Work Done in Pulling the Chain Against Gravity
Category : JEE Main & Advanced
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 \[W=\frac{MgL}{2{{n}^{2}}}\]
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