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Homework Help: Mechanical energy of sliding beads

  1. Jan 20, 2007 #1

    lzh

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    1. The problem statement, all variables and given/known data
    A 0.5 kg bead slides on a curved wire, starting
    from rest at point A as shown in the figure.
    The segment from A to B is frictionless, and
    the segment from B to C is rough. The point
    A is at height 6.9 m and the point C is at
    height 1.6 m with respect to point B.
    The acceleration of gravity is 9.8 m/s^2 :
    http://img376.imageshack.us/img376/2720/motion8pj.gif [Broken]
    the image for this problem is fig.2
    If the bead comes to rest at C, find the change
    in mechanical energy due to friction as it
    moves from B to C. Answer in units of J.


    2. Relevant equations
    mgh=energy of grav.
    Force*displacement=diss. energy


    3. The attempt at a solution
    First, I set mgh at point A to mgh at C + diss energy:
    (.5)(9.8)(6.9)=.5(9.8)+diss.
    diss=26J
    the answer above is not right, am i misunderstanding the question?
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jan 20, 2007 #2

    arildno

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    Science Advisor
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    Well, isn't the height at C 1.6m above B??

    Remember, though that the change of mechanical energy is NEGATIVE.
     
    Last edited: Jan 20, 2007
  4. Jan 20, 2007 #3

    lzh

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    oh, thanks alot!
     
  5. Jan 20, 2007 #4

    lzh

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    since i have fig.1 as well, I might as well post another one of my problem:
    1. The problem statement, all variables and given/known data
    A bead slides without friction around a loop-
    the-loop. The bead is released from a height
    of 9.4 m from the bottom of the loop-the-loop
    which has a radius 2 m.
    The acceleration of gravity is 9.8 m/s2 :
    What is its speed at point A? Answer in
    units of m/s.
    3.Work
    this is basically the same as the first problem, and mass isn't needed because it ends up canceling.
    so:
    mgh=.5mv^2+mgh <-there is also grav. energy, because A is not the lowest
    gh=.5v^2+gh
    (9.8)(9.4)=.5v^2+9.8(4)
    v=10.3m/s
    but that answer is wrong. I'm sure that i have the equation set out right, because at point A kinetic energy is transferred to grav. potential energy.
     
  6. Jan 20, 2007 #5
    you did some thing that does not make logical sense. take a look at your plus/minus signs.
     
  7. Jan 20, 2007 #6

    lzh

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    so you are saying that one of the value that i plugged in should've been negative?
     
  8. Jan 20, 2007 #7
    yep. But don't just change it without understanding why. after it falls, it goes up 4 meters--not down.
     
  9. Jan 21, 2007 #8

    lzh

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    i figured out what i did wrong... I rounded wrong. I was off by more than 1% from the real answer, so Utexa's homework service rejected it.
     
  10. Jan 21, 2007 #9
    Those drawings are down in Latex?
     
  11. Jan 21, 2007 #10

    lzh

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    no, I copied and edited them on paint :)
     
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