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Work and Energy

  1. Dec 16, 2013 #1
    ImageUploadedByPhysics Forums1387228165.386829.jpg

    Okay so in number a) I got the work just fine
    In number b) pushing horizontally meaning that cos angle is 0 which would give a larger force?
    The correct answer is smaller
    In c) the work done on the cart by the shopper how would it be the same if the force is smaller let us say?
     
  2. jcsd
  3. Dec 16, 2013 #2
    pushing the cart at a 0° angle would not produce a greater force--it would increase the work, if you calculate it with the original displacement and applied force: W = (50m)(35N)cos(0°). but the question is asking how the applied force would have to change, if at all, for the speed to stay the same? you can assume the displacement is the same, since it says she goes down "the next aisle."
     
  4. Dec 16, 2013 #3
    So you are saying for the speed to stay the the same the force will be smaller ? Is that correct?
    What about the pushing horizontally part wont it affect at all the force?

    Secondly in c) the work will remain the same how is it possible if the force is smaller ?

    Please Help!
     
  5. Dec 16, 2013 #4

    haruspex

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    The speed doesn't matter. As long as the speed is constant, the forces are in balance.
    In each scenario, there are essentially four forces on the cart, right? The gravitational force, Mg, the shopper's applied force Fs, the frictional force Ff, and the normal force N from the ground.
    Which direction does Ff act in?
    What is the horizontal component of Fs equal to?
    What is the vertical component of Fs equal to?
     
  6. Dec 17, 2013 #5
    In b)

    The friction force acts opposite to the direction of motion

    Vertical component will be equal to mg

    Horizontal will be equal to Fs
     
  7. Dec 17, 2013 #6

    haruspex

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    Since we're trying to compare (a) with (b), best is to answer for an unknown angle of applied force, downwards at angle θ say, then see how the magnitude of Fs depends on θ.
    So horizontal, yes.
    The vertical component of the shopper's push will equal mg?? The shopper is pushing the cart along the ground, not carrying it.
    When pushing horizontally, yes, but I meant in terms of the frictional force.
     
  8. Dec 17, 2013 #7
    Then the frictional force is equal to his horizontal force true. Please see attachment and tell me if this is correct

    ImageUploadedByPhysics Forums1387276263.270333.jpg
    Is this where it comes to say that the force is smaller?
    ImageUploadedByPhysics Forums1387276295.453962.jpg
     
  9. Dec 17, 2013 #8

    haruspex

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    Yes.
    Now what about the work? Remember work done ( by a constant force) = force x what?
     
  10. Dec 17, 2013 #9
    Work = Force x Displacement
    Force is smaller now as we agreed
    And the displacement is the same

    ImageUploadedByPhysics Forums1387278129.579551.jpg
     
  11. Dec 17, 2013 #10
    This is not correct the work should be the same whyyyyyy???
     
  12. Dec 17, 2013 #11

    haruspex

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    But what displacement, exactly? Think about the directions.
     
  13. Dec 17, 2013 #12
    It is displaced by 50 meters in the next aisle right?
     
  14. Dec 17, 2013 #13

    haruspex

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    No, I mean in the statement "work = force x distance", it isn't just any old distance. The distance has a specific relationship to the force. It's to do with the direction of each.
     
  15. Dec 17, 2013 #14
    Okay I see that please elaborate more how will it be smaller?
     
  16. Dec 17, 2013 #15

    haruspex

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    It's the distance in the direction of the force. In vector terms, and allowing for a variable force, it's ∫F.ds, i.e. the integral of the dot product of the force vector with the distance element vector. For a constant force you can simplify that to F.s, again that's the dot product of vectors. If the angle between the force vector and the distance vector is θ then you can write it as |F||s|cos(θ).
    So, it doesn't matter whether you consider the distance component in the direction of the force or the force component in the direction of the distance moved - it comes to the same thing.
    In the present problem, the movement is horizontal. So you need to compare the horizontal components of the forces for the two cases.
     
  17. Dec 17, 2013 #16

    But the magnitude of the force is smaller?
     
  18. Dec 17, 2013 #17

    haruspex

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    The overall magnitude, yes, but what about the horizontal components of each? Are they different?
     
  19. Dec 17, 2013 #18
    Ohhh yes I got it finally the the force(a) horizontal component = force (b)

    Thank you very much !
     
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