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 the vertical position. The work done by the man in this process is?

Answer 1

According to work energy theorem,

$W_{mg}+W_{man}=0$

$\because △K.E$ (changes in kinetic energy)

as according to question it was placed slowly on the table.

$W_{mg}=-mg\frac{l}{2}$

$W_{man}=\frac{mgl}{2}$