OnlineCBT
Physics · PYQ NEET UG 2022
हिन्दीAn electric lift with a maximum load of \( 2000\text{ kg} \) (lift + passengers) is moving up with a constant speed of \( 1.5\text{ ms}^{-1} \). The frictional force opposing the motion is \( 3000\text{ N} \). The minimum power delivered by the motor to the lift in watts is: (\( g = 10\text{ ms}^{-2} \))
Correct answer · Option 1
\( 34500 \)
Explanation
Total upward force \( F = mg + f = (2000 \times 10) + 3000 = 23000\text{ N} \). Power \( P = F \cdot v = 23000 \times 1.5 = 34500\text{ W} \).
