Explain the elastic potential energy of spring and obtain an expression for this energy.

Consider an elastic spring, obeying Hook's law with negligible mass whose one end is tied rigidly to a wall as shown in figure. At the other end of the spring a block has been tied and it restring on a smooth horizontal surface.

We shall for the sake of simplicity restrict the motion of the block in the $X$-direction.

In the normal position of the spring, the position of the block is taken as $x=0$, this is shown, as in figure$ (a)$.

When the block is pulled and the length of spring $+x$ is increased a restoring force $\mathrm{F}_{\mathrm{S}}$ is produced in the spring which tries to bring the spring back to its normal position. The restoring force (spring force) is produced when the spring is compressed. This is shown in figure $(b)$ and $(c)$. The restoring force is directly proportional to the change in the length of the spring and is in the direction opposite to the change in the length. This law of force for the spring is called Hook's law. $\therefore \mathrm{F}_{\mathrm{S}}=-k x$

where $k$ is the spring constant or force constant of the spring.

Spring constant $k=\frac{\mathrm{F}_{\mathrm{S}}}{x}$ (magnitude), its unit is $\mathrm{Nm}^{-1}$.

The spring is said to be stiff if $k$ is large and soft if $k$ is small.

Suppose that the block is pull outwards as in figure (b). If the extension is $x_{m}$, the work done by the spring force is,

$\mathrm{W}_{\mathrm{S}} =\int_{0}^{x_{m}} \mathrm{~F}_{\mathrm{S}} d x=-\int_{0}^{x_{m}} k x d x \quad\left[\because \mathrm{F}_{\mathrm{S}}=-k \Delta x\right]$

$=-k\left[\frac{x^{2}}{2}\right]_{0}^{x_{m}}$

$=-k\left[\frac{x_{m}^{2}}{2}-0\right]$

$=-\frac{k x_{m}^{2}}{2}$