$A$ body weighing $20 \ kg$ just slides down a rough inclined plane that rises $5$ in $12$. The coefficient of friction is

The force required just to move a body up an inclined plane is double the force required just to prevent the body from sliding down. If $\mu$ is the coefficient of friction,the inclination of the plane to the horizontal 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 $g = 10 \ m/s^2$)

There are two inclined surfaces of equal length $(L)$ and the same angle of inclination $45^{\circ}$ with the horizontal. One of them is rough and the other is perfectly smooth. $A$ given body takes $2$ times as much time to slide down the rough surface than on the smooth surface. The coefficient of kinetic friction $(\mu_k)$ between the object and the rough surface is close to:

$A$ particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is $0.5$,the acceleration is ($g=$ Acceleration due to gravity).

The minimum force required to start pushing a body up a rough (frictional coefficient $\mu$) inclined plane is $F_{1}$,while the minimum force needed to prevent it from sliding down is $F_{2}$. If the inclined plane makes an angle $\theta$ with the horizontal such that $\tan \theta = 2\mu$,then the ratio $\frac{F_{1}}{F_{2}}$ is: