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Newton's Laws and the coefficient of kinetic friction

  1. Oct 29, 2007 #1
    Here is the problem:
    A 12.5 Kg wooden crate with an initial velocity of 2.5m/s slides across a rough cement floor for 1.7 m before coming to rest. Find the coefficient of kinetic friction. (Use GUESS method.) The answer is 0.19.

    The equation you use is: coefficient = kinetic friction/normal force

    I worked backwards so I got 0.19 = kinetic friction/(9.8 gravity x 12.5 kg)

    That means the kinetic friction has to be 23.275 N.

    Can someone show me how to work this one correctly. I don't know where the 23.275 came from.
     
  2. jcsd
  3. Oct 29, 2007 #2

    Doc Al

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    Staff: Mentor

    Start by finding the crate's acceleration. Then apply Newton's 2nd law.
     
  4. Oct 29, 2007 #3
    OK, so I figured out that acceleration is -.075 m/s^2. So friction is equal to m x a. This would mean the plugged in values would be: coefficient = (12.5 kg x -.75 m/s^2)/(12.5 kg x 9.8 m/s^2). I got the answer -.077 but this is wrong because the correct coefficient is .19. What am I doing wrong?

    EDIT: I just realized that I forgot to square the velocity as I was solving for a. So the acceleration should have actually been -1.838 m/s^2. Thank you so much for the help. I finally figured it out.
     
    Last edited: Oct 29, 2007
  5. Oct 29, 2007 #4
    I do have one last question, is the working equation (v^2/x)(m) / (g)(m)?
     
  6. Oct 29, 2007 #5
    is the mew exactly .19?
     
    Last edited: Oct 29, 2007
  7. Oct 29, 2007 #6
    i havent done these types of problems in awhile but this is what i have.

    to find acceleration use:
    v[tex]^{2}_{f}[/tex]=v[tex]^{2}_{i}[/tex]+2a[tex]\Delta[/tex]x//solve for a
    a=(v[tex]^{2}_{f}[/tex]-v[tex]^{2}_{i}[/tex])/2[tex]\Delta[/tex]x//substitute known values
    a=-1.838m/s[tex]^{}2[/tex]

    now use:
    [tex]\Sigma[/tex]F=ma//the sum of the forces=ma;the only force acting here is friction
    -[tex]\mu[/tex]mg=ma//m's cancel out and you get
    [tex]\mu[/tex]=-a/g
    [tex]\mu[/tex]=.187
     
    Last edited: Oct 29, 2007
  8. Oct 29, 2007 #7
    Yeah, I got that exact answer too. Thanks for the help.
     
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