Verified Solution • 2025–26 Session moderate

QJEEM16Ph4608: JEE Main 2016 Physics - Class 11 JEEM16Ph4608 Solution: A car of weight W is on an inclined road that rises by 100 m…

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 W20{W \over 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 p2{p \over 2} while moving downhill at speed v then value of υ\upsilon is :

Multiple Choice Options

  • A 20 ms −- 1
  • B 15 ms −- 1
  • C 10 ms −- 1
  • D 5 ms −- 1

Step-by-Step Verified Solution

Here, tan θ\theta = 1001000=110{{100} \over {1000}} = {1 \over {10}}

∴\therefore    sin θ\theta = 110{1 \over {10}} (as    θ\theta   is very small),

when car is moving uphill :

JEE Main 2016 (Online) 9th April Morning Slot Physics - Work Power & Energy Question 105 English Explanation 1

P = f ×\times u

=  (wsin θ\theta + f) ×\times u

=   (w10+w20)×10\left( {{w \over {10}} + {w \over {20}}} \right) \times 10

P = 3w20×10{{3w} \over {20}} \times 10 = 3w2{{3w} \over 2}

When car is moving down hill :

JEE Main 2016 (Online) 9th April Morning Slot Physics - Work Power & Energy Question 105 English Explanation 2

∴\therefore     P2{P \over 2} = (wsin θ\theta −- f) ×\times v

⇒\Rightarrow     3w4{{3w} \over 4} = (w10−w20)\left( {{w \over {10}} - {w \over {20}}} \right) ×\times v

⇒\Rightarrow     w20×{{w \over {20}} \times } v = 3w4{{3w} \over 4}

⇒\Rightarrow    v = 15 m/s

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