Question

A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruis- ing speed of 1.75 m/s. (a) What is the average power of the elevator motor during this period

Answer 1

Answer:
`P=5573.43\ W`

Step-by-step explanation:

Given

Mass of the elevator is
`M=650\ kg\\\`

Time period of ascension
`t=3\ s`

cruising speed
`v=1.75\ m/s`

Distance moved by elevator during this time

Suppose Elevator starts from rest

`\Rightarrow v=u+at\\\Rightarrow 1.75=0+a(3)\\\Rightarrow a=0.583\ s`

Distance moved

`\Rightarrow h=ut+0.5at^2\\\Rightarrow h=0+0.5* 0.5833* (3)^2\\\Rightarrow h=2.62\ m`

Gain in Potential Energy is

`\Rightarrow E=mgh\\\Rightarrow E=650* 9.8* 2.62\\\Rightarrow E=16,720.3\ N`

Average power during this period is

`\Rightarrow P=(E)/(t)\\\\\Rightarrow P=(16,720.3)/(3)\\\\\Rightarrow P=5573.43\ W`

Answer 2

The power is 331.7 W.

Step-by-step explanation:

mass, m = 650 kg

time, t= 3 s

initial velocity, u = 0 m/s

final velocity, v = 1.75 m/s

(a) The power is defined as the rate of doing work.

Work is given by the change in kinetic energy.

W = 0.5 m (V^2 - u^2)

W = 0.5 x 650 x 1.75 x 1.75 = 995.3 J

The power is given by

P = W/t = 995.3/3 = 331.7 W