# Answer to Question #104976 in Mechanics | Relativity for Ceaser

Question #104976
A 6.50 3 102-kg elevator starts from rest and moves upward for 3.00 s with constant acceleration until it reaches its cruis- ing speed, 1.75 m/s. (a) What is the average power of the elevator motor during this period? (b) How does this amount of power compare with its power during an upward trip with constant speed
1
2020-03-10T11:17:32-0400

Data:

"m = 6.5 \\times 10^{2} kg"

"t=3s"

"\\nu = 1.75m\/s"

The acceleration of elevator at first time is:

"a=\\frac{\\nu - \\nu0}{t}"

Motor force is:

"F = m\\times(a+g) = m\\times(\\frac{\\nu - \\nu0}{t} + g)"

when the elevator moves upward with constant speed, "a = 0"

So force of motor is equal:

"Fconst = m \\times g"

Pressure on its power is:

"P = Fconst \\times\\nu"

"F = 6.5 \\times10^{2} \\times (\\frac{1.75 - 0}{3} + 9.8)\n= 650 \\times (0.583 + 9.8) = 6749N"

"Fconst = 6.5 \\times10^2\\times9.8=6370N"

"P = 6370 \\times1.75 = 11147.5W"

1) 6749N

2) 11147.5W

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