Question

A uniformed chain of length $l$, and mass $M$ overcharge a horizontal table with its two third part on the table. The friction cover between the table & the chain is $\mu$. What is the work done by the friction during the period the chain slips off the table?

Answer 1

Force of friction is against the movement of the chain on the table

$d{F}_f$ on the element $dx$ of mass $dm$ due to friction $=\mu .dm.g$

$=\mu \left(\frac{m}{c}\right)g$

Element $dx$ gets displaced by $x$ when chain sides off

So work done on $dx$ by friction $=d{F}_F\times x$

$=\mu g\left(\frac{m}{L}\right)dx.x$

Total woek $={\int }_0^{2L/3}d{F}_f\times x$

$=\frac{\mu gm}{c}{\int }_0^{2L/3}x.dx=\frac{\mu gm}{c}{\left[\frac{x^2}{2}\right]}_0^{2L/3}$

Total work $=\frac{2}{9}\mu mgL$