Elastic Potential Energy Equation | GCSE Physics

The elastic potential energy equation is

$$E_e=\frac{1}{2}k{e}^2$$

It calculates the energy stored when a spring is stretched or compressed, provided the limit of proportionality has not been exceeded. Here, $E_e$ is elastic potential energy in joules ($\mathrm{J}$), $k$ is the spring constant in newtons per metre ($N/m$), and $e$ is the extension or compression in metres ($\mathrm{m}$). The spring constant tells you how stiff the spring is: a larger $k$ means more force is needed for the same extension.

How to use it

Extension means the change from the spring’s original length, not its total stretched length. For example, if a spring goes from $15\mathrm{cm}$ to $25\mathrm{cm}$, its extension is $25-15=10\mathrm{cm}$, or $0.10\mathrm{m}$. Using the final length of $25\mathrm{cm}$ would give the wrong energy.

Find the extension first, convert it to metres if necessary, then substitute it into the equation. Square the extension before multiplying by the spring constant and $\frac{1}{2}$. Squaring means multiplying the extension by itself, so $(0.10{)}^2=0.01$, not $0.20$.

Suppose that spring has a spring constant of $200N/m$. Its stored energy is

$$E_e=\frac{1}{2}\times 200\times (0.10{)}^2=1\mathrm{J}.$$

Assuming the spring has not been permanently deformed, the work done stretching it equals the elastic potential energy stored. The limit of proportionality is the point beyond which extension is no longer directly proportional to force, so do not use this equation beyond that limit. In the proportional region, the force increases as the spring stretches, which helps explain why stored energy depends on the extension squared.

Common exam mistake: using the spring’s final length in place of its extension, or forgetting to square $e$. If the extension doubles while $k$ stays the same, the energy becomes four times as large, not twice as large.

This explanation assumes AQA GCSE separate Physics and applies to both Foundation and Higher tier.