Solve Work Problems: Elevator with Mass 1000kg on Single Cable

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SUMMARY

The discussion centers on calculating the work done on a 1000 kg elevator rising with an acceleration of 2 m/s² for 3 seconds. The correct total work done on the elevator is established as 108 kJ, derived from the tension in the cable. The tension force is calculated as 12,000 N, leading to the work formula W = Fd, where F is tension and d is the distance of 9 m. The confusion arises from the distinction between total work and net work, with the net work calculated as 18 kJ, accounting for gravitational forces.

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johnj7
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Hello, I'm currently studying for the mcat and ran across this problem in one of my practice tests.

An elevator used in a building has a mass of 1000 kg when empty
the elevator hangs on a single cable

step 1. the elevator rises for 2m/s^2 for 3 seconds,
if there is no friction on the elevator or cable, what is the work done on the elevator?

the correct answer is 108 kJ

I understand how to get the answer
T - mg = ma
T = ma + mg = 12000

W = Fd = 12000 X 9 m = 108 kJ

but why in this case do we only use the force in the Tension, I thought we always use NET force, or in this case Fnet = T - mg = 1000 X 2 = 2000
w = fd = 2000 X 9 = 18 kJ

so why just tension and not Net force?

thank you!
 
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You are quite right, the net or total work done on the elevator is 18 kJ; the work done by the cable is 108 kJ and the work done by gravity is -90 kJ. I guess the problem was not properly worded.
 

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