Physics - W.E.P. Question with Solution | TestHub

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 :

$\text{W}=-{\int }_{\text{L/n}}^0\text{mgydy}=\frac{{\text{MgL}}^2}{{\text{2n}}^2}$

$\text{or}\text{W}={\text{MgL/2n}}^2\left[\text{as M} = \text{mL}\right]$