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Work Energy Theorem Problem

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data
    I do not know how to find the height. I am thinking of using mgh=delta k but I am not getting the correct answer. I know the answer is 6.28m because my teacher gave it to me.
    Below is the picture and question:
    2hcm8uh.jpg

    2. Relevant equations
    mgh = delta k


    3. The attempt at a solution
    ep=ek
    mgh=785000
    98000h=785000
    h = 8.01m
     
  2. jcsd
  3. Feb 8, 2013 #2

    Doc Al

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    Staff: Mentor

    Since the truck moves at constant speed, ΔKE = 0. So that won't work.

    Instead, consider the work done by the engine and where that work goes.
     
  4. Feb 8, 2013 #3
    Take truck as the system. What force acts during its ascension ? How many are non conservative ? What is work done by non conservative force ?

    Negative work done by non conservative force= Change in truck's total mechanical energy.
     
  5. Feb 8, 2013 #4
    The non conservative force will be the friction right? so if I set up the equation it will be like this
    -85000=k1+u1
     
  6. Feb 8, 2013 #5
    I should have told you to take truck+earth as the system to account for external forces.

    Yes. Now total change in mechanical energy is change in kinetic energy plus change in potential energy. As change in kinetic energy is zero, so what is the change in potential energy? It will be better to account for minus sign to gravity.
     
  7. Feb 8, 2013 #6
    mgh, Cant work it out, need the height.
     
  8. Feb 8, 2013 #7
    Wfriction = M.Ef-M.Ei

    M.E. is mechanical energy..

    Wfriction = ΔK.E + ΔP.E.

    As ΔK.E=0 , so

    ΔPE=mgh

    Now you know, Wfriction= -85000 J.. Can you work it out from here ?
     
  9. Feb 8, 2013 #8

    Doc Al

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    The height is what you are supposed to find! Just call it h and you will solve for it.
     
  10. Feb 9, 2013 #9
    Okay I think I got the answer:
    Fapp-Ffric=mgh
    7*10^5-8.5*10^4=10000*9.8*h
    h = 6.28m

    Correct?
     
  11. Feb 9, 2013 #10

    Doc Al

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    Staff: Mentor

    Yes, correct. (Where Fapp is the work done by the applied force, the engine in this case. Better to call it Wapp, right? And the work done against friction would be Wfric.)
     
  12. Feb 9, 2013 #11
    Yeah, sorry its work not force. Anyway,for 3.2 what do they mean by the instantaneous power delivered by the truck. I know P=w/t and I got the time to be 4.14s using t=d/s but what should I use for work, must I subtract Wapp-Wfrict and put it into the equation?
     
  13. Feb 9, 2013 #12

    Doc Al

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    Staff: Mentor

    Good.
    To find the power delivered by the engine, just use the work done by the engine.
     
  14. Feb 9, 2013 #13
    Okay so,
    p=w/t
    = 7*10^5/4.14
    = 169082.13 W

    Correct?
     
  15. Feb 9, 2013 #14

    Doc Al

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    Staff: Mentor

    Good. (I suggest rounding off your final answer and using exponential notation.)
     
  16. Feb 9, 2013 #15
    Will do, and thanks for your help
     
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