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Physics · WORK, ENERGY AND POWER

हिन्दी

An elevator can carry a maximum load of \( 1800\text{ kg} \) (elevator + passengers) and moves upwards with a constant speed of \( 2\text{ m s}^{-1} \). A constant frictional force of \( 4000\text{ N} \) opposes its motion. Taking \( g = 10\text{ m s}^{-2} \), the minimum power delivered by the motor to the elevator is:

  1. A
    \( 36000\text{ W} \)
  2. B
    \( 44000\text{ W} \)
  3. C
    \( 40000\text{ W} \)
  4. D
    \( 48000\text{ W} \)
View Answer & Solution

Correct answer · Option B

\( 44000\text{ W} \)

Explanation

Total downward force \( F = mg + f = (1800 \times 10) + 4000 = 22000\text{ N} \). Minimum power \( P = F \cdot v = 22000 \times 2 = 44000\text{ W} \).
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