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Kinetic and Potential Energy

  1. Apr 17, 2008 #1
    A 215 g particle is released from rest at point A inside a smooth hemispherical bowl of radius 30.0 cm. Point B to C have Height= 2/3(R)
    Calculate the following: A. The gravitational potential energy at A relative to B
    B. The particle's kinetic energy at B
    c. Th particle's speed at B
    D. The potential energy and kinetic energy at C

    I solved A,B,C and D(The potential Energy but not the kinetic energy)

    Here is my work:

    A. Changed 215 g into Kg so it will be 0.215 kg
    30.0 cm into m so it will be 0.30m
    PE=mgh
    PE=0.215*9.81*0.30
    PE=0.633J



    B. KE= 1/2mv^2
    KE=1/2*0.215*2.4^2
    KE=0.62J


    C. V^2f=2gh
    V^2f=2*9.81*030
    V^2f=5.9 Apply square root
    Vf= 2.4m/s

    D. PE=mgh
    PE=0.30*9.8(2/3*0.30)
    PE=0.422

    Can someone please help me find the KE at C, the answer is 0.211J according to the book. Well, what I got is 0.43J..



    Thanks in advance.
     
  2. jcsd
  3. Apr 17, 2008 #2
    You missed the point of the problem. Use conservation of energy.
     
  4. Apr 18, 2008 #3
    That is true, you need to use the conservation of energy between C and B.
    mgR= KEc + mg(2/3)R
    KEc=1/3mgR=0.211J
     
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