Question

When a block of mass M is suspended by a long wire of length L, the length of the wire becomes L+l. The elastic potential energy stored in the extended wire is

• A. $Mgl$
• B. $MgL$
• C. $\frac{1}{2}Mgl$
• D. $\frac{1}{2}MgL$

Answer 1

$energy=\frac{1}{2}stress\left(strain\right)\left(volume\right)$

$=\frac{1}{2}\left(\frac{Mg}{A}\right)\frac{l}{L}\left(A\right)\left(L\right)$

$=\frac{1}{2}Mgl$

The correct option is C