Write down the expression for the elastic potential energy of a stretched wire .

Explanation

The work done in stretching the wire by dl,
`dW = F . dl`
The total work done in stretching the wire from 0 to l is
`W = int_0^1 F . dl`
From Young.s modulus of elasticity, force becomes,
`F = (Y Al)/(L) " " …..(1)`
Substituting equation (2) and (1) we get,
`W = int_0^l (Y Al.)/(L) dl = (YA)/L [(l.^2)/(2)]_(0)^(l) = (YA)/L (l^2)/2 = 1/2 [(YAl)/(L)] = 1/2 Fl`
W = 1/2 Fl`
This work done is known as the elastic potential energy of a stretched wire.