OnlineCBT
Physics · WORK, ENERGY AND POWER
हिन्दीAn 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:
Correct answer · Option 2
\( 44000\text{ W} \)
Explanation
Total downward force \( F = mg + f = (1800 \times 10) + 4000 = 22000\text{ N} \). Minimum power \( P = F \cdot v = 22000 \times 2 = 44000\text{ W} \).
