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Work done

  1. Oct 3, 2004 #1
    ____
    | B |
    |__|
    | A |
    |__|

    Suppose that there is a net force from left hand side pushing the 'A' box to right hand side. During the pushing, 'B' box moves in right same as 'A'. According to my textbook, there is zero work done on 'B' because the force(weight) on the 'B' is at right angle to the motion. However, friction(reaction) should be existed from 'A' on 'B' so that 'B' moves from at rest.This friction is along the motion direction. Since friction is a force, so work done should exist. But why does it wrong?

    Anyone can help me?
     
    Last edited: Oct 3, 2004
  2. jcsd
  3. Oct 3, 2004 #2

    Tide

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    I think your textbook is referring to a situation where the boxes are being pushed but maintain constant speed. If Box B changes speed then work is done on it.
     
  4. Oct 3, 2004 #3

    HallsofIvy

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    The friction force between A and B is "static" friction and doesn't do any work.
     
  5. Oct 3, 2004 #4

    arildno

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    In the initial acceleration phase, A certainly does work on B.
    When the constant final velocity is reached, inertia keeps B moving (i.e, no friction between the objects).

    For the system A+B, static friction never does any work.
     
  6. Oct 3, 2004 #5
    It's a net force...
     
  7. Oct 3, 2004 #6

    arildno

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    Well, then A does work on B all the time, through the static friction.
    Static friction doesn't, however, do work on the system A+B
     
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