question_answer

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

Answer:

                Here \[F=200N,l=1mm={{10}^{-3}}m\] Elastic potential energy stored in the wire, \[U=\frac{1}{2}Fl=\frac{1}{2}\times 200\times {{10}^{-3}}=\mathbf{0}\mathbf{.1J}\].