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

Explanation

Here `F = 200 N, l = 1 mm = 10^(-3) m`
Elastic potential energy stored in the wire is
`U = (1)/(2) Fl = (1)/(2) xx 200 xx 10^(-3) = 0.1 J`