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 (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 (P/2) while moving downhill at speed v then value of v is :

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 (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 (P/2) while moving downhill at speed v then value of v is : (A)  20 ms-1 (B)  15 ms-1 (C)  10 ms-1 (D)  5 ms-1

Tardigrade Admin · Accepted Answer

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

$20 m s^{- 1}$

$15 m s^{- 1}$

$10 m s^{- 1}$

$5 m s^{- 1}$

$P = (w s in θ + \frac{w}{20}) 10$
$P = (\frac{w}{10} + \frac{w}{20}) 10 = \frac{3 w}{2}$
$\frac{P}{2} = \frac{3 w}{4} = (\frac{w}{10} - \frac{w}{20}) w$
$\frac{3}{4} = \frac{v}{20}$
$15 m / s = v$

20ms−1

15ms−1

10ms−1

5ms−1

P=(wsinθ+20w​)10 P=(10w​+20w​)10=23w​ 2P​=43w​=(10w​−20w​)w 43​=20v​ 15m/s=v

$20 m s^{- 1}$

$15 m s^{- 1}$

$10 m s^{- 1}$

$5 m s^{- 1}$

$P = (w s in θ + \frac{w}{20}) 10$
$P = (\frac{w}{10} + \frac{w}{20}) 10 = \frac{3 w}{2}$
$\frac{P}{2} = \frac{3 w}{4} = (\frac{w}{10} - \frac{w}{20}) w$
$\frac{3}{4} = \frac{v}{20}$
$15 m / s = v$