Question

. A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed of 1.75 m/s.
(a) What is the average power of the elevator motor during this time interval?
(b) How does this power compare with the motor power when the elevator moves at its cruising speed?

Answer 1

Answer:

(a). The average power of the elevator motor during this time interval is 5903.6 W.

(b).  The power input from motor is $1.115\times10^{4}\ W$

Explanation:

Given that,

Mass of elevator = 650 kg

Time = 3.00 sec

Speed of the elevator = 1.75 m/s

We need to calculate the acceleration of elevator

Using formula of acceleration

$a = \dfrac{v_{f}-v_{i}}{t}$

$a=\dfrac{1.75}{3.00}$

$a=0.58\ m/s^2$

(a). We need to calculate the net force on elevator is

Using formula of force

$T=mg+ma$

$T=m(g+a)$

$T=650(9.8+0.58)$

$T=6747\ N$

We need to calculate average velocity

Using formula of average velocity

$v'=\dfrac{v_{f}+v_{i}}{2}$

$v'=\dfrac{1.75}{2}$

$v'=0.875\ m/s$

We need to calculate the average power

Using formula of power

$P=T\times v'$

$P=6747\times0.875$

$P=5903.6\ W$

(b). We need to calculate the power input from motor

Using formula of power

$P=F\times v$

$P=mg\times v$

$P=650\times9.8\times1.75$

$P=1.115\times10^{4}\ W$

Hence, (a). The average power of the elevator motor during this time interval is 5903.6 W.

(b).  The power input from motor is $1.115\times10^{4}\ W$