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How to use kinetic energy for this problem?

  1. May 9, 2015 #1
    1. The problem statement, all variables and given/known data

    The problem is that my teacher asked to solve this using kinetic energy but i doubt it uses kinetic energy... i thought it uses potential energy.


    You put your little sister (mass m) on a swing whose chains have lenght L and pull slowly back until the swing makes an angle ø with the vertical. Show that the wrok you do is mgL(1 - cos ø).

    2. Relevant equations
    W = integral of (F *dl))

    KE=1/2 m (vf)^2-1/2m(vi)^2

    3. The attempt at a solution
    [tex]\int(Fdl) = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_f^2 [/tex]
    [tex] v_i [/tex] and [tex] v_f [/tex] are 0
    so the integral becomes
    [tex]\int(Fdl) = 0 [/tex]

    ?? what do i do now?

    so my teacher makes us use integral to solve for work.
     
  2. jcsd
  3. May 9, 2015 #2
    also why is everything centered?
     
  4. May 9, 2015 #3
    Indeed, you have to analyze your sister's gravitational potential energy at both the lower position and the upper position in order to solve this problem.

    I am not sure what you mean by this. The swing is pulled back until it makes an angle with the vertical. This seems to imply that the swing was initially making an angle of zero degrees with the vertical.
     
  5. May 9, 2015 #4
    oh no. i was talking about the latex. everything I typed was centered... lol
     
  6. May 9, 2015 #5
    You are using the mathematics inline, or something like that. You can go to the latex website and look at their tutorials and stuff. There is also a primer here on PF that has a good amount of information.

    As far using kinetic energy to solve this problem goes, I'm really not seeing it. If you slowly pull back the swing, I don't see ...your sister... as having much kinetic energy when she reaches the upper position.
     
  7. May 9, 2015 #6
    yes, i think the teacher may have titled the homework set incorectly. this is most probably a potential enregy problem.
     
  8. May 9, 2015 #7

    ehild

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    Your teacher meant Work-Energy Theorem. You also started the solution with it: change of kinetic energy is equal to the work done by the sum of forces, acting on the object. The swing is slowly pulling back, so the kinetic energy can be considered zero during the process, and the net work done is zero, as you have shown.
    Now you have to analyse the forces and their works. What forces act on the swing? Remember, the problem asks your work. Your force is not conservative, it does not have potential energy.
     
  9. May 9, 2015 #8
    the forces are gravitational force, tension force, and the applied force by me. the only thing is how would i represent the applied force? the angle is constantly changing....
     
  10. May 10, 2015 #9

    ehild

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    What can you say about the work of the individual forces? Does the tension do work? What is the work of gravity? (Remember, its work can be determined by the change of potential energy.) As the whole work is zero and you know the other two, you can determine the third one: your work.
     
  11. May 10, 2015 #10
    The work you do is the gravitational potential energy that is now in the girl-swing system (that was not present before). The gravitational potential energy, or the work, put in is equal to mgh, where h is equal to L(1 - cos ø).


    Now, if you have a hard time seeing how h = L(1 - cos ø), that's a trig problem, not a physics problem.
     
  12. May 10, 2015 #11
    how is the whole work 0? also i dont think tension force does work, but im not sure. how do you tell if a certain force does work? are the only 2 ways: if it changes speed or changes the position such that it has pontential energy? i dont think tension force changes either.
     
  13. May 10, 2015 #12

    ehild

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    You have shown
    You have shown in the first post that the whole work is zero as the kinetic energy does not change.
    What do you know about the direction of the force of tension with respect to the displacement? The swing moves along a circle and the chain is the radius...
    Change of potential energy is equal to the negative work done by the conservative force. Gravity is conservative force.
     
  14. May 10, 2015 #13

    ehild

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    According to the rules of this Forum, you must not solve the problem instead of the original poster. Also, you are wrong saying that "The work you do is the gravitational potential energy".
     
  15. May 10, 2015 #14
    its perpendicular so no work is done. the only 2 forces that do work are gravity (which does work in terms of potential energy) and the applied force.
     
  16. May 10, 2015 #15

    ehild

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    Very good! So your work is opposite to the work of gravity. How is the work of gravity related to the change of potential energy?
     
  17. May 10, 2015 #16
    uhm, never learned about potential energy but i just looked it up online. the potential energy equals the displacement times the magnitude of the graviatinal force ( i think) so its more commonly written as mgh. but our teacher said we HAVE to use integrals becuase thats the correct way no matter how easy the work is, so im gonna go ahead and say

    [tex] \int_{x_i }^{x_f}F_g dl [/tex]

    where Fg is constant so you can pull it out of the integral.
    so the new integral becomes:

    [tex] mg \int_{0}^{L-Lsin(\theta)} dl [/tex]

    OMG which simplifies to

    [tex] mgL(1-sin(\theta) [/tex]

    is this the correct thinking? thanks!
     
  18. May 10, 2015 #17
    dang it, why is all my text centered? here is the code i used for the last latex code:

    [tex ] mgL(1-sin(\theta) [/ tex]

    is there anything wrong with that?
     
  19. May 10, 2015 #18

    ehild

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    You have to state what dl and x are. If x means the vertical position of the swing, the integral is OK, but the result is not. Make a drawing and check.
    Remember, theta is the angle the chain makes with the vertical.
     
  20. May 10, 2015 #19
    OHHH sorry, that should only be one variable. change all the x's to l's. our teacher told us to use dl... not dr or dx. would that be correct?
     
  21. May 10, 2015 #20
    also my mistake. the above limit should be the following:

    [tex] l_f = L-Lcos(\theta) [/tex]

    i didnt do a drawing the first time and got confused.
     
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