Question

A chain is held on a frictionless table with $1/n$ th of its length hanging over the edge. If the chain has a length $L$ and a mass $M$ , how much work is required to pull the hanging part back on the table?

• A. $\frac{Mg{L}^{}}{2n^2}$
• B. $\frac{MgL}{n^2}$
• C. $\frac{MgL}{4n^3}$
• D. $\frac{MgL}{3n^{-1}}$

Answer 1

Answer: (A)

If $m$ is the mass per unit length of the chain, the mass of the chain of length $y$ will be $y\times m=ym$ and the force acting on it due to gravity will be $my\times g=mgy$ (assuming that $y$ is the length of the chain hanging over the edge). So, the work done in pulling the $dy$ length of the chain on the table
$dW=F\left(-dy\right)$ [as $y$ is decreasing]
i.e., $dW=mgy(-dy)$ [as $F=mgy$ ]
So, the work done in pulling the hanging portion on the table :
$W=-\underset{L/n}{\overset{0}{\int }}mgydy=\frac{Mg{L}^2}{2n^2}$
orW $=MgL/2n^2[$ as $M=mL]$