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Help! Forces and Friction and Movement! O my!

  1. Oct 19, 2009 #1
    A car is traveling at 40.7 mi/h on a horizontal highway.
    The acceleration of gravity is 9.8 m/s/s.
    If the coefficient of friction between road and tires on a rainy day is 0.12, what is the minimum distance in which the car will stop?
    (1 mi = 1.609 km)

    So given:
    vi(initial velocity): 40.7 mi/h -> 65486.3m/3600s
    vf(final velocity): 0 m/s
    g(gravity): -9.8
    Ff(friction): -0.12
    m(mass): ?
    Fnet(net force): ?
    x(distance): ?

    Equations:
    F = ma
    Fnet = ma - Ff
    Ff = uFn
    Fn = mg
    vf(^2) = vi(^2) + 2ax

    Answer in back of text book: 140.689 m

    I am unsure how to set this problem up. All my attempts have had no success. Please, help would be appreciated.
     
  2. jcsd
  3. Oct 19, 2009 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi kza62! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    i] what is the normal force?

    ii] what is the friction force?

    iii] what is the acceleration? :smile:
     
  4. Oct 19, 2009 #3
    ok, i think i figured it out but i'm not sure how let me try to break it down again
    so if Fgravity is -9.8 then the Fnormal is 9.8?
    so is Ffriction = u * Fnormal?
    which is 1.176 is a?
    and then i used
    vf - vi / 2a
    which gave me 140.689

    did i get it or was this a lucky guess?
     
  5. Oct 19, 2009 #4
    *** vf^2 - vi^2 / 2a = x
    and a = -1.176
     
  6. Oct 20, 2009 #5

    tiny-tim

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    Science Advisor
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    HI kza62kza62 ! :smile:

    (just got up :zzz: …)
    Yes, that's fine!

    (btw, the forces would be mass times g, or times µFn, so what you found were the accelerations, but since the two masses cancel, it makes no difference :wink:)
     
  7. Oct 25, 2009 #6
    WOOHOO! thanx for ur help! i love physics when I get it :D
     
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