Bungee Jumping spring constant

In summary, the conversation discusses the problem of determining the spring constant k for a bungee jumper, Kate, who wants to jump off a bridge and touch the surface of the river on her first downward trip. The conversation mentions that the bungee cord behaves as an ideal spring with a spring constant k and Kate can be treated as a point particle. The attempted solution involves drawing a free body diagram for Kate and using the equation kx - mg = mg to solve for k. However, the correct approach is using the conservation of energy. The conversation also clarifies that the acceleration at the bottom cannot be assumed to be equal to gravity, as the velocity at the bottom is zero.
  • #1
IKonquer
47
0

Homework Statement


Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls.

Assume the following:

* The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k.
* Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward.
* Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle.

If Kate just touches the surface of the river on her first downward trip (i.e., before the first bounce), what is the spring constant k? Ignore all dissipative forces.

Homework Equations





The Attempt at a Solution


I started by trying to draw a free-body diagram on Kate. I had kx from the cord that pulled her up and mg down. At the bottom, her acceleration is g upwards so I wrote the equation:
kx - mg = mg.

k = 2mg / x

x is the distance from the bridge to the water - the length of the unstretched cord.
x = h - L

My answer for the spring constant k = 2mg / (h - L)

I'm not sure why my answer is not right.
 
Physics news on Phys.org
  • #2
IKonquer said:

I started by trying to draw a free-body diagram on Kate. I had kx from the cord that pulled her up and mg down. At the bottom, her acceleration is g upwards so I wrote the equation:
kx - mg = mg.

If you wanted to write a free body diagram for the problem with down being positive then you would have mg-kx = ma where 'a' is the varying acceleration at any time though out the fall. You can't simply say that the acceleration is that of gravity..

Probably the best way to approach this is using the conservation of energy. What energy forms do you have at the top vs any other point?
 
  • #3
I got the correct answer with the conservation of energy, but I am still confused about the acceleration in this problem. Shouldn't the acceleration at the bottom be g pointing upwards?
 
  • #4
You can't say the acceleration at the bottom is g because the acceleration is related to the motion equation from the FBD of mg-kx=ma...all you can say is that the velocity at the bottom is zero.

What you did was solve for what k "would" be if the acceleration at the bottom was equal to gravity but in the positive direction. This is a possible solution for k, but one that doesn't fit to the problem description.
 
  • #5


Your approach and equations are correct, but there may be a mistake in your calculation. Double check your units and make sure they are consistent. The spring constant should have units of force/length, so make sure you are using the correct units for mass, acceleration due to gravity, and distance. Also, keep in mind that the mass m in the equation represents the total mass of Kate and the bungee cord.
 

1. What is the spring constant in bungee jumping?

The spring constant in bungee jumping refers to the stiffness of the bungee cord or rope used in the jump. It is measured in units of force per unit of length, such as N/m or lbf/in.

2. How is the spring constant calculated for bungee cords?

The spring constant for bungee cords can be calculated by dividing the force of the bungee cord by the length change, also known as the extension, of the cord. This calculation can be done using Hooke's Law, which states that the force applied to a spring is directly proportional to the amount of extension or compression of the spring.

3. Does the spring constant affect the safety of bungee jumping?

Yes, the spring constant plays a crucial role in the safety of bungee jumping. A higher spring constant indicates a stiffer bungee cord, which can result in a smoother and safer jump. However, if the spring constant is too high, it can lead to a harsh landing and potentially cause injury. Therefore, it is essential to calculate and use an appropriate spring constant for bungee jumping.

4. Can the spring constant be adjusted for different weights in bungee jumping?

Yes, the spring constant can be adjusted for different weights in bungee jumping. This is because the amount of force needed to stretch the bungee cord varies based on the weight of the jumper. Therefore, the spring constant can be adjusted to ensure a safe and enjoyable jump for individuals of different weights.

5. How does the spring constant affect the height of a bungee jump?

The spring constant can affect the height of a bungee jump by determining how much the bungee cord will stretch during the jump. A higher spring constant will result in a shorter stretch of the cord, leading to a shorter jump. On the other hand, a lower spring constant will result in a longer stretch of the cord, allowing for a higher jump. The spring constant, along with other factors such as the length of the cord and the weight of the jumper, can be adjusted to determine the desired height of the bungee jump.

Similar threads

Replies
44
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
Replies
26
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
367
  • Introductory Physics Homework Help
Replies
14
Views
5K
Replies
4
Views
1K
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top