Question

A wire suspended vertically from one end of its ends is stretched by attaching a weight of $200\text{N}$ to the lower end. The weight stretches the wire by $1\text{mm}$. Then the elastic energy stored in the wire(in J) is

Answer 1

Given $F=200N,x=1\text{mm}={10}^{-3}\text{m}$

Elastic energy $=\frac{1}{2}\times F\times x$
$\therefore E=\frac{1}{2}\times 200\times 1\times {10}^{-3}=0.1\text{J}$