$A$ block is projected up an inclined plane of angle $\theta = 30^o$ with an initial velocity of $5 \, m/s$. It comes to rest in $0.5 \, s$. What is the coefficient of friction?

$A$ box is placed on an inclined plane and has to be pushed down. The angle of inclination is

$A$ homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is:

$A$ 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:

$A$ body takes $1\frac{1}{3}$ times as much time to slide down a rough inclined plane as it takes to slide down an identical but smooth inclined plane. If the angle of the inclined plane is $45^{\circ}$,the coefficient of friction is:

An inclined plane is bent in such a way that the vertical cross-section is given by $y = \frac{x^2}{4}$,where $y$ is in the vertical direction and $x$ is in the horizontal direction. If the upper surface of this curved plane is rough with a coefficient of friction $\mu = 0.5$,the maximum height in $cm$ at which a stationary block will not slip downward is............$cm$.