A motor must lift a 1000-kg elevator cab. The cab\'s maximum occupant capacity i

Question

A motor must lift a 1000-kg elevator cab. The cab's maximum occupant capacity is 400 kg, and its constant "cruising" speed is 1.5 m/s. The design criterion is that the cab must achieve this speed within 2.0 s at constant acceleration, beginning from rest, (a) When the cab is carrying its maximum capacity, at what rate must the motor deliver energy to get the cab up to cruising speed? (b) At what constant rate must the motor supply energy as the fully loaded cab rises after attaining cruising speed?

Explanation / Answer

accelaration of the lift in the first 2 s is

a = (v-u)/t) = (1.5/2) = 0.75 m/s^2

A) power P = w/t = chnage in kinetice enrgy/t =0.5*m*v^2/t = 0.5*1400*1.5*1.5/2 = 787.5 W

B) Power P = F*v = m*g*v = 1400*9.81*1.5 = 20601 W

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