Calculating the Spring Constant for Bungee Jumping

Click For Summary
SUMMARY

The discussion focuses on calculating the spring constant (k) for a bungee jumping scenario using Hooke's Law, specifically the equation F = -kX. A jumper weighing 735N is noted, and it is established that the bungee cord doubles in length under this load. The formula k = F/L is derived, where L represents the original length of the rope. Additionally, the relationship between elastic potential energy (EPE) and gravitational potential energy (GPE) is explored, emphasizing the importance of understanding impulsive forces that can occur during bungee jumps.

PREREQUISITES
  • Understanding of Hooke's Law and its application in physics
  • Knowledge of elastic potential energy (EPE) and gravitational potential energy (GPE)
  • Familiarity with the concept of forces and weight in physics
  • Basic understanding of rope rigging and associated safety considerations
NEXT STEPS
  • Research the application of Hooke's Law in real-world scenarios
  • Study the effects of impulsive forces in bungee jumping and other dynamic systems
  • Learn about safety measures and engineering standards for bungee cord design
  • Explore the relationship between energy conservation and mechanical systems in physics
USEFUL FOR

Physics students, bungee jumping enthusiasts, safety engineers, and anyone involved in the design and safety of bungee jumping equipment.

wangking
Messages
6
Reaction score
0
Hey, i need some help here
I'm trying to figure out the spring constant for bungee jump

Say, I'm 735N and the rope will apparently double in length when i jump off it.
I don't know the length of the rope thought.
but i do know it does have a spring constant.
How to find out is the spring constant using the Hooke's Law F=-KX

Please help :)
 
Physics news on Phys.org
oh, we are trying to find the N/m ratio
how could i do that
 
F= -k/x --> |k| = F/x you cannot find it without some information on its stretching. O I just read that it doubles in length.
 
The rope doubles in length when 735N is added, so does that mean that
k would be equal to 735N/the length of the rope.
k = 735 / L
 
yes. When it has doubled its length it has displaced L from it equilibrium position.
 
Thanks homie, so is the
(the change in F) proportional to (the change in x) = k
 
If i were using the f = k x equation, and then use
EPE formula = half k x squared
Which is also equal to GPE so could i say
mgh = half k x squared,
and is the x still L
 
wangking said:
If i were using the f = k x equation, and then use
EPE formula = half k x squared
Which is also equal to GPE so could i say
mgh = half k x squared,
and is the x still L

Yes, this is the point where all the kinetic energy was converted into elastic potential energy.
 
If this is more that just a thought exercise you should be aware that at the bottom of the jump impulsive forces equal to approximately twice your weight act on the bungee cord.

There have been fatalities due to cord breakages from this cause.

Anyone who has had dealings with rope rigging will be familiar with snatch forces.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 67 ·
3
Replies
67
Views
6K
  • · Replies 27 ·
Replies
27
Views
4K
  • · Replies 23 ·
Replies
23
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
1
Views
9K
  • · Replies 6 ·
Replies
6
Views
3K