Elastic Potential Energy

When you elongate or contract a rod or wire by exerting a force, you do work. This work is stored as elastic potential energy in the rod or wire. The elastic potential energy per unit volume is easy to remember and is equal to ½ stress×strain. Since the Young’s modulus Y = stress/strain, we can modify the above expression as ½ Y×strain². Another form of the same expression is ½ (stress)²/Y.

Consider the following M.C.Q. which appeared in the Kerala Engineering Entrance test paper of 2006:

A work of 2×10^-2 J is done on a wire of length 50cm and area of cross section 0.5mm². If the Young’s modulus of the material of the wire is 2×10¹⁰ N/m², then the wire must be

(a) elongated to 50.1414cm (b) contracted by 2mm (c) stretched by 0.707mm (d) of length changed to 49.293cm (e) of length changed to 50.2cm

Since the total work done is involved in this prblem, we write,

½ (Y×strain²) ×volume = 2×10^-2

Here, strain = δ/0.5 where ‘δ’ is the elongation (or contraction) and volume = AL = (0.5×10^-6) ×0.5

Substituting proper values, δ works out to 1.414×10^-3 m = 1.414mm = 0.1414cm. The wire therefore elongates to 50.1414cm [option(a)].

An expression you should remember for obtaining the total work done in stretching or contracting a rod or wire through ‘l’ by exerting a force ‘F’ is, W = ½ F × l.