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Homework Help: Energy and conservative forces

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data
    I've attached the problems in a PDF file.

    2. Relevant equations

    3. The attempt at a solution
    I've honestly made an attempt to all of these problems, and I've simply demonstrated to myself that I have no idea what I'm doing. I already posted part a of the first problem in another thread, so don't worry about that one...

    For 1 b I began by calculating the distances for each route, but I'm unsure how to calculate work if I am not given a force. This is a little peculiar, isn't it?

    For problem 2 I am a tad puzzled by the force of the spring. Since the spring is not strained, I'm going to have to guess that the force of the spring is completely irrelevant. This would be than gravity is the only force acting on the system.

    I then treated gravity as an internal forces, including it as potential energy. But I have too many variables to possible solve force velocity. I have both change in distance as well as velocity, and there does not appear to be another equation that I can use to eliminate a variable.

    What should I set the change in energy side of the equation to. I am guessing 0... But I can't understand why work would be zero. It makes sense that the change in energy is 0, because KE increases at the same rate that PE decreases, but there is only one force, and it does act over some distance on the mass. Wouldn't this imply a net work?

    Attached Files:

    Last edited: Oct 25, 2008
  2. jcsd
  3. Oct 25, 2008 #2


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    Hi zoner7,

    I don't see an attachment. I don't think a powerpoint attachment is allowed (it says valid file extensions for attachments are: bmp doc gif jpe jpeg jpg m ms mw mws nb pdf png psd txt xls zip.) Can you upload something else, or type it in?
  4. Oct 25, 2008 #3
    I've edited the original post significantly
  5. Oct 25, 2008 #4


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    Okay, but it may take a while for the new attachment to be approved.
  6. Oct 26, 2008 #5
    What's that mean exactly?...

    does that mean hours or days? or by tomorrow?
  7. Oct 26, 2008 #6


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    I think you are needing to show that the result is the same for any constant force. Pick a general (constant) force, and then caculate the work done along each path.

    I don't think this is right. The spring is only unstressed at the beginning, before the object is released. After that, the spring is stretching and putting a force on the object, so you'l have to account for spring potential energy.
  8. Oct 26, 2008 #7
    ahhh, I see what you mean. I got the impression that the spring was in equilibrium while the block was hanging, implying that the spring would contract as the block was released; thus, the force of the spring would not influence the block...

    Your way makes a lot more sense
  9. Oct 26, 2008 #8
    so when I solve for the force of the spring I get mg. That means the work is mgx. Since x is equal mg/-k, the work of the spring is mg^2. Meanwhile, the work due to gravity is mgh.
    So if the work of the spring and the work of gravity are treated as internal forces,
    w = KEf - (FsX - FgH)
    w = .5mv^2 - mg^2/-k +mgh
    w = .5mv^2 + mg^2/k + mgh

    For some reason I was thinking that since the KE is a direct product of the potential energy, the KE must be equal to PE. This would imply that work is 0.

    -> .5mv^2 = -mg^2/k - mgh
    The masses cancel
    -> .5v^2 = -mg^2/k - mgh
    so V = sqrt(-2(mg^2/k + mgh)) ... now this is all messed up seeing as I cant divided by a negative... and even more so, there are two variables. So my answer is wrong. Any thoughts? actually, on second thought, seeing as h is downward, if the force of gravity is greater than the force of the spring, I'm in luck.
  10. Oct 26, 2008 #9


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    That sounds like what you want to do here.

    No, the spring force is not constant, and so that is not the work done. If you need to find the work done by the spring (and you don't need to here) there is a formula for it.

    That's right, and that will let you find x.

    There are some problems with the rest of the equations you have. A starting point for the energy approach is the equation:

    W_{\rm nc} = E_f - E_i

    So what is the total initial energy? And what is the total final energy (for the point you are studying in part a)? You have probably only studied three types of energy (kinetic, gravitational potential, and spring potential) so you only have to consider three terms for each point.

    Finally is there any non-conservative work being done here?

    Using the formulas for kinetic, gravitational potential, and spring potential energies you can write the above equation to apply to part a. What do you get?
  11. Oct 26, 2008 #10
    well, there aren't any non-conservative forces, because the only two forces are the force of the spring and gravity, and they are both conservative forces.

    I understand what you mean by the spring not exerting a constant force; clearly it varies based on how far it is stretched from its equilibrium position. So I would use 1/2kx^2 from xf to xi to find work, but you said that this isn't necessary. You suggested that I alternatively find the potential energy of the spring... but dosen't the work-energy theorem assert that work is simply a change in energy? So, what exactly is the difference between these two ideas?
  12. Oct 26, 2008 #11


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    That's right, so Wnc=0, and the equation becomes:

    0=E_f - E_i


    E_i = E_f

    With the three types of energy you had in your previous post, this is:


    where PE is gravitational potential energy, SE is spring potential energy, and some of these terms are zero based on which two points you pick. You have formulas for these three types of energy, so you can plug those in and apply this equation to the two points that are important in part a.

    The point about potential energy is that when you are writing down an equation, you either have terms for the spring potential energy or the work done by the spring, but not both. I was referring to the fact that your eventual equations (and words) were referring to potential energies, so you would not have to deal with the work done by the conservative forces--their effects are handled by the potential energy terms.
  13. Oct 26, 2008 #12
    So I set the equation up, solved and ended up with a square root of a negative again. My units aren't canceling properly either...

    I set the equation up like the following:

    KEi = 0 PEi = -mg SEi = 0
    KEf = 1/2mg^2 PEf = 0 SEf = mg

    so, -mg = 1/2mv^2 + mg.

    I think there is something wrong with the signs, but there must be something more, because if I take the square root of m/2^2, I won't end up with m/s (the units of velocity).

    I have no idea why I'm struggling so much with this... first issue I've had in physics all year.
  14. Oct 26, 2008 #13


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    This term is correct, since the objects starts out with zero velocity.

    This is not correct. (-mg) is the gravitational force. The gravitational potential energy has the formula mgy ( or perhaps mgh), so you need to know the initial height.

    This is right, since at the beginning the spring is not stretched or compressed, it has no stored energy.

    This should be [itex] KE_f = \frac{1}{2} mv_f^2[/itex]; where [itex]v_f[/itex] is the speed you are looking for.

    This is okay if you are saying that the height of the object at this point is at y=0. The decision of what level is at y=0 is up to you, so this is okay. Once you have set either the initial or final height equal to zero, then that would determine the other one.

    This is not correct. The formula for spring potential energy here is:

    SE_f = \frac{1}{2} kx_f^2

    and the important thing is that you have already determined the numerical value of x (when you set the spring and gravitational forces equal in post #8).

    Using these six terms should give you the total energy equation. The last thing is to determine what to put in for hi (in the initial gravitational potential energy term) and you can solve for the speed.
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