Starting from rest,a body slides down a $45^o$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is:

$A$ board is balanced on a rough horizontal semicircular log. Equilibrium is obtained with the help of the addition of a weight to one of the ends of the board when the board makes an angle $\theta$ with the horizontal. The coefficient of friction between the log and the board is:

$A$ block slides down an inclined plane of slope angle $\theta$ with a constant velocity. It is then projected up the plane with an initial velocity $u$. The distance up to which it will rise before coming to rest is

$A$ 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.

$A$ small block starts sliding down an inclined plane forming an angle $45^{\circ}$ with the horizontal. The coefficient of friction $\mu$ varies with distance $s$ as $\mu = C s^2$,where $C$ is a constant of appropriate dimensions. The distance covered by the block before it stops is:

Two inclined planes are placed as shown in the figure. $A$ block is projected from point $A$ of the inclined plane $AB$ along its surface with a velocity just sufficient to carry it to the top point $B$ at a height of $10 \ m$. After reaching point $B$, the block slides down the inclined plane $BC$. The time it takes to reach point $C$ from point $A$ is $t(\sqrt{2}+1) \ s$. The value of $t$ is........ (use $g = 10 \ m/s^2$)