IGCSEPhysicsOxford AQARevision NotesEnergyForces & EnergyElastic Potential Energy

Elastic Potential Energy (Oxford AQA IGCSE Physics): Revision Note

Exam code: 9203

Written by: Leander Oates

Reviewed by: Caroline Carroll

Updated on 3 April 2024

Elastic Potential Energy

What is elastic potential energy?

Energy in the elastic potential store of an object is defined as:

The energy stored in an elastic object when work is done on the object

This means that any object can change shape by stretching, bending or compressing (eg. springs, rubber bands)
- When a spring is stretched (or compressed), work is done on the spring which results in energy being transferred to the elastic potential store of the spring
- When the spring is released, energy is transferred away from its elastic potential store

The extension of a spring when a force is applied

Stretching a spring, for IGCSE & GCSE Physics revision notes — **When a force is applied to either end of a spring, it stretches. Work is done on the spring as it extends, therefore energy is transferred**

How to calculate elastic potential energy

The amount of energy in the elastic potential store of a stretched spring can be calculated using the equation:

E_e = ½ × k × e²

Where:
- E_e = elastic potential energy in joules (J)
- k = spring constant in newtons per metre (N/m)
- e = extension in metres (m)

The above elastic potential energy equation assumes that the spring has not been stretched beyond its limit of proportionality

A spring stretched beyond its elastic limit

Elastic limit of a spring, for IGCSE & GCSE Physics revision notes — **When a spring is stretched beyond its elastic limit it will no longer return to its original length**

Worked Example

A mass is attached to the bottom of a hanging spring with a spring constant of 250 N/m. It stretches from 10.0 cm to 11.4 cm.

Calculate the energy in the elastic potential store of the stretched spring.

Step 1: Determine the extension of the spring

The extension of the spring is its extended length minus its original length

$e = 11.4 - 10.0$

$e = 1.4 cm$

Convert cm to m

$e = \frac{1.4}{100}$

$e = 0.014 m$

Step 2: List the known quantities

Spring constant, $k = 250 N / m$
Extension, $e = 0.014 m$

Step 3: Write out the elastic potential energy equation

$E_{e} = \frac{1}{2} \times k \times e^{2}$

Step 4: Calculate the elastic potential energy

$E_{e} = \frac{1}{2} \times 250 \times {(0.014)}^{2}$

$E_{e} = 0.0245 J$

Step 5: Round the answer to 2 significant figures

$E_{e} = 0.025 J$

Examiner Tips and Tricks

Look out for units! If the question gives you units of cm for the length you MUST convert this into metres for the calculation to be correct.

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