The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction,when the angle of inclination of the plane is $60^{\circ}$ is

The upper half of an inclined plane of inclination $\theta$ is smooth,while the lower half is rough. If a block released from the top comes to rest at the bottom,what is the coefficient of friction between the block and the rough surface?

$STATEMENT-1$: $A$ block of mass $m$ starts moving on a rough horizontal surface with a velocity $v$. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of $30^{\circ}$ with the horizontal and the same block is made to go up on the surface with the same initial velocity $v$. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. because
$STATEMENT-2$: The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.

$A$ 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 $g = 10 m/s^2$)

Find the reading of the weighing machine $(WM)$ in the following case. The system is in equilibrium.

The time taken by an object to slide down a $45^{\circ}$ rough inclined plane is $n$ times the time it takes to slide down a perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the inclined plane is: