$A$ block is at rest on an inclined plane making an angle $\alpha$ with the horizontal. As the angle $\alpha$ of the incline is increased,the block starts slipping when the angle of inclination becomes $\theta$. The coefficient of static friction between the block and the surface of the inclined plane is:

$A$ rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is $\mu$. Let the mass of the box be $m$.
$(a)$ At what angle of inclination $\theta$ of the plane to the horizontal will the box just start to slide down the plane?
$(b)$ What is the force acting on the box down the plane,if the angle of inclination of the plane is increased to $\alpha > \theta$?
$(c)$ What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?
$(d)$ What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration $a$?

The acceleration of a body of mass $m$ sliding down an inclined plane with an angle of inclination $\theta$ and a coefficient of kinetic friction $\mu$ 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 $2 \, kg$ rests on a plane inclined at an angle of $30^o$ with the horizontal. The coefficient of friction between the block and the surface is $0.7$. The frictional force acting on the block is . . . . . . $N$.

The force required just to move a body up an inclined plane is double the force required just to prevent the body from sliding down. If the coefficient of friction is $0.25$,the angle of inclination of the plane is ...... $^o$.