Question

An uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $\left(\frac{L}{n}\right)$ th part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be

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

Answer 1

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


Given, mass of the cable is $M$
So, mass of $\frac{1}{n}$ th part of the cable i.e hanged part of the cable is = $M/n$ ...... (i)
Now, centre of mass of the hanged part will be its middle point.
So, its distance from the top of the table will be $L/2n$.
$\therefore$ Initial potential energy of the hanged part of cable,
$U_i=\left(\frac{M}{n}\right)(-g)\left(\frac{L}{2n}\right)$
$\Rightarrow U_i=-\frac{MgL}{2n^2}$....(ii)
When whole cable is on the table, its potential energy will be zero.
$\therefore U_f=0....\left(iii\right)$
Now, using work-energy theorem,
$W_{net}=\Delta U=U_f-U_i$
$\Rightarrow W_{net}=0-(-\frac{MgL}{2n^2})$ [Using Eqs. (ii) and (iii)]
$\Rightarrow W_{net}=\frac{MgL}{2n^2}$