An electric lift with a maximum load of $2000 ext{ kg}$ (lift + passengers) is moving up with a constant spee

Question

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

Accepted Answer

The total upward force required to move the lift at constant velocity is $F = mg + F_{friction} = (2000 	imes 10) + 3000 = 20000 + 3000 = 23000	ext{ N}$. The power delivered is $P = F 	imes v = 23000 	imes 1.5 = 34500	ext{ W}$.

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