Answer:
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 :

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 :
Also , displacement is :

Therefore , work done is :

SO , power P is :

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