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Calculate the coefficient of kinetic friction between the slide and the child

  1. Sep 28, 2004 #1
    A child slides down a slide with a 28° incline, and at the bottom her speed is precisely one half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.

    here's the two equations I used:

    sin28*m*g=m(v2/t) since v2=v1+at

    the other one is with friction:

    sin28*m*g-u*cos28*m*g=m((1/2*v2)/t)

    but when I solve for u i am not getting the correct answer...
     
  2. jcsd
  3. Sep 29, 2004 #2

    Tide

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    Calculate the work done by friction (coefficient times distance traveled) then apply energy conservation to both situations.
     
  4. Feb 26, 2005 #3
    :confused: Ok, so I used the same equations as the first post. But I don't know what Tide means by using the coefficient*distance (we have no distance).

    I'm totally lost now
     
  5. Feb 26, 2005 #4

    dextercioby

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    Use what Tide said (namely to apply the theorem of variation of KE),but with the correct (i'm sure it was a typo) expression for the friction force.

    Answer:[tex] \mu=\frac{1}{2}\tan 28[/tex]°

    Daniel.
     
  6. Feb 26, 2005 #5
    I have no clue how you figured out the u=(.5)tan28 I usually don't waste a submission on an answer that I don't understand how to get to, but I tried your formula, and it still said I was wrong. In my problem the incline is 26degrees and the speed is 1/4 instead of 1/2. So obviously, I substituted my values, but its not working.

    I thought maybe you were solving for u using the 2 equations in the first post, but there are 3 variables (Vf, a, and t), but only 2 equations. This is driving me nuts! :cry:
     
  7. Feb 26, 2005 #6

    dextercioby

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    Okay.For your numbers:In the first case (no friction),the theorem reads:

    [tex]m\frac{v^{2}}{2}=mgl\sin 26[/tex]°--------->[tex] v=\sqrt{2gl\sin 26}[/tex](*)

    For the second case:

    [tex] m\frac{v^{2}'}{2}=-\mu mgl\cos 26+mgl\sin 26[/tex](**)

    Now the ratio is 1/4,which means:

    [tex] \frac{v^{2}'}{v^{2}}=\frac{1}{16} [/tex] (***)

    Combine the 3 relations marked with stars to find the final correct answer.

    Daniel.

    EDIT:Final answer:[tex] \mu=\frac{7}{8}\tan 26[/tex]°
     
    Last edited: Feb 26, 2005
  8. Feb 26, 2005 #7
    ok, well now it makes sense how you got to that...but the damn thing is still telling me that 0.427 answer is wrong. Wow, what a pain in the...
     
  9. Feb 27, 2005 #8

    dextercioby

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    That's simply weird.I'm sure that's the ways to do it and the signs for the gravity rok and the friction force work are the ones i've chosen.

    Daniel.
     
  10. Feb 27, 2005 #9
    well, the assignment is past due...the answer key says that the correct answer was 0.457. And of course, there is no explanation about how they got that answer. :mad:
     
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