The time taken by a block to slide down a rough inclined plane of angle $30^{\circ}$ is $n=2$ times the time taken to slide down a frictionless inclined plane of the same angle $30^{\circ}$. The coefficient of kinetic friction between the block and the plane is:

$A$ block of mass $5 \text{ kg}$ is placed on a rough inclined surface as shown in the figure. If $\vec{F}_1$ is the force required to just move the block up the inclined plane and $\vec{F}_2$ is the force required to just prevent the block from sliding down,then the value of $|\vec{F}_1|-|\vec{F}_2|$ is: [Use $g=10 \text{ m/s}^2$]

$A$ uniform chain of total length $L$ is at rest,partially on an incline of angle $\theta = 30^{\circ}$ and partially hanging vertically. The coefficient of friction between the chain and the incline is $\mu = \frac{1}{2\sqrt{3}}$. Find the ratio $\frac{L_{\max}}{L_{\min}}$,where $L_{\max}$ is the maximum length of the chain on the incline and $L_{\min}$ is the minimum length of the chain on the incline such that the chain remains at rest.

The upper half of an inclined plane with an angle of inclination $\phi$ is smooth,while the lower half is rough. $A$ body starting from rest at the top of the inclined plane comes to rest at the bottom of the inclined plane. Then the coefficient of friction for the lower half is

$A$ block of mass $5 \ kg$ is moving on an inclined plane which makes an angle of $30^{\circ}$ with the horizontal. The coefficient of friction between the block and the inclined plane surface is $\frac{\sqrt{3}}{2}$. The force to be applied on the block so that the block will move down without acceleration is . . . . . . $N$.

$A$ block of mass $15\, kg$ is resting on a rough inclined plane as shown in the figure. The block is tied by a horizontal string which has a tension of $50\, N$. The coefficient of friction between the surfaces of contact is $(g = 10\, m/s^2)$.