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How much total work is done in raising the ball?

  1. Dec 6, 2009 #1
    A 0.075Kg ball in a kinetic sculpture is raised 1.33m above the ground by a motorized vertical conveyor belt. A constant frctional force of 0.350N acts in the direction opposite the conveyor belt. How much total work is done in raising the ball?

    This confuses me quite a bit. I understand some of it. Could someone help me please? :)
     
  2. jcsd
  3. Dec 6, 2009 #2

    rock.freak667

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    How would you normally find work done? What is the total energy of the sculpture?
     
  4. Dec 6, 2009 #3
    Use the formula W=Fd. And thats the problem I'm having is trying to find the total energy of the sculpture.
     
  5. Dec 6, 2009 #4

    rock.freak667

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    the question says

    how much energy does it have at 1.33m above the ground?
     
  6. Dec 6, 2009 #5
    .099?
     
  7. Dec 6, 2009 #6
    I'm sorry I'm not trying to sound stupid. I'm just trying to understand this question!
     
  8. Dec 6, 2009 #7

    rock.freak667

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    I don't have a calculator but it is around that. So this is the total amount of energy needed to lift the object 1.33m above the ground.

    To lift the object, the total work done is the sum of the work done in lifting the object and what else?
     
  9. Dec 6, 2009 #8
    So then .345N force is acting on it and therefore since there is a force acting the opposite way there is no work that is being done on the ball in the kinetice sculpture. But there is work acting on the conveyor belt?
     
  10. Dec 6, 2009 #9

    rock.freak667

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    right but the 0.345N does frictional work.

    So the total work done = work by conveyor + work done by friction.

    What is the work done by friction equal to? (see your 2nd post)
     
  11. Dec 6, 2009 #10

    Use the equation W=Fd or .099?

    So what your saying is to find the work by the conveyor and then add that to work done by friction (.350N) which then will equal the total work done in raising the ball?
     
  12. Dec 6, 2009 #11

    rock.freak667

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    You have the right idea, just jumbled.

    Total work done= work done by conveyor+ work done by friction.

    You know the left side is 0.099. The work done by friction is Fd. So what is 'd' and the frictional work?
     
  13. Dec 6, 2009 #12
    Ah, so 0.099 = ______ + .350N x 1.33m?

    0.099 + .465 = .564J?
     
  14. Dec 6, 2009 #13

    rock.freak667

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    yes, just change '+' to '-', I forgot to compensate with signs.
     
  15. Dec 6, 2009 #14
    so .099 - .465 = 366J? And tha would be how much total work is done in raising the ball.
     
  16. Dec 6, 2009 #15
    Right?
     
  17. Dec 6, 2009 #16

    rock.freak667

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    No the previous answer was correct. It was supposed to be this

    0.099 = ______ - .350N x 1.33m
     
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