Question

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

Answer 1

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:

$P = Fconst \times\nu$

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

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

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

Answer:

1) 6749N

2) 11147.5W