# Elastic Potential Energy

#### Back to Different Types of Forces

## Description

This is a basic physics tutorial that is targeted at GCSE (grade 9 and grade 10) standard briefly reviewing the energy transfer that occurs when a spring is deformed. The video for this written tutorial is on its way.

## Energy Transfer

From our brief introduction to work and energy (with more to come in future modules), you'll know that when a force is applied to something, energy is transferred. In the case of a spring, this energy becomes stored in the spring. Within its elastic limit, the energy stored is exactly the same as the work done on the spring.

We refer to this stored energy as *elastic potential energy*. It is referred to as *potential* energy because it gives the spring the potential to do work when released - but when it's stored in the spring it's it is not actually doing anything.

We can calculate the elastic potential energy using the formula:

### E_{e} = half x k x e^{2}

Where:

- E
_{e}refers to elastic potential energy, measured in joules (J) - k refers to the spring constant, reported in newtons per meter (N/m)
- e refers to the deformation (extension in this case), reported in metres (m)

## Practice Problem

Using the graph below, let's calculate the spring constant, and then the elastic potential energy that is stored in the spring when it is extended to 13 cm.

#### Spring Constant Calculation

We know that the spring constant is the same as the gradient of a force-extension graph. *Remember*: This only the case if the y-axis represents force and the x-axis represents extension.

Gradient = rise ÷ run |
Gradient = force ÷ extension |
k = force ÷ extension |

**gradient = k**

So let's calculate the gradient.

*= 60 N ÷ 0.13 m*

*= 461.5 N/m*

Therefore the spring constant is 4,61.5 N/m.

#### Elastic Potential Energy Calculation

*E _{e} = half x k x e^{2}*

*E*

_{e}= 0.5 x 461.5 N/m x (0.13 m)^{2}*E*

_{e}= 0.5 x 461.5 N/m x 0.0169 m^{2}*E*

_{e}= 0.5 x 7.8 Nm*E*

_{e}= 3.9 Nm or 3.9 joulesBecause the work done in stretching the spring is the same as the elastic potential energy stored in the spring, we can say that it takes 3.9 joules to stretch the spring in question by 13 cm.