$A$ 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 .............. $cm$.

$A$ 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 $g=10 \,ms^{-2}$)

$A$ $2 \ kg$ brick begins to slide over a surface which is inclined at an angle of $45^{\circ}$ with respect to the horizontal axis. The coefficient of static friction between their surfaces is:

$A$ 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

When a body is placed on a rough plane inclined at an angle $\theta$ to the horizontal,its acceleration is

$A$ body is placed on a rough inclined plane of inclination $\theta$. As the angle $\theta$ is increased from $0^o$ to $90^o$,the contact force between the block and the plane