Solids: Practice Problem & Solution
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 : (2019)
Solution Explained:
To solve this problem, we apply the core principles of Solids. Understanding the underlying formula is key to arriving at the correct answer below:
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, $U = \frac{1}{2} Mgl$.
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