Max Stretch of Bungee

  • #1
purpleperson1717
12
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Homework Statement:
A 4kg box is attached to a bungee cord of spring constant 20N/m and dropped off a 52m platform. It is in free fall for 3s, until the bungee catches. Will the box hit the ground?
Relevant Equations:
Conservation of energy
This is more of a check that I solved this assignment correctly. I got to an answer but I’m not sure it’s correct.

First, I decided that I needed to solve for the maximum stretch of the bungee. To do that I think I needed the length of the bungee (which is also the initial compression).

So next, I used d=v1t + 1/2at^2 to find the length of the bungee (this is the part I’m not so sure about). I used a=9.8m/s^2, t=3 and v1=0 and got d=44.1m. So I think this was the magnitude of the initial compression of the bungee.

Finally I used conservation of energy: mgh1 + 1/2kx1^2=1/2kx2^2
I used m=4, g=9.8, h1=44.1+x2, k=20, x1=44.1. After using the quadratic formula I got x2=48.02. I added that max stretch to 44.1 and got 92.12m, which is way more than 52 (the height of the platform). So my answer is that yes the box will hit the ground. But I’m not really sure that I did this right because 92.12 is not at all close to 52 so I would really appreciate some help.
 

Answers and Replies

  • #2
Doc Al
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So next, I used d=v1t + 1/2at^2 to find the length of the bungee (this is the part I’m not so sure about). I used a=9.8m/s^2, t=3 and v1=0 and got d=44.1m. So I think this was the magnitude of the initial compression of the bungee.
Careful. The mass falls for 3 seconds and you correctly found the distance fallen to be 44.1 m below its starting point. (So how far above the ground is it then?) How fast is it moving?

This is the point where the bungee cord just begins to start stretching. (Up to that point, it's just been slack.)
 
  • #3
purpleperson1717
12
1
Careful. The mass falls for 3 seconds and you correctly found the distance fallen to be 44.1 m below its starting point. (So how far above the ground is it then?) How fast is it moving?

This is the point where the bungee cord just begins to start stretching. (Up to that point, it's just been slack.)
Do you mean the 44.1m doesn’t count as initial compression because the bungee was slack?
 
  • #4
Doc Al
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Do you mean that the 44.1m doesn’t count as the initial compression because the bungee was slack?
Correct.

That's what they mean when they say "the bungee catches".
 
  • #5
purpleperson1717
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image.jpg

Oh ok. So this is the diagram I was using, should I have no x1?
 
  • #6
Doc Al
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If you measure everything from the point where the bungee catches, then you can start with an unstretched "spring" (no X1). Note that the mass is moving; be sure to include that in your energy calculation.
 
  • #7
purpleperson1717
12
1
If you measure everything from the point where the bungee catches, then you can start with an unstretched "spring" (no X1). Note that the mass is moving; be sure to include that in your energy calculation.
If you measure everything from the point where the bungee catches, then you can start with an unstretched "spring" (no X1). Note that the mass is moving; be sure to include that in your energy calculation.
Oh I see. So I calculated the speed just when the bungee begins to stretch as 29.4m/s. Then I used conservation of energy as mgh1 + 1/2mv1^2 = 1/2kx2^2. I used h1=x2 and v1=29.4m/s, and got x2=15.3m. 15.3+44.1 is 59.4, so the box would still hit the ground but the answer is more reasonable. Does that sound right?
 
  • #8
Mister T
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Finally I used conservation of energy
Why not just use conservation of energy for the entire motion?
 
  • #9
Doc Al
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Oh I see. So I calculated the speed just when the bungee begins to stretch as 29.4m/s. Then I used conservation of energy as mgh1 + 1/2mv1^2 = 1/2kx2^2. I used h1=x2 and v1=29.4m/s, and got x2=15.3m. 15.3+44.1 is 59.4, so the box would still hit the ground but the answer is more reasonable. Does that sound right?
I didn't do the calculation (too lazy!) but your method is good. But note (as @Mister T suggests) that you could solve the whole thing with conservation of energy. It's easier that way. So do it that way and compare answers.
 
  • #10
purpleperson1717
12
1
I didn't do the calculation (too lazy!) but your method is good. But note (as @Mister T suggests) that you could solve the whole thing with conservation of energy. It's easier that way. So do it that way and compare answers.
Great thanks for all your help!
 
  • #11
purpleperson1717
12
1
Why not just use conservation of energy for the entire motion?
I’m a bit confused on how to do that since there’s a few variables I don’t know. I would use m=4, g=9.8, h1=52, v1=0, k=20, x1=0 (I think), and h2=0. But then I don’t know v2 or x2.
 
  • #12
Doc Al
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I’m a bit confused on how to do that since there’s a few variables I don’t know. I would use m=4, g=9.8, h1=52, v1=0, k=20, x1=0 (I think), and h2=0. But then I don’t know v2 or x2.
The mass starts and ends at rest. I would just solve for "how far below the starting point" does the mass reach. Call that distance D (if you want). Measure the gravitational potential energy from the lowest point and solve for D. (You'll still need that calculation of how far the mass falls in 3 seconds, so that part's done.)
 
  • #13
purpleperson1717
12
1
The mass starts and ends at rest. I would just solve for "how far below the starting point" does the mass reach. Call that distance D (if you want). Measure the gravitational potential energy from the lowest point and solve for D. (You'll still need that calculation of how far the mass falls in 3 seconds, so that part's done.)
Oh ok, that makes sense! Thanks again for your help, I really appreciate it.
 

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