Respuesta :
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 :
[tex]P=\dfrac{Work\ Done}{Time }[/tex]
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 :
[tex]v-u=at\\a=\dfrac{v-u}{t}\\\\a=\dfrac{2-0}{3}\\\\a=\dfrac{2}{3}\ m/s^2[/tex]
Also , displacement is :
[tex]d=ut+\dfrac{at^2}{2}\\\\d=0+\dfrac{2 \times 3^2}{2\times 3}\\\\d=3\ m[/tex]
Therefore , work done is :
[tex]W=m\times a\times d\\\\W=675\times \dfrac{2}{3}\times 3\\\\W=1350\ J[/tex]
SO , power P is :
[tex]P=\dfrac{1350\ J}{3\ s}\\\\P=450\ W[/tex]
Therefore , the average power of the elevator motor during this period is 450 W .
