$A$ block of mass $m$ rests on a rough inclined plane. The coefficient of friction between the surface and the block is $\mu$. At what angle of inclination $\theta$ of the plane to the horizontal will the block just start to slide down the plane?

$A$ small block starts slipping down from a point $B$ on an inclined plane $AB$,which is making an angle $\theta$ with the horizontal. The section $BC$ is smooth and the remaining section $CA$ is rough with a coefficient of friction $\mu$. It is found that the block comes to rest as it reaches the bottom (point $A$) of the inclined plane. If $BC = 2AC$,the coefficient of friction is given by $\mu = k \tan \theta$. The value of $k$ is

$A$ heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and table surface is $0.25$,then the maximum fraction of length of the chain that can hang over one edge of the table is ...... $\%$

$A$ given object takes $n$ times more time to slide down a $45^o$ rough inclined plane as it takes to slide down a perfectly smooth $45^o$ incline. The coefficient of kinetic friction between the object and the incline is

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

The force required to just move a body up an inclined plane is double the force required to just prevent the body from sliding down the plane. If the coefficient of friction is $\mu$,then the inclination $\theta$ of the plane is: