Question

A uniform rod of length L and mass m is placed on rough horizontal surface.If coefficient of friction is $\mu =kx$,where k is constant and $\mu$ exist for $0\le x\le L$ (i.e for $x>L$ friction is zero.). If rod moves towards the right, magnitude of work done by friction is:

• A. $\frac{kmg{L}^2}{9}$
• B. $\frac{kmg{L}^2}{6}$
• C. $\frac{kmg{L}^2}{3}$
• D. $\frac{2kmg{L}^2}{3}$

Answer 1

The correct option is B $\frac{kmg{L}^2}{6}$


$df=\mu dN=kx\frac{m}{L}dxg$
$f={\int }_0^{L}kx\frac{m}{L}gdxworkdone=force\times diplacementdw=(kx\frac{m}{L}dx.g)(L-x)W={\int }_0^L\frac{kmg}{L}x(L-x)dx=\frac{kmg{L}^2}{6}$