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Finding the pan's distance from the ceiling

  1. Oct 24, 2008 #1
    1. The problem statement, all variables and given/known data
    A massless pan hangs from a spring that is suspended from the ceiling. When empty, the pan is 53 below the ceiling. If a 100 clay ball is placed gently on the pan, the pan hangs 72 below the ceiling. Suppose the clay ball is dropped from the ceiling onto an empty pan.

    What is the pan's distance from the ceiling when the spring reaches its maximum length?

    2. Relevant equations
    F = ma
    Fspring = -k*(sf-si)
    Kinetic energy = (1/2)m*v^2
    spring potential energy = (1/2)k*s^2
    gravitational potential energy = mgh

    3. The attempt at a solution

    k = mg/sf-si = (.1meters*9.8)/.19meters = 5.158 N/m (J)
    potential energy of clay ball going onto pan: .1*9.8*.53 = .5194 N/m (J)
    spring potential energy: (1/2)(5.158)(s)^2

    taking potential energy = spring potential energy
    solve for delta s = .448 m. though, this doesn't seem to be right because when i add the .53 initial "height" to that, I get the wrong answer.
  2. jcsd
  3. Oct 24, 2008 #2


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    Hi jheld,

    This is an error. Since you have chosen the initial height as 53cm, the h=0 level is 53cm below the ceiling. So when the spring reaches its maximum length, the gravitational potential energy of the ball is not zero.
  4. Oct 25, 2008 #3
    So the potential energy of the clay ball going onto the pan should be:
    (.1)(9.8)(0)= 0J...correct?

    Where would I go from there?

    Sorry, it is just taking me awhile to understand this problem.
  5. Oct 25, 2008 #4


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    The general approach in your first post was correct. You calculated the energy at two points: initial energy when the ball was at the ceiling, and final energy when the ball was at the lowest point. You then set those equal to each other.

    I believe your mistake was that you assumed that gravitational potential energy was zero at the final point.

    Gravitational potential energy here is mgy. But where is the origin of your y axis? It's up to you where to set it, and you have decided to make y=0 to be where the end of the spring was initially.

    This means the ball begins at y=53cm (which you have in your first post). But at the lowest point, it is below the y=0 level, so gravitational potential energy is not zero, and you need two terms for the final energy. What would it be?
    Last edited: Oct 25, 2008
  6. Oct 25, 2008 #5
    I need the gravitational potential energy and the elastic (spring) potential energy.
    So, I take the beginning distance + the distance the spring stretches = the total distance from the ceiling.

    If I set mgh = (1/2)k*s^2 and solve for s, i get:
    x = sqrt((2(.1)(9.8)(.53))/5.158) = .448 m

    and the potential energy for the spring at the absolute beginning is:
    (1/2)k*.53^2 = .72 m

    therefore adding the distance from the spring potential energy at the beginning with the spring distance once the ball is dropped onto it, i get:
    .72 m + .448 m ~~ 1.21 m.
    I think that's right.
  7. Oct 25, 2008 #6


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    I think that answer is about right, but that is not the correct way to approach it.

    You are right that energy is conserved, so this is correct:

    E_i = E_f

    You have also seen that there is no kinetic energy at the initial and final points; the only energy terms you have to worry about is gravitational potential and spring potential. Putting these in:

    PE_i + SE_ i = PE_f + SE_f
    where PE is gravity potential energy and SE is for the spring. We have formulas for these that we can plug into the above formula:

    m g y_i + \frac{1}{2} k x_i^2 = m g y_f + \frac{1}{2} kx_f^2

    This is what you essentially had in your first post, but you were making a few bad assumptions. Comparing the above equation to the forms that you had in your post:

    It was fine to say that yi=+0.53m, as you had in your first post.

    What is xi? In other words, when the ball was at the ceiling, how much was the spring compressed or stretched from its natural length?

    You set xf, the final stretch of the spring, to the unknown variable s. That's fine.

    Now, what is yf? In other words, if the ceiling is at y=+0.53m, what would you set yf equal to? Hint: you're not setting it to a numerical value here.

    Once you get that, you should have three terms in your equation, and you can solve for s.
  8. Nov 15, 2008 #7
    Well, now that I've had time to redo this problem, I have stumbled upon the answer. At first it made sense, but now, I don't think I began it correctly, and therefore, it works, but I want to go about it a different way.

