$A$ 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:

An inclined plane is bent in such a way that the vertical cross-section is given by $y = \frac{x^2}{4}$,where $y$ is in the vertical direction and $x$ is in the horizontal direction. If the upper surface of this curved plane is rough with a coefficient of friction $\mu = 0.5$,the maximum height in $cm$ at which a stationary block will not slip downward is............$cm$.

$A$ brick of mass $2 \, kg$ begins to slide down on a plane inclined at an angle of $45^\circ$ with the horizontal. The force of friction will be

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 $1 \, kg$ is projected from the lowest point up along an inclined plane with an angle of inclination $45^{\circ}$ and coefficient of friction $\mu = 0.5$. If $g = 10 \, ms^{-2}$,the retardation experienced by the block is ............. $ms^{-2}$.

Starting from rest,the time taken by a body sliding down on a rough inclined plane at $45^{\circ}$ with the horizontal is twice the time taken to travel on a smooth plane of the same inclination and same distance. Then the coefficient of kinetic friction is