Question

A uniform chain of length L and mass M overhangs a horizontal table with its two third part on the table. The friction coefficient between the table and the chain is μ. Find the work done by the friction during the period the chain slips off the table.

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Mass of chain = $M$

Length of chain = $L$

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

Let's take a small dx element at a distance x from the edge of the table mass of that element = $dm$ = $\frac{M}{L}dx$

Friction force on that element = $\mu dmg\left(leftward\right)$

Work done by friction on dx till dx falls off table = $\left(\frac{\mu M}{L}gdx\right)\times cos{180}^{\circ }$

$\underset{0}{\overset{w}{\int }}d{W}_{f}r=\underset{0}{\overset{\frac{2L}{3}}{\int }}\frac{-\mu M}{L}gxdx$

$W_f=\frac{-\mu M}{L}g\frac{x^2}{2}{|}_0^{\frac{2L}{3}}$

= $\frac{-\mu Mg}{2}\frac{4L}{9x}$

$W_f=\frac{4\mu MgL}{18}=\frac{-2\mu MgL}{9}$