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  1. Oct 23, 2012 #1
    A 1.8 x 102-g particle is released from rest at point A on the inside of a smooth hemispherical bowl of radius R = 37.0 cm (figure below).

    5-p-073.gif

    (a) Calculate its gravitational potential energy at A relative to B.

    Energy of system at point A: PEA + KEA
    EA: PEA + KEA
    EA: PEA + 0
    EA: PEA

    So,

    PEgravitational = mgh
    PEgravitational = (0.18)(9.81)(0.37)
    PEgravitational = 0.653


    Got this one correct...

    (b) Calculate its kinetic energy at B.

    Energy of system at point B: PEB + KEB
    EB: PEB + KEB
    EB: mg(0) + KEB ***height is 0 at point B, so PE = 0***
    EB: KEB


    The solution says that KEB = mgR
    Howwww????????!!!!!!!! Isn't it supposed to be (1/2)mv2???
    Please help!
     
  2. jcsd
  3. Oct 23, 2012 #2
    A 1.8 x 102-g particle is released from rest at point A on the inside of a smooth hemispherical bowl of radius R = 37.0 cm (figure below).

    5-p-073.gif

    (a) Calculate its gravitational potential energy at A relative to B.

    Energy of system at point A: PEA + KEA
    EA: PEA + KEA
    EA: PEA + 0
    EA: PEA

    So,

    PEgravitational = mgh
    PEgravitational = (0.18)(9.81)(0.37)
    PEgravitational = 0.653


    Got this one correct...

    (b) Calculate its kinetic energy at B.

    Energy of system at point B: PEB + KEB
    EB: PEB + KEB
    EB: mg(0) + KEB ***height is 0 at point B, so PE = 0***
    EB: KEB


    The solution says that KEB = mgR
    How?? Isn't it supposed to be (1/2)mv2???
    Please help!
     
  4. Oct 23, 2012 #3

    TSny

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    How does the total energy at B compare to the total energy at A?
     
  5. Oct 23, 2012 #4

    phinds

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    yes ...
     
  6. Oct 23, 2012 #5
    I have no idea!!! That's why I'm asking...

    PEA + KEA = PEB + KEB

    ????
     
  7. Oct 23, 2012 #6

    TSny

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    Have you covered the law of conservation of mechanical energy?
     
  8. Oct 23, 2012 #7
    Well kind of...

    But my teacher never really elaborated on it enough.
     
  9. Oct 23, 2012 #8

    TSny

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    Total mechanical energy is conserved in a problem like this because there is no friction to convert some of the mechanical energy into heat energy. You actually set up the equation that expresses this idea when you wrote:
    So, apply this equation to your problem and see what it gives you for the KE at B.
     
  10. Oct 23, 2012 #9
    PEA + 0 = 0 + KEB
    PEA = KEB
    mgR = KEB
    (0.18)(9.81)(0.37) = KEB
    0.635 = KEB

    Is that right?
     
  11. Oct 23, 2012 #10

    TSny

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    Yes, that's right.

    I noticed that you posted this question a second time because you had not received an answer within an hour of your first post. We ask that you not do that. Sometimes it just takes a while for someone to respond to your question. Please read https://www.physicsforums.com/showpost.php?p=4021232&postcount=4

    especially the part "How long did you wait for a reply". Thanks.
     
    Last edited: Oct 23, 2012
  12. Oct 23, 2012 #11
    What about part d though???

    (d) Calculate its potential energy at C relative to B.

    I did...

    PEB + KEB = PEC + KEC
    0 + KEB = PEC + 0
    KEB = PEC
    (1/2)mv2 = PEC

    But after I plugged in all the values and got PEC = 0.651, it said it was wrong!

    So is it supposed to be:

    PEB + KEB = PEC + KEC
    PEB + 0 = PEC + 0
    mgh = PEC
    mg(2R/3) = PEC

    Is that right??
     
  13. Oct 23, 2012 #12
    Ohhh ok!
     
  14. Oct 23, 2012 #13

    haruspex

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    Re: Law of Conservation of Energy - Urgent help needed!

    How will its total (KE+PE) at B compare with that at A?
     
  15. Oct 23, 2012 #14

    cepheid

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    Duplicate threads merged.
     
  16. Oct 23, 2012 #15

    TSny

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    The potential energy at C relative to B just means to calculate PEC - PEB. You won't need to use the idea of conservation of energy to do that. Just use the formula for PE.

    [EDIT: Maybe a better way to say it would be that the potential energy at C relative to B means to find the potential energy at C using h as the height measued from point B.]
     
    Last edited: Oct 23, 2012
  17. Oct 23, 2012 #16

    TSny

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    The reason you didn't get the right answer was because you assumed the KE at C is zero. But it isn't. The particle is still sliding upward at that point.
     
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