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mkwok
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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)
[tex]\sum[/tex]F = 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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