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Homework Help: Conservation of energy, potential energy function

  1. Nov 12, 2014 #1
    1. The problem statement, all variables and given/known data
    The potential energy for a mss m = 1.0 kg moving in one dimension is given by U(x) = (2.5J) sin πx.
    The mass starts at x = 0 with an initial velocity v = 0.71m/s. (The plus sign means the motion is
    in the positive x-direction.) Describe the subsequent motion of the mass. Suppose the initial velocity
    were 3.0 m/s. What would the subsequent motion look like?

    2. Relevant equations
    1/2mv2, mgh

    3. The attempt at a solution
    Looking at the initial conditions when x=0, U(x)=0.

    Considering the initial velocity as 0.71m/s, substituting this and the mass into Ek=1/2mv2 we obtain
    Ek=0.25205 J

    We know that sin maximizes at 1/2 pi, thus, at x=0.5 our U(x) function will be maximized.
    U(0.5)=(2.5J)sin π(0.5) = 2.5 J

    Since Ek=0.25205 J

    So, Ek<U(0.5) and we conclude that the object will not reach the top of the peak defined by our Potential energy function.

    Moving ahead early (I'll explain why in a moment)
    for part b, we see that Ek=1/2(1kg)(3m/s)2
    so Ek=4.5 J and therefore now Ek>U(0.5) so it will pass over the crest.

    My question for this problem is as follows:

    In part a), will the object fall into negative values of x? and therefore also negative potential energy?
    In part b) will the object continue over the crest and again, proceed into negative potential energy?

    I recall my professor explaining that potential energy is set by the person, so negative values are able to be obtained. In this situation, I see part a oscillating back and forth, while in part b it continues on with its motion indefinitely.

    What is the best way to "Describe the subsequent motion". This seems like a loose statement with a fair amount of points attached to it vulnerable to misinterpretation on my part.

    Thank you for any insight to this problem.
  2. jcsd
  3. Nov 12, 2014 #2

    Simon Bridge

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    ... it may help you to think about it in terms of forces to start with.
    The force on the mass is the gradient of the potential energy function.
    At x=0, which direction does the force act?
    Initially it travels in the +x direction - it moved into an area of increasing potential energy - so what happens to the speed?
    When the potential energy is exactly equal to the initial kinetic energy - what is the speed?
    At that position, which way does the resultant force act?

    "describe subsequent motion" is actually a pretty precise request - you work out how the object moves for t>0 and then you describe it. Does it speed up, get slower,stop? Are their any critical points where something special happens?

    i.e. if you were to describe the subsequent motion of a ball fred from a cannon, you'd say that it follows a parabolic arc, reaching a maximum height, and coming back to the ground; right? If you had details about the initial velocity, then you'd be able to say what the maximum height was and where it lands.
  4. Nov 12, 2014 #3
    I have considered this in terms of forces. I am unsure how your reply has much to do with my actual question. Your answers are: at x=0 the force acts in the negative x direction since the slope of the tangent to the curve of the function U(x) is positive. This remains the same answer for your later asking of it. Around x=0.016 the kinetic energy and potential energy will be equal. The speed will be approximately 0.5m/s

    So once it doesnt make it over the crest, and begins to return to x=0.

    I'm not sure how this relates to my question about it achieving negative energies. (Honestly, I have no clue what this has to do with any of that). I may be missing something important. Thanks for trying to help though.
  5. Nov 12, 2014 #4
    Also, I am not trying to be rude with what I wrote above, I honestly do not make the connection between my question and your response :/ Looking back it seems rude to me, but I never meant it like that. I am exhausted and pushed to my limit and have no clue what to do. The more I struggle with this, the more lost I become. I feel like I've reached a point where I have no clue whats even happening, and thats a really crappy feeling.
    Last edited: Nov 12, 2014
  6. Nov 12, 2014 #5

    Simon Bridge

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    You actual question was - does the object travel into negative values of x?
    Please bear with me: the reply are designed to help you figure out the solution for yourself, in accord with PF rules and guidelines.
    It also helps me figure out where you are up to. For eg.

    ... careful: if the final potential energy is equal to the initial kinetic energy at some point x=0.016m (I didn't check if this was correct.)
    And energy is conserved ... then what is it's speed at x=0.016m? - hint: nowhere near 0.5m/s.

    Well done - so what is it's velocity once it has returned to position x=0m?
    Which direction does the force point at x=0m?

    Did I misread?
    We are getting there.

    Note: the gravitational potential energy function about the Earth is U=-GMm/r ... notice how the potential energy is always negative?
    Your question about energy translates to: can masses move into the Earth's gravity: i.e. a region of negative potential energy?
    But I'm hoping to get you thinking about what the periodic potential energy function is telling you.

    No worries - you are encouraged to be direct.
  7. Nov 12, 2014 #6
    Good, I just didn't want to come off as an arse. And spoon-fed answers do NOT help me on my exam!

    I have poked at this some more, and I think I have come to a good description of this motion. (Still need to compute values...)

    Since we can define U=0 at any place on the vertical plane for convenience, then we are able to agree that the U value CAN in fact have a negative value, depending on this predefined location. In this situation, we have U=0 cutting the sine curve clean in half (I know theres a technical term for this but it is eluding me). So with our initial velocity being 00.71m/s, as stated previously the object will rise up to a point, which cannot clear the crest and turn around. This will then move in the negative x direction, into the trough, where it will then again rise until it reaches the same condition as it initially met to reach a zero velocity and change directions. The object then turns around and begins to head back in the positive x direction. It will again return to the same location in the positive x range, to repeat this motion over and over and over.

    For part b, again, since we mentioned it would clearly hop over that saddle point, it would then decrease in velocity as it climbed up, and as it went over begin to gain velocity until it reached the bottom of the trough. Where it would then have adequate kinetic energy to again hop over the next peak, to repeat this motion ... forever...

    I plan to compute all of these specific values, but I think now that this is the way to look at the system. It is nonsense to me to believe that a function like this would "hit a wall". and suddenly stop. That information would have been provided, right? I am though concerned about "describe the subsequent motion". It just feels so loose, as if they truly wanted a mathematical representation of it.

    However, I am not sure how to describe an entire motion any better than defining a potential energy function for it, such as was already provided... I'm certain that this would not generate me a respectable grade to bring home, so I have to lean on what you're saying that it wants a written description with key points indicated on a graph, with computed values for these key points. I found in my notes a nice quadratic that had a U(x) function, and how we drew a horizontal line to identify the key points of intersection with U(x) to visually show the permitted values for the given motion.

    If this all makes sense... HAH!

    Is this a good way to treat the system?

    Edit: Also, When I computed 2.5sin pi (.016) the result from my calculator is 0.1256. I am in radians too. This is just reflecting from before, however if my initial kinetic energy was 0.25J i expected for when U=K it would be then half of 0.25J, so 0.125. That was my train of thought.
  8. Nov 12, 2014 #7

    Simon Bridge

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    The problem itself has defined where U=0 can be.

    ... only if you define the +x direction as "upwards".
    If you want to call "increasing potential energy" as "upwards" - by analogy with gravity - then there are places where the +x direction is "down".
    But there is no reason to make any such interpretation.

    I'm concerned you are thinking a tad too literally about the potential energy function being like hills and valleys.
    The "up and down" of the U vs x graph is potential energy, not space. The object moves back and forth along the x axis.

    However your description is good - in (a) the object has periodic motion. In (b) it slows and speeds up - but keeps going in the +x direction.
  9. Nov 12, 2014 #8
    Right, thats exactly what I'm trying to say, my words aren't as clear as I'd like them to be. I'm not thinking of it as a literal object rolling around in a bucket against gravity, but that the potential energy being conserved, will increase and decrease throughout the motion showing that it is conserved. If energy were not conserved, this motion would NOT continue forever and diminish over time. Since it is conserved however, the potential energy will continue to oscillate up and down (maybe oscillate is not the proper term) forever, governed by the amount of total energy in the system (kinetic convers to potential, then back to kinetic, rinse, repeat.).
  10. Nov 12, 2014 #9

    Simon Bridge

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    ... if something changes, then it is not being conserved.
  11. Nov 12, 2014 #10
    Correct, since energy is conserved, the motion will continue back and forth across x. (part 1) and will continue across x in the positive direction (part 2). If energy was not conserved, it would at some point in time stop, depending on the loss of energy.
  12. Nov 13, 2014 #11

    Simon Bridge

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    I think you have a handle on things - you just have to be a bit careful about how you say it.
  13. Nov 13, 2014 #12
    Indeed, I do! The good news is that I scored 5/5 on that question. Thanks for all of your help with it. The good news is that my Thursday lecture next week is canceled so I get an extra period of time to work on my problem set (5 extra days!). The bad news is that this caused the prof to only be able to give us 9 problem sets this semester instead of his intended 10, so the problems are quite a bit trickier. I'm sure with the extra time it will all work out.

    Thanks again for your help Simon, its greatly appreciated. :)
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