Question

A wire suspended vertically from one of Its ends is stretched by attaching a weight of $\text{200}$N to the Lower End. The weight stretches the wire by $1$mm. Then the elastic energy stored in the wire is ?

Answer 1

Step 1: Given parameters

Force, F = $\text{200}$N.

Stretching of wire, x = $1$mm.

Step 2: Formula used

Elastic energy$=\frac{1}{2}\times \mathrm{F}\times \mathrm{x}$ where F is the force and x is the extension produced.

Step 3: Calculating the elastic energy stored in the wire.

By substituting the given values, we get

$$\begin{gathered} =\frac{1}{2}\times 200\times 1\times {10}^{-3} \\ =0.1\mathrm{J} \end{gathered}$$

Therefore, the elastic energy will be $0.1\mathrm{J}$.