Question

A block of mass $m$ slides down the plane inclined at an angle $60$ with an acceleration $\frac{g}{2}$. Find the coefficient of kinetic friction?

Answer 1

At an inclined normal force on the block or surface is

$N=mg\cos 60=\frac{mg}{2}$

also, friction, $F_k={\mu }_k\times N$

$\therefore F_k={\mu }_k\times \frac{mg}{2}$

Force responsible for downward motion of block is $mg\sin \theta$

$\therefore F_{net}=mg\sin \theta -\mu mg\cos \theta$

or $mg(\sin \theta -\mu \cos \theta )=m\times \frac{g}{2}$

$\sin \theta -\mu \cos \theta =\frac{1}{2}$

or $\frac{\sqrt{3}-\mu }{2}=\frac{1}{2}$

or $\mu =\sqrt{3}-1=0.73$