Posted by **heeeeeeeeeelpppp!!!** on Sunday, December 11, 2011 at 12:06pm.

A 2.7 × 10

3

kg elevator carries a maximum

load of 888.9 kg. A constant frictional force

of 2.1 × 10

3

N retards 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

- physics -
**Henry**, Monday, December 12, 2011 at 4:26pm
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.

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