SATHEE: Work Energy And Power Question 281

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

Options:

A) $20 m s^{- 1}$

B) $5 m s^{- 1}$

C) $15 m s^{- 1}$

D) $10 m s^{- 1}$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] While moving downhill power $P = (w s i n θ + \frac{w}{20}) l 0$

$P = (\frac{w}{10} + \frac{w}{20}) l 0 = \frac{3 w}{2}$

$\frac{P}{2} = \frac{3 w}{4} = (\frac{w}{10} + \frac{w}{20}) V$

$\frac{3}{4} = \frac{v}{20} \Rightarrow v = 15 m / s$

$∴$ Speed of car while moving downhill $v = 15 m / s .$

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Work Energy And Power Question 281

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