Work Energy And Power Question 281
Question: 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 $ 10m{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:
Options:
A) $ 20m{s^{-1}} $
B) $ 5m{s^{-1}} $
C) $ 15m{s^{-1}} $
D) $ 10m{s^{-1}} $
Show Answer
Answer:
Correct Answer: C
Solution:
While moving downhill power $ P=( w\sin\theta +\frac{w}{20} )l $
$ P=( \frac{w}{10}+\frac{w}{20} )l=\frac{3w}{20} $
$ \frac{P}{2}=\frac{3w}{4}=( \frac{w}{10}+\frac{w}{20} )V $
$ \frac{3}{4}=\frac{v}{20}\Rightarrow v=15m/s $
$ \therefore $ Speed of car while moving downhill $ v=15\ \text{m/s}. $
BETA
