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Help with the work-energy principle

  1. May 27, 2016 #1
    1. The problem statement, all variables and given/known data
    IEFChsO.jpg

    2. Relevant equations
    PE at A = 3mgx
    WD = Fs
    KE = 1/2mv^2

    3. The attempt at a solution
    The question I am stuck on is part ii.

    I worked out from part i that the PE at A is 3mgx, so therefore all this must go towards the KE and the sound, and doing work against friction, etc...

    I said that the change in energy must equal the work done so -3mgx = μmgd, and the mg must cancel so d = -3x/μ

    However the mark scheme says d = 3x/μ, without the negative. So have I done something wrong or is this just to do with the direction I have chosen?
     
  2. jcsd
  3. May 27, 2016 #2
    The right side of the expression is negative because the displacement vector and the force vector are in opposite directions
     
  4. May 27, 2016 #3
    The mark scheme has said: 3mgx = μmgd.

    So if I had taken the directions the other way I would have said -3mgx = -μmgd?
     
  5. May 27, 2016 #4
    The way I like to think of it is the loss (negative) in potential energy is equal to the loss of energy due to heat (friction). Both are negative values but algebraically they can be simplified to be positive.
     
  6. May 27, 2016 #5
    How is that possible? Doesn't some of the energy go towards movement and sound etc?
     
  7. May 27, 2016 #6

    NascentOxygen

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    Staff: Mentor

    Merely relocating something from X to Y takes no nett energy. The body starts from rest and ends up at rest so gains no KE, and in the absence of force fields it gains/loses no PE, and in the absence of friction it produces no heat (and no sound).

    Certainly, to get it initially to begin moving you must give it energy, but as it nears destination Y you must arrange to take back all that kinetic energy so it comes to a standstill at Y. You get back the energy you put in, so the relocation itself has consumed zero nett energy. :smile:
     
  8. May 27, 2016 #7
    Surface CD is rough so there is friction.
     
  9. May 27, 2016 #8

    NascentOxygen

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    Staff: Mentor

    That's the method provided here for taking back the energy. Once all KE has been removed by friction, the body is at standstill.
     
  10. May 27, 2016 #9
    So WD against friction = μmgd. Therefore since all PE is lost to this then 3mgx = μmgd, so d = 3x/μ. That makes a little more sense right now.

    But what I'm trying to get my head around is that fact that ΔE = WD, so I would have thought work done is actually -3mgx, so -3mgx = μmgd? Forgive me if I am being slow. I just see that it has lost energy so the change is negative.
     
  11. May 27, 2016 #10

    NascentOxygen

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    Staff: Mentor

    A positive force produces a positive acceleration, (F=m·a), so whichever direction you choose to consider positive for the friction force must likewise be the direction you use for positive displacement.
     
  12. Jun 1, 2016 #11
    I understand it now. -3mgx = -μmg * d as friction acts in the opposite direction. :)
     
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