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Kinetic energy and potential energy concept question

  1. Oct 12, 2011 #1
    i need a review. why does lifting a book at a constant speed result in a constant KE but changing PE? obviously KE = 1/2 mv^2 but im looking for a conceptual answer. doesnt KE need to equal PE for energy to be conserved?

    im in e&m right now and theres a statement in my book saying that the book situation is analogous to potential within an emf source. so W = 0 (work done on a test charge within an ideal emf source), U is increased, and KE is constant. but in the book analogy doesnt the book have to be moving in the +/- x direction for W = 0?

    so to summarize i have two questions, (1) concerns how energy is conserved with a constant KE but changing U and (2) concerns how W = 0 if there is a changing U

    please answer both questions in terms of the book analogy AND the emf source! thanks in advance :)
  2. jcsd
  3. Oct 13, 2011 #2


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    Staff: Mentor

    Lifting a book is a quite different situation to a falling book. :smile: In lifting it, you are doing work all the while, so you are increasing its potential energy by moving it higher.

    For something falling, its PE is converted into KE. No extra work is being done on the system. One form of energy is being converted into the other. And that KE all gets wasted when the book hits the floor.
  4. Oct 13, 2011 #3
    that makes sense but how is there zero work being done in an emf source yet there is a changing KE and static PE? :/ its almost like the falling book situation and lifting book situation combined...
  5. Oct 13, 2011 #4
    strike that, reverse it :) changing PE, static KE
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