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 $\mu$. The work done by the friction during the period, the chain slips off the table is :

• A. $-\frac{2}{9}\mu MgL$
• B. $-\frac{6}{7}\mu MgL$
• C. $-\frac{1}{4}\mu MgL$
• D. $-\frac{4}{9}\mu MgL$

Answer 1

The correct option is A $-\frac{2}{9}\mu MgL$

The linear mass density is $\frac{M}{2}$. The small work done for the slippage the small distance is given by $dW=-\mu \left[\begin{array}{c}\frac{M}{L}\end{array}\right]gldl$

Now, total work done

$W={\int }_0^{2L/3}-\frac{\mu Mg}{L}ldl$

$=-\frac{\mu Mg}{L}{\left[\begin{array}{c}\frac{l^2}{2}\end{array}\right]}_0^{2L/3}=-\frac{\mu Mg}{2L}\left[\begin{array}{c}\frac{4L^2}{9}-0\end{array}\right]$

$\Rightarrow W=-\frac{2}{9}\mu MgL$