When a block of mass $M$ is suspended by a long wire of length $L$, the length of the wire becomes $(L+l)$. Th

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:

Accepted Answer

1. **Identify Extension:** The initial length is $L$ and the final length is $(L+l)$. Therefore, the extension (elongation) produced in the wire is $\Delta L = (L+l) - L = l$.
2. **Identify Force:** The force stretching the wire is the weight of the suspended block, so $F = Mg$.
3. **Formula for Potential Energy:** The elastic potential energy ($U$) stored in a stretched wire is equal to the work done in stretching it. For a wire behaving elastically, this is given by:
   $$ U = \frac{1}{2} 	imes 	ext{Load} 	imes 	ext{Extension} $$
   
4. **Calculation:** Substituting the values:
   $$ U = \frac{1}{2} 	imes (Mg) 	imes (l) = \frac{1}{2}Mgl $$

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