    I made kf + Ugf + Usf = Ki + Ugi + Usi

    Ki = 0, and Kf = 0. Usi = (1/2)(5.158 (spring constant))(.19). Usf = (1/2)k(ds)^2. Ugi = mg(.72), Ugf = 0.

    ds = sqrt((Usi + Ugi)2/k) = .6808.

    so .6808 + .53 = 1.21. this is the correct answer.
    but, Ugi should be : mg(.53).
    Usi = 0
    Usf = (1/2)kds^2
    and solve for ds in the conservation of energy equation

    but, with the ds i get, it isn't the correct length.
    i got the right answer, but it wasn't the right way of doing it (as much as I want it to be)
    any help would be great, thanks.
  9. Nov 15, 2008 #8


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    Hi jheld,

    I don't believe your terms are correct (as you point out below), but even then I don't think you get this answer. I think you neglected to square the x value for Usi, for example.

    It's because you are missing the final gravitational potential; it's not zero.

    Look at Ugi: you have mg(0.53). That means the initial height of the putty is at 53cm.

    Now the putty drops, and the spring stretches by a distance s. What is the final height of the putty? It's not zero, because the putty fell by more than 53 cm.

    In other words, the spot where the pan is initially is at y=0. The initial height of the putty is yi=+53cm, because the putty is initially 53cm above that spot. When the spring stretches by a distance s, how far below that y=0 spot is the putty?

    Since it is below the y=0 level, it will be a negative height. You can then solve for s and you should get 0.68; do you get that answer?
  10. Nov 15, 2008 #9
    Haha, now I truly get it.
    Well, if you noticed above, I did get the right answer and I got .68 for the ds, which was correct.

    my problem was that I already assumed something without understanding where i was getting it.

    i had set the initial Ug's height to .72, which worked, of course, but in the concept of physics, it isn't technically correct. so, of course in the final Ug, the height would be (as you have explained very well) -.19 m. and when you derive for ds, it becomes a positive .19 m, and adds to the .53 already there (being that both are Ug and have mg as the same). thanks for the help, now I get it. :)
  11. Nov 16, 2008 #10


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    Okay, I was just saying that I think you were using:

    Usi = (1/2)(5.158 (spring constant))(.19)

    to get the 0.677, when it should be something more like

    Usi = (1/2)(5.158 (spring constant))(.19)^2

    (But of course the 0.19 is wrong in the first place as you pointed out. And maybe I'm just misunderstanding what you did. As long as you understand it now, then this other stuff is not important.)

    Let me just mention one minor change I would personally make when doing energy problems that I think makes things a bit easier. I would have said that the initial height of the putty is zero (that is, the ceiling is at y=0).

    Then the final height would be yf= - (s + 0.53), giving:

    Ei = E f

    0 = \frac{1}{2} k s^2 + m g [ - (s+0.53) ]

    Of course there is no real difference at all, but I like having only to keep track of one height in the problem.
  12. Nov 16, 2008 #11
    Well, I've back to this problem, just making sure that I can do it on command, you know?
    But now it seems that I can't do it your way, no my way either.

    If Ei = Ef (which it does),
    0 = (1/2)ks^2 + mg[-(s+.53)]
    s = .2 m
    s should equal .68 m
    I understand setting Ei when y = 0, so therefore Ei equals 0.
    Yet that way of doing it doesn't seem to work.
  13. Nov 16, 2008 #12


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    From that equation I get that s=0.68m.

    Also, if I plug back s=0.2m into the equation:

    (1/2)k s^2 + mg[-(s+.53)]= (1/2) (5.158) (0.2)^2 + (0.1)(9.8)[ -(0.2+0.53)]= 0.10316 - 0.7154

    that does not give zero, so I don't think s=0.2m is a solution.

    If you are still getting that answer, can you show the steps that gave you 0.2m?
  14. Nov 16, 2008 #13
    I have:
    -mg(-(s + .53)) = (1/2)ks^2
    divide s both sides....
    -mg(-.53) = (1/2)ks
    s = mg(.53)(2)/k
    this levies 0.2 m.

    I can't get .68 from that.
    I guess you could say that my earlier version (the one where I assumed things without realizing where they were coming from) was the perfect version, because it gave the right answer.
  15. Nov 16, 2008 #14


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    This line is incorrect. If you divide by s, you'll get:

    -mg \frac{-(s+0.53)}{s} = \frac{1}{2} ks

    which is not helpful. Instead, write out the left side as two terms and then I think you have to use the quadratic equation formula for this one.
  16. Nov 16, 2008 #15
    Haha, right you are.

    I didn't even see to use the quadratic formula.

    I still don't quite understand why what I was trying to do doesn't work out algebraically.
    though I suppose if i had foiled out the mg(-(s+.53)) from the start, then I would have seen it. so, i suppose I just answered my own question right there.

    And, this (as you have eluded) is the correct way of doing this problem (at least from this way).
    My original way worked as well, but I don't think I was doing it correctly (though it worked).
    But this way I know I can do on command.
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