$A$ block of mass $10 \,kg$ is kept on a fixed rough $(\mu=0.8)$ inclined plane of angle of inclination $30^{\circ}$. The frictional force acting on the block is ........... $N$.

$A$ 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 blocks are in equilibrium. The friction force acting on the $10 \, kg$ block is:

$A$ block of mass $m$ is placed on a surface with a vertical cross-section given by $y = \frac{x^3}{6}$. If the coefficient of friction is $0.5$,the maximum height above the ground at which the block can be placed without slipping is:

The upper half of an inclined plane of inclination $\theta$ is perfectly smooth,while the lower half is rough. $A$ body starting from rest at the top comes back to rest at the bottom. The coefficient of friction $\mu$ for the lower half is given by:

$A$ block of mass $5\,kg$ and surface area $2\,m^2$ just begins to slide down an inclined plane when the angle of inclination is $30^{\circ}$. Keeping the mass the same,the surface area of the block is doubled. The angle at which this starts sliding down is: