Asked by Michelle
A 500 kg elevator starts from rest. It moves upward for 4.00 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?
(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?
Answers
Answered by
Mallory
work:
acceleration = velocity/time = 1.75/4 = .4375
distance = .5*acceleration*t^2 = .5*.4375*16 = 3.5
energy = mass*gravity*distance + (mass/2)*v^2
=500*9.8*3.5+(500/2)*(1.75)^2
=17150.0+765.625 = 17915.625
power = work/time = 17915.625/4 = 4478.9063
answer for part a) 4478.9063
do not know how to do part b) could not figure out for myself
acceleration = velocity/time = 1.75/4 = .4375
distance = .5*acceleration*t^2 = .5*.4375*16 = 3.5
energy = mass*gravity*distance + (mass/2)*v^2
=500*9.8*3.5+(500/2)*(1.75)^2
=17150.0+765.625 = 17915.625
power = work/time = 17915.625/4 = 4478.9063
answer for part a) 4478.9063
do not know how to do part b) could not figure out for myself
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