First slide

Power

Question

A car of weight w is on inclined road, that rises by 100m 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 , the car needs power p. If it needs power $\frac{p}{2}$ while moving downhill at speed v, then value of v is

Moderate

A

20 m/s
B

5 m/s
C

15 m/s
D

10 m/s

Solution

   $i n c l i n a t i o n o f t h e r o a d i s \tan θ = \frac{100}{1000}$
$\begin{array}{rcl} ∴ \tan θ & ≃ & \sin θ = \frac{1}{10} (f o r v e r y s m a l l v a l u e o f θ) \\ s i t u a t i o n (1) w h e n c a r i s m o v i n g u p h i l l \\ P o w e r P & = & F o r c e . v e l o c i t y \\ P & = & (W \sin θ + f) . v e l o c i t y \\ s u b s t i t u t e g i v e n v a l u e s h e r e \\ P & = & (\frac{w}{10} \cdot + \frac{w}{20}) 10 \\ P & = & \frac{3 w}{2} - - - (1) \end{array}$




$\begin{array}{rcl} s i t u a t i o n (2) w h e n c a r i s m o v i n g d o w n h i l l \\ P o w e r P & = & F o r c e . v e l o c i t y \\ g i v e n p o w e r & = & \frac{p}{2} d o w n h i l l \\ \Rightarrow & \frac{p}{2} = F' . v \\ \Rightarrow & \frac{p}{2} = (w \sin θ - f) v \\ s u b s t i t u t e e q n (1) h e r e \end{array}$
$\frac{3 w}{4} = (\frac{w}{10} - \frac{w}{20}) v$

$\frac{3 w}{4} = \frac{w}{20} \cdot v \Rightarrow v = 15 m / \sec$

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