$A$ block slides down an inclined plane with an acceleration $g/2$ as shown in the figure. Then the coefficient of kinetic friction is

$A$ 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 of mass $m$ is at rest with respect to a lift when placed on an inclined plane of inclination $\theta$ inside the lift. If the lift moves upward at a constant velocity $v$,then the work done by the friction force on the block in time $t$ is:

$A$ block of mass $5 \text{ kg}$ starts up a $45^{\circ}$ inclined plane with an initial kinetic energy of $100 \text{ J}$. If the coefficient of friction between the block and the plane is $0.5$,then the distance covered by the block before it stops is (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)

$A$ rough inclined plane is placed on a car moving with a constant velocity $u$ on horizontal ground. $A$ block of mass $M$ rests on the inclined plane. Is any work done by the force of friction between the block and the inclined plane? Is there any dissipation of energy?

An engine of mass $1$ metric ton is ascending an inclined plane,at an angle $\theta = \tan^{-1}(1/2)$ with the horizontal,with a speed of $36 \; km/h$. If the coefficient of friction of the surface is $1/\sqrt{3}$,then the power developed by the engine is: