$A$ $2 \ kg$ brick begins to slide over a surface which is inclined at an angle of $45^{\circ}$ with respect to the horizontal axis. The coefficient of static friction between their surfaces is:

$A$ body weighing $20 \ kg$ just slides down a rough inclined plane that rises $5$ in $12$. The coefficient of friction is

$A$ cone of radius $r$ and height $h$ rests on a rough horizontal surface,the coefficient of friction between the cone and the surface being $\mu$. $A$ gradually increasing horizontal force $F$ is applied to the vertex of the cone. The largest value of $\mu$ for which the cone will slide before it topples 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?

When a body slides down an inclined plane with a coefficient of friction $\mu$,its acceleration will be:

$A$ particle of mass $m$ slides from rest down a plane inclined at $30^{\circ}$ to the horizontal. The force of resistance acting on the particle during motion is $ms^2$,where $s$ is the displacement of the particle from its initial position. The velocity (in $m/s$) of the particle when $s = 1\,m$ is $v$. The value of $\frac{3v^2}{14}$ is: