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Work Power and Energy question

2016 · 9 Apr · Shift 1 · Q64
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Work Power and Energy question

2016 · 9 Apr · Shift 1 · Q64

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
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}20W​ 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}2p​ while moving downhill at speed v then value of υ\upsilonυ is :
  1. A
    20 ms −-− 1
  2. B
    15 ms −-− 1
  3. C
    10 ms −-− 1
  4. D
    5 ms −-− 1
View written solutionFree

Correct answer: B

  1. Find the slope of the road

The road rises by 100 m100\,\text{m}100m over a distance of 1 km=1000 m1\,\text{km} = 1000\,\text{m}1km=1000m.

So,

sin⁡θ=1001000=110\sin\theta = \frac{100}{1000} = \frac{1}{10}sinθ=1000100​=101​

Hence the component of weight along the incline is

Wsin⁡θ=W10W\sin\theta = \frac{W}{10}Wsinθ=10W​

The frictional force is given as

f=W20f = \frac{W}{20}f=20W​
  1. Power needed while moving uphill

When the car moves uphill at speed 10 m s−110\,\text{m s}^{-1}10m s−1, the resisting forces are:

  • component of weight down the incline =W10= \dfrac{W}{10}=10W​
  • friction =W20= \dfrac{W}{20}=20W​

Total resisting force:

Fup=W10+W20=3W20F_{\text{up}} = \frac{W}{10} + \frac{W}{20} = \frac{3W}{20}Fup​=10W​+20W​=203W​

Power required is

P=Fup⋅v=3W20×10=3W2P = F_{\text{up}} \cdot v = \frac{3W}{20} \times 10 = \frac{3W}{2}P=Fup​⋅v=203W​×10=23W​
  1. Power needed while moving downhill

While moving downhill, gravity helps the motion and friction opposes it.

So along the downhill direction:

  • driving component of weight =W10= \dfrac{W}{10}=10W​
  • opposing friction =W20= \dfrac{W}{20}=20W​

Net force helping downhill motion:

W10−W20=W20\frac{W}{10} - \frac{W}{20} = \frac{W}{20}10W​−20W​=20W​

To move downhill at constant speed, the engine must supply negative power (or equivalently braking effect is needed). The magnitude of power involved is

Pdown=W20 vP_{\text{down}} = \frac{W}{20}\,vPdown​=20W​v

Given this is P2\dfrac{P}{2}2P​,

W20v=P2\frac{W}{20}v = \frac{P}{2}20W​v=2P​

Using P=3W2P = \dfrac{3W}{2}P=23W​,

W20v=12⋅3W2=3W4\frac{W}{20}v = \frac{1}{2}\cdot \frac{3W}{2} = \frac{3W}{4}20W​v=21​⋅23W​=43W​

Cancel WWW:

v20=34\frac{v}{20} = \frac{3}{4}20v​=43​ v=15 m s−1v = 15\,\text{m s}^{-1}v=15m s−1
  1. Check options
v=15 m s−1\boxed{v = 15\,\text{m s}^{-1}}v=15m s−1​

So the correct option is B.

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