$A$ 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 ....... $^o$

$A$ car is going at a speed of $6\, m/s$ when it encounters a $15\, m$ slope of angle $30^o$. The friction coefficient between the road and tyre is $0.5$. The driver applies the brakes. The minimum speed of the car with which it can reach the bottom is ........ $m/s$. $(g = 10\, m/s^2)$

$A$ 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

The upper half of an inclined plane with an angle of inclination $\phi$ is smooth,while the lower half is rough. $A$ body starting from rest at the top of the inclined plane comes to rest at the bottom of the inclined plane. Then the coefficient of friction for the lower half is

The coefficient of friction between a body and the surface of an inclined plane at $45^o$ is $0.5$. If $g = 9.8\,m/s^2$,the acceleration of the body downwards in $m/s^2$ is

If $250 \,J$ of work is done in sliding a $5 \,kg$ block up an inclined plane of height $4 \,m$,the value of work done against friction is ............. $J$ $\left(g=10 \,ms^{-2}\right)$.