Hooke's Law applied to a bungee cord

In summary, the daredevil plans to bungee-jump from a hot-air balloon 65.0 m above a carnival midway using a uniform elastic cord tied to a harness. In a preliminary test, his weight stretched a 5.0 m section of the cord by 1.5 m. Using Hooke's law and energy equations, it is possible to determine the length of cord needed and the maximum acceleration the daredevil will experience. The mass term ultimately cancels out, and the cord does not begin to stretch until he has fallen the length of the cord. However, the practical problem of ensuring the end attached to the balloon stays stationary must also be considered.
  • #1
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Homework Statement



"A Daredevil plans to bungee-jump from a hot-air balloon 65.0 m above a carnival midway. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 10.0 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, hanging at rest from a 5.00-m length of the cord, the daredevil finds his body weight stretches the cord by 1.50 m. He intends to drop from rest at the point where the top end of a longer section of the cord is attached to the stationary hot-air balloon.

(a) What length of cord should he use?
(b) What maximum acceleration will he experience?

Homework Equations



Hooke's Law:
[tex]F=-kx[/tex]

Most likely other equations.

The Attempt at a Solution



For part (a) I know I will be using Hooke's law, but my problem is I was not given the mass of the "daredevil". I know I need to use the fact that the daredevil's weight stretched a 5.0m section of the cord by 1.5m, but I do not know what the correlation is without being given the mass.

I have not yet attempted part (b), but I am guessing that he will experience the maximum acceleration when the cord is fully stretched and is just about to reverse his downward direction. If that is true, then I would probably use Hooke's law again, substituting the distance the cord will stretch for x.

I know the rules state that you guys are not supposed to give me the answer, but I just want to reiterate, I only want help with this, if I cannot yet solve a problem like this on my own then I need to realize that now and not during an exam.

Thank you very much in advance!
 
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  • #2
What is the energy stored in the cord for a displacement x from equilibrium? What is the potential energy of the daredevil in the balloon compared to his energy at 10 m above the ground. Equate those two energies.

Use the fact that his weight stretches the cord by 1.5 m. to set up an expression for k in terms of his weight. Be careful. The k of the cord changes with length. For example, his weight would stretch a 10 m. cord by 3 m. (this is like two 5 m. lengths in series). Then use that expression for k in the energy equation. You will see that the mass term ultimately cancels out.

That should give you an expression for the displacement from equilibrium that the cord must have in order to absorb the change in potential energy of the daredevil. This will be a function of length of the original cord.

Note: The cord does not begin to stretch until he has fallen the length of the cord. It then stretches the amount you determined it needs to stretch until he stops.

[PS: the practical problem that is overlooked here is: how do you know that the end attached to the balloon will stay stationary? It would have to be a very big balloon! ]

AM
 
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  • #3


I would like to point out that Hooke's Law is a fundamental principle in physics that describes the relationship between the force applied to a spring (or an elastic cord) and the resulting displacement. It states that the force is directly proportional to the displacement, with a constant of proportionality known as the spring constant (k). In this case, the daredevil's weight is the force applied to the cord, and the displacement is the amount the cord stretches.

To solve for the length of the cord (part a), we can use Hooke's Law and the given information about the daredevil's weight and the amount the cord stretched. We can assume that the daredevil's weight is evenly distributed along the length of the cord, so we can use the formula F = mg (where m is the mass of the daredevil and g is the acceleration due to gravity) to calculate the force applied to the cord. Then, we can use Hooke's Law to solve for the length of the cord.

For part (b), we can use the formula for acceleration (a = F/m) and the maximum force on the cord (which occurs when the cord is fully stretched) to calculate the maximum acceleration the daredevil will experience.

However, it is important to note that this model assumes the cord and the daredevil's body behave as a point particle, which may not be the case in reality. Factors such as air resistance, the mass and shape of the daredevil's body, and the elasticity of the cord may affect the results. As a scientist, it is important to consider these limitations and potential sources of error in any experiment or calculation.
 

FAQ: Hooke's Law applied to a bungee cord

1. What is Hooke's Law?

Hooke's Law is a principle in physics that states the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. This means that the more the spring is stretched or compressed, the greater the force it exerts.

2. How is Hooke's Law applied to a bungee cord?

In the case of a bungee cord, the elastic material of the cord behaves like a spring. As the bungee cord is stretched, it exerts a force that is proportional to the amount of stretching. This allows it to provide a cushioning effect for the person attached to it, preventing them from hitting the ground with too much force.

3. What factors affect the application of Hooke's Law to a bungee cord?

The factors that affect the application of Hooke's Law to a bungee cord include the length and thickness of the cord, the material it is made of, and the weight and velocity of the person attached to it. These factors can influence the amount of stretching and the force exerted by the bungee cord.

4. Is Hooke's Law applicable to all types of bungee cords?

No, Hooke's Law is only applicable to bungee cords that behave like ideal springs. This means that they have a linear relationship between the force and displacement, and that they return to their original length once the force is removed. Bungee cords made of non-ideal materials may not follow Hooke's Law.

5. How is Hooke's Law used in bungee jumping?

In bungee jumping, Hooke's Law is used to calculate the length of the bungee cord needed for a specific jump. The length of the cord is determined by the weight and velocity of the person jumping, as well as the desired amount of stretch to provide a safe and enjoyable experience. This calculation ensures that the bungee cord will provide enough force to slow the person down before they reach the ground.

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