© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Physical Situation
A block of mass M is suspended from a long wire of original length L. Due to the weight of the block, the wire extends by an amount $l$, making its new total length $(L + l)$. We wish to find the elastic potential energy stored in the wire due to this extension.
Step 2: Identify the Tension in the Wire
In equilibrium, the tension T in the wire equals the weight of the block:
$$T = Mg\,.$$
Although the tension starts from nearly 0 when the wire is unextended and increases to Mg when fully extended, the extension process can be treated as having an average tension of $\frac{T}{2} = \frac{Mg}{2}$ over the stretch.
Step 3: Apply the Work-Energy Principle
The elastic potential energy stored in the wire can be found by the work done in stretching it. Since the tension increases uniformly from 0 to Mg, the average force is $\frac{Mg}{2}$. Hence, the work done (and stored as potential energy) is:
$$U = \text{Average force} \times \text{Extension} = \left(\frac{Mg}{2}\right) \times l = \frac{1}{2} Mgl\,.$$
Step 4: Final Answer
Thus, the elastic potential energy stored in the wire is:
$$\frac{1}{2} Mgl\,.$$
