$A$ block of mass $m$ is pressed against a vertical wall and is in equilibrium. An external upward force of $\frac{mg}{2}$ is applied to the block. The minimum coefficient of friction $\mu$ required is:

$A$ block is at rest on an inclined plane making an angle $\alpha$ with the horizontal. As the angle $\alpha$ of the incline is increased,the block starts slipping when the angle of inclination becomes $\theta$. The coefficient of static friction between the block and the surface of the inclined plane is:

$Assertion$: Mountain roads rarely go straight up the slope.
$Reason$: The slope of mountains is large,therefore there are more chances for a vehicle to slip from the roads.

$A$ 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 of mass $2 \ kg$ is at rest at the bottom of an inclined plane of length $8 \ m$ and height $1 \ m$ as shown in the figure. If the coefficient of friction is $0.2$,what is the work done in moving the body from the bottom to the top of the incline (in $J$)? (Take $g = 10 \ m/s^2$)

$A$ block of mass $M$ is sliding down an inclined plane. An external force $F$ is applied vertically downwards on the block. The coefficient of static friction is $\mu_s$ and the coefficient of kinetic friction is $\mu_k$. The friction force acting on the block is: