$A$ small bar starts sliding down an inclined plane forming an angle $\theta$ with the horizontal. The friction coefficient depends on the distance $x$ covered as $\mu = kx$,where $k$ is a constant. Find the distance covered by the bar until it stops.

The force required to just move a body up the inclined plane is double the force required to just prevent the body from sliding down the plane. The coefficient of friction is $\mu$. The inclination $\theta$ of the plane 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.

$A$ body starts from rest on a long inclined plane of slope $45^o$. The coefficient of friction between the body and the plane varies as $\mu = 0.3x$,where $x$ is the distance travelled down the plane. The body will have maximum speed (for $g = 10 \ m/s^2$) when $x = $ ........ $m$.

$A$ body of mass $2 \ kg$ slides down with an acceleration of $4 \ m/s^2$ on an inclined plane having a slope of $30^{\circ}$. The external force required to take the same body up the plane with the same acceleration will be (Acceleration due to gravity $= 10 \ m/s^2$) (in $N$)

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 ........ $^o$.