A man places a vertical uniform chain (of mass $m$ and length $l$ ) on a table slowly. Initially the lower end of the chain just touches the table. The man drops the chain when half of the chain is in vertical position. Then, the work done by the man in this process is :

- \frac{m g l}{2}

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- \frac{m g l}{4}

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- \frac{3 m g l}{8}

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- \frac{m g l}{8}

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Solution

The correct option is C $- \frac{3 m g l}{8}$
The magnitude of the work done by man is equal to the magnitude of change of potential energy of the chain.
The initial potential energy of the chain, $U_{i} = m g \frac{L}{2}$ as the center of mass of mass $m$ is at height $L / 2$ above the ground as shown in the figure.
The final potential energy of the chain, $U_{f} = \frac{m}{2} g \frac{L}{4}$ as center of mass of mass $m / 2$ is at height $L / 4$ above the ground as shown in the figure.
$Δ U = U_{f} - U_{i} = m g \frac{L}{8} - m g \frac{L}{2} = - 3 m g \frac{L}{8}$

$∴ W = Δ U = - \frac{3 m g l}{8}$

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A man places a vertical uniform chain (of mass m and length l ) on a table slowly. Initially the lower end of the chain just touches the table. The man drops the chain when half of the chain is in vertical position. Then, the work done by the man in this process is :

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