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Work done by friction on potted plant

  1. Nov 17, 2012 #1
    1. The problem statement, all variables and given/known data
    A 2.75-kg potted plant rests on the floor. Determine the work required to move the
    plant at a constant speed along the shelf for 1.07 m where the coefficient of kinetic friction is 0.549

    3. The attempt at a solution
    I found the Fn:

    Fn = mg
    Fn = 26.95 N

    Ff = Fnμ
    Ff = (26.95N)(0.549)
    Ff = 14.79555 N

    W = Ff x d
    W = 14.79555 N x 1.07 m
    W = - 15.8312 J

    This would be negative correct?
  2. jcsd
  3. Nov 17, 2012 #2


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    Well the work done by the frictional force would be positive due to sign conventions. So your answer would be W = + 15.8312 J, however the work done on the plant due to friction is W = -15.8312 J.

    Your answer is correct, but the sign depends on the convention used.
  4. Nov 17, 2012 #3
    I don't think you need a sign convention for this.

    The angle between the applied force and the displacement is 0, so the work done by the applied force must be positive.

    The angle between the frictional force and the displacement is 180, so the work done by the frictional force must be negative.
  5. Nov 17, 2012 #4
    would u like to expand?
  6. Nov 17, 2012 #5


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    Gold Member

    You are asked for the work required to move the plant, i.e. the work done by the force that moves it. That force and the distance moved are in the same direction, so the product must be positive, no matter which way you assign the positive direction in the co-ordinate system. That is, either F and d are both negative or they're both positive, but Fd will always be positive.
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