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Homework Help: Find the coefficient of friction

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A box weighing 3.0 kN is placed in the bed of a pick-up truck. The pick-up accelerates uniformly from rest to 40 km/h in a distance of 40 m. In that time the box slides 1.0 m back toward the end of the bed. Find the coefficient of friction between the bed and the box.

    2. Relevant equations

    v^2 = v0^2 + 2a (x-x0)

    F = ma

    w = mg

    Fk = micro k * N
    3. The attempt at a solution

    Given:
    v0 = 0 km/h
    v = 40 km/h
    x-xo = Delta x = 40 m = 0.04 km
    weight = 3 kN.

    First, find a:

    v^2 = v0^2 + 2 a (x-x0)
    40^2 = 0 + 2 a * 0.04
    => a = 2*10^4 km/s^2.

    My imagine that we have 3 forces in Free Body Diagrams: Friction Forces, Newton's law and weight of the box.

    We have:

    m*g = 3 kN => m = 3000/9.8 = 306.1 grams = 0.306 kg.

    N = ma = 0.306 * (2*10^4) = 6120 kN

    but the box slides 1.0 m back toward the end of the bed, so

    (0.039km * 6120) / 0.04 = 5967 kN

    Therefore:

    coefficient of friction between the bed and the box is 5967 N / 6120 N = 0.975 = final answer => INCORRECT

    Please help! Thanks!
     
    Last edited: Feb 23, 2010
  2. jcsd
  3. Feb 23, 2010 #2
    My answer is NOT correct. Please help! Anyone?
     
  4. Feb 23, 2010 #3
    Im no expert, i worked it out and this is what i got: You deduced the first part right, so
    a = 1.5m/s2, then the weight of the box = 3000N, so mg = 3000N, so the force of friction = uk(3000N). If you use Newtons third law the foward force = the reaction force, so it is constant motion. So, the force of friction = Fnet, uk(3000N)=ma, a = the above acceleration and m is obtained from the weight equation Fg=mg, m = 306Kg, plug it into above equation and u shud get uk = 0.15.

    Hope this helps:)
     
  5. Feb 23, 2010 #4
    How about "In that time the box slides 1.0 m back toward the end of the bed" ???
     
  6. Feb 23, 2010 #5
    Its neglected since the same acceleration is needed to move the box 1.0m.
     
  7. Feb 23, 2010 #6
    You mean the final answer is 0.15 ???
     
  8. Feb 23, 2010 #7
    Yup, looks right:)
     
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