To reduce the power requirement of elevator motors, elevators arecounterbalanced
ID: 1673378 • Letter: T
Question
To reduce the power requirement of elevator motors, elevators arecounterbalanced with weights connected to the elevator by a cablethat runs over a pulley at the top of the elevator shaft. Neglectany effects of friction in the pulley. Assume a 1000 kg elevator that carries a maximum load of 800 kgis counterbalanced with a mass of 1400 kg. (a) What is the power provided by the motorwhen the elevator ascends fully loaded at a speed of 2.6 m/s.kW
(b) How much power is provided by the motor when the elevatorascends at 2.6 m/s without a load?
kW (a) What is the power provided by the motorwhen the elevator ascends fully loaded at a speed of 2.6 m/s.
kW
(b) How much power is provided by the motor when the elevatorascends at 2.6 m/s without a load?
kW
Explanation / Answer
A)
(1000+800)-1400 = 400 kg
mgh = 400* 9.8 *2.6 = 10192 J = 10.192 kJ
b)1000-1400 = -400
-400*9.8*2.6 = -10192 J = -10.192kJ (generated)
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