Question

When a block a mass M is suspended by a long wire of length L, the elastic potential potential energy stored in the wire is $\frac{1}{2}$ × stress × strain × volume. Show that it is equal to $\frac{1}{2}$ Mgl, where l is the extension. The loss in gravitational potential energy of the mass earth system is Mgl. Where does the remaining $\frac{1}{2}$ Mgl energy go?

Answer 1

Let the CSA of the wire be A.

$$\begin{gathered} \mathrm{Stress}=\frac{\mathrm{Force}}{\mathrm{Area}}=\frac{Mg}{A} \\ \mathrm{Strain}=\frac{l}{L} \\ \mathrm{Volume}=AL \\ \mathrm{We}\mathrm{need}\mathrm{to}\mathrm{calculate}\mathrm{the}\mathrm{elastic}\mathrm{potential}\mathrm{energy}\mathrm{stored}\mathrm{in}\mathrm{the}\mathrm{wire}\mathrm{which}\mathrm{is}\mathrm{given}\mathrm{to}\mathrm{be}\mathrm{equal}\mathrm{to}\frac{1}{2}\times \mathrm{Stress}\times \mathrm{Strain}\times \mathrm{Volume}. \\ \mathrm{Elastic}\mathrm{potential}\mathrm{energy}=\frac{1}{2}\times \mathrm{Stress}\times \mathrm{Strain}\times \mathrm{Volume} \\ =\frac{1}{2}\times \frac{Mg}{A}\times \frac{l}{L}\times AL \\ =\frac{1}{2}Mgl \end{gathered}$$

The other $\frac{1}{2}Mgl$ is converted into kinetic energy of the mass.

When the mass leaves its initial point on the spring, it acquires a velocity as it moves down. The velocity reaches its maximum at the end point. The spring oscillates. Finally, when the kinetic energy is dissipated into heat, the spring comes to rest.