Question
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 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 :
$$ - \frac{{3mgL}}{8}$$
Solution :
The work done by man is negative of magnitude of decreases in potential energy of chain
$$\eqalign{ & \Delta U = mg\frac{L}{2} - \frac{m}{2}g\frac{L}{4} = 3mg\frac{L}{8} \cr & \therefore W = - \frac{{3mgL}}{8} \cr} $$
The work done by man is negative of magnitude of decreases in potential energy of chain
$$\eqalign{ & \Delta U = mg\frac{L}{2} - \frac{m}{2}g\frac{L}{4} = 3mg\frac{L}{8} \cr & \therefore W = - \frac{{3mgL}}{8} \cr} $$