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Pulling a box by a 10N force

  1. Mar 8, 2007 #1
    1. The problem statement, all variables and given/known data
    A 2.0 kg box is pulled along a horizontal plank by a 10N force parallel to the direction of motion. The rate of acceleration is 0.50 m/s^2. Find the coefficient of kinetic friction.

    I think we use the formula: coefficient of friction = F (so 10) / N.
    But I am not sure how to use the acceleration and the N variable.


    2. Relevant equations
    F - μk mg = ma
    10 - 2(9.8)μk = 2 x 0.5

    Would this be the correct way to go?

    Thanks!
     
  2. jcsd
  3. Mar 8, 2007 #2

    PhanthomJay

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    Yes, but it contradicts what you said in the problem statement.
     
  4. Mar 9, 2007 #3
    Yes, so I use the second equation? Is the data correctly plugged in?
     
  5. Mar 9, 2007 #4

    cristo

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    Yes it is.
     
  6. Mar 9, 2007 #5
    I am getting -0.005 as the answer... does it sound right? Where should the acceleration be used?

    Thanks for all the help! :)
     
  7. Mar 9, 2007 #6

    PhanthomJay

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    you have the correct equation, with the given acceleration entered correctly, but your math is off.
    [tex]10 -19.6\mu_k = 2(.5) = 1[/tex]
    [tex] -19.6\mu_k = -9[/tex]
    [tex]\mu_k = ?[/tex]
     
  8. Mar 9, 2007 #7
    oops, it's 0.46, right?
     
  9. Mar 9, 2007 #8

    PhanthomJay

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    yes, correct.
     
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