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Helium Balloon and spring

  1. Feb 6, 2008 #1


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    1. The problem statement, all variables and given/known data
    A light spring of constant k= 90.0N/m is attatched to a table vertically.
    A 2.00g balloon is filled with helium (d= 0.180kg/m^3) to a volume of 5.00m^3 and is then connected to the spring, causing the spring to stretch as shown. Determine the distance L when the balloon is in equillibrium.


    2. Relevant equations

    I think...not sure ...

    [tex]K_i + U_i = K_f + U_f [/tex]
    [tex] K_i + (U_g + Ui_s) = K_f + (U_g + Ui_s) [/tex]

    B= \rho Vg
    (B= buoyant force of air)

    I know that if it is in equillibrium it would be:

    [tex] \sum Fy= 0 [/tex]

    3. The attempt at a solution

    I'm not sure how to put all this that I compiled together...good grief

    first of all I think that
    the equation would sort of look like this:

    [tex]K_i + U_i = K_f + U_f [/tex]
    [tex] K_i + (U_gi + Ui_s) = K_f + (U_g + Ui_s)[/tex]

    BUT AM I Supposed to use this equation? or...F= kx
    and how do I incorperate the buoyant force into that?
    [I do know that the weight of the balloon will be a force downward and the spring will be a downward pulling force and the buoyant force is facing up but other than that....well I tried.]

    And now I'm officially confused and lost here...:cry:
    (I need help setting up the equation)

    help please
    Last edited: Feb 6, 2008
  2. jcsd
  3. Feb 6, 2008 #2

    Shooting Star

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    What are the vertical forces acting upward on the balloon? What are the vertical forces acting downward on the balloon? The sum must be zero for equilibrium.

    For example, you know that one vert force acting downward is the weight of the balloon. Try to think of all the forces.

    (BTW, what are these U terms -- Ui etc? Potential energy?)
  4. Feb 6, 2008 #3


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    well. yes the pot e = Ui (initial potential E [it's a general equation])

    forces acting upward on balloon => Buoyancy force of air

    Forces acting downward=> weight of balloon & the force of the spring (not sure how to represent that except by F= -kx)
    so would it be ...

    [tex] \Sum F= F_b - Mg - 1/2 kx^2 = 0 [/tex]

    somehow this equation looks funny to me.. I'm not sure about that.

  5. Feb 6, 2008 #4

    Shooting Star

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    Restoring force of spring = 1/2 kx^2???? Write the correct expression and eqn will become correct.

    Note that Mg would be weight of helium + the weight of the material of the balloon itself. You'll need the density of air.
  6. Feb 6, 2008 #5


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    [tex] \Sum F= F_b - Mg - F_s = 0 [/tex]

    [tex] \Sum F= F_b - g - kx = 0 [/tex]

    [tex] \Sum F= (1,000kg/m^3)(9.8m/s^2)(5.00m^3) - (0.9kg + 0.002kg)(9.8m/s)- (90.0N/m)(x) = 0 [/tex]

    I think I'd then solve for x and that should be it right?

  7. Feb 6, 2008 #6

    Shooting Star

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    Where did the M go from the 1st eqn? Anyay, you've included that in the last eqn.

    What have you put for density of air? Put the correct density of air and that'll give you the answer.
  8. Apr 17, 2009 #7
    hi guys i'm new to the forum so hello, I'm a bit stuck on the same question. I didn't use the potential energy method above. But went with Forces.

    The only force up is the boyant force which i think is the the volume of the balloon X air density...

    5 x 1.29 = 6.45N

    this must be equal and opposite to the combined forces of the spring kx, and the weight of the balloon and the helium within it.

    with this i get 6.45=0.0196 + 0.9 + 90x
    solving for x i get = 0.06144, which is way off the answer is .604m

    where am i going wrong?
  9. Apr 17, 2009 #8


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    5 x 1.29 = 6.45N
    This is mass of air displaced by the balloon in kg, not N.
    with this i get 6.45=0.0196 + 0.9 + 90x
    Check this step also.
    Last edited: Apr 17, 2009
  10. Apr 18, 2009 #9
    yep forgot to times by 9.8 cheers
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