$A$ mass of $4\; kg$ rests on a horizontal plane. The plane is gradually inclined until at an angle $\theta = 15^{\circ}$ with the horizontal,the mass just begins to slide. What is the coefficient of static friction between the block and the surface?

$A$ block of mass $2\,\,kg$ is placed on a rough inclined plane as shown in the figure $(\mu = 0.2)$ so that it just touches the spring. The block is allowed to move downwards. The spring will be compressed to a maximum of

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 . . . . . . .

The minimum force required to start pushing a body up a rough (frictional coefficient $\mu$) inclined plane is $F_{1}$,while the minimum force needed to prevent it from sliding down is $F_{2}$. If the inclined plane makes an angle $\theta$ with the horizontal such that $\tan \theta = 2\mu$,then the ratio $\frac{F_{1}}{F_{2}}$ is:

$A$ block slides down an inclined plane with an acceleration $g/2$ as shown in the figure. Then the coefficient of kinetic friction is

$A$ body is made to move up along an inclined plane of inclination $30^{\circ}$ and the coefficient of friction is $0.5$. Then its retardation is ($g$ = acceleration due to gravity).