$A$ block of mass $2 \, kg$ slides down an inclined plane of inclination $30^{\circ}$. The coefficient of friction between the block and the plane is $0.5$. The contact force between the block and the plane is:

$A$ 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:

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

The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is $\frac{1}{2 \sqrt{3}}$, then the angle of the inclined plane is (in $^{\circ}$)

When a body is placed on a rough plane (coefficient of friction $= \mu$) inclined at an angle $\theta$ to the horizontal,its acceleration is (acceleration due to gravity $= g$)

$A$ block moves down a smooth inclined plane of inclination $\theta$. Its velocity on reaching the bottom is $v$. If it slides down a rough inclined plane of the same inclination,its velocity on reaching the bottom is $v/n$,where $n$ is a number greater than one. The coefficient of friction $\mu$ is given by: