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Energy Problem, Confused.

  1. Dec 31, 2006 #1
    1. The problem statement, all variables and given/known data
    A child's pogo stick stores energy in a spring (k=2.5 x 10^4 N/m). At position A (x_1=-0.1m), the spring compression is a maximum and the child is momentarily at rest. At position B (x=0), the spring is relaxed and the child is moving upward. At position C, the child is again momentarily at rest at the top of the jump. There is a displacement (x_2) above the x=0 mark. Assuming that the combined mass of the child and pogo stick is 25kg, a) Calculate the total energy of the system if both potential energies are zero at x=0, b) determine x_2, c) calculate the speed of the child at x=0, d) determine the value of x for which the kinetic energy of the system is a maximum, and e) obtain the child's maximum upward speed.


    2. Relevant equations
    E(before)=E(after)--- Both Potential and Kinetic


    3. The attempt at a solution

    I wasn't able to find a) right off the bat. The answer I got (125J) does not match the answer in the back of the book (101J). What I did to find a) was put: U(spring)= KE(after) , and to find U(spring) I used:
    1/2kx^2 --> 1/2(2.5 x 10^4 N/m)(-0.1m)^2

    That should give you 125J but again, it doesn't match the answer given. Am I doing something wrong? Something about the displacement of the spring?

    I got b) right I think. To find x_2 I used KE(before)=U_g(after), so:
    1/2mv^2=mgh ---> 101J(Used book's answer)= (25)(9.8)h

    I solved for h and i got .41m, which is supposedly correct. =]

    I got c) by using U(spring)=KE(after) so: 101J=1/2(25)v^2; v=2.8425

    So overall, I don't know how to find a), d) or e).

    For both d) and e), how do you know where the KE and speed is maximum?
     
  2. jcsd
  3. Dec 31, 2006 #2

    AlephZero

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    Think about the potential energy of the child and stick (relative to the datum position x = 0) as well as the potential energy in the spring.

    For (d) and (e), the KE+PE is constant, so the KE is a maximum when the PE is a minumum. You can write down the PE as a function of x.
     
  4. Dec 31, 2006 #3
    Okay, I get how KE+PE is always constant. But how do you mean either one a maximum/minimum?? make it equal to .00000000000001 or something? Can you show me how to make it "a function of x"?
     
  5. Dec 31, 2006 #4

    radou

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    The answer is in your question, since, if the sum is constant, minimizing one summand maximized the other, and vice versa.
     
  6. Dec 31, 2006 #5
    d) determine the value of x for which the kinetic energy of the system is a maximum, and e) obtain the child's maximum upward speed.

    You need to find numbers. For d) its -9.8mm and e) 2.85 m/s?

    How do you find those exactly? I'm gonna try something.
     
  7. Dec 31, 2006 #6
    Using conservation of energy, you can find the answer.

    We can help you more easily if you show us your conservation of energy expression. Once you figure out how to write that, the rest is just arithmetic.
     
  8. Jan 1, 2007 #7

    AlephZero

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    When the pogo stick is below x = 0, the PE is (1/2)Kx^2 - mgx (you left out the mass term in your attempt at the solution). At the minimum (or maximum), d(PE)/dx = 0.
     
  9. Jan 1, 2007 #8
    Ohh.. alright I got the first part. So any time x is not zero you have to use the gravitational potential energy also? I thought the only potential energy it had was from the spring. Anyways.. Isn't the total energy in the system always the same? Unless it is nonconservative... Because if you make the potential energy 0 the Kinetic energy would have 101J, the same as potential energy before. What am I doing wrong? Thanks. Also, I don't know how to use calculus or any of those derivations.
     
  10. Jan 1, 2007 #9
    Potential energy is completely relative. Gravitational potential energy (our mgh approximation) has no meaning except in terms of a certain reference point with zero gravitational potential.

    Let's set the point where the spring is fully compressed as zero gravitational potential energy. You can also set it to any other point - but you have to be able to write the gravitational potential energy expression. They'll all come out to be the same thing in the end, regardless of where you have your reference point.

    In this particular system, yes.

    But we have to take into account gravitational potential energy, like this:

    [tex]
    \left (PE_\rmmath{spring} + PE_\rmmath{gravity} + KE \right)_{initial} = \left ( PE_\rmmath{spring} + PE_\rmmath{gravity} + KE \right )_{final}
    [/tex]

    Why can we say this? Let's think about it. Energy is conserved - as you said - so all the initial forms of energy must equal the final forms of energy. It remains to be seen which ones are zero.

