A chain of mass M is placed on a smooth table with $\large\frac{1}{n}$$^{th} of its length L hanging over the edge. The work done in pulling the hanging portion of chain back to the surface of table is $(a)\;\frac{MgL}{n}\quad (b)\;\frac{MgL}{2n} \quad (c)\;\frac{MgL}{n^2} \quad (d)\;\frac{MgL}{2n^2}$ 1 Answer Answer: W=\large\frac{MgL}{2n^2} Work done= Change in potential energy of the center of mass of hanging part of chain. Mass of hanging part of the chain =\large\frac{M}{n} The height through which the center of mass rises =\large\frac{L}{2n} Work W = mgh = \large\frac{M}{n}$$ \times g \times \large\frac{L}{2n}$$= \large\frac{MgL}{2n^2}$
edited Aug 25, 2014