$A$ 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:

$A$ cube of side length $a$ is resting on an inclined plane. What must be the value of the coefficient of friction $\mu$ between the cube and the plane so that the cube topples before sliding?

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

Consider a uniform cubical box of side $a$ on a rough floor that is to be moved by applying a minimum possible force $F$ at a point $b$ above its centre of mass (see figure). If the coefficient of friction is $\mu = 0.4$,the maximum possible value of $100 \times \frac{b}{a}$ for a box not to topple before moving is

The time taken by an object to slide down a $45^{\circ}$ rough inclined plane is $n$ times the time it takes to slide down a perfectly smooth $45^{\circ}$ inclined plane. The coefficient of kinetic friction between the object and the inclined 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 $g=10 \text{ m/s}^2$]