If the coefficient of friction between the block of mass $m$ and the wedge is $\mu $,then the maximum horizontal acceleration of the wedge for which the block will remain at rest with respect to the wedge is

$A$ 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

$A$ block of mass $5\,kg$ and surface area $2\,m^2$ just begins to slide down an inclined plane when the angle of inclination is $30^{\circ}$. Keeping the mass the same,the surface area of the block is doubled. The angle at which this starts sliding down is:

Two inclined planes are placed as shown in the figure. $A$ block is projected from point $A$ of the inclined plane $AB$ along its surface with a velocity just sufficient to carry it to the top point $B$ at a height of $10 \ m$. After reaching point $B$, the block slides down the inclined plane $BC$. The time it takes to reach point $C$ from point $A$ is $t(\sqrt{2}+1) \ s$. The value of $t$ is........ (use $g = 10 \ m/s^2$)

$A$ block of mass $10\, kg$ is placed on an inclined plane. When the angle of inclination is $30^\circ$,the block just begins to slide down the plane. The force of static friction is ....... $kg\, wt$.

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 $^{\circ}$)