question_answer

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\[10\text{ }m{{s}^{-1}}\], the car needs power P. If it needs power \[\frac{P}{2}\] while moving downhill at speed v then value of v is:

A) \[20m{{s}^{-1}}\]

B) \[5m{{s}^{-1}}\]

C) \[15m{{s}^{-1}}\]

D) \[10m{{s}^{-1}}\]

Correct Answer: C

Solution :

[c] While moving downhill power \[\operatorname{P}=\left( w\,sin\theta +\frac{w}{20} \right)l0\] \[\operatorname{P}=\left( \frac{w}{10}+\frac{w}{20} \right)l0=\frac{3w}{2}\] \[\frac{P}{2}=\frac{3w}{4}=\left( \frac{w}{10}+\frac{w}{20} \right)V\] \[\frac{3}{4}=\frac{v}{20}\Rightarrow v=15m/s\] \[\therefore \] Speed of car while moving downhill\[v=15\text{ }m/s.\]