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Nonconservative work

  1. Jan 10, 2014 #1
    This isn't homework. I'm reviewing physics after many years of neglect. As with most of my posts, I made this problem up.

    Let object A have mass [itex]m_A[/itex] and object B have mass [itex]m_B[/itex]. One of A's surfaces is flat, as is one of B's. These flat surfaces are in contact and slide relative to each other in a straight line for a distance, x, experiencing kinetic friction. The kinetic friction, of magnitude f, is constant and is the only force acting on A or B in the direction of motion, and the only other forces acting on A and B are the normal forces pressing the surfaces together. The normal forces are assumed to be equal and opposite, so that the only acceleration of the objects is in the direction of friction. No net force acts on the center of mass.

    During a finite time interval, the speed of each surface relative to the other decreases from [itex]v_i[/itex] to [itex]v_f[/itex] due to friction. The nonconservative work done by each surface on the other is [itex]W_{noncons} = -fx[/itex] since each surface experiences the same displacement and the magnitude of the frictional force exerted by each surface on the other is the same since the two frictional forces comprise a force/reaction force pair.

    Therefore the change in kinetic energy of A, [itex]\Delta K_A[/itex] as seen from B's restframe is equal to the change in kinetic energy of B, [itex]\Delta K_B[/itex], as seen from A's restframe, since we are assuming that no other forces accelerate the objects.

    Therefore [itex]\frac{1}{2}m_A v_{f}^{2} - \frac{1}{2}m_A v_{i}^{2} = \Delta K_A = \Delta K_B = \frac{1}{2}m_B v_{f}^{2} - \frac{1}{2}m_B v_{i}^{2} \Rightarrow m_A = m_B[/itex], an absurd result.

    Something's wrong and it's driving me crazy! Please help me find the error in this reasoning, thanks.
  2. jcsd
  3. Jan 10, 2014 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Try analyzing things from an inertial frame. Since the objects are accelerating, their rest frames are non-inertial.
  4. Jan 11, 2014 #3
    Thank you! It was staring me in the face.
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