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Different cases of work done done by 3 conservative forces

  1. Sep 1, 2015 #1

    Titan97

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    Gold Member

    1. The problem statement, all variables and given/known data
    WP_20150901_23_50_24_Pro.jpg

    2. Relevant equations
    W=-ΔU for conservative force.

    3. The attempt at a solution
    Let all three forces be conservative.
    Since particle comes back to where it was, ΔU=0.
    Hence ΔUAB+ΔUBC+ΔUCA=0
    Hence WCA=-WAB-WBC
    For case A, if both WAB, WBC>0, then WCA<0. Else its greater than zero. It is not possible for WCA=WBC because it will contradict the given statement. (Let WAB=WBC=x and other be equal to y. Then 2x=y and x=y which is not possible).
    Hence for A: q,s.
    Given answer: q,s

    Moving onto case C, again q is true. But r can't be true since they have opposite signs. s can also be true.
    Hence for C: q,s
    Given answer: q,r,s
    Are they considering only the magnitude?
     
  2. jcsd
  3. Sep 1, 2015 #2
    If you are a student at school, this is not addressed to you.
    Strictly, a force does not do work: whatever applies the force F does work and loses some of its available energy. The specified force always acts in the same direction so it cannot be responsible for the return of the particle from C to A. Some other force must be applied by some other agency to bring this about. This agency does the work on the return leg and whatever applies F has work done on it. "Doing negative work" is an unnecessary oxymoron. A negative calculated value indicates that the force in the calculation is not involved in doing work but that work is being done on whatever applies the force. One possible situation is that a spring is suspended with vertically with a mass m at the lower end. At rest, the mass is at B. The mass m is displaced through h to A and released. F = mg where g N/kg is the strength of the gravitational field. The work done by gravity to reach B is mgh and to reach C is 2mgh. This is the gravitational potential energy lost by the mass-Earth system. Beyond B the extended spring applies an increasing upward force on m and gains energy at the expense of the kinetic energy of the mass until, at C, the motion is reversed. The spring loses elastic potential energy while doing work on the mass. At B, the mass has kinetic energy and is able to do work to compress the spring and to move away from the Earth until point A is reached. Between C and A the force F, the weight of the body, does no work. Instead, the interaction that gives rise to the weight, gravity, has work done on it and gains energy. Between A and B the spring does work and loses energy. Between B and C it has work done on it and gains energy. Between C and B the spring does work and loses energy. Between B and A it has work done on it and gains energy.
    Incidentally, I dislike the notion of 'conservative' and 'non-conservative' forces. All interactions conserve energy and there is nothing special about ones involving only kinetic and one or two forms of potential energy. I would rather express the question in terms of situations where only KE and PE were involved and those where other forms of energy were involved. This is slightly more wordy but much more concrete and comprehensible.
     
  4. Sep 1, 2015 #3

    mfb

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    Staff: Mentor

    WCA = WBC = 2 J, WAB = - 4 J is an example for (r).
     
  5. Sep 1, 2015 #4

    Titan97

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    How do I approach case B and D?
     
  6. Sep 1, 2015 #5
    I don't understand how in case B ,the given condition is satisfying q ?
     
  7. Sep 1, 2015 #6

    Titan97

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    For case B and D, apply W=ΔKE. The particle can be moved such that kinetic energy may not change.
    All answers p,q,r,s satisfy B and D.
    Hence
    A: q,s
    B: p,q,r,s
    C: q,r,s
    D: p,q,r,s
     
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