A car of weight W is on an inclined road that rises by 100 m over a distance of 1km and applies a constant frictional force 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 P 2 while moving downhill at speed υ then value of υ is:

P = w sin ⁡ θ + w 20 10 P = w 10 + w 20 10 = 3 w 2 P 2 = 3 w 4 = w 10 - w 20 v 3 4 = v 20 15 m / s = v

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A car of weight W is on an inclined road that rises by 100 m over a distance of 1km 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 $υ$ then value of $υ$ is:

Options:

Answer:

Solution:

$P = (w \sin θ + \frac{w}{20}) 10$

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

$\frac{P}{2} = \frac{3 w}{4} = (\frac{w}{10} - \frac{w}{20}) v$

$\frac{3}{4} = \frac{v}{20}$

$15 m / s = v$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Power

⏱ 2mℹ️ Source: QB

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