Help with IB Physics IA about bungee jumping....

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  • #1
Hi, guys, this is my first time on PF!

My IA is about bungee jumping where I increase the mass and look at the efficiency of energy at the lowest point of the jump.

My question is: How do I find the spring constant k for an elastic cord? I need it so I can plug it into the elastic potential energy formula.

My thoughts:

I was thinking to do this by mgh=1/2kx^2
but by doing this I get a different k value each time as the mass and therefore the extension, x is changing but isn't k always a constant and cannot change unless the cord deforms... so is this wrong?

I was also wondering if I can find the the k constant using Hooke's law (F=kx). I was thinking to graph force (mg) on the Y-axis and the extension on the X-axis and the slope of that would be the k constant. Will this be a valid approach?

Please help me!
Thanks a looooot in advance!
 

Answers and Replies

  • #2
Chandra Prayaga
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Your experimental approach is quite correct.
I am not sure what you mean by "efficiency of energy at the lowest point of the jump." Could you explain what exactly you want to do, once you get the spring constant?
 
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  • #3
efficiency of energy is = (useful energy out/total energy in) x 100

In my case, the energy out will be in the form of elastic potential energy and the total energy in will be in the form of gravitational potential energy.

so the efficiency of energy will be: 1/2kx^2/mgh (I need k for this)

By the lowest point, I meant the height the person will reach after the maximum extension of the bungee cord. (like the point of the jump before rebound)
...Do you understand? Any suggestions on how can I word it better?

So I can graph Force and extension, x to find the spring constant right?

Thank you so much for helping me out!

I just wasn't sure if Hooke's law can be applied to elastic cords..
 
  • #4
CWatters
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How are you measuring X? Normally the bungee doesn't start stretching until the mass has fallen some distance already.
 
  • #5
I'm measuring X using a motion sensor. Yes, I modeled a bungee jump using a teddy bear.

My question though is: Can I find the the k constant using Hooke's law (F=kx)? (by finding the slope of the graph) I just wasn't sure if Hooke's law can be applied to elastic cords.
 
  • #6
ZapperZ
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I'm measuring X using a motion sensor. Yes, I modeled a bungee jump using a teddy bear.

My question though is: Can I find the the k constant using Hooke's law (F=kx)? (by finding the slope of the graph) I just wasn't sure if Hooke's law can be applied to elastic cords.

And this is where you have to test your assumption FIRST before applying something that may or may not work.

Why not test out if your elastic cord does follow Hooke's law before you use it as your description for that cord? You might even find a bonus when you do this by actually getting the spring constant falling onto your lap! SURPRISE!

Zz.
 
  • #7
So you mean not all elastic cords follow Hooke's law?
 
  • #8
ZapperZ
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So you mean not all elastic cords follow Hooke's law?

Those common rubber bands often don't. So why would ALL elastic cords be assumed to follow such a law?

Zz.
 
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  • #9
CWatters
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The spring constant of a rubber band increases dramatically just before it fails. Try stretching one with your hands.
 
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  • #10
Thank you for the replies.

With some research, I have learned that the larger the displacement of the rubber band the higher the spring constant.

I'm changing the mass of my bungee jumper. So the extension of the cord increases as the mass increases which means the spring constant should increase with it as well.

SO my question: The 'k' (Hooke's law constant) does not have to be a constant value? whaaaat

Thank you!
 
  • #11
ZapperZ
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Thank you for the replies.

With some research, I have learned that the larger the displacement of the rubber band the higher the spring constant.

I'm changing the mass of my bungee jumper. So the extension of the cord increases as the mass increases which means the spring constant should increase with it as well.

SO my question: The 'k' (Hooke's law constant) does not have to be a constant value? whaaaat

Thank you!

It is no longer "Hooke's law"!

The elastic band follows a different set of rules now.

Zz.
 
  • #12
kuruman
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You should also know that elastic bands often display hysteresis effects. That means that the length by which they stretch under a given load depends on what was done to them previously. Try this, hang masses in increasing order measuring the displacement each time. Once you reach the highest mass, reverse the order to decreasing mass. You will find that the displacement for a given mass is not the same when you go up in mass increments as when you go down. You might also discover that it takes some time before the band fully stretches under a fixed load so describing the spring "constant" as k(x) may be too simplistic.
 
  • #13
CWatters
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From my model aircraft flying days I can confirm that the performance of a twisted rubber motor can vary with use. Bit like "running in" a car engine.
 
  • #14
kuruman
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From my model aircraft flying days I can confirm that the performance of a twisted rubber motor can vary with use. Bit like "running in" a car engine.
Not to mention becoming progressively brittle with age.
 
  • #15
hmm okay

If I want to use the formula of elastic potential energy for my bungee cord: 1/2kx^2

How do I find k?
 
  • #16
ZapperZ
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hmm okay

If I want to use the formula of elastic potential energy for my bungee cord: 1/2kx^2

How do I find k?

If your cord does not obey Hooke’s law, that expression for PE isn’t valid either. The value that you find will be meaningless.

Zz.
 
  • #17
I was thinking to graph force (mg) on the Y-axis and the extension on the X-axis

Do this. The area under the curve is the stored energy in the bungee cord. The 1/2KX^2 equation is the area under the force / extension curve when the curve is a straight line. That equation only applies when the curve is a straight line. Plus the hysteresis and creep effects mentioned above.

Spend some time doing the experiment recommended by kuruman above.
 

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