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Q.

A man places a chain (of mass m and length l) on a table slowly. Initially, the lower end of the chain just touches the table. The man brings down the chain by length l/2. Work done by the man in this process is

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a

$- \frac{m g l}{4}$

b

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

c

$- m g \frac{l}{2}$

d

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

answer is C.

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Detailed Solution

Change in potential energy = work done by force

$W = Δ U = (\frac{m}{2} g \frac{l}{4}) - (m g \frac{l}{2}) = - 3 m g \frac{l}{8}$

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