$A$ car of weight $W$ is on an inclined road that rises by $100 \, m$ over a distance of $1 \, km$ and applies a constant frictional force $\frac{W}{20}$ on the car. While moving uphill on the road at a speed of $10 \, m/s$,the car needs power $P$. If it needs power $\frac{P}{2}$ while moving downhill at speed $v$,then the value of $v$ is ........ $m/s$.

$A$ block kept on a rough inclined plane,as shown in the figure,remains at rest up to a maximum force $2 \ N$ down the inclined plane. The maximum external force up the inclined plane that does not move the block is $10 \ N$. The coefficient of static friction between the block and the plane is: [Take $g = 10 \ m/s^2$]

Starting from rest,a body slides down a $45^o$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is:

$A$ block rests on a fixed wedge inclined at an angle $\theta$. The coefficient of friction between the block and the plane is $\mu$. The maximum value of $\theta$ for the block to remain motionless on the wedge is

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 $a$?