Energy Stored in Stretched Spring

How stretching stores energy

When you pull a spring to make it longer, you apply a force while its end moves. You therefore do work on the spring: energy is transferred into its elastic potential store. The stretched spring stores energy even when it is held still.

When released, the spring can transfer energy out of this store as it returns towards its original length. For example, a stretched spring can make an attached object move.

For a spring stretched within its limit of proportionality, the work done in stretching it equals the elastic potential energy stored. Calculating one therefore gives you the other.

Extension is the change in length

The extension is how much longer the spring becomes, not its total length:

$$e = \text{stretched length} - \text{original length}$$

Both lengths must be measured in the same units before subtracting. The extension then needs to be expressed in metres for the energy calculation.

Calculating the stored energy

The elastic potential energy of a stretched spring is:

$$E_e = \frac{1}{2}ke^2$$

Here:

• $E_e$ is elastic potential energy in joules (J).
• $k$ is the spring constant in newtons per metre (N/m).
• $e$ is extension in .

This equation is given on the Physics equations sheet. It applies provided the limit of proportionality has not been exceeded: the spring's extension must still be directly proportional to the applied force.

The spring constant describes how stiff the spring is. At the same extension, a spring with a larger spring constant stores more energy because more work is needed to stretch it.

The extension is squared, so its effect is especially important. For the same spring, doubling the extension stores four times as much energy, as long as both extensions remain within the proportional range.

From spring lengths to energy and work done

Suppose a spring has a spring constant of $250\text{ N/m}$ and stretches from an original length of $10.0\text{ cm}$ to $11.4\text{ cm}$. Assume its limit of proportionality has not been exceeded.

First find the extension:

$$e = 11.4 - 10.0 = 1.4\text{ cm} = 0.014\text{ m}$$

Then substitute into the energy equation, squaring the extension:

$$E_e = \frac{1}{2}\times 250\times (0.014)^2 = 0.0245\text{ J}$$

The spring stores approximately $0.025\text{ J}$ of elastic potential energy, to two significant figures. The work done in stretching it from its original length is also approximately $0.025\text{ J}$: this is the energy transferred into the spring's elastic potential store.

Energy and work done

Stretching a spring transfers energy into its elastic potential store.

Within the limit of proportionality:

$$W = E_e = \frac{1}{2}ke^2$$