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Question

Please help me with this problem!

Aa 683 kg elevator starts from rest and moves upward for 2.60 s with constant acceleration until it reaches its cruising speed, 1.82 m/s.

(a) What is the average power of the elevator motor during this period?
(b) What is the average power during an upward cruise with constant speed?

Answer 1

Power = Work / Time
Work = Force * distance
Force = mass * acceleration

Step 1: Find Force.
F = ma + mg (because you must overcome gravity
a = Δv/Δt = 1.82 m/s / 2.60 s = 0.7 m/s^2
F = 683 * .7 + 683 * 9.8 = 7171.5N

Step 2: Find Work
W = Fd
d = vt + 1/2at^2
d = 0 + .5 * .7 * 2.6^2 = 2.366m
W = Fd = 7171.5 * 2.366 = 16967.8 J

Step 3: Find Power
P = ΔW/Δt
P = 16967.8 / 2.60 = 6526 J (solution for a)

b.) To find the power during an upward cruise at constant speed:
F = mg (because velocity does not change)
W = Fd
d = v * t, so W = mg * vt
P = W/t = mgvt/t = mgv
P = 683 * 9.8 * 1.82 m/s = 12182 J