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An elevator can carry a maximum load of `1800 kg` (celevator + passengers) is moving up with a constant speed of `2 ms^(-1)`. The friction force oppposite the motion is `4000 N`.What is minimum power delivered by the motor to the elevator?A. `22 kW`B. `44 kW`C. `66 kW`D. `88 kW` |
Answer» Correct Answer - B Here, `m = 1800 kg` Frictional force `f = 4000 N` Uniform speed, `v = 2ms^(-1)` Downward force on elevator is `F = mg + f` `= (1800 kg xx 10 ms^(-2)) + 4000 N = 22000N` The motor must supply enough power to balanced this force hence, `P = Fv = (22000 N) (2 ms^(-1)` `= 44000 W = 44 xx 10^(3)W = 44 kW`. |
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