Question

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

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

Answer 1

The correct option is A $\frac{MgL}{2n^2}$

The chain is pulled slowly without acceleration.
Let $\lambda$ is the mass per unit length of the chain.
$\lambda =\frac{M}{L}$


Now considering a differential element of the chain of length $dy$ at a depth $y$ from the surface of the table. The force acting on the element due to gravity is equal to $\lambda \left(dy\right)g$.

$dF=\lambda \left(dy\right)g$

So the work done in pulling this element of the chain on the table is

$dW=dFy\cos {180}^{\circ }=-ydF$

$dW=-y\left(\lambda gdy\right)$

For total workdone, integrating both side

$W=-\lambda g\underset{0}{\overset{-L/n}{\int }}ydy=-\lambda g[\frac{y^2}{2}{]}_0^{-L/n}$

$W=-\frac{\lambda g{L}^2}{2n^2}=-\frac{M}{L}\frac{g{L}^2}{2n^2}=-\frac{MgL}{2n^2}$

So, magnitude of work is $\frac{MgL}{2n^2}$.

Hence, option $\left(a\right)$ is correct.

Alternative approach (using center of mass): Let us take the tabletop as our reference level where potential energy is zero. Initially $\frac{L}{n}$ length is hanging whose mass is $\frac{M}{L}\frac{L}{n}=\frac{M}{n}$ and center of mass is $\frac{L}{2n}$ below the table. Initial $PE=-\frac{M}{n}g\frac{L}{2n}$ Finally, no length is hanging, and the center of mass is on the table. So, Final $PE=0$ As we know that $W_{ext}=\Delta (U+K)$ Looking at initial and final situation, kinetic energy is zero in both cases. $\therefore W_{ext}=P{E}_f-P{E}_i$ $W_{ext}=0-(-\frac{M}{n}g\frac{L}{2n})=\frac{MgL}{2n^2}$