A car of weight W is on an inclined road that rises by 100 m over a… | JEE Main 2016 (9 Apr, Shift 1)

Q35JEE Main 2016 · 9 APR · SHIFT 1Difficult

Power

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 ms−1, the car needs power P. If it needs power $\frac{P}{2}$ while moving downhill at speed v then value of $v$ is:

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Answer: (b)

$\sin\theta = \frac{100}{1000} = \frac{1}{10}$ , so the gravity component along the road is $\frac{W}{10}$ .
Uphill: $P = \left(\frac{W}{10} + \frac{W}{20}\right)(10) = 1.5W$
Downhill (magnitude of net resisting force): $\frac{P}{2} = \left(\frac{W}{10} - \frac{W}{20}\right)v = \frac{W}{20}v$
$0.75W = \frac{W}{20}v \Rightarrow v = 15\text{ m/s}$
So the answer is (b) 15 m/s.

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