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Homework Help: Coefficient of Friction and Energy

  1. Feb 20, 2014 #1
    1. The problem statement, all variables and given/known data

    A 30kg sled is sliding on a frictionless sheet of ice at a velocity of 4m/s. The sled encounters a rough patch of ice and begins to slow down. After traveling on the rough patch of ice for 3m, the sled's velocity is 2m/s. Determine the coefficient of friction between the rough ice and sled.

    2. Relevant equations

    KE = (1/2)mv2

    FF = uk * FN

    W = f * d

    v2 = vi2 + 2a(x - xi)

    3. The attempt at a solution

    I begin by finding the Kinetic energy when the sled is on the frictionless sheet of ice:

    KE = (1/2)mv2

    KE = (1/2) * (30) * 42

    KE = 240J

    Then I find the Kinetic Energy when the sled is on the rough patch:

    KE = (1/2)mv2

    KE = (1/2) * (30) * 22

    KE = 60J

    I now note the amount of energy released due to friction:

    240J - 60J = 180J

    I continue to evaluate the force acting on the sled when on the rough patch:

    W = fd
    60J = f * 3m
    f = 20N

    Then I solve for acceleration in the rough patch:

    v2 = vi2 + 2a(x - xi)

    4 = 0 + 2*a*3
    a= (2/3)

    Now I sum the forces:

    20N - (uk * (30kg * 9.8)) = 30kg * (2/3)
    uk = 0

    This is obviously not correct as it states friction is present. I don't know where I'm making a mistake. Presumably my logic is incorrect, not the math itself. I think finding the kinetic energy for the frictionless surface was useless with the approach I'm taking.

    Could anyone please help me understand what's incorrect?

    Last edited: Feb 20, 2014
  2. jcsd
  3. Feb 20, 2014 #2


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    Hi Blakeasd! :smile:
    No, W = 180J. :wink:

    (and what is the relevance of acceleration? :confused:)
  4. Feb 20, 2014 #3
    Thank you for your response.

    When I sum up the forces, I set them equal to ma. I need acceleration so I can solve for the coefficient of friction.

    Can you please explain how you found work to be 180J .

  5. Feb 20, 2014 #4


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    Hi Blakeasd! :smile:

    The work energy theorem says that the change in mechanical energy equals the work done

    That change is 180J :wink:

    (and you can get µ from W = Fd and F = µmg)
  6. Feb 20, 2014 #5
    Ok, I see why it is 180J, thanks! Here is my revamped work:

    W = fd
    180J = f * 3m
    f = 60N

    60N - (u * (30*9.8)) = (2/3) * 30
    u = -.14

    This is not a choice, though:

    a.) 0.07
    b.) 0.12
    c.) 0.20
    d.) 0.27
    e.) 0.60

    When I use W = Fd and F = umg , I get .204 (an answer choice).

    So why doesn't this work when I sum forces? If I were to assume constant velocity, then summing forces would work, but how could I tell it is constant velocity? Why is it appropriate to assume this?
  7. Feb 20, 2014 #6


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    No, that should be
    16 = 4 + 2*a*3
    a= 2 :wink:

    corrected for a, that's 60N - (µ * (30*9.8)) = 2 * 30

    which is F - µmg = ma

    that's wrong because you've used the same force twice

    there is only one force (= ma), and it's either F or µmg :smile:
  8. Feb 20, 2014 #7
    I don't understand how I'm using the same force twice. I take the force pushing the object forward and subtract the force of friction pushing in the other direction. When summing forces I thought it is necessary to set their sum equal to ma. Am I missing something fundamental?

  9. Feb 20, 2014 #8


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    there is no force pushing the sled!!

    if it wasn't for the rough patch, the sled would go on for ever at 4m/s

    the only force is the friction! :smile:
  10. Feb 20, 2014 #9
    Ah... so if I were to sum the forces:

    0 - (u*30*9.8) = 60

    u = .204

    It works!

    If you don't mind me asking another question, can you explain you reasoning for: v2 = vi2 + 2a(x - xi)

    16 = 4 + 2*a*3

    Why don't you use 2m/s for the final velocity?

  11. Feb 20, 2014 #10


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    ohhh, that should have been

    4 = 16 + (-2)*a*3
    a= -2 …​

    the acceleration from the friction is actually -2 :wink:
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