1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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 .

    Thanks!
     
  5. Feb 20, 2014 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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?

    Thanks!
     
  9. Feb 20, 2014 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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?

    Thanks!
     
  11. Feb 20, 2014 #10

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    ohhh, that should have been

    4 = 16 + (-2)*a*3
    a= -2 …​

    the acceleration from the friction is actually -2 :wink:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted