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Homework Help: How to get the coefficient of kinetic friction?

  1. Jan 2, 2018 #1
    1. The problem statement, all variables and given/known data
    0-02-05-1603638271dcdb24f0d995ff49536e57b62af5140df45f1a646612c4.jpg
    2. Relevant equations
    W = Fk d cos θ
    Fk = FN μκ
    3. The attempt at a solution
    Here's my work :
    m = 1 kg vi = 10 m/s vf = 0 m/s xi = 0 m xf = 0.1 m d = 0.1 m k=60 N/m μκ=?
    The work done be the frictional force :
    W = Fκ d cos θ = FN*μκ*d*cosθ = mg μκ d cos 180 = - 0.1*10*1 *μκ = - μκ
    - μκ = W = ΔKE + ΔUg + ΔUs
    ΔUg = 0 , because the height doesn't change
    - μκ = 0.5 m vf^2 - 0.5 m vi^2 + 0.5 k ( xf )^2 - 0.5 k ( xi )^2
    - μκ = 0 - 0.5 * 1 * 10^2 + 0.5 * 60 *( 0.1 )^2 - 0
    - μκ = - 49.7
    μκ = 49.7
    I don't know why I've got unreasonable answer . Could somebody help me to solve this problem ?
    Thanks for your help
     
  2. jcsd
  3. Jan 2, 2018 #2

    haruspex

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    It is large, but not impossible. I see no error in your working.

    Edit: 60 N/m is a fairly weak spring.
     
    Last edited: Jan 2, 2018
  4. Jan 2, 2018 #3

    jbriggs444

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    When in doubt, sanity-check by another approach.

    We have an object moving at 10 m/s coming to a stop over a distance of 0.1 meters. If the deceleration were constant (it's not), that would take 0.02 seconds = 20 milliseconds.

    10 m/s in 0.02 seconds is 500 meters/sec2 = 50 g's.

    The spring, when compressed by 0.1 meters, will produce 6 N. On average (distance-weighted) that would be 3 N. On a one kg mass, that's only 0.3 g.

    So a back of the envelope estimate is 49.7 g's from kinetic friction - a coefficient of kinetic friction of 49.7, just as you obtained with the proper analysis.

    Edit: [with more significant figures than are warranted for a "g" that is approximated as 10 m/s2].
     
  5. Jan 3, 2018 #4

    scottdave

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