$A$ 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 base $10 \ cm \times 10 \ cm$ and height $15 \ cm$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$. The inclination $\theta$ of this inclined plane from the horizontal plane is gradually increased from $0^{\circ}$. Then

The force required to just move a body up the inclined plane is double the force required to just prevent the body from sliding down the plane. The coefficient of friction is $\mu$. The inclination $\theta$ of the plane is

$A$ block of mass $m$ slides down an inclined plane at an angle of $30^{\circ}$ with an acceleration of $\frac{g}{4}$. The value of the coefficient of kinetic friction will be:

$A$ block slides down on an inclined surface of inclination $30^{\circ}$ with the horizontal. Starting from rest,it covers $8 \ m$ in the first two seconds. Find the coefficient of friction. (Take $g = 10 \ m/s^2$)

$A$ block is placed on a rough inclined plane of inclination $\theta = 30^o$. If the force required to drag it up along the plane is to be smaller than the force required to lift it vertically,the coefficient of friction $\mu$ should be less than: