$A$ box when dropped from a certain height reaches the ground with a speed $v$. When it slides from rest from the same height down a rough inclined plane inclined at an angle $45^{\circ}$ to the horizontal,it reaches the ground with a speed $v/3$. The coefficient of sliding friction between the box and the plane is:

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

$A$ 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 $N$)

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

When a body slides down from rest along a smooth inclined plane making an angle of $30^{\circ}$ with the horizontal,it takes time $T$. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance,it takes time $\alpha T$,where $\alpha$ is a constant greater than $1$. The coefficient of friction between the body and the rough plane is $\frac{1}{\sqrt{x}}\left(\frac{\alpha^{2}-1}{\alpha^{2}}\right)$ where $x = .....$.

The upper half of an inclined plane with inclination $\phi$ is perfectly smooth,while the lower half is rough. $A$ body starting from rest at the top will again come to rest at the bottom,if the coefficient of friction for the lower half is given by-