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Car, Coffee, and static friction

  1. Feb 25, 2004 #1
    Here is a problem from a worksheet I have..I'm having a little trouble with it. My teacher neglected to explain most of it:

    Now, I know everything is done in m/s, so that's 100/9 m/s. I know what the problem is asking and everything..but I don't have any formulas that I can use (I don't think)...in this case, can acceleration be used the same as deceleration? Is it okay to just put it in an equation as a negative?
  2. jcsd
  3. Feb 25, 2004 #2


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    What is the question even asking? An even better question is, what is the question? What you quoted was a statement.

    Acceleration and deceleration can be used the same way.

    Here are some formulas you can try using:
    F = ma
    Fd = (1/2)mv^2
    F = uN
  4. Feb 25, 2004 #3
    Sorry. It's a very outdated book. That's what the question says...I'm 99% sure we are supposed to figure the coefficient of static friction.

    Is it impossible? That just kinda came to me...we don't know the mass of the cup.
  5. Feb 25, 2004 #4


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    Ok that makes things easier

    setup a formula for equilibrium and the answer writes itself.

    F = F
    uN = ma
    umg = ma
    ug = a
    u(9.8) = (40km/h * 1h/3600s * 1000m/km)/3.5s
    u = 0.324
    Last edited: Feb 25, 2004
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