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Change of kinetic energy problem

  1. Feb 28, 2012 #1
    Hi Forum :)

    This is not a specific homework problem, just something I tried to solve myself.

    1. The problem statement, all variables and given/known data

    A 10 kg object is moving at 10 m/s.

    The object is losing 2v^2 J of kinetic energy per second.

    Determine the time it takes for the speed to decrease from v_1 = 10 m/s to v_2 = 5 m/s.

    EDIT: or even better, how do I obtain an expression for velocity as a function of time?

    2. Relevant equations

    [tex]KE=0.5*m*V^2[/tex]

    3. The attempt at a solution

    Alright, so the difference in kinetic energy between v_1 and v_2 is -375 J. I also noticed that the derivative of KE with respect to v is -2v^2. I somehow need to get the time variable in there.. How do I continue to solve the problem? I'm not interested in a numerical answer, more in the way of doing it.

    Thanks in advance!
     
    Last edited: Feb 28, 2012
  2. jcsd
  3. Feb 28, 2012 #2

    tiny-tim

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    welcome to pf!

    hi marjuna! welcome to pf! :wink:

    you have d(mv2/2)/dt = 2v2

    ok, now separate the variables :smile:
     
  4. Feb 28, 2012 #3
    Thanks tiny-tim!

    So you get:

    d(mv2/2)/dt = -2v2

    d(mv2/2) = -2v2 dt

    integrate both sides: d(mv2/2) was -375 J

    -375 = -2v2*t

    Uhm, how exactly do I continue from here?
     
  5. Feb 28, 2012 #4

    tiny-tim

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  6. Feb 28, 2012 #5
    oooh, sorry

    write it as:

    1/(-2v2) * d(mv2/2) = dt

    THEN integrate both sides right?

    But how do I integrate with respect to mv2/2 when the variable on the left side is just v?
     
  7. Feb 28, 2012 #6

    tiny-tim

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    two ways:

    i] treat v2 as the variable, intead of v

    ii] write d(v2) = 2v dv :wink:
     
  8. Feb 28, 2012 #7
    Okay, then:

    case i]

    Write 1/(-2v2) * d(mv2/2) = dt as

    1/(-4v2/m) d(v2) = dt

    Then integrate the left side from v_1 to v_2 (substitute values) and the right side becomes t, obviously and then you've got an answer, right?


    and if you go for ii] then the equation becomes:

    m/-2v dv = dt

    and then also integrate both sides


    but one of them must be wrong since i get different expressions after integrating.. hmm
     
  9. Feb 28, 2012 #8

    tiny-tim

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    no, d(v2)/v2 = d(ln(v2)) …

    try substitution if you can't see it immediately :smile:
     
  10. Feb 28, 2012 #9
    Uhm, I don't get it :(

    How do you get rid of the m/2 inside d(mv2/2) so you're left with just d(v2)?


    But if you do it the other way:

    write d(mv2/2) as mv dv..

    then the equation becomes m/-2v dv = dt


    is that correct? oh man I really should start polishing up those calculus skills :)
     
    Last edited: Feb 28, 2012
  11. Feb 28, 2012 #10

    tiny-tim

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    well, yes, you do have to keep the m/2 also
    that's right! :smile:

    and now integrate …​
     
  12. Feb 28, 2012 #11
    :smile: Alright then!

    After integration the equation becomes:

    t = -m/2 * ( ln[v_2] - ln[v_1] )

    Substituting, m = 10 kg, v_1 = 10 m/s and v_2 = 5 m/s gives t ≈ 3.47 s

    Seems plausible, right?
     
  13. Feb 28, 2012 #12

    tiny-tim

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    looks good! :smile:
     
  14. Feb 28, 2012 #13
    Alright then.

    tiny-tim, thank you a lot for your time and help, I really appreciate it :biggrin: I hope you have a great day!
     
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