A 2.6x10^3 kg elevator carries a maximum load of 888.1 kg. A constant frictional force of 4.9x10^3 N s the elevator's motion upward. The acceleration of gravity is 9.81 m.s^2.
What minimum power must the motor deliver to lift the fully loaded elevator at a constant speed 2.49 m/s?
Answer in units of kW
1 answer
[(Fully loaded mass)*g + (Friction force]*V = Power required
5 answers