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Total Mechanical Energy Swing Problem

  1. Jan 15, 2012 #1
    A 30 kg child is playing on a swing. Another child is pushing. That second child does 75 J of work on the swing. Ignore friction. The swing has zero mass compared to the child on the swing. There is a diagram. So basically: Postion A Height = .6 m. Position B Height = .2 m and Postion Height C = ?

    a. How much kinetic energy does the child on the swing have at position A just after the push?
    Work = change in kinetic energy. So I did 75=.5(30)(v2)-.f(30)(0) to get velocity (I assumed the initial velocity is 0) And then I was gonna plug that into KE = .5mv2. Is that right?

    c. What is the maximum speed of the child on the swing?
    d.What is the maximum height off the ground the child on the swing reaches following the initial push.

    Part b is the total energy at positions B and C. I can figure that out. I'm really just confused about c, and I want to know if that is how to do a.

    Please help!
     
  2. jcsd
  3. Jan 15, 2012 #2

    BruceW

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    The kid on the swing has two different forms of energy. You've already said about kinetic energy, but there is another.
     
  4. Jan 15, 2012 #3
    Potential energy.
     
  5. Jan 15, 2012 #4

    BruceW

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    Yep. The kid also has potential energy. Do you know the equation for it?
     
  6. Jan 15, 2012 #5
    Yup! It's PE = mgh
     
  7. Jan 15, 2012 #6

    BruceW

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    That's the one. You are also given the initial height and kinetic energy, so from this you can calculate the total energy (which is a conserved quantity).
     
  8. Jan 15, 2012 #7
    Would the initial kinetic energy be the answer I have for part a? In that case, part a is right?
     
  9. Jan 15, 2012 #8

    BruceW

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    I think you were right that the work done is equal to the initial KE of the kid. So I think you did get a) right. (and you don't need to find the velocity, since it only asks for the KE).
     
  10. Jan 15, 2012 #9
    But KE=.5mv^2, so I need to find velocity before I can calculate kinetic energy.
     
  11. Jan 15, 2012 #10

    Doc Al

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    But you're starting with the kinetic energy. So taking the extra step of solving for the velocity, then plugging it back into the same expression can only give you back what you started with.
     
  12. Jan 15, 2012 #11
    But I never had the kinetic energy to begin with. I had work and mass, but I was never given any velocity.
     
  13. Jan 15, 2012 #12

    Doc Al

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    But the work equals the kinetic energy. That's your first equation.
     
  14. Jan 15, 2012 #13
    Oh I see what you're saying. Sorry! Thanks! So the kinetic energy is 75 J.
     
  15. Jan 16, 2012 #14

    BruceW

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    yep. That's the initial KE.
     
  16. Jan 16, 2012 #15
    Ok, but now how do I solve part c? And also if Work = the change in kinetic energy, but 75 joules is only the initial energy, how does that work?
     
  17. Jan 16, 2012 #16

    BruceW

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    At least give it a try yourself. What physics principles/laws do you think might help?
    Yeah, the problem is maybe a bit badly worded. It is most likely that they are saying the kid was initially at rest, then work was done on him to give him kinetic energy. So the best bet is to assume that the work done at the start is the initial KE of the kid.
     
  18. Jan 16, 2012 #17
    The only equation that has "speed" is kinetic energy. So would I just say that 75 = .5(30)v62 and solve for v? If that's the case then I have to assume 75 J is the change in kinetic energy, not the initial kinetic energy.
     
  19. Jan 16, 2012 #18

    BruceW

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    No, that's not right. The question, as you wrote it in the first post, implies that the kid is pushed, and it is during this time that work is done on him. And at the end of this push, the kid is at position A.

    Now I can only assume that the work done on the kid is equal to the kinetic energy of the kid at position A. This is my own interpretation of the question, but I think it is most likely what the question did mean.

    So this gives you the initial KE and position of the kid. From now on, no work is being done on the kid. So how would you work out the relationship between KE and position? and how can you use this to find the max KE which the kid will have, once he goes through pendulum motion?
     
  20. Jan 16, 2012 #19
    Well position implies height, right? Which is in the equation for PE (PE = mgh). I really just don't know. I appreciate the help, but I guess I'll just go see the teacher about this.
     
  21. Jan 17, 2012 #20

    BruceW

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    You're on the right track there. seeing the teacher is a good idea though.
     
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