Question

A uniform chain of mass m and length $l$ is kept on a smooth horizontal surface such that 1/4th of its length is hanging from the edge of the table. If it slides down, the change in kinetic energy when hanging length becomes $\frac{3l}{4}$ will be

• A. $\frac{mgl}{4}$
• B. $\frac{mgl}{2}$
• C. $\frac{mgl}{32}$
• D. $\frac{9mgl}{32}$

Answer 1

Answer: (A)

$\frac{mgl}{4}$

The center of mass of chain has descended through $\frac{l}{8}$ to $\frac{3l}{8}$

The increase in potential energy will be gain in kinetic energy :

$\Delta K.E.=\Delta P.E.=\frac{3m}{4}\times \frac{3l}{8}\times g-\frac{m}{4}\times \frac{l}{8}\times g$

$=\frac{8mgl}{32}$

$=\frac{mgl}{4}$