$A$ block of mass $5 kg$ is at rest on a rough inclined surface. If the angle of inclination of the plane is $60^{\circ}$,then the total force applied by the surface on the block is .......... $N$. (Take $g = 10 m/s^2$)

If a block moving up an inclined plane at $30^{\circ}$ with a velocity of $5 \, m/s$ stops after $0.5 \, s$,then the coefficient of friction will be nearly:

$A$ 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?

$A$ 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:

$A$ block is at rest on an inclined plane making an angle $\alpha$ with the horizontal. As the angle $\alpha$ of the incline is increased,the block starts slipping when the angle of inclination becomes $\theta$. The coefficient of static friction between the block and the surface of the inclined plane is:

$A$ block of mass $m$ slides down a plane inclined at an angle $\theta$. Which of the following will $NOT$ increase the energy lost by the block due to friction?