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Homework Help: Conservation of Mechanical Energy in rope swing

  1. Feb 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Tarzan and Jane, whose total mass is 130.0 kg, start their swing on a 5.0 m long vine when the vine is at an angle of 30 degrees with the horizontal. At the bottom of the arc, Jane, whose mass is 50.0 kg, releases the vine. What is the maximum height at which Tarzan can land on a branch after his swing continues? (Hint: Treat Tarzan's and Jane's energies as separates quanitities.)

    2. Relevant equations

    How to I calculate this?

    3. The attempt at a solution


    mgh=.5mv(f)^2 + .5 mv(f)^2
    (130)(9.81)(2.5)=.5(50)v(f)^2 + .5(80)v(f)^2
    3188.25=25v(f)^2 + 40v(f)^2

    not sure if this is correct or where to go from here??
  2. jcsd
  3. Feb 26, 2008 #2
    Try this out.

    When the swing starts, there is only potential energy.


    At the bottom, there is only kinetic

    1/2(m1+m2)v^2. Therefore, equate these 2 quantities and solve for v. Now, Jane (m2) releases off. So the energy on the bottom becomes:

    1/2(m1+m2)v^2 - 1/2(m2)v^2 = Total

    Finally, this total goes into how high he can go alone.

    Total = m1 g h, solve for h. Good luck!
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