A 650-kg elevator starts from rest and moves upward for 2.80 s with constant acceleration until it reaches its cruising speed, 1.64 m/s.

(a) What is the average power of the elevator motor during this period?
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(b) How does this amount of power compare with its power during an upright trip with constant speed?
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To find the average power of the elevator motor during the acceleration phase, we will use the formula:

Power = (Change in energy) / (Time)

For Part (a):

Step 1: Find the change in energy of the elevator motor during the acceleration phase.
To find the change in energy, we need to calculate the final kinetic energy and the initial kinetic energy.

Final kinetic energy (Kf):
We are given the cruising speed of the elevator, which is 1.64 m/s. Using the kinetic energy formula (K = 0.5 * m * v^2), we can calculate the final kinetic energy:
Kf = 0.5 * (650 kg) * (1.64 m/s)^2

Initial kinetic energy (Ki):
The elevator starts from rest, so initially its kinetic energy is zero. Ki = 0 J.

Change in energy (ΔK):
The change in energy is the difference between the final and initial kinetic energies:
ΔK = Kf - Ki

Step 2: Calculate the average power.
The average power is the change in energy divided by the time:
Power = ΔK / t
where t is the time taken for the acceleration phase (2.80 s).

Finally, calculate the average power of the elevator motor.

For Part (b):

To compare the power during an upright trip with constant speed, we need to consider that during a trip with constant speed, there is no change in kinetic energy. Therefore, the power required will be only to overcome the forces of friction and other losses. Typically, this power will be much lower than the power required during acceleration since there is no need to change the kinetic energy significantly.

So, the power during the upright trip with constant speed will be much less than the power during the acceleration phase.