# Strain energy

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u6-strain-energy/

This module covers strain energy, the elastic potential energy stored in deformed solids. You will learn to calculate strain energy for linear and non-linear materials and solve common CIE exam problems.

**Prerequisites:** [Hooke's Law and force-extension graphs](https://www.owlsprep.com/study/cie-9702-u6-hookes-law/); [Stress and strain definitions](https://www.owlsprep.com/study/cie-9702-u6-stress-strain/)

## Learning objectives

- Define strain energy and relate it to work done on deformed solids
- Derive and use strain energy formulae for Hooke's law materials
- Calculate strain energy from force-extension graphs for any material
- Solve exam problems involving stored strain energy in deformed solids

## Definition & Basic Strain Energy Formula

**Strain Energy** — The elastic potential energy stored in a material when work is done to deform it against internal restoring forces. It is fully recoverable if the material stays within its elastic limit.

*Notation:* E, U

*Example:* A stretched catapult stores strain energy, which is released to launch a projectile.

When you stretch or compress a material, the applied force moves through a displacement, so work is done. This work is stored as strain energy if no permanent deformation occurs. For a material that obeys Hooke's law ($F = kx$), the strain energy formula can be derived from the area of the triangle under the linear force-extension graph.

**Worked example:** A spring with spring constant $k = 80 \text{ Nm}^{-1}$ is stretched by 0.25 m within its elastic limit. Calculate the stored strain energy.

1. Use the standard strain energy formula for Hooke's law materials:

   $$E = \frac{1}{2} k x^2$$
2. Substitute the given values for $k$ and extension $x$:

   $$E = \frac{1}{2} \times 80 \times (0.25)^2$$
3. Calculate the final result:

   $$E = 2.5 \text{ J}$$

> **Exam tip**
>
> Always confirm the material is within its elastic limit: if it has undergone plastic deformation, not all work is stored as strain energy.

## Strain Energy from Force-Extension Graphs

The relationship $\text{Work done} = \int F \, dx$ tells us that for any material (linear or non-linear), strain energy equals the total area under the force-extension graph up to the current extension. This rule applies to all deformation types.

> **info**
>
> Only the area under the graph up to the elastic limit counts as recoverable strain energy. Work done beyond the elastic limit goes into permanent plastic deformation, not stored strain energy.

**Worked example:** A non-linear elastic rubber band has an area of $0.08 \text{ N·m}$ under its force-extension graph when stretched by 4 cm. What strain energy is stored?

1. Strain energy equals the area under the force-extension graph, by definition. Units of N·m are equivalent to Joules.
2. Therefore the stored strain energy is:

   $$E = 0.08 \text{ J}$$

**Check your understanding**

Test your understanding:

1. Which of the following statements is true for strain energy in a non-linear elastic material?

   - A: Strain energy cannot be calculated
   - B: Strain energy equals the area under the force-extension graph
   - C: Strain energy is always equal to $\frac{1}{2}Fx$
   - D: Strain energy is zero below the elastic limit

   *Why:* The area rule applies to all materials. $\frac{1}{2}Fx$ only works for linear (Hooke's law) materials.

## Alternative Strain Energy Formulae

**Derivation:** Derive $E = \frac{1}{2} Fx$ from $E = \frac{1}{2}kx^2$ for Hooke's law materials

*Starting from:* Hooke's law: $F = kx \implies k = \frac{F}{x}$

1. Substitute $k = \frac{F}{x}$ into the original strain energy formula:
2. $$E = \frac{1}{2} \left(\frac{F}{x}\right) x^2$$
3. Simplify the expression to get the final result:
4. $$E = \frac{1}{2} F x$$

*Conclusion:* For any linear elastic material, strain energy equals half the product of maximum applied force and total extension.

Another common form for uniform deformation is written in terms of stress $\sigma$, strain $\varepsilon$, and total volume $V$ of the material: $E = \frac{1}{2} \sigma \varepsilon V$. This is useful for problems where you are given stress and strain instead of force and extension.

**Worked example:** A force of 200 N stretches a Hooke's law wire by 3.0 mm. Calculate the stored strain energy.

1. Convert extension to SI units: $3.0 \text{ mm} = 0.0030 \text{ m}$
2. Use $E = \frac{1}{2}Fx$ for Hooke's law materials:

   $$E = \frac{1}{2} \times 200 \times 0.0030$$
3. Calculate the final result:

   $$E = 0.30 \text{ J}$$

> **Exam tip**
>
> If you are given force and extension for a Hooke's law material, $E = \frac{1}{2}Fx$ is faster to use than calculating $k$ first.

## Common pitfalls

- **Wrong:** Using $E = \frac{1}{2}Fx$ for non-linear materials
  - Why it fails: This formula only applies to materials with a linear force-extension relationship (obeying Hooke's law). It underestimates or overestimates strain energy for non-linear materials.
  - Correct: Always use the area under the force-extension graph to find strain energy for non-linear materials like rubber.
- **Wrong:** Forgetting to convert extension from millimetres to metres
  - Why it fails: SI unit calculations for energy (Joules) require extension in metres. Leaving it in millimetres gives an answer 1000× too large.
  - Correct: Divide extension in mm by 1000 to convert to m before substituting into any formula.
- **Wrong:** Counting all work done after plastic deformation as strain energy
  - Why it fails: Work done beyond the elastic limit causes permanent deformation and is not stored as recoverable strain energy.
  - Correct: Only the area under the force-extension graph up to the elastic limit counts as strain energy.
- **Wrong:** Using inconsistent units for $k$ and $x$
  - Why it fails: If $k$ is given in Ncm⁻¹ and $x$ in m, the resulting energy value will be incorrect due to unit mismatch.
  - Correct: Convert all values to SI units: $k$ in Nm⁻¹, $x$ in m, $F$ in N to get energy in Joules.

## Cheatsheet

| Scenario | Formula | Key Notes |
| --- | --- | --- |
| Linear (Hooke's law) | E = \frac{1}{2} k x^2 | k = spring constant, x = extension |
| Linear (Hooke's law) | E = \frac{1}{2} F x | F = maximum force, x = extension |
| Any material (all types) | E = \text{Area under } F-x \text{ graph} | Works for linear and non-linear materials |
| Uniform linear deformation | E = \frac{1}{2} \sigma \varepsilon V | σ = stress, ε = strain, V = total volume |

## What's next

Understanding strain energy is a core foundation for further topics in solid mechanics, including energy transformations in springs, elastic collisions, and the behavior of materials under dynamic load. It connects directly to work and energy concepts from earlier mechanics units, and is regularly tested in both multiple choice and structured questions in CIE A-Level Physics. Mastering strain energy calculations will help you access full marks on all deformation of solids questions.

- [Waves](https://www.owlsprep.com/study/cie-9702-u7-overview/)
- [Progressive waves](https://www.owlsprep.com/study/cie-9702-u7-progressive-waves/)
- [Transverse and Longitudinal Waves](https://www.owlsprep.com/study/cie-9702-u7-transverse-and-longitudinal-waves/)

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