Question

A 674-kg elevator starts from rest and moves upward for 2.90 s with constant acceleration until it reaches its cruising speed, 1.80 m/s.

(a) What is the average power of the elevator motor during this period?______________ hp

(b) How does this amount of power compare with its power during an upward trip with constant speed? (Give the power during an upward trip with constant speed.)____________________ hp

Answer 1

Given

Mass of the elevator , m = 674 kg

Initial velocity of elevator is , vi = 0 m/s

Time taken , t = 2.90 s

Speed of the elevator is, v = 1.80 m/s

Acceleration of elevator is , a = 1.80 m/s / 2.90 s

                                           a = 0.6206 m/s^2

a) Net force on elevator is

         T = mg + ma = m (g +a)

         T = 674 kg( 9.8 m/s^2 +0.6206 m/s^2)

          T =7023.4844 N

   Average velocity is , v' = (1.80 m/s + 0) /2 = 0.90 m/s

Average power , P = T v' = 7023.4844 N * 0.90 m/s

                            P = 6321.135 W

b) Power input from motor is

          P = F * v = mg * v

              = 674kg *9.8 m/s^2 * 1.80 m/s

         P = 1.1889 *10^4 W