = kx, straight line through the origin):
Eₑ = (1/2)Fx = (1/2)kx²
  • If the graph is not a straight line, find the area by splitting it into simple shapes (triangles/trapezia/rectangles).
Force–extension graph with proportional and elastic limitsForce is proportional to extension from the origin to point P. The material then remains elastic through a curved region until point E; beyond E, unloading may leave permanent extension.Extension, xForce, FPEF = kxlinear; gradient = knon-linear butstill elasticpermanent extensionmay remain
Scroll diagram horizontally to read all labels.
The proportional limit P ends the straight Hooke’s-law region. The elastic limit E is a different condition: deformation remains reversible only up to E.

3. Detailed Explanations

A. Why “area under the graph” gives energy

For a constant force parallel to the displacement, work done is W = Fx.

If the force changes with extension, then:

  • split the extension into small parts Δ x,
  • work done on each part is approximately FΔ x,
  • total work is the sum of those areas.

So the work done stretching the object is the area under the F–x curve. If the deformation is elastic (no energy lost), this work becomes elastic potential energy stored.

B. Hooke’s law case (straight line)

When F = kx, the F–x graph is a straight line through the origin, so the area is a triangle:

Eₑ = 1/2 × (base x) × (height F) = (1/2)Fx = (1/2)kx²

Linear and non-linear force–extension responses

One line shows a Hooke's-law response through the origin. A second response curves upward, so its work must be obtained from the actual area under the data.

Scroll across the graph to read all labels.

One line shows a Hooke's-law response through the origin. A second response curves upward, so its work must be obtained from the actual area under the data.One line shows a Hooke's-law response through the origin. A second response curves upward, so its work must be obtained from the actual area under the data.
The area under either curve is the work transferred during loading. It equals stored elastic potential energy only when that loading is elastic and losses are negligible.
Open full-size graph
View figure data
Values and uncertainty for Linear and non-linear force–extension responses
SeriesExtension, x (m)Extension, x uncertaintyForce, F (N)Force, F uncertainty
Hooke’s law00
Hooke’s law0.0510
Non-linear00
Non-linear0.011.4
Non-linear0.023.2
Non-linear0.035.5
Non-linear0.048.4
Non-linear0.0512

4. Common Mistakes

  • Using the total length instead of the extension.
  • Using Fx with F taken at the final extension (a rectangle) instead of the area under the curve.
  • Forgetting conversions (e.g. cm → m).
  • Forgetting the 1/2 factor when the graph is a straight line from the origin.

5. Exam Tips

  • Always write the unit check: N m = J.
  • Do not confuse the two limits: a response may be non-linear beyond the proportional limit yet remain elastic until the elastic limit.
  • Area under a loading curve is work input. Call it stored elastic energy only when deformation is elastic and dissipative losses are negligible.

6. Worked Examples

Modelled example 1

Use Eₑ = (1/2)kx²

Core

Problem

A spring of force constant 250 N m⁻¹ is stretched by 0.060 m. Find its stored elastic potential energy.
Study the worked solution

  1. Check the model

    Method

    Use the Hooke-law triangular-area expression Eₑ = (1/2)kx².

    Reason

    The stated force constant represents a linear force–extension response over this extension.

    Working

    Eₑ = (1/2)kx²

  2. Calculate stored energy

    Method

    Eₑ = 0.45 J.

    Reason

    Extension is already in metres, consistent with the force-constant unit.

    Working

    Eₑ = (1/2)(250)(0.060)² = 0.45 J

Guided practice 2

Use area under a straight-line graph

About 4 min

Problem

A force–extension graph is a straight line from (0,0) to (0.080 m,12 N). Find the energy stored at 0.080 m, assuming elastic loading with negligible losses.

Try this before viewing the solution

Unit: J

Hints

Hint 1: identify the graph shape
Use one half times the 0.080 m base times the 12 N height.
View solution step by step

  1. Identify the work area

    Method

    The area is triangular.

    Reason

    Force increases linearly from zero to its final value.

    Working

    Eₑ = (1/2)Fx

  2. Calculate the area

    Method

    Eₑ = 0.48 J.

    Reason

    Area under the loading graph equals work input and, under the stated assumptions, stored elastic energy.

