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Work Power and Energy question

2020 · 7 Jan · Shift 2 · Q48
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Work Power and Energy question

2020 · 7 Jan · Shift 2 · Q48

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
An elevator in a building can carry a maximum of 10 persons, with the average mass of each person being 68 kg, The mass of the elevator itself is 920 kg and it moves with a constant speed of 3 m/s. The frictional force opposing the motion is 6000 N. If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator (g = 10 m/s2) must be at least :
  1. A
    48000 W
  2. B
    62360 W
  3. C
    56300 W
  4. D
    66000 W
View written solutionFree

Correct answer: D

  1. Find the total mass being lifted

The elevator carries:

  • Mass of elevator =920 kg= 920\,\text{kg}=920kg
  • Number of persons =10= 10=10
  • Average mass of each person =68 kg= 68\,\text{kg}=68kg

So, total mass of persons: 10×68=680 kg10 \times 68 = 680\,\text{kg}10×68=680kg

Hence total mass: m=920+680=1600 kgm = 920 + 680 = 1600\,\text{kg}m=920+680=1600kg

  1. Find the total downward resisting force

Since the elevator is moving upward at constant speed, acceleration is zero. So motor force must balance all opposing forces:

  • Weight of elevator + passengers: mg=1600×10=16000 Nmg = 1600 \times 10 = 16000\,\text{N}mg=1600×10=16000N
  • Frictional force opposing motion: 6000 N6000\,\text{N}6000N

Therefore total opposing force is: F=16000+6000=22000 NF = 16000 + 6000 = 22000\,\text{N}F=16000+6000=22000N

  1. Calculate power delivered by the motor

For constant speed, P=FvP = FvP=Fv

Given speed v=3 m/sv = 3\,\text{m/s}v=3m/s, P=22000×3=66000 WP = 22000 \times 3 = 66000\,\text{W}P=22000×3=66000W

  1. Match with options

66000 W66000\,\text{W}66000W corresponds to Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So the derived answer agrees with the stored answer.

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