$A$ body is sliding down an inclined plane (angle of inclination $45^{\circ}$). If the coefficient of friction is $0.5$ and $g = 9.8\, m/s^2$,then the acceleration of the body downwards in $m/s^2$ is

$A$ body is sliding down a rough inclined plane. The coefficient of friction between the body and the plane is $0.5$. The ratio of the net force required for the body to slide down and the normal reaction on the body is $1:2$. Then the angle of the inclined plane is (in $^{\circ}$)

The time taken by an object to slide down a $45^{\circ}$ rough inclined plane is $n$ times the time it takes to slide down a perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the inclined plane is:

$A$ block of mass $2 \, kg$ rests on a plane inclined at an angle of $30^o$ with the horizontal. The coefficient of friction between the block and the surface is $0.7$. The frictional force acting on the block is . . . . . . $N$.

$A$ box of mass $2 \ kg$ is placed on an inclined plane that makes $30^{\circ}$ with the horizontal. The coefficient of friction between the box and the inclined plane is $0.2$. $A$ force $F$ is applied on the box perpendicular to the incline to prevent the box from sliding down. The minimum value of $F$ is (acceleration due to gravity $= 10 \ ms^{-2}$) (in $N$)

$A$ body is placed on a rough inclined plane of inclination $\theta$. As the angle $\theta$ is increased from $0^o$ to $90^o$,the contact force between the block and the plane