$A$ block of mass $10 \, kg$ is moving down an inclined plane with a constant velocity of $10 \, m/s$. The angle of inclination is $37^{\circ}$. The coefficient of kinetic friction between the inclined plane and the block is:

$A$ body takes just twice the time to slide down a plane inclined at $30^\circ$ to the horizontal as it would if the plane were frictionless. The coefficient of friction between the body 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 length $L$ which hangs partially from a table,is kept in equilibrium by friction. The maximum length that can hang without slipping is $l$. Then,the coefficient of friction between the table and the chain is:

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

$A$ body is sliding down a rough inclined plane. The coefficient of friction between the body and the plane is $0.5$. The ratio of the net force required for the body to slide down and the normal reaction on the body is $1:2$. Then the angle of the inclined plane is (in $^{\circ}$)