Question

A box of mass 10 kg slides down a rough inclined plane of inclination ${30}^0$ with uniform speed. The coefficient of kinetic friction in this case is

Answer 1

$N=log\cos 30$

$=log\frac{\sqrt{3}}{2}$

$N=5\sqrt{3}g$

$f_s=N\mu$

$=5\sqrt{3}g\mu$

Now Since the block is sliding down with constant velocity i acceleration = 0

$\therefore f_{net}=0$

So $f_s=mg\sin {30}^{\circ }$

$5\sqrt{3}g\mu =10g\left(\frac{1}{2}\right)$

$5\sqrt{3}\mu =5$

$\mu =\frac{1}{\sqrt{3}}$