Question

A 700-kg elevator starts from rest and moves upward for 3.70 s with constant acceleration until it reaches its cruising speed, 1.76 m/s. (a) What is the average power of the elevator motor during this period

Answer 1

Answer:

$P_{avg}=11488.4\ Watts$

Explanation:

Given that,

Mass of the elevator, m = 700 kg

Time taken, t = 3.7 s

Initial velocity, u = 0

Final velocity, v = 1.76 m/s

Let a is the acceleration of the elevator. Using the first equation of motion to find it as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{1.76}{3.7}$

$a=0.475\ m/s^2$

The net force acting on the elevator is :

$F_{net}=m(a-g)$

$F_{net}=700(0.475-9.8)$

$F_{net}=-6527.5\ N$

Let P is the average power of the elevator motor during this period. It is given by :

$P_{avg}=F_{net}\times v$

$P_{avg}=6527.5\times 1.76$

$P_{avg}=11488.4\ Watts$

The average power of the elevator is 11488.4 watts. Hence, this is the required solution.