$A$ body is sliding down an inclined plane having a coefficient of friction $0.5$. If the normal reaction is twice that of the resultant downward force along the incline,the angle between the inclined plane and the horizontal is ....... $^o$

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

$A$ small block starts slipping down from a point $B$ on an inclined plane $AB$,which is making an angle $\theta$ with the horizontal. The section $BC$ is smooth and the remaining section $CA$ is rough with a coefficient of friction $\mu$. It is found that the block comes to rest as it reaches the bottom (point $A$) of the inclined plane. If $BC = 2AC$,the coefficient of friction is given by $\mu = k \tan \theta$. The value of $k$ is

$A$ 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 $m$ is stationary on a rough inclined plane making an angle $\theta$ with the horizontal. Find the contact force between the block and the plane.