When a body is placed on a rough plane inclined at an angle $\theta$ to the horizontal,its acceleration 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 upper half of an inclined plane with an angle of inclination $\phi$ is smooth,while the lower half is rough. $A$ body starting from rest at the top of the inclined plane comes to rest at the bottom of the inclined plane. Then the coefficient of friction for the lower half is

Two touching blocks $1$ and $2$ are placed on an inclined plane forming an angle $60^{\circ}$ with the horizontal. The masses are $m_1$ and $m_2$ and the coefficients of friction between the inclined plane and the two blocks are $1.5 \mu$ and $1.0 \mu$, respectively. The force of reaction between the blocks during the motion is ($g=$ acceleration due to gravity).

Find the work done by friction if a $1 \, kg$ block reaches the end of an inclined plane of length $10 \, m$ and inclination $30^{\circ}$ with constant velocity.

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