$A$ block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient of friction is $\mu$. The force required to just push it up the inclined plane is $3$ times the force required to just prevent it from sliding down. If we define $N=10 \mu$,then $N$ is

$A$ block of mass $m$ is at rest with respect to a lift when placed on an inclined plane of inclination $\theta$ inside the lift. If the lift moves upward at a constant velocity $v$,then the work done by the friction force on the block in time $t$ is:

$A$ block of mass $10 \, kg$ is released on a rough inclined plane. The block starts descending with an acceleration of $2 \, m/s^2$. The kinetic friction force acting on the block is ..... $N$ (take $g = 10 \, m/s^2$).

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:

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:

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