Question 41:
moderateWhen 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 :
(2019)
The elastic potential energy stored in a stretched wire is given by $U = \frac{1}{2} \times \text{Force} \times \text{Extension}$. Here, the applied force is the weight of the block $Mg$ and the extension is $l$. Therefore, $U = \frac{1}{2} Mgl$.