When a cube of side length $a$ is in limiting equilibrium on an inclined plane and it is about to topple,what is the coefficient of friction between the cube and the plane?

Starting from rest,the time taken by a body sliding down on a rough inclined plane at $45^{\circ}$ with the horizontal is twice the time taken to travel on a smooth plane of the same inclination and same distance. Then the coefficient of kinetic friction is

If a block moving up an inclined plane at $30^{\circ}$ with a velocity of $5 \, m/s$ stops after $0.5 \, s$,then the coefficient of friction will be nearly:

$A$ 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 . . . . . . $N$.

Consider a block kept on an inclined plane (inclined at $45^{\circ}$) as shown in the figure. If the force required to just push it up the incline is $2$ times the force required to just prevent it from sliding down,the coefficient of friction between the block and inclined plane $(\mu)$ is equal to

$A$ body starts from rest on a long inclined plane of slope $45^o$. The coefficient of friction between the body and the plane varies as $\mu = 0.3x$,where $x$ is the distance travelled down the plane. The body will have maximum speed (for $g = 10 \ m/s^2$) when $x = $ ........ $m$.