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Work Done Incline

  1. Jul 11, 2015 #1
    1. The problem statement, all variables and given/known data
    screenshot_22.png

    2. Relevant equations
    W = F d
    F = ma


    3. The attempt at a solution
    Constant Speed ==> a = 0 ==> Net Force = 0 ==> F cos 37 = mg sin 37 ==> F = 37.6 N
    sin 37 = h/d ==> 6/10 = 0.5/d ==> d = 5/6
    W = F d = 5/6 * 37.6 = 31.3 N [Not In Choices] - Any Help?
     
  2. jcsd
  3. Jul 11, 2015 #2
    Perhaps it would be better to consider the work done as the change in potential energy and approach it that way.
     
  4. Jul 11, 2015 #3
    yup! W = mgh = 5 * 10 * 0.5 = 25 J
    In my previous attempt, my mistake was not writing cos 37 in finding the work
     
  5. Jul 11, 2015 #4
    Sir,
    I applied the method you mentioned on this questions but it does not work with me!
    screenshot_23.png
    Confused now on when to use that method

    That method find work done by gravity, is it same as work done by F ? Why?
    Thanks in advance
     
  6. Jul 12, 2015 #5

    SammyS

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    The original problem had constant speed. I think you will find that this one doesn't.

    Your initial method (after correcting by including cos(37°) in that case) should work here.
     
  7. Jul 12, 2015 #6
    How "constant speed" differ?
     
  8. Jul 12, 2015 #7

    CWatters

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    KE is constant.
     
  9. Jul 12, 2015 #8
    Sorry, but how is that relevant?
    The way I think of the work done is final energy - initial energy
    In final energy I have potential energy (mgh) only
    In initial energy there is nothing in this case
    So,
    W = mgh

    Where is my point of confusion?
     
  10. Jul 12, 2015 #9
    The final energy will not be only in the form of potential energy.
     
  11. Jul 12, 2015 #10
    Without thinking too much, why don't you attempt this question using your previous knowledge on basics...
     
  12. Jul 12, 2015 #11
    ok, let say the speed is not constant..
    E initial is zero because potential is zero and it is at rest at the begining
    E final is potential and kinetic energy will be ..

    That also the same with if speed is constant.

    So, why it differ
     
  13. Jul 12, 2015 #12
    I did attempt the question and got the answer of 52 J but I am asking why I can't use that method
     
  14. Jul 12, 2015 #13
    If the speed is constant to a direction, it means it has no external unbalancrd force towards that direction..
     
  15. Jul 12, 2015 #14
    When there's an unbalanced force and the object shows a displacement, there's a work done.
     
  16. Jul 12, 2015 #15
    That's right, but it does not answer my question in that why I can't use the concept of conservation of energy
     
  17. Jul 12, 2015 #16
    We don't know the final kinetic energy. Since there is an external unbalanced force there is an acceleration so the initial KE cannot be taken as the final KE. So to apply conservation of energy, we don't have enough data.
     
  18. Jul 12, 2015 #17
    OK Thanks, what about if it is constant speed. Initially velocity = 0 but final is not 0. So how we only used W = change in potential energy, without considering hte KE
     
  19. Jul 12, 2015 #18
    If there is a motion I think there should be a velocity.;)
     
  20. Jul 12, 2015 #19

    CWatters

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    That is correct for #1 (constant velocity) but not #4 (constant force)

    In #4 the block is accelerating so the final KE is not the same as the initial KE.

    You can use conservation of energy to solve this but you would need to calculate the final velocity.

    It is easier to use work = force * displacement.
     
  21. Jul 12, 2015 #20
    Thanks, but why you said it is correct for #1 ?
    It is constant velocity so we still have velocity in Ef but we don't have it in Ei (before applying the force)
     
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