$A$ brick of mass $2 \, kg$ begins to slide down on a plane inclined at an angle of $45^\circ$ with the horizontal. The force of friction will be

$A$ block of mass $10 \,kg$ is kept on a fixed rough $(\mu=0.8)$ inclined plane of angle of inclination $30^{\circ}$. The frictional force acting on the block is ........... $N$.

$A$ 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

$A$ cubic block of mass $m$ is sliding down on an inclined plane at $60^{\circ}$ with an acceleration of $\frac{g}{2}$. The value of the coefficient of kinetic friction is:

When a body slides down from rest along a smooth inclined plane making an angle of $45^{\circ}$ 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 is seen to take time $pT$,where $p$ is some number greater than $1$. Calculate the coefficient of friction between the body and the rough plane.

$STATEMENT-1$: $A$ block of mass $m$ starts moving on a rough horizontal surface with a velocity $v$. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of $30^{\circ}$ with the horizontal and the same block is made to go up on the surface with the same initial velocity $v$. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. because
$STATEMENT-2$: The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.