Question

A chain of uniform mass m and length l is hanging from the end of a table , find work done in pulling the chain up the table

Answer 1

The work done is $\frac{mgl}{2}$.

Answer 2

The work done in pulling the chain up is equal to the change in its gravitational potential energy. The initial potential energy of the chain, with its center of mass at $-l/2$, is $PE_{initial} = m \cdot g \cdot (-l/2) = -\frac{mgl}{2}$. When the chain is pulled onto the table, its center of mass is at $0$, so the final potential energy is $PE_{final} = 0$. The work done is $W = PE_{final} - PE_{initial} = 0 - (-\frac{mgl}{2}) = \frac{mgl}{2}$.