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Work Done with Friction

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    What is the minimum work needed to push a 900 kg car 880 m up along a 9.5° incline? Assume the effective coefficient of friction retarding the car is 0.20.

    2. Relevant equations
    W = F * d * cos (θ)
    Normal Force - mg * cos (9.5°)
    Force of kinetic friction = μ * Normal Force

    3. The attempt at a solution
    I drew out a free body diagram. The first thing I did:

    Normal Force - y component of gravity = 0 because it is not accelerating in the y direction.
    Normal Force - mg * cos (9.5°).
    Normal Force = mg * cos (9.5°).
    Normal Force = (900 kg)(9.8 m/s^2) * cos (9.5°)
    Normal Force = 1.73*10^3 N

    Second: Force of Kinetic Friction = (0.20)(8.69*10^3 N)
    Force of Kinetic Friction = 1.73*10 ^3N

    Work done by Friction : (1.73*10^3)(880 m) * sin 9.5°
    W = 2.5*10^5 J

    2.5*10^5 J + 1.28 *10^6 J (This is the minimum work required without friction. The first part of the question asked this and it confirmed this was the correct answer) =

    1.53 * 10^6 J
    Last edited: Feb 28, 2012
  2. jcsd
  3. Feb 28, 2012 #2


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    I get 1.74 x 10^3 rather than your 1.73 for the friction force; don't know if that is important.

    In finding the work, why multiply by sin (9.5) ? The force of friction and the distance are in the same direction so no sine factor should be used.

    The rest of your calc looks good!
  4. Feb 28, 2012 #3
    Oops! I forgot - friction acts over the entire distance, not just the height as I was thinking.
  5. Feb 28, 2012 #4


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    You bet.
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