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Homework Help: Work, Stairstep Ball

  1. Feb 15, 2004 #1
    Okay, I have a ball balanced on a pole. The P.E. is given as 30 J. At the bottom of the pole, if it were to fall off, what would the P.E. be? I say zero. Okay, but now the ball is setting on top of a series of steps. The topmost step is the same height as the pole was, so I am guessing the ball has 30 J P.E. again. But, if it goes to the next step (three steps total, it is setting on the top one), I mean falls down to that step and sticks, is the P.E. one-third as much since it is one-third of the way down, or two-thirds as much, seeing as it is still 2/3 of the way up? It is going down, yes, but it's still 2/3 of the way off the ground. Does the 5dt^2 thing have anything to do the the ball now?
  2. jcsd
  3. Feb 15, 2004 #2
    I would say that the ball has a PE of two thirds it's original PE once it has lost one third of it's height. This is assuming that we are referencing it's PE with respect to the floor (bottom of the stairs.
    Remember that PE=mgh where mg is the weight (mass*gravity) and h is the height. Since it is only two third the original height, it should have only two thirds the original PE.

    As for the "5dt^2 thing," I have no idea what you are talking about. Can you elaborate?
  4. Feb 15, 2004 #3
    doesn't it depend on the place you take to be 'zero potential energy'? if you set it to the center of the earth, for instance, it has a long ways to go. further still if you make it the center of the galaxy =P . but if you're assuming that the bottom of the stairs is ZPE then when it is a third of the way down, it would have two thirds of the energy remaining as compared to when it started. since the equation i'm guessing youd use is Ep = mgh, m and g are constant [cuz i doubt the changes from e = mc^2 and g = Gm/r^2 would make a difference :) ] so h is the only factor that really makes a difference here. h' is two thirds h, [or, it still has two thirds of the distance to fall] the energy at that point is also 2/3. did that help any? ^^

    and i don't know about that 5dt^2 equation you're referring to, sorry :(
  5. Feb 15, 2004 #4


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    that seems quite reasonable to me
    it has 2/3 as much altitude so it must
    have 2/3 as much PE

    anyone see a flaw in that?

    Personally I dont see how the 5dt^2 thing is relevant since yr not being
    asked about the time it takes to fall
    only the change in PE

    edit: I think what holly means by the 5dt^2 thing is that
    the distance something falls in time dt is
    about 5 meters per second2 times the square of the time it has been falling----(dt)2

    but the problem does not involve calculating the falling time, dt, or calculating anything else from the falling time. so that formula plays not role.
    Last edited: Feb 15, 2004
  6. Feb 15, 2004 #5


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    paul, vodka,

    what do you bet she will be back with another question
    I think it is better than even odds.
  7. Feb 15, 2004 #6
    i think i did more difficult stuff in grade 10 and should move my question in the k-12 forum into this one so it'll get attention [honors physics class, cover first/2nd year university stuff anyways] :wink:
  8. Feb 15, 2004 #7
    Youse are getting mean.

    Thank you for the help given.:frown: I won't ask more questions.
  9. Feb 15, 2004 #8


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    I was hoping you would!
    didnt realize we were mean sounding
    Last edited: Feb 15, 2004
  10. Feb 15, 2004 #9


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    for goodness sake holly, nobody meant to sound mean

    we all like getting physics problems questions

    if you have more it would be shame not to ask them!

    BTW where did you find the remake of the quote from
    a Famous Physicist, it is a very funny take-off of what
    he said (in all his false modesty) about seeing more
    than others because he stood on giants shoulders.
    Last edited: Feb 15, 2004
  11. Feb 15, 2004 #10
    indeed, no offence intended sorry [b(]

    and even if we did, by listening to us you'd just be hurting yourself since you wouldn't be using such a valuable asset. i think its about as much fun to give help as it is to get it and i'll be damned if i don't need help right now...
  12. Feb 15, 2004 #11
    hmph...I wouldn't blame anyone for being tired of my stupid questions. I am so sick of not understanding, embarrassed, too.

    I made up the take-off on Newton's remark. Ha ha, it's how I feel.

    Thanks for answering the questions.
  13. Feb 15, 2004 #12
    Glad to see you are back.

    Don't be embarrassed about not understanding. If everyone understood, this forum would not exist. We all have something we get stuck on. I am glad to have found this forum, and I have had a few questions answered so far.

    So, is this question you originally posted answered for you? If not, what are you still not clear about? I just completed my own physics HW (for now anyway), and I will check back here later (after the kids are asleep).[zz)]
    Catch you later.
  14. Feb 15, 2004 #13
    Thx, I think I understand about the ball's P.E. now. I guess gravity didn't have anything to do with the problem, not sure why I thought it did. Oh, well.
  15. Feb 15, 2004 #14


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    holly the problem has to do with gravity
    the potential energy you are dealing with is called gravitational potential energy

    gravitational PE = mgh = mass x 10 m/s^2 x height

    it is proportional to the height

    there is more than one formula involving gravity

    distance fallen = (1/2)g (time)^2

    formula is one that involves falling-time
    but it is not the only one
  16. Feb 15, 2004 #15


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    in every introductory physics course
    there is a certain goal of what to teach

    out of several possible formulas about gravity, the teacher and the t textbook writers may have chosen just two gravity-formulas for you to learn how to use

    (or they may have three, with one about a cannonball flying in a parabolic arc, or they may have more---there are several they can choose)

    let's you and me suppose the simplest, that they just want you to learn a mere TWO gravity formulas

    those two would be the same the world over but the symbols vary,
    the two formulas (partly in my crude personal symbols) would be




    and this little gee symbol stands for g = 10 meter per second2

    so naturally one half of gee = (g/2) = 5 meter per second2

    If the problem involves work done to rais something, measured in joules, then you use the second formula.

    If the problem involves the time it takes (measured in seconds) for something to fall a certain distance (measured in meters) then you use the first formula.

    You must make a list of all the formulas you are expected to know how to use. For a one-semester class this will normally take one side of one sheet of typing paper.
    (However if it is CalTech or MIT it may take both sides)

    You must memorize, together with the formula, the circumstances in which you use it.

    You must say this over to your self at least once a week until the final exam is over.

    You can use this PF message board to develop your list, with us to help, if you wish.

    the list that a freshman in college physics accumulates in one semester is apt to be much the same wherever you go even like
    Romania or Bangladesh but certainly in Texas

    we have all been thru this, and it ultimately turns out to be fun
    Last edited: Feb 15, 2004
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