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:

[c] While moving downhill power $ P=( wsin\theta +\frac{w}{20} )l0 $

$ P=( \frac{w}{10}+\frac{w}{20} )l0=\frac{3w}{2} $

$ \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=15m/s. $



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