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2008 F=ma Olympiad Problems # 10 + 11

  1. Jan 19, 2009 #1
    These two problems are based one some data from an experiment. Of course, experiment often don't carry out perfectly matching theory... And I'm having some difficulty understanding the answer.


    The following information applies to the next two problems

    An experiment consists of pulling a heavy wooden block across a level surface with a spring force meter, the constant force for each try is recorded, as is the acceleration of the block. The data are shown below.

    Force F in Newtons:_____3.05__|__3.45__|__4.05__|__4.45__|__5.05
    Acceleration a in m/s2:___0.095_|__0.205_|__0.295_|__0.405_|__0.495

    10) Which is the best value for the mass of the block?

    11) Which is the best value for the coefficient of friction between the block and the surface?

    Answer: 10)b 11)a

    Would someone please explain the answer? THank you very much~
    Last edited: Jan 20, 2009
  2. jcsd
  3. Jan 19, 2009 #2


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    What are your thoughts on this so far?

    EDIT: I'd check your data as well.
    Last edited: Jan 19, 2009
  4. Jan 19, 2009 #3
    I saw that they gave F and a, so F = ma, I tried to find m, except that the m i obtained is different every time, varying from 30 to 10kg.

    #11 i guessed it correctly, but i didn't have much reasoning. Static friction is the force that the spring has to overcome in order to accelerate the block. So max static friction force is mg*mus, which equals Ffr and Fspring - Ffr = F of the spring.
  5. Jan 20, 2009 #4


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    If it is accelerating it will be kinetic friction. If friction is present how will you have to modify the F=ma equation?

    Again, check your data. I managed to find the paper online and the data is slightly different on there.
  6. Jan 20, 2009 #5
    Thank you for noticing the typo in the data!

    Well for #10:
    If Kinetic friction exists, then F = ma = Fspring - Ffriction
    F = Fspring - m*g*mukinetic?

    Since there are two unknowns: muk and m, just take two cases to set up two equations?

    AH HAH! I took the first and the last trial and got m = 5kg!

    And i'm assuming that they are asking for the kinetic friction?
    nd yes, i got 0.0485!

    THank you very much :D
  7. Jan 20, 2009 #6


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    If it was a true experiment the results would have been graphed and the gradient is the mass and the intercept would give you the term with mu in it. Well done!
  8. Jan 20, 2009 #7
    hm... Don't quite understand how you relate the graph of F vs. a to gradient (you mean the vector gradient?).

    Also another question in 2008 Test:
    #18 The answer is c

    Is it because that... the gravitational force is exerted by the Center of Mass of the ring? Is that why the new maximum speed is the same? But then since the two rings have the same mass density, wouldn't they have different mass?

    I tried to apply [tex]F = \frac{Gm_{particle}M_{ring}}{r^2} [/tex], does M increase by a factor of 4? But I'm not sure how.

    And number 19... I have no idea where to start...

    I listed out some equations that relates to Power: P = dW/dt = Fv...
  9. Jan 23, 2009 #8


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    Sorry for the late reply. The PDF doesn't appear to be available any more. As for the force on a particle given by a ring of mass, you can't just use Newton's law for point masses. Do you know how to treat extended masses?

    If the rings have the same mass density but different sizes then they will have different total masses.
  10. Jan 24, 2009 #9
    does anyone happen to have an answer key for last year's exam that is posted?
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