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

The minimum force required to start pushing a body up a rough (frictional coefficient $\mu$) inclined plane is $F_{1}$,while the minimum force needed to prevent it from sliding down is $F_{2}$. If the inclined plane makes an angle $\theta$ with the horizontal such that $\tan \theta = 2\mu$,then the ratio $\frac{F_{1}}{F_{2}}$ is:

$A$ solid hemisphere of weight $P$ rests with its curved surface in contact with a rough inclined plane. $A$ weight $Q$ is placed at some point on the rim of the hemisphere to keep its plane surface horizontal. Then its minimum coefficient of friction is

The upper half of an inclined plane of inclination $\theta$ is smooth,while the lower half is rough. If a block released from the top comes to rest at the bottom,what is the coefficient of friction between the block and the rough surface?

$A$ 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 $5 \, kg$ is placed on a rough inclined plane with an angle of $30^\circ$. If the block moves down with a constant velocity,what is the coefficient of friction? (Take $g = 10 \, m/s^2$)