Question regarding work.

  • Thread starter ssb
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  • #1
ssb
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My TA gave us a conceptual example today that I dont understand. She was explaining how to derive work if you are given acceleration.

She wrote on the board that an object accelerates with 3t^2. the object weighs 3kg, starts at t=0. She asked us how much work is done on the object durring the first 2 seconds and then she said the answer was zero.

Is this right? what am I not getting:confused: :confused:
 

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  • #2
arildno
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No, it is not.
The net work done equals the change in the object's kinetic energy, which certainly is non-zero.
 
  • #3
ssb
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No, it is not.
The net work done equals the change in the object's kinetic energy, which certainly is non-zero.
Thanks I knew she was crazy. That makes me feel better.

Do you know how to calculate this in joules by the way so I can go back to her and talk to her about this?
 
  • #4
Doc Al
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You have the acceleration as a function of time. What would you need to do to get the velocity as a function of time?
 
  • #5
ssb
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take the derivative right? so it would be 6x right? sorry im new to this stuff
 
  • #6
ssb
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i tried solving for the position function which is just 6. graphicly speaking, a graph of y = 6 from 0 to 2 would cover 12 meters.
since work is force x distance and force is mass x acceleration:

work = (3kg * 3x^2) * (12)
so at time = 2, the work would be 432 joules?

Im sorry but im thinking i did something wrong.
 
  • #8
ssb
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ahhhh cool it is starting to make sence. Thanks!
 
  • #9
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i tried solving for the position function which is just 6. graphicly speaking, a graph of y = 6 from 0 to 2 would cover 12 meters.
since work is force x distance and force is mass x acceleration:

work = (3kg * 3x^2) * (12)
so at time = 2, the work would be 432 joules?

Im sorry but im thinking i did something wrong.
That would work if your force was constant, but it acceleration is changing and so is force. The best way is to figure out the change in kinetic energy
 

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