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Friction of a stone sliding on a pond

  1. Nov 28, 2004 #1
    This isn't really homework because I'm not doing it for school. I'm on my holidays at the moment, and I'm trying to do a bit of Physics on my own. As you can imagine, I have not gotten far, and am stuck on this problem:

    A stone slides in a straight line across a frozen pond. Given that the initial speed of the stone is 5m/s and that it slides 20m before coming to rest, calculate the coefficient of friction between the stone and the surface of the frozen pond.

    The deceleration of the stone is (20*2)/5 = 8 m/s2.

    How should I proceed next?

    EDIT: I'm really confused. Is the resultant force = 8 X mass?

    Is the resultant force = frictional force - forward force?

    I know that frictional coefficient = Frictional force/contact force.

    However, I've done the algebra and it comes out very messy.
    Last edited: Nov 28, 2004
  2. jcsd
  3. Nov 28, 2004 #2
    vf^2 = vi^2 + 2ad

    a = -1.6

    a = (mu)(g)

    mu = .194
  4. Nov 28, 2004 #3


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    Use the relation

    [tex] v^2 = v^2_{0} + 2a \Delta x [/tex]

    and the fact that the only force acting on the stone with a nonzero horizontal component is friction.

    Newton's 2nd Law

    [tex] \mu mg = ma [/tex]
  5. Nov 28, 2004 #4
    I don't really understand the variables that the two of you have used above. Is there a site that you can refer me to for this?

    OK, I see my careless mistake. The deceleration is not 8 m/s2, that is the time taken for it to decelerate. So the deceleration is 5/8 = 0.625 m/s2, is it not?
    Last edited: Nov 28, 2004
  6. Nov 28, 2004 #5


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    Newton's Laws:


    By the way:

    [tex] v^2 = v^2_{0} + 2a \Delta x [/tex]

    It covers 20 meters then it stops so Final speed is 0.

    [tex] 0 = v^2_{0} + 2a \Delta x [/tex]

    [tex] a = \frac{-v^2_{0}}{2\Delta x} [/tex]
    Last edited: Nov 28, 2004
  7. Nov 28, 2004 #6
    OK, I have the answer now guys. Thanks a lot.
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