Question

A uniform chain of length $\text{L}$ and mass $\text{M}$ is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If $\text{g}$ is acceleration due to gravity, the work required to pull the hanging part on the table is

• A. $\text{MgL}$
• B. $\frac{MgL}{3}$
• C. $\frac{MgL}{9}$
• D. $\frac{MgL}{18}$

Answer 1

The correct option is D $\frac{MgL}{18}$

Mass $\frac{M}{3}$ has its centre of mass $\frac{L}{6}$ below the table surface.
$\therefore \text{W = mgh}=\left(\frac{M}{3}\right)\left(g\right)\left(\frac{L}{6}\right)$
$=\frac{MgL}{18}$