    When the spring is fully compressed, there is just spring potential energy. So we write that down. Let's take the fully compressed point to be our zero gravitational potential energy point. Also, when the spring is fully compressed, the stick and the boy aren't moving - no initial kinetic energy. Let's strike those terms, then.

    [tex]
    \left (PE_\rmmath{spring})_{initial} = \left ( PE_\rmmath{spring} + PE_\rmmath{gravity} + KE \right )_{final}
    [/tex]

    Now it's just a matter of deciding what happens next. Is your final state the one where the spring is fully decompressed? If so, there's no spring potential energy in the end. Or is it the one where the stick and boy have reached the maximum height? Then there's no kinetic energy or spring energy on the right side.

    In summary, here's how you want to approach conservation of energy problems:
    1. Determine if there is any nonconservative work being done. If so, take that into account (that's not the case here).
    2. Write out all of the possible forms of energy as "initial" and "final".
    3. Determine what forms of energy exist at the beginning. Strike out those that are zero.
    4. Determine what forms of energy are there in the final state. Strike out the ones that are zero.
    5. Solve for whatever you are solving for.

    Be careful! It's tricky to get the gravitational potential energy expressions correct.
     
  11. Jan 1, 2007 #10
    Thanks for the awsome reply, but I'm still confused about a) calculate the total energy of the sytem if both potential energies are zero at x=0.. The equation that I used is 1/2kx^2+mgh=K, I just solved for the left side and I got 101J, which is correct as state in the back of the book. Why do you need to consider the mgh??

    or mabye I wrote it wrong. it should be 1/2kx^2=mgh right? because at first the spring is fully compressed, then when the child reaches his max height, it is gravitational potential energy?

    I still don't get how you would solve for the maximum of something, do you just make something 0?
     
  12. Jan 1, 2007 #11
    They're telling you that you should use x_0 as the zero gravitational potential energy point.

    So the total energy would be all potential when the spring is fully compressed, which is equal to:

    [tex]
    PE_{grav} + PE_{spring}
    [/tex]

    Right?

    So now we have
    [tex]
    PE_{grav} + 125J
    [/tex]

    Gravitational potential energy is negative, because we are below the reference point, so that would be -mgh, which is -24.5J.

    No - you have to think about what becomes zero (from the complete initial-final energy expression) when something is at a maximum. There's a lot of thinking in these energy problems, and once you're done thinking the rest is just arithmetic.

    (EDIT: I sent you a PM about the other thing.)
     
    Last edited: Jan 1, 2007
  13. Jan 1, 2007 #12
    Hmm, well if the energy is conserved and all the energy before is the same as after. U(before)=K(after) if U was 101J before, then wouldn't K(after) be 101J if there was no potential energy left and all of it was K?

    Thanks a lot for all your help btw. You pm too =D
     
  14. Jan 1, 2007 #13
    No, because all of it will never be KE.

    Think about what I just said, and how it can be true. (Hint: Gravitational potential energy.)
     
  15. Jan 1, 2007 #14
    hmm I think I have the equation:

    1/2kx^2+mgh=101J

    x and h are supposedly the same. To get the x of max KE, I just solve for x right?
     
  16. Jan 1, 2007 #15
    No, you would want

    1/2 kx^2 + mgx_0 = KE_max + mgx = 101J

    where x is the distance above the x=0 line. Do what AlephZero said and differentiate.

    To find maximum velocity, use the above equation.
     
  17. Jan 1, 2007 #16

    AlephZero

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    Hmm... I can't think of a way to do part (d) without using calculus - or at least, without using the same general ideas you would learn in a first calculus course. I can't imagine you are supposed to work out those ideas for yourself, without taking the course - and this isn't the right place to write a textbook about calculus... :confused:
     
  18. Jan 1, 2007 #17
    wow really?? yeah I got only the first few a) b) c), but I can't get d) or c) I tried using the equation but I end up with .55 m which is 550mm, the book says -9.8mm =P I don't know what I'm doing wrong... Seriously, you need calculus to solve this??
     
  19. Jan 1, 2007 #18

    Doc Al

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    You don't need calculus to do part d (or e). Hint: Consider the net force on the system as the spring decompresses.
     
  20. Jan 3, 2007 #19
    So for d) determine the balue of x for which the kinetic energy of the system is maximum..

    Do I use PE=KE and just substitute all of the correct numbers in?
     
  21. Jan 4, 2007 #20

    Doc Al

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    Staff: Mentor

    d)

    No. You need to determine the position at which the KE is maximum. Do that by considering the net force on the system as it moves from its lowest point. Hint: What will the net force be when KE is maximum?
     
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