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Help velocity and mass given

  1. Dec 18, 2011 #1
    1. The problem statement, all variables and given/known data

    With an initial speed of 5.00 m/s a 1.00 kg disk is slid on a horizontal steel plate. How long will it take for the disk to stop, and how far does it travel? ( do not neglect friction )

    2. Relevant equations
    I don't know how to use the mass in any equations. help.


    3. The attempt at a solution
    something with xo= vo + 1/2at^2 ?? how is the mass used?
     
  2. jcsd
  3. Dec 18, 2011 #2

    gneill

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    The question appears to be incomplete. You need to know the coefficient of dynamic friction between the disk and steel plate in order to proceed. You'll use the coefficient of friction and the mass to determine the force that retards the motion, hence the acceleration.
     
  4. Dec 19, 2011 #3
    ok so I checked and yeah the coefficient is given as 0.04 what do I do now? thanks
     
  5. Dec 19, 2011 #4

    gneill

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    What is the relationship between the coefficient of friction and the frictional force?
     
  6. Dec 19, 2011 #5
    weight X coefficient of friction = friction force?
    how does that tie in with the velocity given?
     
  7. Dec 19, 2011 #6

    gneill

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    Yes.
    Draw the free body diagram for the disk. What forces are acting? What is the acceleration of the disk (hint: Newton's second law).
     
  8. Dec 20, 2011 #7
    Ff= force of friction
    U= coefficient of friction
    Fn= force normal
    m=mass
    g=acceleration due to gravity =9.81
    a= acceleration
    d=distance

    Ff=UxFn
    which is the same as
    m x a= U x m x -g
    divide out the m and you get
    a=U x -g
    and so
    a=.04 x -9.81
    a= -.3924
    so is it slowing down by .3924 m/s^2?
    how do I link the velocity, time, and acceleration now?
     
  9. Dec 20, 2011 #8

    gneill

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    What's the relationship between acceleration and velocity? How about if there's an initial velocity?
     
  10. Dec 20, 2011 #9
    the initial velocity is 5.00 m/s and since the acceleration is (I think) -.3924 and I'm looking for the time would this equation work?
    a = (V2 - V1)/T
    where
    (V2 - V1) = change in velocity
    so it would say
    -.3924 = (V2 - 5.00)/T
    but what is V2? or how do I find it?
     
  11. Dec 20, 2011 #10

    gneill

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    That will work. It's a rearrangement of the usual equation, v2 = v1 + a*t .
    What is the speed of the puck when it stops?
     
  12. Dec 20, 2011 #11
    so V2 = 0
    and now I just have to find T
    so according to your equation it would look like this:
    v2 = v1 + a*t
    and if I fill it in I get
    0 = 5.00 + (-.3924) * t
    -5 = -.3924 * t
    (-5/-.3924) = t
    t = 12.42 sec (is that correct?)
    thanks a lot for helping, but there's a second part to the problem
    How long will it take for the disk to stop, and how far does it travel? ( do not neglect friction )
    how do I begin the second part now that I have the time? Is there some equation?
     
  13. Dec 20, 2011 #12

    gneill

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    Probably a typo, t = 12.74 should be your value.
    I'm rather surprised that you don't seem to have had any introduction to the basic kinematic formulas. Yes there is an equation (there's always an equation :smile ). In this case you want one that relates distance traveled to initial velocity, acceleration, and time. Any ideas as to which of the standard kinematic equations might fit?
     
  14. Dec 20, 2011 #13
    would it be one like this?
    X - Xo = .5(Vo + V)t
    which could be simplified to
    X = .5(5+V)(12.74)
    how do I find the velocity though?
     
  15. Dec 20, 2011 #14

    gneill

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    That equation will work. It incorporates the initial velocity (Vo) and the final velocity (V = 0) in order to take care of the acceleration. Remember that you previously wrote: a = (V2 - V1)/T.

    So plug in your V=0 and find distance X.

    Another equation that you should become familiar with is:

    [itex]d = v_o t + \frac{1}{2} a t^2 [/itex]

    See the Physics Formulary
     
  16. Dec 20, 2011 #15
    ok so I got like
    X = 31.86 meters?
    is that correct?
     
  17. Dec 20, 2011 #16

    gneill

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    Staff: Mentor

    Yes, that is correct.
     
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