$A$ body of $5 \,kg$ weight kept on a rough inclined plane of angle $30^{\circ}$ starts sliding with a constant velocity. Then the coefficient of friction is (assume $g=10 \,ms^{-2}$)

$A$ 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 . . . . . . $N$.

$A$ 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:

The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is $\frac{1}{2 \sqrt{3}}$, then the angle of the inclined plane is (in $^{\circ}$)

$A$ block slides down an incline of angle $30^o$ with an acceleration $\frac{g}{4}$. Find the coefficient of kinetic friction.

$A$ 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