Finding the Spring Constant for Bungee Jumping

Click For Summary
SUMMARY

The discussion focuses on calculating the spring constant for a bungee jumping scenario using Hooke's Law, specifically the equation F = -KX. The jumper's weight is given as 735N, and it is noted that the rope doubles in length upon jumping. The user attempts to derive the spring constant (K) by equating the potential energy stored in the spring to the gravitational potential energy, leading to the equation 2mg = -kx. The solution requires understanding the relationship between force, displacement, and energy in elastic systems.

PREREQUISITES
  • Understanding of Hooke's Law and its application in elastic systems.
  • Knowledge of gravitational potential energy (GPE) and elastic potential energy (EPE).
  • Basic algebra for manipulating equations and solving for variables.
  • Familiarity with the concepts of force, mass, and acceleration due to gravity.
NEXT STEPS
  • Study the derivation and application of Hooke's Law in various contexts.
  • Explore the relationship between potential energy and kinetic energy in mechanical systems.
  • Learn about the principles of energy conservation in bungee jumping scenarios.
  • Investigate real-world examples of spring constants in different materials and applications.
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in understanding the mechanics of bungee jumping and elastic systems.

wangking
Messages
6
Reaction score
0

Homework Statement


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.
we are trying to find its N/m ratio.
How to find out is the spring constant using the Hooke's Law F=-KX

Please help :)


Homework Equations


F=-KX
EPE=1/2KX^2
EPE=GPE
mgh=1/2kx^2


The Attempt at a Solution



i can't find the spring constant
 
Physics news on Phys.org
Fall of potential - mgx.
PE stored in the spring is 1/2kx^2
-mgx = 1/2kx^2
2mg = -kx
 

Similar threads

Replies
26
Views
3K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
3
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
4
Views
2K