Question

An elevator is designed such that when traveling upward, it has a constant acceleration from rest to its cruising speed of 2.00 m/s in a time of 3.00 s. The elevator has a mass of 675 kg and you may ignore friction. What is the average power of the elevator motor during this period?

Answer 1

Answer:

The average power of the elevator motor during this period is 450 W .

Explanation:

Given :

Initial velocity , u = 0 m/s .

Final velocity , v = 2 m/s .

Time taken , t = 3 s .

Mass of elevator , M = 675 kg .

We need to find the average power of the elevator motor during this period.

Now , we know power is given by :

$P=\dfrac{Work\ Done}{Time }$

Now , work done is given by :

W=F.d

Here , F is the force applied and d is the distance covered .

Now , acceleration is given by :

$v-u=at\\a=\dfrac{v-u}{t}\\\\a=\dfrac{2-0}{3}\\\\a=\dfrac{2}{3}\ m/s^2$

Also , displacement is :

$d=ut+\dfrac{at^2}{2}\\\\d=0+\dfrac{2 \times 3^2}{2\times 3}\\\\d=3\ m$

Therefore , work done is :

$W=m\times a\times d\\\\W=675\times \dfrac{2}{3}\times 3\\\\W=1350\ J$

SO , power P is :

$P=\dfrac{1350\ J}{3\ s}\\\\P=450\ W$

Therefore , the average power of the elevator motor during this period is 450 W .