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Change in kinetic energy of a box

  1. Oct 1, 2006 #1
    A 41 kg box initially at rest is pushed 4 m along a rough, horizontal floor with a constant applied force of 121 N. If the coefficient of kinetic friction between the box and the floor is 0.29, find
    (a) the work done by the applied force W= FΔx
    Ans: 4.84e2 J
    b) the increase in internal energy of the box + floor system due to friction ΔEint= fkd fk= ukn= 116.5N
    ΔEint= 4.66e2 J
    c)the work done by the normal force on the block due to the floor
    =0.00 J
    d) the work done by the gravitation force on the block due to the earth =0.00 J
    e)the change in kinetic energy of the box
    ΔK = -fkd + ∑W This is where I'm stuck. Is ∑W= to 4.84e2J because I thought that the applied force was included in the change in kinetic energy but this according to my homework is wrong. Can someone explain what I'm missing here?
  2. jcsd
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