As shown in the figure,a body of mass $m$ moving vertically downward with a speed of $3\, m/s$ hits a smooth fixed inclined plane and rebounds with a velocity $v_f$ in the horizontal direction. If the angle of the inclined plane is $30^{\circ}$,the velocity $v_f$ will be:

$A$ particle of mass $m$ slides from rest down a plane inclined at $30^{\circ}$ to the horizontal. The force of resistance acting on the particle during motion is $ms^2$,where $s$ is the displacement of the particle from its initial position. The velocity (in $m/s$) of the particle when $s = 1\,m$ is $v$. The value of $\frac{3v^2}{14}$ is:

$A$ 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

The wedge is fixed and the mass of the block is $m$. The minimum value of $F$ so that the block is in equilibrium is (neglect the dimension of the block).

The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is $\frac{1}{2 \sqrt{3}}$, then the angle of the inclined plane is (in $^{\circ}$)

For the equilibrium of a body on an inclined plane of inclination $45^{\circ}$, the coefficient of static friction must be: