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Solving for the friction without mass

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data

    A car is traveling at 48.0 km/h on a flat highway.
    (a) If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?

    (b) What is the stopping distance when the surface is dry and the coefficient of friction is 0.600?

    2. Relevant equations

    F = ma
    Ff = u*Fn
    kinematics equations

    3. The attempt at a solution

    I attempted to get the F in F = ma to equal the friction equation because friction is the only force acting on the car.
    so i got:
    F = Ff
    ma = μ*Fn
    ma = μ*m*g
    a = μ*g
    a = .1*-9.8
    a = -.98

    Vf^2 = Vi^2 + 2aX
    0 = 48^2 + 2(-.98)(X)
    -2304 = -1.96 (X)
    1175.5 = X

    But this isn't correct... So i was hoping someone would be able to show me the error of my ways so i could fix part a and do part b
     
  2. jcsd
  3. Oct 22, 2012 #2

    SteamKing

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    Check your units. Do you think it is reasonable for a car travelling at a moderate speed to take over 1000 meters to stop?
     
  4. Oct 22, 2012 #3
    it doesnt seem reasonable...
     
  5. Oct 22, 2012 #4
    could i get a hint potentially?
     
  6. Oct 23, 2012 #5

    ehild

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    What is the unit of your x?


    ehild
     
  7. Oct 23, 2012 #6
    meters
     
  8. Oct 23, 2012 #7

    ehild

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    Why? Substitute the data together with their units.

    ehild
     
  9. Oct 23, 2012 #8
    What?
     
  10. Oct 23, 2012 #9

    ehild

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    You calculated x from the equation vf^2=vi^2-2ax. What are the units of vf, vi and a?

    ehild
     
  11. Oct 23, 2012 #10
    v = m/s
    g = m/s/s
     
  12. Oct 23, 2012 #11
    v = km/h
    my bad
     
  13. Oct 23, 2012 #12

    ehild

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    Do you get the right values when substituting v is m/s?:biggrin:

    ehild
     
  14. Oct 23, 2012 #13
    could you explain that a little more...?
     
  15. Oct 23, 2012 #14

    ehild

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    X=vi^2/(2a). If you substitute vi in km/h, you get x in (km/h)^2s^2/m. Does it have sense? You need to convert km/h to m/s to get X in meters.

    [tex]X=\frac{(V (m/s))^2}{2*0.1*9.8 (m/s^2)}=\frac{(V (m/s))^2}{2*0.1*9.8 (m/s^2)}[/tex]

    ehild
     
  16. Oct 23, 2012 #15
    i used this equation to get 1175.51 km.
    to get meters i would divide by 1000, right?
    to get a stopping distance of 1.17551 meters?
     
  17. Oct 23, 2012 #16

    SteamKing

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    The car is travelling 48 km/hr. Are these units consistent with a g = 9.81 m/s/s?
     
  18. Oct 23, 2012 #17
    I got it!
    i wasn't doing the conversion right.
    right away when the velocity is 48 km/hr i needed to change that to meter/sec. so 48 km/hr = 13.3 m/s.
    then plug 13.3 into the equation X=vi^2/(2μa)
    resulting in
    A = 90.25 m
    B = 15.041 m
    THANK YOU VERY MUCH!!!
     
  19. Oct 23, 2012 #18
    they are not consistent...
    soo i would convert 48 km/hr to 13.3 m/s
    so using x=v^2/(2ua)
    for (a) i got 90.25
    and (b) i got 15.041

    These seem accurate... thank you very much for your help!!!!
     
  20. Oct 23, 2012 #19

    ehild

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    Good work at last!

    ehild
     
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