$A$ piece of ice slides down a rough inclined plane at $\theta=45^{\circ}$ inclination in twice the time that it takes to slide down an identical but frictionless inclined plane. What is the coefficient of kinetic friction $\mu$ between the ice and the incline?

$A$ block of mass $m$ is lying on an inclined plane. The coefficient of friction between the plane and the block is $\mu$. The force $(F_1)$ required to move the block up the inclined plane will be

$A$ 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 ........ $m/s$.

There are two inclined surfaces of equal length $(L)$ and the same angle of inclination $45^{\circ}$ with the horizontal. One of them is rough and the other is perfectly smooth. $A$ given body takes $2$ times as much time to slide down the rough surface than on the smooth surface. The coefficient of kinetic friction $(\mu_k)$ between the object and the rough surface is close to:

$A$ body is moving up an inclined plane of angle $\theta$ with an initial kinetic energy $E$. The coefficient of friction between the plane and the body is $\mu$. The work done against friction before the body comes to rest is

$A$ block of mass $1 \ kg$ is pushed up a surface inclined to the horizontal at an angle of $60^{\circ}$ by a force of $10 \ N$ parallel to the inclined surface as shown in the figure. When the block is pushed up by $10 \ m$ along the inclined surface,the work done against the frictional force is: $\left[g=10 \ m/s^2, \mu_s = 0.1\right]$