Asked by Casey
A 645-kg elevator starts from rest and reaches a cruising speed of 1.47 m/s after 3.13 seconds. It moved 2.75 m during that time.
What average power (W) is delivered by the motor during the initial acceleration of the elevator during the first 3.13 seconds?
The correct solution is 5780 W but I am no where near this number. The equation I am using is P = (mv^2)/2t + mgv/2
(645 kg x 1.47 m/s)/2(3.13) = 151.46
(645 kg x 9.8 x 1.47)/2 = 4645.93
4645.93 + 151.46 = 4797 W
Should I be incorporating the distance traveled that is given (2.75 m )?
What average power (W) is delivered by the motor during the initial acceleration of the elevator during the first 3.13 seconds?
The correct solution is 5780 W but I am no where near this number. The equation I am using is P = (mv^2)/2t + mgv/2
(645 kg x 1.47 m/s)/2(3.13) = 151.46
(645 kg x 9.8 x 1.47)/2 = 4645.93
4645.93 + 151.46 = 4797 W
Should I be incorporating the distance traveled that is given (2.75 m )?
Answers
Answered by
bobpursley
energy used= Force*distance+final KE
= m(g+a)*2.75+1/2 m*1.47^2
=645(9.8+1.47/3.13)*2.75 +1/2*645*1.47^2
That should give you the energy used, then divide it by 3.13 to get watts.
= m(g+a)*2.75+1/2 m*1.47^2
=645(9.8+1.47/3.13)*2.75 +1/2*645*1.47^2
That should give you the energy used, then divide it by 3.13 to get watts.
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