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Mastering physics again

  1. Oct 11, 2005 #1
    Again another work/kinetic energy problem I need help with.
    A small rock with a mass of 0.240 kg is released from rest at point A, which is at the top edge of a large, hemispherical bowl with a radius R = 0.500 m. Assume that the size of the rock is small compared to the radius of the bowl, so that the rock can be treated as a particle, and assume that the rock slides rather than rolling. The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl is - 0.160 J. What is the speed of the rock when it reaches point B?
    I really don't know where to start.
    picture is attached.
     

    Attached Files:

  2. jcsd
  3. Oct 11, 2005 #2
    Use energy methods.

    Point A:

    Whats the associated energy at that point a? Gravitational? Kinetic? Electric?

    Whats the associated energy at point b?

    Show some work and well go from there.

    Thats a very nice picture by the way, how did you get it?
     
    Last edited: Oct 12, 2005
  4. Jun 30, 2007 #3
    hey i am not sure abot the answer. is this correct

    mgh=mv^2/r
    (.24*10*.5)-.16=(.24*v^2)/2
    v=2.94m/s
     
  5. Jun 30, 2007 #4

    HallsofIvy

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    It is very difficult to help you when you wont at least indicate what you DO know! Why do you think that formula will help you? Are you using kinetic energy and potential energy? But what about "The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl is - 0.160 J"?
     
  6. Jun 30, 2007 #5
    i hav reduced it from the potential energy. so it equalls to the kinetic energy
     
  7. Jun 30, 2007 #6
    mgh=mv^2/r...

    So PE = Fc?

    Fc being the centripetal force. I don't think that's right. PE is in joules and Fc is in newtons.

    I think it might actually be:

    PE=KE

    mgh = [(1/2) mv^2] + Wf,

    where height (h) is your radius and W is the work done by friction (-.160J). When you use -.160J make it positive in the equation since the KE plus the work done by friction should be equal to the PE.

    Solve for v.
     
  8. Jul 1, 2007 #7
    thanx for pointing out my mistakes.
     
  9. Oct 22, 2009 #8
    I would approach the problem using energy as well. Since there is frictional force then energy conservation must involve work. so Kinetic energy(final)+potential energy(final)+Work=potential energy inital+kinetic energy initial. Where potential initial=mgR and kinetic initial=potential final=0 and kinetic final=.5mv^2. Im not entirely sure someone correct my mistakes ^_^
     
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