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 1.7 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 average 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

Total weight of elevator and occupants, m = 1000 kg + 400 kg = 1400 kg

Time taken to achieve cruising speed, tc = 2 s

Initial speed, u = 0

FInal speed to achieve, v = 1.7 m/s

required acceleration, a = (v - u)/tc = (1.7 - 0)/2 = 0.85 m/s2

Total distance lifted during the acceleration phase, s = (v2 - u2)/2a = (1.72 - 02)/(2*0.85) = 1.7 m

Average speed during the acceleration phase, vavg = s/tc = 1.7/2 = 0.85 m/s

A) Constant force applied by the motor, F = m(a + g) = 1400(0.85 + 9.81) = 14924 N

Average rate of delivering energy by motor, Eavg = F * vavg = 14924 * 0.85 = 12685.4 J/s

B) Constant force applied by the motor, F = mg = 1400 * 9.81 = 13734 N

Constant rate of supplying energy at cruising speed, E = F * v = 13734 * 1.7 = 23347.8 J/s

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