Question

A chain is held on a frictionless table with ${\left(\frac{1}{n}\right)}^{th}$ of its length hanging over the edge. If the chain has 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. $3\frac{MgL}{2n^2}$
• D. $\frac{MgL}{6n^2}$

Answer 1

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

The chain is pulled very slowly. Let $\lambda$ is the mass per unit length of the chain. Consider 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 $\lambda dyg$.
So the work done in pulling this element of the chain on the table is dW = F(y)

mass of this element is $dm=\lambda y$
$dW=\lambda gydy=\frac{M}{L}gydyW=\frac{Mg}{L}{\int }_0^{L/n}ydy=\frac{MgL}{2n^2}$