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Friction, speed & distance question

  1. Nov 4, 2003 #1
    How far will a person slide if they are sliding at 3m/s and the coefficient of friction between the floor and their feet is 0.5
  2. jcsd
  3. Nov 5, 2003 #2


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    There's not enough information. You would have to know the person's weight (or mass) to calculate the actual friction force (friction coefficient times weight). That would determine the distance.
  4. Nov 5, 2003 #3


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    Sorry, Halls, that's wrong. The mass cancels.

    See this thread:


    - Warren
  5. Nov 6, 2003 #4

    Yes, the mass was given. 120kg

    With that i found the Ff by uk*mg.

    Then i used the equation F= m [(vf^2-vi^2)/2d] and rearranged it to d=[m(vf^2-vi^2)]/2F
    The force that i subbed into that equation was the Ff=uk*mg, is that right???

    Is there another way to do this, or did i do this wrong???
  6. Nov 6, 2003 #5

    Doc Al

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    Not wrong, just a little wasted effort. Looking at your own equations, you can see that the mass cancels (as chroot pointed out) so it really is extraneous information.

    Resist the temptation to start plugging in numbers prematurely. Figure things out symbolically as much as you can, then plug in the numbers.
  7. Nov 6, 2003 #6
    I am not sure about my approach, but here is what i think:

    Δx = ( v^2 - v(initial)^2 ) / 2a

    ΣFx = max = Ff

    max = μkFN * cos 180deg

    since cos 180deg = -1, Ff is a negative quantity.
    and the mass m cancels out.

    ax = - 0.5 * 9.8 = -5 m/s

    plug back in Δx equation:

    Δx = abs(-9/10) = 9/10 m.

    The sliding guy will travel 0.9 meters.
    Last edited: Nov 7, 2003
  8. Nov 6, 2003 #7

    What does the mass cancel out with?
  9. Nov 6, 2003 #8


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    F = [mu] N = [mu] m g

    F = m a

    [mu] m g = m a

    a = - [mu] g (negative sign because the slider is slowing down)

    v(t) = v0 + a t

    v(t) = v0 - [mu] g

    When he stops, his velocity is zero:

    0 = v0 - [mu] g

    t = v0 / ([mu] g)

    s(t) = v0 t + 1/2 a t2

    s(t) = v0 (v0 / ([mu] g) ) - 1/2 [mu] g * (v02 / ([mu] g)2 )

    s(t) = 1/2 v02 / ([mu] g)

    Does this help?

    - Warren
  10. Nov 6, 2003 #9
    mass cancels out with mass. it's a mathematical "trick".
  11. Nov 6, 2003 #10

    Doc Al

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    With itself, of course! (D'oh!)

    Using conservation of energy:

    (all the KE goes into work against friction)

    Fd=1/2m v^2

    umgd = 1/2mv^2 (the mass cancels here)

    d= v^2/(2ug)
  12. Dec 3, 2003 #11


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    The answers posted seem to be in nice accordance with the physics I was taught in high school. However, in practice they fail dismally, unless one at least brings the area of contact between the two surfaces into the equation. There is a reason formula one cars have tyres as wide as a barn door, and that the trimming of cars usually involves changing to wider tyres. I think there is some relationship which dictates that the greater the area, the greater the friction becomes. Does anyone know anything about this?
  13. Dec 3, 2003 #12


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    Tires are quite special, because the deform, heat up, and become greasy as they are used. That does not mean that the area is important in determining friction -- in general, it is not.

    - Warren
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