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

`P_(avg)=11488.4\ Watts`

Step-by-step 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=(v-u)/(t)`

`a=(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)* v`

`P_(avg)=6527.5* 1.76`

`P_(avg)=11488.4\ Watts`

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