Question

A wire is suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The wire stretches the wire by 1 mm. The elastic energy stored in the wire is$:$

Answer 1

The elastic energy stored $U=\frac{1}{2}\times F\times \Delta I$

$==\frac{1}{2}\times 200\times {10}^{-3}J$

$=0.1J$

We get, $U=0.1J.$