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Work Power and Energy question

2020 · 7 Jan · Shift 1 · Q62
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Work Power and Energy question

2020 · 7 Jan · Shift 1 · Q62

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to : (1 HP = 746 W, g = 10 ms-2)
  1. A
    1.5 ms-1
  2. B
    1.7 ms-1
  3. C
    2.0 ms-1
  4. D
    1.9 ms-1
View written solutionFree

Correct answer: D

  1. Given data
  • Motor power =60 HP=60\ \text{HP}=60 HP
  • 1 HP=746 W1\ \text{HP}=746\ \text{W}1 HP=746 W
  • Total load mass m=2000 kgm=2000\ \text{kg}m=2000 kg
  • Frictional force f=4000 Nf=4000\ \text{N}f=4000 N
  • g=10 m s−2g=10\ \text{m s}^{-2}g=10 m s−2
  1. Convert power into watts

P=60×746=44760 WP=60\times 746=44760\ \text{W}P=60×746=44760 W

  1. Find total resisting force while lifting at constant speed

The elevator must overcome:

  • its weight: mgmgmg
  • friction: fff

So,

F=mg+f=2000×10+4000=20000+4000=24000 NF=mg+f=2000\times 10+4000=20000+4000=24000\ \text{N}F=mg+f=2000×10+4000=20000+4000=24000 N

  1. Use power relation

For constant speed,

P=FvP=FvP=Fv

Hence,

v=PF=4476024000=1.865 m s−1v=\frac{P}{F}=\frac{44760}{24000}=1.865\ \text{m s}^{-1}v=FP​=2400044760​=1.865 m s−1

  1. Closest option

v≈1.9 m s−1v\approx 1.9\ \text{m s}^{-1}v≈1.9 m s−1

So the correct option is D.

  1. Check all options
  • A: 1.5 m s−11.5\ \text{m s}^{-1}1.5 m s−1 ❌ too low
  • B: 1.7 m s−11.7\ \text{m s}^{-1}1.7 m s−1 ❌ not closest
  • C: 2.0 m s−12.0\ \text{m s}^{-1}2.0 m s−1 ❌ slightly high
  • D: 1.9 m s−11.9\ \text{m s}^{-1}1.9 m s−1 ✅ closest to 1.8651.8651.865

Therefore, the elevator speed at full load is 1.9 m s−11.9\ \text{m s}^{-1}1.9 m s−1.

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