Work elevator problem

  • Thread starter johnj7
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  • #1
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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!
 

Answers and Replies

  • #2
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,171
509
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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