3.2 Work done on a spring by an external force

1.1 Work done by area under force displacement curve

2.1 Properties of work done
2.2 Work done by a variable force

3.1 Work done by a spring force
3.2 Work done on a spring by an external force

6.1 Potential energy $\left( {\Delta U} \right)$ is negative of the work done by conservative forces.
6.2 Types of potential energy
6.3 Law of conservation of mechanical energy

When an external force ${\overrightarrow F _{Ext}}$ is applied on a spring and it undergoes an elongation of distance $x$,

From FBD, when the spring is elongated by a distance $x$ we get, $${\overrightarrow F _{Ext}} = kx$$

Work done by an external force to displace the spring by an infinitesimal distance $dx$ we get, $$\begin{equation} \begin{aligned} {W_{Ext}} = \int\limits_0^x {{{\overrightarrow F }_{Ext}}.d\overrightarrow x } \\ {W_{Ext}} = \int\limits_0^x {kxdx} \\ {W_{Ext}} = \frac{1}{2}k\left[ {{x^2}} \right]_0^x \\ {W_{Ext}} = \frac{1}{2}k{x^2}...(v) \\\end{aligned} \end{equation} $$

From equations $(iii)$ & $(v)$ we get, $$\begin{equation} \begin{aligned} {W_S} = - \frac{1}{2}k{x^2} \\ {W_{Ext}} = \frac{1}{2}k{x^2} \\\end{aligned} \end{equation} $$ or $${W_S} = - {W_{Ext}}$$

Question 7. Find the work done by a spring $(k=10N/m)$, when it elongates by a distance of $0.3m$. Also, find the potential energy stored in the spring.

Given,$$\begin{equation} \begin{aligned} k = 10N/m \\ x = 0.3m \\\end{aligned} \end{equation} $$

Work done by a spring is given by, $$\begin{equation} \begin{aligned} {W_S} = - \frac{1}{2}k{x^2} \\ {W_S} = - \frac{1}{2}\left( {10} \right){\left( {0.3} \right)^2} \\ {W_S} = - 0.45J \\\end{aligned} \end{equation} $$

Potential energy is the negative of the work done by spring, $$\Delta U = - {W_S} = 0.45J$$

Question 8. Find the potential energy stored in the spring in its equilibrium position.

Solution: At equilibrium position, the spring is elongated by a distance $x_0$.

So, from FBD, we get, $$\begin{equation} \begin{aligned} k{x_0} = mg \\ {x_0} = \frac{{mg}}{k}...(i) \\\end{aligned} \end{equation} $$

Potential energy stored in the spring is given by,$$\begin{equation} \begin{aligned} \Delta U = \frac{1}{2}k{x^2} \\ \Delta U = \frac{1}{2}k{\left( {\frac{{mg}}{k}} \right)^2} \\ \Delta U = \frac{{{{\left( {mg} \right)}^2}}}{{2k}} \\\end{aligned} \end{equation} $$