Question

Derive the expression for elastic potential energy stored in a stretched wire under stress

Answer 1

let the initial length of the wire be L1 and final length after stretching be L2

Now during a small elongation dx.

Work done by Tension will be -Fdx

Y=stress /strain = (F*L1)/Ax

F= (YA/L1)*x

thus total work done = integration of -Fdx from 0 to L2-L1
=-(YA/L1)[x²/2]
=-1/2(YA/L1)(L2-L1)²

potential energy stored = negative of work done = 1/2*(k)*x²
where
k=YA/L1
x=(L2-L1)

Answer 2

Explanation:

F = -kx, where F is the force and x is the elongation.

The work done = energy stored in stretched string = F.dx

The energy stored can be found from integrating by substituting for force,

and we find,

The energy stored = kx2/2, where x is the final elongation.

The energy density = energy/volume

= (kx2/2)/(AL)

=1/2(kx/A)(x/L)

= 1/2(F/A)(x/L)

= 1/2(stress)(strain)