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Friction force, and external forces

  1. Oct 21, 2012 #1
    If we have a box with mass m placed on a horizontal tabletop. Then we push the box a distance of d. THe coefficient of kinetic friction between the box and tabletop is .35.

    When we're asked to find the external work done on the block-table system, it's not the work done by the friction force at all. Why??? Can someone explain to me??

    My textbook refers the work done by friction is transfered to thermal energy. So they don't call it the external work. But in the previous before, they say friction force is external force.
     
  2. jcsd
  3. Oct 21, 2012 #2
    Of you push the box, what forces are acting?

    Which of these moves its point of application?

    Does our book talk about Work done = energy expended or transferred or some such?
     
  4. Oct 21, 2012 #3
    what they say is : W external = W(me on block) + W (gravity on block) + W(gravity on table) + W (floor on table) = W(me on block) + 0 + 0 +0

    WHy don't they include the work done by friction?? Friction, of course, is not in the system
     
  5. Oct 21, 2012 #4
    I'm not being funny, I'm trying to get you to see for yourself.

    If you push the box what forces are acting (call it P)

    If the block slides along a horizontal table, with or without friction, does the normal reaction or the weight (gravity) force of the block do any work?

    Look at the sketch and suppose you push the block from A to B.

    At A there is a side thrust P pushing the block along the table.

    As the block moves to B the work done by this force is P(xa-xb) since this is in the direction of the line of action of P.

    Force R does no work since there is no movement in its direction of action.

    Ditto Force W.

    Now the friction force, F, is different because, unlike P it moves along with the block so its point of action

    is always under the block

    This is why F does no work.

    You can say if you like, that when the block is at B, F no longer exists at A.

    However if we are pushing with with a stick, then P still exists at A when the block has moves to B since the stick must extend back through A to its source.
     

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    Last edited: Oct 21, 2012
  6. Oct 21, 2012 #5
    but isnt there for a short time and short distance some force Pstatic > P whereby the work done during that short time and short distance is != P(Xb-Xa) ??

    i may be wrong, but the work done on moving a block from rest != P(Xb-Xa) ??
     
  7. Oct 21, 2012 #6

    haruspex

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    I think there are two ways of looking at that, and it depends whether you consider the frictional action to be part of the block or part of the table (or anything between those two extremes).
    The view you express above is taking the frictional action to be part of the block. So the work done in pushing the block remains in the block (as heat).
    Viewing the frictional action as external to the block, the frictional force is in the opposite direction to the distance moved, so the 'work done' by it is negative and exactly cancels the work done in pushing. In this view, the block gains no energy, the heat now being external to it.
     
  8. Oct 21, 2012 #7
    Honestly, I knew the force P does work, and the normal force and weight do no work. It's obvious. My question is only about the friction force.

    I'm not saying that you're being funny or anything. If you're not happy, well, I can find other access.
     
  9. Oct 21, 2012 #8
    Thank you :)...I just can't recognize when the frictional action is part of the system and when it's not. :(
     
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