$A$ body is projected up along a rough inclined plane of inclination $45^{\circ}$. The coefficient of friction is $0.5$. Then the retardation of the block is

The time taken by an object to slide down a $45^{\circ}$ rough inclined plane is $n$ times the time it takes to slide down a perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the inclined plane is:

$A$ block rests on a rough inclined plane making an angle of $30^o$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$. If the frictional force on the block is $10 \, N$,the mass of the block (in $kg$) is (take $g = 10 \, m/s^2$)

$A$ block moves down a smooth inclined plane of inclination $\theta$. Its velocity on reaching the bottom is $v$. If it slides down a rough inclined plane of the same inclination,its velocity on reaching the bottom is $v/n$,where $n$ is a number greater than one. The coefficient of friction $\mu$ is given by:

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 $g = 10 \ m/s^2$)

When a body slides down from rest along a smooth inclined plane making an angle of $30^{\circ}$ with the horizontal,it takes time $T$. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance,it takes time $\alpha T$,where $\alpha$ is a constant greater than $1$. The coefficient of friction between the body and the rough plane is $\frac{1}{\sqrt{x}}\left(\frac{\alpha^{2}-1}{\alpha^{2}}\right)$ where $x = .....$.