$A$ block is placed on a parabolic shape ramp given by the equation $y = \frac{x^2}{20}$. If the coefficient of static friction $\mu_s$ is $0.5$, then what is the maximum height above the ground at which the block can be placed without slipping (in $m$)?

$A$ block starts moving up an inclined plane of inclination $30^{\circ}$ with an initial velocity of $v_{0}$. It comes back to its initial position with velocity $\frac{v_{0}}{2}$. The value of the coefficient of kinetic friction between the block and the inclined plane is close to $\frac{I}{1000}$. The nearest integer to $I$ is......

$A$ block is projected upwards on an inclined plane of inclination $37^{\circ}$ along the line of greatest slope with a coefficient of friction $\mu = 0.5$ and an initial velocity of $5 \ m/s$. The block first stops at a distance of .......... $m$ from the starting point.

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 the coefficient of friction is $0.25$,the angle of inclination of the plane is ...... $^o$.

$A$ body is projected up along a rough inclined plane of inclination $45^{\circ}$. The coefficient of friction is $0.5$. Then the retardation of the block is

$A$ uniform chain of length $L$ which hangs partially from a table,is kept in equilibrium by friction. The maximum length that can hang without slipping is $l$. Then,the coefficient of friction between the table and the chain is: