What is the average power of the elevator motor during this period?

Question

A 680 kg elevator starts from rest. It moves upward for 3.23 s with constant acceleration until it reaches its cruising speed, 1.86 m/s

Help I'm stuck!!!!

Answer 1

First, find the acceleration of the elevator. You know:

a) The elevator started at 0 m/s.
b) The elevator ended at 1.86 m/s.
c) The elevator traveled a constant acceleration for 3.23 seconds.

So the acceleration is simply (1.86 m/s)/(3.23 seconds) = 0.576 m/s/s. Thus, the force applied to the elevator was simply mass*acceleration or:

(680 kg)*( 0.576 m/s/s ) = 391.680 Newtons

AVERAGE power is FORCE*DISTANCE/TIME. Thus, all we need is distance. The distance traveled in 3.23 seconds at 0.576 m/s/s is:

0.5*( 0.576 m/s/s )*(3.23 seconds)^2 = 3.0 meters

So all you're left with is:

( 391.680 Newtons )*( 3.0 meters )/( 3.23 seconds ) = 363.79 Watts

See? 391.680 Newtons was applied for 3 meters. That means that 1175 Joules of energy was used. That energy was used over 3.23 seconds, so we get 1175/3.23 Watts, or 363.79 Watts.

Another way to look at this:

(AVERAGE Power) = Force * (AVERAGE Velocity)

The force applied was 391.680 Newtons. The elevator went up 3 meters in 3.23 seconds, so its average velocity was 3/3.23 m/s.

So again, the AVERAGE power is 363.79 Watts.

Answer 2

First, lets get the force of the elevator.

Force = mass * (delta velocity / delta time)

Starts from rest, so initial velocity is 0 and initial time is 0.

Force = 680 kg * ((1.86 m/s) / (3.23 s)) = 391.58 N (mistype in calculator the first time)

Power = Force * velocity

Power = (391.58 N) * (1.86 m/s) = 728.3 W = 0.728 kW

Answer below neglects units -> Average velocity is not 3 m divided by 1.86 m/s. That is average time -> units end up in seconds.

Also, it is not by coincidence that my answer is approximately two times the answer given below. Below is a pitfall that only half the problem's power was calculated. The power needed to reach the cruise speed given in the problem.