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Kinetic Energy of a Particle in a semi-spherical bowl

  1. Nov 5, 2016 #1
    1. The problem statement, all variables and given/known data
    A 410 g particle in a semi-spherical bowl of radius 0.9 m is released from rest at point A at the level of the center of the bowl, and the surface of the bowl is rough. The speed of the particle at B is 2.6 m/s. The acceleration of gravity is 9.8 m/s 2 . What is its kinetic energy at B? Answer in units of J.

    2. Relevant equations
    KE= (1/2)mv^2

    3. The attempt at a solution
    KE= (1/2)(410)(2.6)^2= 1385.8
  2. jcsd
  3. Nov 5, 2016 #2


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    2017 Award

    Staff: Mentor

    You should work with units, that would have helped you to spot your mistake here.
  4. Nov 5, 2016 #3
    Remember that a joule is equal to 1 N⋅m
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