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

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

Step-by-step 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=(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=(v-u)/(t)\\\\a=(2-0)/(3)\\\\a=(2)/(3)\ m/s^2`

Also , displacement is :

`d=ut+(at^2)/(2)\\\\d=0+(2 * 3^2)/(2* 3)\\\\d=3\ m`

Therefore , work done is :

`W=m* a* d\\\\W=675* (2)/(3)* 3\\\\W=1350\ J`

SO , power P is :

`P=(1350\ J)/(3\ s)\\\\P=450\ W`

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