$A$ block rests on a rough inclined plane making an angle of $30^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$. If the frictional force on the block is $10 \, N$,the mass of the block (in $kg$) is (take $g = 10 \, m/s^2$).

$A$ small bar starts sliding down an inclined plane forming an angle $\theta$ with the horizontal. The friction coefficient depends on the distance $x$ covered as $\mu = kx$,where $k$ is a constant. Find the distance covered by the bar until it stops.

$STATEMENT-1$: $A$ 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.

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 ........ $^o$.

In the given arrangement of a doubly inclined plane,two blocks of masses $M$ and $m$ are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is $0.25$. The value of $m$,for which $M=10 \text{ kg}$ will move down with an acceleration of $2 \text{ m/s}^2$,is: (take $g=10 \text{ m/s}^2$ and $\tan 37^{\circ}=3/4$) (in $\text{ kg}$)

$A$ block is stationary on a rough inclined plane. How many forces are acting on the block?