Helium Balloon and spring

In summary: Thank you all for your help!In summary, the problem involves a spring attached to a table and a balloon filled with helium. The spring has a constant of 90.0N/m and the balloon has a density of 0.180kg/m^3 and a volume of 5.00m^3. The goal is to determine the distance L at equilibrium. The equation used is \sum F= F_b - g - kx = 0, where F_b is the buoyancy force, g is the weight of the balloon and helium, and kx is the restoring force of the spring. The correct density of air must be used to get the correct answer.
  • #1
~christina~
Gold Member
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Homework Statement


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.

http://img222.imageshack.us/img222/9371/93040485ij0.th.jpg

Homework 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]

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
Thanks
 
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  • #2
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?)
 
  • #3
Shooting star said:
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?)

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.

Thanks
 
  • #4
~christina~ said:
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.

Thanks

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.
 
  • #5
Shooting star said:
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.


Oh...um

[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?

Thanks
 
  • #6
~christina~ said:
[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]

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.
 
  • #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?
 
  • #8
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:
  • #9
yep forgot to times by 9.8 cheers
 

Related to Helium Balloon and spring

1) What is the relationship between helium balloons and springs?

Helium balloons and springs have a direct relationship in terms of buoyancy and weight. Helium balloons are able to float because helium gas is lighter than the surrounding air. Similarly, springs are able to compress and stretch due to the force of gravity acting on the weight of the object attached to it.

2) How does a helium balloon behave when attached to a spring?

When a helium balloon is attached to a spring, it will float upward due to the buoyancy force of the helium gas. As it rises, it will also stretch the spring due to the weight of the balloon pulling down on it. Once the upward force of the helium is equal to the downward force of the weight, the balloon will reach equilibrium and stop rising.

3) Can a helium balloon and spring be used as a model for understanding gas laws?

Yes, a helium balloon and spring can be used as a simple model for understanding gas laws. The behavior of the balloon and spring can demonstrate how the volume, pressure, and temperature of a gas are related and how changes in one factor can affect the others.

4) Why do helium balloons eventually lose their buoyancy and start to sink?

Helium balloons eventually lose their buoyancy because the gas inside them slowly diffuses through the latex material of the balloon. This process is known as permeability. As the helium escapes, the balloon becomes less buoyant and will eventually sink.

5) Can the behavior of a helium balloon and spring be affected by external factors?

Yes, the behavior of a helium balloon and spring can be affected by external factors such as temperature, pressure, and humidity. Changes in these factors can impact the volume and pressure of the gas inside the balloon, ultimately affecting its buoyancy and the behavior of the attached spring.

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