$A$ block of mass $M$ slides down a rough inclined plane with constant velocity. The angle made by the inclined plane with the horizontal is $\theta$. The magnitude of the contact force will be:

$A$ piece of ice slides down a rough inclined plane at $\theta=45^{\circ}$ inclination in twice the time that it takes to slide down an identical but frictionless inclined plane. What is the coefficient of friction between ice and incline?

When a cube of side length $a$ is in limiting equilibrium on an inclined plane and it is about to topple,what is the coefficient of friction between the cube and the plane?

$A$ block of mass $m$ is stationary on a rough inclined plane making an angle $\theta$ with the horizontal. Find the contact force between the block and 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

The upper half of an inclined plane with an angle of inclination $\phi$ is smooth,while the lower half is rough. $A$ body starting from rest at the top of the inclined plane comes to rest at the bottom of the inclined plane. Then the coefficient of friction for the lower half is