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Someone pls help me with this qn on energy

  1. Jun 23, 2005 #1
    ok.. guyz, i'm not really sure on how i'm gonna phrase this qn, so juz hope you guyz out there understands

    work done= Change in energy
    = Energy(final) - energy(initial)
    For this eqn, do you have to add in external energy loss eg: heat n sound energy into your energy final..? or do you juz take account the energy change in the body?

    Sry.. not very good at this
  2. jcsd
  3. Jun 23, 2005 #2


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    If there are losses, the net work is
    Work = energy(initial) - energy(losses) - energy(final)
  4. Jun 23, 2005 #3
    That means that energy losses shouldn't be accounted for work done rite?

  5. Jun 24, 2005 #4
    just in case you aren't clear yet, this is what I understand from work-energy theorem:
    there are many kinds of energy. consider kinetic energy. KE of any point mass of mass 'm' at an instant when its velocity is 'v' is KE=(1/2)mv^2
    the outright formula for work, applicable in any case whatsoever, with sound loss, heat loss, etc. is:
    Net work done = Final KE - Initial KE
    remember that this is the work done by all forces, including dissipative ones. The sign of each work done MUST be taken into account. for eg. if a body of mass m is thrown up through a height h with only gravity acting, the force due to gravity=mg, and it acts downwards, but the displacement h is upwards so the work done is
    W=(mg)*(-h)= -mgh
    Now, there are 2 kinds of forces. CONSERVATIVE forces are those whose 'works' DO NOT depend on the path of the particle, for example gravity, a fixed charge etc. NON-CONSERVATIVE forces are those whose works DO depend on the path of the particle, for example friction. these forces generally result in heat loss, sound etc. When only conservative forces act on a body, we assign potential energies related to these forces. the total Mechanical energy (ME) of a body is the sum of its kinetic energy and its potential energies due to all the conservative forces. Then the following equation holds:
    Net work done by all external forces (other than these conservative forces) = ME(final) - ME(initial)
    Let the work done by friction and all the other such dissipative forces be w(nf). Let the net work done on the body due to all other forces be W. So,
    W + w(nf) = ME(final) - ME(initial)
    therefore, W = ME(final) - w(nf) - ME(initial)
    but the work done by the dissipative forces causes a reduction (assumed to be reduction, if I took it as increase, I would simply have some changes in sign in the final formula) in ME of the body, ME(losses).
    Thus, w(nf) = ME(losses)
    so, the final formula is
    W = ME(final) - ME(losses) - ME(initial)
    as said before.
    IF you wanted to find the NET work done by ALL forces other than those for which you have assigned potential energy(all, including dissipative and non-dissipative), you can write this straight:
    W = ME(final) - ME(initial)
    If you still don't get it, I recommend the book "Fundamentals of Physics" by Resnick, Halliday and Walker (6th ed) or "Fundamentals of Physics" by Resnick and Halliday(2nd ed).
  6. Jul 4, 2005 #5
    All formulas must contain all resistance to the energy input.Ergo,resistance of a 'bearing' of a particular motor must be involved along with loads of that motor.
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