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 s 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?

Question

Please show all work :) Thanks!

Answer 1

F=(m+M) g +F(fr)= (1000+800)9,8 + 4000 = 5440 N
P=Fv = 5440•3=16320 W