Question

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

Answer 1

Elastic potential energy U = \(\frac{1}{2}\) x Stress x Strain x Volume

U = \(\frac{1}{2}\times\frac{F}{A}\times\frac{\Delta l}{l}\times AI = \frac{1}{2}F\Delta I\)

U = 1212 x 200 x 10^-3

= 0.1 Joule