# What is the maximum stretch of the bungee and will the mass hit the ground?

• purpleperson1717
In summary, the box will hit the ground because the mass falls for 3 seconds and the initial compression of the bungee was 92.12m.
purpleperson1717
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.

purpleperson1717 said:
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.)

Doc Al said:
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?

purpleperson1717 said:
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".

Oh ok. So this is the diagram I was using, should I have no x1?

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.

Doc Al said:
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.
Doc Al said:
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?

purpleperson1717 said:
Finally I used conservation of energy
Why not just use conservation of energy for the entire motion?

purpleperson1717 said:
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.

Doc Al said:
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!

Mister T said:
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.

purpleperson1717 said:
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.)

Doc Al said:
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.

## 1. What is the maximum stretch of the bungee?

The maximum stretch of a bungee cord depends on several factors such as the length and elasticity of the cord, the weight of the attached mass, and the force of gravity. It is typically calculated using mathematical equations and can vary depending on the specific situation.

## 2. Will the mass hit the ground?

The mass attached to the bungee cord will not hit the ground if the maximum stretch of the cord is greater than the distance between the attachment point and the ground. However, if the maximum stretch is less than this distance, the mass will hit the ground.

## 3. How is the maximum stretch of a bungee cord calculated?

The maximum stretch of a bungee cord can be calculated using the equation: Lmax = (2 * m * g) / k, where Lmax is the maximum stretch, m is the mass attached to the cord, g is the force of gravity, and k is the spring constant of the cord. This equation takes into account the weight of the mass and the elasticity of the cord.

## 4. Can the maximum stretch of a bungee cord be changed?

Yes, the maximum stretch of a bungee cord can be changed by altering the length or elasticity of the cord, or by adjusting the weight of the attached mass. These factors can be manipulated to achieve a desired maximum stretch and ensure the safety of the bungee jumper.

## 5. Is the maximum stretch of a bungee cord the same for all situations?

No, the maximum stretch of a bungee cord can vary depending on the specific situation. Factors such as the length and elasticity of the cord, the weight of the attached mass, and the force of gravity can all affect the maximum stretch. Therefore, it is important to calculate the maximum stretch for each bungee jumping scenario to ensure safety.

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