# Elastic potential energy

Elastic potential energy (${E}_{\text{e}}$) is the energy stored in an object when work is done to compress or stretch it. Like kinetic and gravitational energy, elastic energy is simply defined as the amount of work it took to get the system in that state. If we treat objects as springs, then the deformation of the spring is represented by $x\text{,}$ the displacement of the free end of the spring from its equilibrium position. For example, if a 5 m spring is squished to half its size, $x$ will be 2.5 m.

*Hooke’s Law* tells us that the deformation of the spring is proportional to the force that is applied to it. That is,

$F=kx\text{,}$

where $k$ is a constant representing the stiffness of the spring, measured in newtons per metre (N/m). According to Newton’s third law, we also have a reaction force of $-kx$ that is “trying” to restore the spring to equilibrium.

A change in elastic potential energy represents work being done. We can derive an equation for change in elastic energy using our definition of work. We begin with

$W=F\mathrm{\Delta}d\text{.}$

If the spring deformation changes from ${x}_{1}$ to ${x}_{2}\text{,}$ we have our displacement. The force is a bit more complicated—it won’t be constant since $x$ is changing, so we want to use the *average* force:

$W=\frac{{F}_{1}+{F}_{2}}{2}({x}_{2}-{x}_{1})\text{.}$

Now we can substitute the forces according to Hooke’s law:

$W=\frac{k{x}_{1}+k{x}_{2}}{2}({x}_{2}-{x}_{1})=\frac{1}{2}k({x}_{2}+{x}_{1})({x}_{2}-{x}_{1})\text{.}$

Recognizing a difference of squares here, we can make a little simplification. Now we have another formula to go along with our equations for changes in kinetic and gravitational energy:

$W=\mathrm{\Delta}{E}_{\text{e}}=\frac{1}{2}k\mathrm{\Delta}{x}^{2}\text{.}$