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Kinetic friction homework

  1. Jul 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A block of mass m is at rest at the origin at t=0. It is pushed with constant force F0 from x=0 to x=L across a horizontal surface whose coefficient of kinetic friction is μk=μ0(1−x/L). That is, the coefficient of friction decreases from μ0 at x=0 to zero at x=L.

    2. Relevant equations

    ax = (dvx/dx)vx
    a = (F0 - mg(1-x/L))/m

    3. The attempt at a solution

    and it wrong
  2. jcsd
  3. Jul 30, 2013 #2
    What is the problem?
  4. Jul 30, 2013 #3
    find an expression for the block's velocity when it reaches position x=L.
    Express your answer in terms of the variables L, F0, m, μ0, and appropriate constants.
  5. Jul 30, 2013 #4
    Please show your work. It looks like you made some math errors, and you also left out a factor of μ0 in your equation for the acceleration.
  6. Jul 30, 2013 #5
    a = (F0 - mg(1-x/L))/m
    vdx = (Fo/m)dx - uogdx + (uog/L)xdx
    (1/2)v^2 = (Fo/m)x - uogx + (uog/2L)x^2
    and solve for v and plug x = L
    I got
  7. Jul 30, 2013 #6


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    Right so far
    That's certainly wrong. For others to see where you went wrong you will need to post every step of your working.
  8. Jul 30, 2013 #7
    (1/2)v^2 = (Fo/m)x - uogx + (uog/2L)x^2
    and solve for v and plug x = L
    v^2 = 2 (Fo/m)x - 2 uogx+ 2(uog/2L)x^2
    v^2 = 2x ((Fo/m)- uog) + (uog/L)x^2
    v^2 = 2L((Fo/m)- uog) + (uogL)

    v= √(2L((Fo/m)- uog) + (uogL))

    is it right?
  9. Jul 30, 2013 #8


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    Yes, but you can simplify it a little.
  10. Jul 30, 2013 #9
    Where can I simplify it? I thought I already simplified it to the most simplest form.
    Can you give some hint?
  11. Jul 30, 2013 #10
    Check out the terms under the radical (which extends over the entire rhs), and see if you can combine anything.
  12. Jul 30, 2013 #11
    okay.. got it
    Last edited: Jul 30, 2013
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