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:

$A$ 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:

$A$ block of mass $m$ is stationary on a rough inclined plane making an angle $\theta$ with the horizontal. Find the contact force between the block and the plane.

$A$ 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

$A$ 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 $N$)

$A$ body takes just twice the time to slide down a plane inclined at $30^\circ$ to the horizontal as it would if the plane were frictionless. The coefficient of friction between the body and the plane is: