# Question #b79c6

Nov 27, 2016

Let
$m \to \text{mass of the block}$
$\theta \to \text{angle of inclination of the inclined plane } = {29.7}^{\circ}$
$g \to \text{acceleration due to gravity}$
${\mu}_{k} \to \text{coefficient of kinetic friction }$

Since in this case the block moves downward with constant speed the net force acting on it zero,
Here downward gravitational pull along the plane $\textcolor{b l u e}{m g \sin \theta}$ acting on the block is just balanced by the upward frictional force $\textcolor{red}{{\mu}_{k} m g \cos \theta}$ along the plane

So $m g \sin \theta = {\mu}_{k} m g \cos \theta$

$\implies {\mu}_{k} = \tan \theta = \tan {29.7}^{\circ} = 0.57$