Question

A chain is held on a friction-less table with one third of its length hanging over the edge. The total length of the chain is $1$ and its mass is $m$. Find the work required to pull the hanging part back to the table.

• A. $2mgl$
• B. $\frac{mgl}{5}$
• C. $\frac{mgl}{9}$
• D. $\frac{mgl}{18}$

Answer 1

Answer: (D)

$\frac{mgl}{18}$

The mass of hanging chain is $\frac{m}{3}$, length is $\frac{l}{3}$ .

Now the the center of mass of hanging chain is at a distance of $\frac{l}{6}$ from table top.

The mass hanging over the edge is $\frac{mg}{6}$, so initial force required to pull the hanging mass on the table =$\frac{mg}{3}$.

The final force required to pull the remaining mass = $0$.

So, Average force =$\frac{mg}{6}$

Displacement in the chain = $\frac{l}{3}$.

Thus, work done by the variable force

= force x distance = $\frac{mgl}{2}{x}^2$

Where $\frac{1}{x}$ is the fraction of the length of the chain which is initially hanging

Here, $x=3$

So, work done $=mg\frac{l}{2\times 3^2}=mg\frac{l}{18}=\frac{mgl}{18}$