JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Work Done in Pulling the Chain Against Gravity

A chain of length L and mass M is held on a frictionless table with (1/n)th 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.

\[W=-\int_{L/n}^{0}{mgy\,dy}=-mg\,\left[ \frac{{{y}^{2}}}{2} \right]_{L/n}^{0}\]\[=\frac{mg\,{{L}^{2}}}{2{{n}^{2}}}\]

\[\therefore \] \[W=\frac{MgL}{2{{n}^{2}}}\]                       [As m = M/L]

Alternative method :

If point mass m is pulled through a height h then work done W = mgh  

Similarly for a chain we can consider its centre of mass at the middle point of the hanging part i.e. at a height of L/(2n) from the lower end and mass of the hanging part of chain \[=\frac{M}{n}\]

So work done to raise the centre of mass of the chain on the table is given by

\[W=\frac{M}{n}\times g\times \frac{L}{2n}\]                      [As W = mgh]

or \[W=\frac{MgL}{2{{n}^{2}}}\]