A 690 kg elevator starts from rest. It moves upward for 3.09 s with constant acc

Question

A 690 kg elevator starts from rest. It moves upward for 3.09 s with constant acceleration until it reaches its cruising speed, 1.86 m/s. What is the average power of the elevator motor during this period? in W) A: 90/ B: 12061 -C: 1604|| D: 2133 E: 283-F: 3773/ -G: 50191 H: 6675 Submit Answer Tries 0/3 What is the power of the elevator motor during an upward cruise with a constant speed of 1.86 m/s? in W) A: 9457| : 12577| C 167281 Submit Answer Tries 0/3 D 22248|| E 29590 F: 39355 G 52342H: 69614

Explanation / Answer

Mass m=690kg

Initial speed u =0

Final speed v =1.86m/s

Time t=3.09s

(a).average Power P=m(a+g) (v/2)

Where a=(v-u) /t=0.6091m/s^2

P=690(0.6091+9.8)(1.86/2)

=6675 W

(b).force F=mg=690*9.8=6762N

Power p=Fv=6762*1.86=12577. 32W

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