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Conservation of Energy Question

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data

    In the absence of friction, how fast will the masses move after the 2kg mass has dropped 25 cm? [answer: v = 1.57 m/s]

    2. Relevant equations

    Conservation of Energy

    3. The attempt at a solution

    KE2F + KE1F = PE2I + PE1I

    .5*m2*vf2^2 + .5*m1*vf1^2 = m2*g*h2 + m1*g*h1

    .5 *2*vf2^2 + .5*1*vf1^2 = 2*9.8*.25 + 1*9.8*???

    vf2 = vf1

    3/2*vf^2 = ???

    Attached Files:

  2. jcsd
  3. Nov 18, 2009 #2


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    Hi Souleater, welcome to PF. Conservation of energy is the right approach, but
    is not correct. The correct form is
  4. Nov 18, 2009 #3
    But shouldn't I combine the total energy from both masses?
  5. Nov 18, 2009 #4


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    Yes, and you should also combine the total potential energy for both masses.

    Mechanical energy conservation says that

    Kinetic plus potential energy of the system at point A is the same as kinetic plus potential energy at point B.

    When a mass moves from point A to point B it trades one form of energy for the other form in such a way as to keep the sum the same at all times.

    It's like taking money out of your left pocket and putting it in your right pocket. As you do this, you have varying amounts of money in each pocket, but the sum on your person does not change.
  6. Nov 18, 2009 #5
    But initially, isn't it all potential energy? Therefore the initial kinetic energy for both masses would be 0.
    And at the end of the movement, wouldn't both of the masses have converted that potential energy to kinetic energy? Therefore the potential energy at the end would be 0.
  7. Nov 18, 2009 #6
    The 1kg block will be gaining gravitational potential energy as it rises.
  8. Nov 18, 2009 #7


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    Welcome to PF!

    Hi SoulEater! Welcome to PF! :wink:

    :smile: KE + PE = constant. :smile:
  9. Nov 18, 2009 #8
    Thank you. :smile:

    So how would the equation look like?
  10. Nov 19, 2009 #9


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    ke1f + ke2f + pe1f + pe2f = ke1i + ke2i + pe1i + pe2i
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