Question

A chain of mass $M$ is placed on a smooth table with $1/n$ of its length $L$ hanging over the edge. The work done in pulling the hanging portion of the chain back to the surface of the table is:

• A. $MgL/n$
• B. $MgL/2n$
• C. $MgL/n^2$
• D. $MgL/2n^2$

Answer 1

Answer: (D)

Mass per unit length $=\frac{M}{L}$

mass of dx element $=\frac{M}{L}dx$

$dm=\frac{M}{L}dx$

Work done by pulling's dx element over the table $=dw=\left(dmx\right)g$

$dw=\frac{Mg}{L}xdx$

$w={\int }_0^{\left(\frac{L}{n}\right)}\frac{Mg}{L}xdx=\frac{Mg}{L}\stackrel{\frac{L}{n}}{\underset{0}{\left[\frac{x^2}{2}\right]}}=\frac{Mg}{L}\frac{L^2}{2n^2}=\frac{mgL}{2n^2}$