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A 1.0 e 3 kg elevator carries a maximum load of 800.0 kg. A constant frictional force of 4.0 e 3 N retards the elevator's motio...Asked by K
A 1.0 x 10^3 kg elevator carries a maximum load of 800.0 kg. A constant frictional force of 4.0 x 10^3 N retards the elevator's motion upward. What minimum power, in kilowatts, must the motor deliver to lift the fully loaded elevator at a constant speed of 3.0 m/s?
Please show all work :) Thanks!
Please show all work :) Thanks!
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Answered by
Elena
F=(m+M) g +F(fr)= (1000+800)9,8 + 4000 = 5440 N
P=Fv = 5440•3=16320 W
P=Fv = 5440•3=16320 W
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