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 $^{\circ}$)

$A$ block starts moving up an inclined plane of inclination $30^{\circ}$ with an initial velocity of $v_{0}$. It comes back to its initial position with velocity $\frac{v_{0}}{2}$. The value of the coefficient of kinetic friction between the block and the inclined plane is close to $\frac{I}{1000}$. The nearest integer to $I$ is......

In the given arrangement of a doubly inclined plane,two blocks of masses $M$ and $m$ are placed. The blocks are connected by a light string passing over an ideal pulley as shown. The coefficient of friction between the surface of the plane and the blocks is $0.25$. The value of $m$,for which $M=10 \text{ kg}$ will move down with an acceleration of $2 \text{ m/s}^2$,is: (take $g=10 \text{ m/s}^2$ and $\tan 37^{\circ}=3/4$) (in $\text{ kg}$)

$A$ 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

$A$ block of mass $4 \ kg$ is at rest on a rough inclined plane making an angle of $\theta$ with the horizontal. The coefficient of static friction between the block and the plane is $0.5$ and the frictional force on the block is $14.14 \ N$. Find the value of $\theta$. (in $^{\circ}$)