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

$A$ box of mass $2 \ kg$ is placed on an inclined plane that makes $30^{\circ}$ with the horizontal. The coefficient of friction between the box and the inclined plane is $0.2$. $A$ force $F$ is applied on the box perpendicular to the incline to prevent the box from sliding down. The minimum value of $F$ is (acceleration due to gravity $= 10 \ ms^{-2}$) (in $N$)

$A$ 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 $g = 10 \, m/s^2$).

$A$ block of mass $8\, kg$ is at rest on a rough inclined plane as shown in the figure below. The magnitude of the net force exerted by the surface on the block will be ........ $N$ $(g = 10\, m/s^2)$.

$A$ block of mass $5 \ kg$ is moving on an inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The coefficient of friction between the block and the inclined plane surface is $\frac{\sqrt{3}}{2}$. The force to be applied on the block so that the block will move down without acceleration is . . . . . . $N$.

$A$ small block starts sliding down an inclined plane forming an angle $45^{\circ}$ with the horizontal. The coefficient of friction $\mu$ varies with distance $s$ as $\mu = C s^2$,where $C$ is a constant of appropriate dimensions. The distance covered by the block before it stops is: