A car of weight w is on inclined road, that rises by 100m over a distance of 1km 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 value of v is 

  $inclination of the road is \tan \theta =\frac{100}{1000}$

$\begin{array}{rcl}\therefore \tan \theta & \simeq & \sin \theta =\frac{1}{10}\left(forverysmallvalueof\theta \right)\\ & & situation\left(1\right) when car is moving uphill\\ Power P& =& Force.velocity\\ P & =& (W\sin \theta +f).velocity \\ & & substitute given values here\\ P& =& (\frac{w}{10}\cdot +\frac{w}{20})10\\ P& =& \frac{3w}{2} ---\left(1\right) \end{array}$   

 

      
  

   $\begin{array}{rcl} & & situation\left(2\right) when car is moving down hill\\ Power P& =& Force . velocity\\ given power & =& \frac{p}{2} down hill\\ & \Rightarrow & \frac{p}{2}=F\text{'}.v\\ & \Rightarrow & \frac{p}{2}=(w\sin \theta -f)v\\ & & substitute eqn\left(1\right) here\end{array}$

$\frac{3w}{4}=(\frac{w}{10}-\frac{w}{20})v$
     
   $\frac{3w}{4}=\frac{w}{20}\cdot v\Rightarrow v=15m/\sec$