$A$ block is lying on an inclined plane which makes $60^\circ$ with the horizontal. If the coefficient of friction between the block and the plane is $0.25$ and $g = 10\,m/s^2$,then the acceleration of the block when it moves along the plane will be ........ $m/s^2$.

$A$ 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:

If for an inclined plane the coefficient of static friction is $\mu_s = \frac{3}{4}$,then the angle of repose for the inclined plane will be ........ $^o$.

If the coefficient of friction between the block of mass $m$ and the wedge is $\mu $,then the maximum horizontal acceleration of the wedge for which the block will remain at rest with respect to the wedge is

$A$ brick of mass $2 \, kg$ begins to slide down on a plane inclined at an angle of $45^\circ$ with the horizontal. The force of friction will be

An inclined plane of length $5.60 \ m$ making an angle of $45^{\circ}$ with the horizontal is placed in a uniform electric field $E = 100 \ V \ m^{-1}$. $A$ particle of mass $1 \ kg$ and charge $10^{-2} \ C$ is allowed to slide down from rest from the maximum height of the slope. If the coefficient of friction is $0.1$,the time taken by the particle to reach the bottom is . . . . . . .