Bungee Jumping Physics: Hooke's Law & Potential Energy

In summary, a daredevil plans to bungee jump from a balloon 59.0 m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 13.0 m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke's law. In a preliminary test he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.65 m. He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon.
  • #1
Abid Rizvi
20
0

Homework Statement


A daredevil plans to bungee jump from a balloon 59.0 m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 13.0 m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke's law. In a preliminary test he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.65 m. He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon.

Homework Equations


Hookes Law = 1/2 kx^2
Potential Energy = mgx

The Attempt at a Solution


Okay so for starters I said when he falls out of the balloon he falls a distance x at which he is now not falling.
x = 59-13 = 46
Then I set out to find k for Hookes law. I said mg = ky where y is the 1.65 m. So
k = (mg)/y
I then said r is the actual length of the chord, and L is the length that the chord stretches. So
r + L = x
I then said the total potential energy of the devil is mgx, and the amount of work in opposition to him by the chord is 1/2KL^2 = 1/2 * (mgL^2)/y
I set this equal to mgx: 1/2 * (mgL^2)/y = mgx
= (L^2)/(2y) = x
= L = sqrt(2yx)
solving for L and subtracting it from x to get r (the actual length of the chord) I get 37.3 m. Which apparently is wrong. What am I missing? Thanks in advance
 
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  • #2
k=mg/y Is the spring constant of an 5 m long chord. What is K for an L m long chord of the same kind?

Also, the potential energy of a stretched spring depends on the change of length (r) instead of the unstretched length (L) as you wrote.

ehild
 
  • #3
Hi and thank you for responding
I did set L as the change of length and r as the unstretched length.

Also I just assumed K = mg/y. Does K change because of length?
 
  • #4
Imagine a 0.5 m length of chord. Would it stretch to 2.15 m if the fellow hung from it ?
 
  • #5
Lol no I suppose not. Ok so K =mg/(stretched length) for a cord. I will try it with this.
 
  • #6
Ok so I put K = mg/L instead of mg/y for the chord, and I get L=2x which is obviously incorrect. I'm lost on what to do now...
 
  • #7
Abid Rizvi said:
Ok so I put K = mg/L instead of mg/y for the chord, and I get L=2x which is obviously incorrect. I'm lost on what to do now...
No, it is still the case that for a given piece of elastic K = mg/(length of stretch). Your problem is how to convert the K for one length of material to the K for a greater length. It's the innate property of the material that's constant. If a given load stretches a rope length L by x, by how much would it stretch a rope of twice the length?
 
  • #8
So k=mg/1.65 for the 5 m long chord. Assuming it is r m long, what is the spring constant?

Think: 5 m chord stretches 1.65 m. You bind two 5 m chords together. How much does that 10 m chord stretch?
If the chord is r m long, how much does it stretch by the weight of the man?

ehild
 
  • #9
YES! I got it. Okay so from what I understood, if a 5 m chord stretches 1.65 m, a 10 m chord should stretch 3.3 m (=1.65*2). So to do the conversion I need to find the factor to multiply by and that factor is given by f = (r/5) where 5 is the length of the small cord. So mg/fy = K. Putting that back into the original stuff, I get 2.5L^2 = yx(x-L), and therefore L = 25.156 m. So to find the length of the cord r, x- l = r which is about 20.8 which the website accepts as correct. Thank you!
 
  • #10
Well done! :)
 

FAQ: Bungee Jumping Physics: Hooke's Law & Potential Energy

How does Hooke's Law apply to bungee jumping?

Hooke's Law states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed. In bungee jumping, the bungee cord acts as a spring, and the jumper's weight causes it to stretch. As the cord stretches, it exerts an increasing force on the jumper, until it reaches its maximum length and the force remains constant. This is what allows the jumper to bounce up and down repeatedly.

What is the relationship between potential energy and bungee jumping?

Potential energy is the energy an object has due to its position or state. In bungee jumping, the jumper's potential energy increases as they are lifted to a higher point before the jump. As they fall, this potential energy is converted into kinetic energy, which allows them to bounce back up. This cycle continues until all of the potential energy has been converted into kinetic energy and the jumper comes to a stop.

How do factors like cord length and jumper weight affect bungee jumping physics?

The length of the bungee cord and the weight of the jumper both play a role in determining the force exerted on the jumper during the jump. A longer cord will stretch more, allowing for a longer free fall and a more intense bounce. A heavier jumper will experience a greater force due to their weight, leading to a more intense bounce as well. These factors should be carefully considered by bungee jump operators to ensure the safety and enjoyment of their customers.

Can Hooke's Law and potential energy be applied to other extreme sports?

Yes, Hooke's Law and potential energy can be applied to other extreme sports such as zip lining and skydiving. In zip lining, the tension in the cable follows Hooke's Law as the rider's weight pulls it down. Potential energy also plays a role as the rider is lifted to a higher point before the ride begins. In skydiving, potential energy is converted into kinetic energy as the person falls, and a parachute uses Hooke's Law to slow their descent.

How do engineers use bungee jumping physics to design safe and thrilling experiences?

Engineers use their understanding of Hooke's Law and potential energy to design bungee jumping experiences that are both safe and thrilling. They carefully calculate the length and strength of the bungee cord, as well as the height and location of the jump, to ensure that the jumper experiences a controlled and enjoyable bounce. They also take into account the weight and physical capabilities of the jumper to ensure that the bungee cord can handle the force exerted on it. By applying these principles, engineers can create a bungee jumping experience that is both exhilarating and safe.

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