$A$ block of mass $m$ rests on a rough inclined plane. The coefficient of friction between the surface and the block is $\mu$. At what angle of inclination $\theta$ of the plane to the horizontal will the block just start to slide down the plane?

$A$ body is sliding down an inclined plane (angle of inclination $45^{\circ}$). If the coefficient of friction is $0.5$ and $g = 9.8\, m/s^2$,then the acceleration of the body downwards in $m/s^2$ is

$A$ block of mass $2 \, kg$ 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 $0.7$. The frictional force on the block is ....... $N$.

$A$ block of mass $m$ is placed on a surface having a vertical cross-section given by $y = x^2 / 4$. If the coefficient of friction is $0.5$,the maximum height above the ground at which the block can be placed without slipping is:

An inclined plane of length $5.60 \ m$ making an angle of $45^{\circ}$ with the horizontal is placed in a uniform electric field $E = 100 \ V \ m^{-1}$. $A$ particle of mass $1 \ kg$ and charge $10^{-2} \ C$ is allowed to slide down from rest from the maximum height of the slope. If the coefficient of friction is $0.1$,the time taken by the particle to reach the bottom is . . . . . . .

$A$ uniform ladder of length $5 \ m$ is placed against a wall as shown in the figure. If the coefficient of friction $\mu$ is the same for both the wall and the floor,what is the minimum value of $\mu$ for it not to slip?