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: $P=5573.43\ W$

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

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

Gain in Potential Energy is

$\Rightarrow E=mgh\\\Rightarrow E=650\times 9.8\times 2.62\\\Rightarrow E=16,720.3\ N$

Average power during this period is

$\Rightarrow P=\dfrac{E}{t}\\\\\Rightarrow P=\dfrac{16,720.3}{3}\\\\\Rightarrow P=5573.43\ W$

Answer 2

Answer:

The power is 331.7 W.

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