Question

A(n) 603 kg elevator starts from rest. It moves upward for 3.88 s with a constant acceleration until it reaches its cruising speed of 1.78 m/s. The acceleration of gravity is 9.8 m/s 2 . Find the average power delivered by the elevator motor during this period. Answer in units of kW.

Answer 1

Given
Mass of the elevator , m = 603 kg
Initial velocity of elevator is , vi = 0 m/s
Time taken , t = 3.88 s
Speed of the elevator is, v = 1.78 m/s
Acceleration of elevator is , a = 1.78 m/s / 3.88s
a = 0.458762887 m / s^2
a) Net force on elevator is
T = mg + ma = m (g +a)
T =603 *( 9.8 +0.458762887 )
T = 6186.03402N
Average velocity is , v' = (1.78 m/s + 0) /2 = 0.89 m/s
Average power , P = T v' = 6186.03402 * 0.89 m/s
P = 5505.57028 W

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