A 2.7 × 10

Question

3
kg elevator carries a maximum
load of 888.9 kg. A constant frictional force
of 2.1 × 10
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 2.10 m/s?
Answer in units of k

Answer 1

m = m1 + m2 = 2700 + 888.9 = 3589 kg. =
Total mass.

F = mg = 3589 * 9.81 = 35,208 N.

Fn = Fap - F - Fr = 0, a = 0.
Fap - 35208 - 2100 = 0,
Fap = 37,208 N. = Force applied by motor.

Power = F * V = 37,208 * 2 m/s = 74,416
Joules/s = 74,416 Watts. = 74.416 Kilowatts.