A cubic block of mass $m$ is sliding down on an inclined plane at $60^{\circ}$ with an acceleration of $\frac{\mathrm{g}}{2}$, the value of coefficient of kinetic friction is

Question

Aicharya · Accepted Answer

$\mathrm{mg} \sin 60^{\circ}-\mu \mathrm{mg} \cos 60^{\circ}=\mathrm{ma}$ $g \sin 60-\mu g \cos 60=\frac{g}{2}$ $\frac{\sqrt{3}}{2}-\frac{\mu}{2}=\frac{1}{2}$ $\mu=\sqrt{3}-1$

A cubic block of mass $m$ is sliding down on an inclined plane at $60^{\circ}$ with an acceleration of…

Solution