    Working

    Eₑ = (1/2)(12)(0.080) = 0.48 J

Common misconception 3

Final force is not the average force

Find and correct the mistake

Learner claim

A spring’s force rises linearly from zero to 8.0 N as extension reaches 0.050 m. A learner calculates the stored energy as Fx = 0.40 J. Diagnose the method and find the correct energy.

Try this before viewing the solution

Unit: J

View solution step by step

  1. Identify the false constant-force assumption

    Method

    The force is not 8.0 N throughout loading.

    Reason

    It rises linearly from zero, so Fx draws a rectangle larger than the area under the graph.

    Working

    F_average = (1/2)F_final = 4.0 N

  2. Calculate the triangular area

    Method

    Eₑ = 0.20 J.

    Reason

    Stored energy equals the force–extension area for elastic loading.

    Working

    Eₑ = (1/2)(8.0)(0.050) = 0.20 J

Examiner practice 4

Piecewise graph (triangle + rectangle)

4 marks

Examination question

Force rises linearly from zero to 10 N as extension increases to 0.040 m, then remains at 10 N until 0.060 m. Find the loading work and hence the stored energy, assuming elastic deformation with negligible losses. [4 marks]

Try this before viewing the solution

Unit: J

View solution step by step

  1. Calculate the triangular area

    1 mark

    Method

    0.20 J.

    Reason

    Force rises linearly over the first 0.040 m.

    Working

    (1/2)(10)(0.040) = 0.20 J

  2. Calculate the rectangular area

    1 mark

    Method

    0.20 J.

    Reason

    Force remains constant over the final 0.020 m.

    Working

    (10)(0.060-0.040) = 0.20 J

  3. Add the loading work

    1 mark

    Method

    W = 0.40 J.

    Reason

    Total work is the full area under both graph regions.

    Working

    W = 0.20 + 0.20 = 0.40 J

  4. State stored energy

    1 mark

    Method

    Eₑ = 0.40 J.

    Reason

    The question states elastic deformation and negligible losses.

    Working

    Eₑ = W = 0.40 J

Challenge 5

Find k from the gradient and then find energy

Minimal support

Independent transfer

A Hooke-law spring extends 6.0 cm under an 18 N force. Find its force constant and the elastic potential energy stored at that extension.

Try this before viewing the solution

Unit: N m^-1
Unit: J

Hints

Hint 1: convert the graph coordinate
6.0 cm = 0.060 m; then the same force–extension point gives both gradient and area.
View solution step by step

  1. Convert extension

    Method

    x = 0.060 m.

    Reason

    Force constant is expressed per metre.

    Working

    6.0 cm = 6.0 × 10⁻² m

  2. Find the force constant

    Method

    k = 3.0 × 10² N m⁻¹.

    Reason

    The Hooke-law force–extension gradient is F/x.

    Working

    k = 18/0.060 = 300 N m⁻¹

  3. Find stored energy

    Method

    Eₑ = 0.54 J.

    Reason

    The straight-line graph area is triangular.

    Working

    Eₑ = (1/2)Fx = (1/2)(18)(0.060) = 0.54 J

7. Mind Stretchers

Mind stretcher 1: Estimate energy from a non-linear graph (trapezium rule)Extension

The table gives force vs extension:

x (m)0.000.020.040.06
F (N)04915

Estimate the elastic potential energy stored at x = 0.06 m.

Show Answer

Use trapezia of width Δ x = 0.02 m:

Eₑ ≈ ∑ ((F₁ + F₂)/2)Δ x; = [(0 + 4)/2 + (4 + 9)/2 + (9 + 15)/2](0.02); = (2 + 6.5 + 12)(0.02); = 0.41 J

Mind stretcher 2: Energy lost in loading/unloading (hysteresis idea)Extension

A spring is stretched and then released. The area under the loading F–x curve is 0.80 J, but the area under the unloading curve is 0.65 J.

  1. How much energy is dissipated in one load–unload cycle?
  2. Where does that energy go?
Show Answer

  1. Energy dissipated: Eₗₒₛₜ = 0.80 - 0.65 = 0.15 J

  2. It is transferred mainly to internal energy (heating) of the spring and surroundings, and sometimes sound.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027