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An electric lift with a maximum load of 2000 kg2000\text{ kg} (lift + passengers) is moving up with a constant speed of 1.5 ms−11.5\text{ ms}^{-1}. The frictional force opposing the motion is 3000 N3000\text{ N}. The minimum power delivered by the motor to the lift in watts is : (g=10 ms−2)(g = 10\text{ ms}^{-2})

1

23000

2

20000

3

34500

4

23500

Step-by-Step Solution

The total upward force required to move the lift at constant velocity is F=mg+Ffriction=(2000×10)+3000=20000+3000=23000 NF = mg + F_{friction} = (2000 \times 10) + 3000 = 20000 + 3000 = 23000\text{ N}. The power delivered is P=F×v=23000×1.5=34500 WP = F \times v = 23000 \times 1.5 = 34500\text{ W}.

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