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Newtons Laws

  1. Nov 20, 2009 #1
    1. The problem statement, all variables and given/known data

    A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.18 and the push imparts an initial speed of 4.1 m/s?

    2. Relevant equations

    MG of the object = mass * 9.8
    Friction = mg * 0.18


    3. The attempt at a solution

    //


    ANY HELP WOULD BE MUCH APPRECIATED
     
  2. jcsd
  3. Nov 20, 2009 #2

    tiny-tim

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    Hi tigerwoods99! :smile:

    Use the work-energy theorem …

    work done = change in energy :wink:
     
  4. Nov 20, 2009 #3
    sorry, im not familiar with this theorem
     
  5. Nov 20, 2009 #4

    tiny-tim

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    oops! :redface:

    In that case, find the acceleration from µ = 0.18, and then use one of the standard constant acceleration equations, with vi = 4.1 and vf = 0. :smile:
     
  6. Nov 20, 2009 #5
    That would make sense. I know the acceleration has to be negative because it comes to a stop, the acceleration is -> (direction) and the friction is <- (direction)

    Are these the formulas I could use to find the acceleration using mu? I have a feeling i have to know the weight of the object though to find the mg and fn
    a = Fnet/m
    Ffriction = Fnormal * mu



    Vi: 4.1 m/s
    Vf: 0 m/s
    D:
    A:
    T:
     
  7. Nov 20, 2009 #6

    ideasrule

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    If a=Fnet/m, Fnet=Fnormal*mu, and Fnormal=mg, then a=?
     
  8. Nov 20, 2009 #7
    yes, i understand the formulas but how do i get the values for the different forces, if the mass of the object is unknown?
     
  9. Nov 20, 2009 #8

    ideasrule

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    Just assume the mass is m and try it. You'll find that m cancels out.
     
  10. Nov 20, 2009 #9
    A = Fnet/m
    A = (Fnormal *mu)/m

    A = (Fnormal *u) ?
    So how would I find the FN
     
  11. Nov 20, 2009 #10

    ideasrule

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    Fn exactly balances gravity, or else the object would accelerate in the y direction. So Fn=mg.

    BTW:

    A = (Fnormal *mu)/m
    A = (Fnormal *u) ?

    Think about that. What's mu?
     
  12. Nov 20, 2009 #11
    mu = mg/9.8 * u
     
  13. Nov 20, 2009 #12

    ideasrule

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    "mu" is a single constant, representing the coefficient of friction. It is not m*u, so mu/m isn't equal to u (which is meaningless).
     
  14. Nov 20, 2009 #13
    thats what i thought, but wasn't sure becuase i am used to seeing it as just u
     
  15. Nov 20, 2009 #14
    a = fnet/m
    a = (fnormal * mu)/ m
    a = (mg * mu)/m
    a = gravity * mu

    but becuase the object is moving vertically there is no acceleration
     
  16. Nov 21, 2009 #15

    ideasrule

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    Yes, that's right
     
  17. Nov 21, 2009 #16

    tiny-tim

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    Hi tigerwoods99! :smile:

    (just got up :zzz: …)
    (oh, if only everybody had a Mac instead of a PC, with a sensible keyboard! :rolleyes:)

    have a mu … µ :wink:
    (try using the X2 tag just above the Reply box :wink:)

    Are you confusing the vertical and horizontal accelerations?

    Vertically, a = 0, and Fnet = N - mg, so N = mg.

    Horizontally, Fnet = µmg, so ma = µmg. :smile:
     
  18. Nov 21, 2009 #17
    thanks i got it!
     
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