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

A $Mgl$

B $MgL$

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

D $\frac{1}{2}MgL$

The answer is C.

Solution

Elastic potential energy stored $=\frac{1}{2}\times\text{load}\times\text{extension}$
$U=\frac{1}{2}Mgl$

Concept energy stored in a stretched wire

15 questions from this chapter have appeared across 11 exam years.