$A$ block rests on a rough inclined plane making an angle of $30^{\circ}$ 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 takes time $t$ to slide down a plane inclined at $45^\circ$ to the horizontal. If the surface is made smooth (frictionless), the block takes time $t/2$ to slide down the plane. The coefficient of friction between the block and the inclined plane is $\frac{\alpha}{100}$. The value of $\alpha$ is . . . . . . .

The upper $\left(\frac{1}{n}\right)^{\text{th}}$ of an inclined plane is smooth and the remaining lower part is rough with coefficient of friction $\mu_k$. If a body starting from rest at the top of the inclined plane will again come to rest at the bottom of the plane,then the angle of inclination of the inclined plane is

$A$ body of mass $100\, g$ is sliding down an inclined plane of inclination $30^\circ$. What is the frictional force experienced if the coefficient of friction $\mu = 1.7$?

$A$ small mass slides down an inclined plane of angle $\theta$ with the horizontal. The coefficient of friction is $\mu = \mu_0 x$,where $x$ is the distance through which the mass slides down and $\mu_0$ is a constant. Then,the distance covered by the mass before it stops is:

The upper half of an inclined plane with an angle of inclination $\phi$ is smooth,while the lower half is rough. $A$ body starting from rest at the top of the inclined plane comes to rest at the bottom of the inclined plane. Then the coefficient of friction for the lower half is