Question

A man places a chain (of mass ${}^{\prime }{m}^{\prime }$ and length ${}^{\prime }{L}^{\prime }$) 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 work done by the man in this process is:

• A. $-mg\frac{L}{2}$
• B. $-\frac{mgL}{4}$
• C. $-\frac{3mgL}{8}$
• D. $-\frac{mgL}{8}$

Answer 1

The correct option is C $-\frac{3mgL}{8}$

The work done by man is negative of magnitude of decrease in potential energy of chain.

Initially the lower end of the chain just touches the table, So COM is initially at the height of $\frac{L}{2}$ from ground.
Potential energy is $\frac{mgL}{2}$

The man drops the chain when half of the chain is in vertical position. So,
Potential energy is $\frac{m}{2}g\frac{L}{4}$
Hence
$W=\Delta U=\frac{m}{2}g\frac{L}{4}-mg\frac{L}{2}=-3mg\frac{L}{8}$
$\therefore W=-\frac{3mgL}{8}$