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Work Energy and Power question

2022 · Q139
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Work Energy and Power question

2022 · Q139

NEETPhysicsWork Energy and PowerMCQ+4 / −1

An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms−-−1. The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : (g = 10 ms−-−2)

  1. A
    23000
  2. B
    20000
  3. C
    34500
  4. D
    23500
View written solutionFree

Correct answer: C

Fup = 2000 g + 3000

= 23000 N

Minimum power Pmin = $\overrightarrow F $ . $\overrightarrow v $

Pmin = Fv = 23000 $\times$ ${3 \over 2}$

= 34500 W

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