A chain of mass $M$ and length $L$ is held on a frictionless table such that a fraction $f$ of its length hangs over the edge. The work done to pull the hanging part back onto the table is:

Solution

1. The mass of the hanging part is $m=fM$.
2. The center of mass of the hanging part is at a distance $h=fL/2$ below the table.
3. The potential energy of the hanging part is $$\begin{array}{rl}U& =-mgh\\ & =-(fM)g(fL/2)\end{array}$$.
4. The work done to bring it to the table level is $$\begin{array}{rl}W& =-U\\ & =MgL{f}^2/2\end{array}$$.
5. Hence the answer is (A).