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A Gymnast on a Trampoline

  1. May 28, 2014 #1
    1. The problem statement, all variables and given/known data
    A gymnast springs vertically upward from a trampoline. The gymnast leaves the trampoline at a height of 1.20 m and reaches a maximum height of 4.80 m before falling back down. All heights are measured with respect to the ground. Ignore air resistance, determine the initial speed v0 with which the gymnast leaves the trampoline.


    2. Relevant equations
    [tex]Wnc = 1/2 \cdot m \cdot V_f^2 + m \cdot g \cdot h_f - (1/2*m \cdot v_0^2 + m \cdot g \cdot h_0)[/tex]


    3. The attempt at a solution
    I know this can be solved with pure kinematics and I know how to do that, but this is in a chapter on Work and Energy and I want to understand the concepts there, so that's how I want to solve it.

    So here's my try:

    [tex]Wnc = (KE_f-KE_0) + (PE_f-PE_0)[/tex]
    We know that KE_f will be 0, so it can be removed from the equation.
    [tex]KE_0 = 1/2 m \cdot V_0^0[/tex]
    [tex]PE_0 = m \cdot g \cdot h_0[/tex]
    [tex]PE_f = m \cdot g \cdot h_f[/tex]
    So we get:

    [tex]Wnc = -1/2 \cdot m \cdot V_0^0 + m \cdot g \cdot h_f - m \cdot g \cdot h_0[/tex]
    Solving for V0, we get:
    [tex]V_0=\sqrt{2 \cdot g (h_f - h_0) - 2Wnc/m}[/tex]
    Correct so far?
    Now, what is Wnc? According to my book, since only the gravitational force acts on the gymnast in the air, it is the net force, and we can evaluate the work by using the relation [tex]W_gravity = m \cdot g (h_0 - h_f)[/tex]

    I don't understand how they get to this? First of all, how is W_gravity related to Wnc? It seems here they are assuming they are the same?

    Secondly, why does the equation they use have the expression [tex](h_0 - h_f)[/tex] instead of [tex](h_f - h_0)[/tex]?

    Please help me clarify this so that I can understand these concepts.
     
  2. jcsd
  3. May 28, 2014 #2

    Doc Al

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    Staff: Mentor

    What does Wnc mean? Do any non-conservative forces act here?
     
  4. May 28, 2014 #3
    No I guess not, only gravitational force. But if you are saying the equation for Wnc doesn't apply. How are KE and PE related to the work being done?
     
  5. May 28, 2014 #4

    Doc Al

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    Staff: Mentor

    So what does Wnc equal if there are no non-conservative forces acting?

    Consider the work-energy theorem.
     
  6. May 28, 2014 #5
    I understand now. Wnc = 0. That's great, I spent half an hour trying to figure out something in a problem when I only missed a minor thing like this.
     
  7. May 28, 2014 #6
    Maybe my second question can be clarified though? Why would they write (h0-hf) instead of (hf-h0)?
     
  8. May 28, 2014 #7

    Doc Al

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    Staff: Mentor

    The work done by gravity is -ΔPE.
     
  9. May 28, 2014 #8
    Interesting. Could you please explain why it's not ΔPE?
     
  10. May 28, 2014 #9

    Doc Al

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    Staff: Mentor

    Figure it out. What's the definition of work?
     
  11. May 28, 2014 #10
    Another relevant question. If there is a non conservative force involved. Let's say the gymnast gets hit by a ball. Is there any difference on the total amount of work being done depending on where/when the external non conservative force acts? If it appears "at" v0/h0 or at vf/hf or in between?
     
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