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

Let's use the work energy theorem to answer this problem. The first case

The force along the incline are component of gravity and frictional force.

The external force should balance both of this forces,

${\mathrm{F}}_{\mathrm{ext}}=\mathrm{mgsin\theta }+\mathrm{f}=\frac{\mathrm{W}}{10}+\frac{\mathrm{W}}{20}=\frac{3\mathrm{W}}{20}$

${\mathrm{P}}_{\mathrm{uphill}}=\frac{3\mathrm{W}}{20}\times 10=\frac{3\mathrm{W}}{2}$....(1)

Gravity will support the motion in the downhill case, so,

${\mathrm{F}}_{\mathrm{ext}}=\frac{\mathrm{W}}{10}-\frac{\mathrm{W}}{20}=\frac{\mathrm{W}}{20}$

${\mathrm{P}}_{\mathrm{downhill}}=\frac{\mathrm{W}}{20}\mathrm{u}$....(2)

from (1) and (2)

$\frac{\mathrm{W}.\mathrm{u}}{20}=\frac{1}{2}\times \left(\frac{3\mathrm{W}}{2}\right)\Rightarrow \mathrm{u}=15\mathrm{m}/\mathrm{s}$

Hence the correct answer is 15m/s.