Question

A 613-kg elevator starts from rest and moves upward for 2.50 s with constant acceleration until it reaches its cruising speed, 1.63 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? (Give the power during an upward trip with constant speed.)

Answer 1

According to the given problem,

Given

Mass of the elevator , m = 613 kg

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

Time taken , t = 2.50 s

Speed of the elevator is, v = 1.63 m/s

Acceleration of elevator is , a = 1.63 m/s / 2.50 s

                                           a = 0.652 m/s²

a) Net force on elevator is

         T = mg + ma = m (g +a)

         T = 613 kg( 9.81 m/s² +0.652 m/s²)

          T = 6413.206 N

   Average velocity is , v' = (1.63 m/s + 0) /2 = 0.815 m/s

Average power , P = T v' = 6413.206 N * 0.815 m/s

                            P = 5226.8 W

b) Power input from motor is

          P = F * v = mg * v

              = 613kg *9.81 m/s² * 1.63 m/s

        P = 9.8 *10³ W.

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