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

Two blocks $A$ and $B$ are released from the top of a rough inclined plane so that $A$ slides along the plane and $B$ falls down freely. Which will have higher velocity on reaching the ground?

$A$ block of mass $m$ is lying on an inclined plane. The coefficient of friction between the plane and the block is $\mu$. The force $(F_1)$ required to move the block up the inclined plane will be

If for an inclined plane the coefficient of static friction is $\mu_s = \frac{3}{4}$,then the angle of repose for the inclined plane will be ........ $^o$.

$A$ small block starts slipping down from a point $B$ on an inclined plane $AB$,which is making an angle $\theta$ with the horizontal. The section $BC$ is smooth and the remaining section $CA$ is rough with a coefficient of friction $\mu$. It is found that the block comes to rest as it reaches the bottom (point $A$) of the inclined plane. If $BC = 2AC$,the coefficient of friction is given by $\mu = k \tan \theta$. The value of $k$ is

$A$ uniform chain of total length $L$ is at rest,partially on an incline of angle $\theta = 30^{\circ}$ and partially hanging vertically. The coefficient of friction between the chain and the incline is $\mu = \frac{1}{2\sqrt{3}}$. Find the ratio $\frac{L_{\max}}{L_{\min}}$,where $L_{\max}$ is the maximum length of the chain on the incline and $L_{\min}$ is the minimum length of the chain on the incline such that the chain remains at rest.