$A$ particle of mass $m$ slides from rest down a plane inclined at $30^{\circ}$ to the horizontal. The force of resistance acting on the particle during motion is $ms^2$,where $s$ is the displacement of the particle from its initial position. The velocity (in $m/s$) of the particle when $s = 1\,m$ is $v$. The value of $\frac{3v^2}{14}$ is:

$A$ block of mass $2 \, kg$ slides down an inclined plane of inclination $30^{\circ}$. The coefficient of friction between the block and the plane is $0.5$. The contact force between the block and the plane 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).

$A$ block 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 $0.8$. If the frictional force on the block is $10 \, N$,the mass of the block (in $kg$) is (take $g = 10 \, m/s^2$).

$A$ rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is $\mu$. Let the mass of the box be $m$.
$(a)$ At what angle of inclination $\theta$ of the plane to the horizontal will the box just start to slide down the plane?
$(b)$ What is the force acting on the box down the plane,if the angle of inclination of the plane is increased to $\alpha > \theta$?
$(c)$ What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?
$(d)$ What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration $a$?

$A$ block of mass $5 \ kg$ is moving on an inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The coefficient of friction between the block and the inclined plane surface is $\frac{\sqrt{3}}{2}$. The force to be applied on the block so that the block will move down without acceleration is . . . . . . $N$.