What Is the Average Power Output of an Elevator Motor During Acceleration?

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The discussion focuses on calculating the average power output of a 750 kg elevator motor during its acceleration phase. The average power is derived from the work done, which involves the tension in the cable and the distance traveled during the acceleration period. The tension is calculated as 7568.75 N, and the work done is approximately 19867.97 J. For part b, the tension at cruising speed is questioned, emphasizing that it differs from the tension during acceleration, as the elevator moves at constant velocity. The key takeaway is that the average power must be calculated using the correct formulas for work and time, and the tension changes when the elevator reaches cruising speed.
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Homework Statement



A 750 kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed, 1.75 m/s.
(a) What is the average power of the elevator motor during this period?(b) How does this power compare with the motor power when the elevator moves at its cruising speed?
Pcruising =

Homework Equations


W=Fd

The Attempt at a Solution



(a)
\sumF = T-mg=ma
Vavg=1.75/2 == > a=1.75/2/3
=> T- 750(9.8) = (750)(1.75/2/3)
T=7568.75N

W=T*d
d=Vavg*t=1.75/2*3
W=7568.75*1.75/2*3=19867.97Am I doing this correctly?

(b) how do I do part b?I don't think I did part a correctly because I got a much different value... and the internet assignment said I am very close
 
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mkwok said:

Homework Statement



A 750 kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed, 1.75 m/s.
(a) What is the average power of the elevator motor during this period?


(b) How does this power compare with the motor power when the elevator moves at its cruising speed?
Pcruising =


Homework Equations


W=Fd



The Attempt at a Solution



(a)
\sumF = T-mg=ma
Vavg=1.75/2 == > a=1.75/2/3
=> T- 750(9.8) = (750)(1.75/2/3)
T=7568.75N

W=T*d
d=Vavg*t=1.75/2*3
W=7568.75*1.75/2*3=19867.97


Am I doing this correctly?

(b) how do I do part b?


I don't think I did part a correctly because I got a much different value... and the internet assignment said I am very close
Part a you calculated the work correctly. But the question asks for the average power, not the work done. You must note the formula for power as a function of work and time.
For part b, it is simplest to use the formula P=Fv, waht is F in this case??
 
I guess F is just the tension? therefore
F = 7568.75N
will v be the final velocity or the average velocity?
 
mkwok said:
I guess F is just the tension? therefore
F = 7568.75N
will v be the final velocity or the average velocity?
F is the tension, but since the elevator is cruising at constant speed, and not accelerating, the F will not be 7568N. What is F at constant velocity?
 
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