Question

Refer the attached picture. ​

Answer 1

assume 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

assume 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

= -1/2(YA/L1)(L2-L1)2

potential energy stored negative of work done 1/2"(k)*x

where

k=YA/L1

x=(L2-L1)