Question

A uniform chain of length $2\text{m}$ is kept on a table such that a length of $60\text{cm}$ hangs freely from the edge of the table. The total mass of the chain is $4\text{kg}$. What is the work done in pulling the entire chain on the table?

• A. $7.2\text{J}$
• B. $3.6\text{J}$
• C. $120\text{J}$
• D. $1200\text{J}$

Answer 1

The correct option is B $3.6\text{J}$

A chain of length $L$ and mass $M$ is held on a frictionless table with $y$ of its length hanging over the edge.

Let $m=\frac{M}{L}=$ mass
per unit length of the
chain and $y$ is the length of the chain hanging over the edge. So the mass of the chain of length $y$ will be $ym$ and the force acting on it due to gravity will be $mgy$.
The work done in pulling the $dy$ length of the chain on the table.
$dW=F(-dy)$ [As $y$ is decreasing]
i.e., $dW=mgy(-dy)$
So, the work done in pulling the hanging portion on the table.
$W=-{\int }_y^{0}mgydy=-mg{\left[\frac{y^2}{2}\right]}_y^0=\frac{mg{y}^2}{2L}[Asm=M/L]\therefore W=\frac{4\times 10\times (0.6{)}^2}{2\times 2}=3.6\text{J}$