Question

A uniform chain has a mass M and length L. It is placed on a frictionless table with length $L_0$ hanging over the edge. The chain begins to slide down. Then, the speed v with which the end slides down from the edge is given by

• A. $v=\sqrt{\frac{g}{L}(L+L_0)}$
• B. $v=\sqrt{\frac{g}{L}(L-L_0)}$
• C. $v=\sqrt{\frac{g}{L}(L^2-L_0^2)}$
• D. $v=\sqrt{2g(L-L_0)}$

Answer 1

The correct option is C $v=\sqrt{\frac{g}{L}(L^2-L_0^2)}$

Decrease in gravitational PE = increase in kinetic energy

or $\{-\frac{M}{L}{L}_{0}g\frac{L_0}{2}\}-\{-\frac{MgL}{2}\}=\frac{1}{2}M{v}^2$
or $v=\sqrt{\frac{g}{L}(L^2-L_0^2)}$