An ice cube is kept on an inclined plane of angle $30^{\circ}$. The coefficient of kinetic friction between the block and the inclined plane is $(1 / \sqrt{3})$. What is the acceleration of the block? $...... \ m / s ^2$

Two blocks $A$ and $B$ are released from the top of a rough inclined plane so that $A$ slides along the plane and $B$ falls down freely. Which will have higher velocity on reaching the ground?

$A$ body is made to move up along an inclined plane of inclination $30^{\circ}$ and the coefficient of friction is $0.5$. Then its retardation is ($g$ = acceleration due to gravity).

$A$ 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 ............. $ms^{-2}$.

$A$ block of mass $5 \text{ kg}$ starts up a $45^{\circ}$ inclined plane with an initial kinetic energy of $100 \text{ J}$. If the coefficient of friction between the block and the plane is $0.5$,then the distance covered by the block before it stops is (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)

When a body slides down from rest along a smooth inclined plane making an angle of $30^{\circ}$ with the horizontal,it takes time $T$. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance,it takes time $\alpha T$,where $\alpha$ is a constant greater than $1$. The coefficient of friction between the body and the rough plane is $\frac{1}{\sqrt{x}}\left(\frac{\alpha^{2}-1}{\alpha^{2}}\right)$ where $x = .....$.