Max Acceleration of a Bungee Jumper

Click For Summary
SUMMARY

The maximum acceleration of a bungee jumper occurs at a point 17 meters below the jump point, which is 20 meters above the ground. The correct formula for calculating acceleration is a = (kx - mg)/m, where k is the spring constant (100 N/m), x is the stretch of the bungee cord (17 m), m is the mass of the jumper (60 kg), and g is the acceleration due to gravity (9.81 m/s²). The confusion arises from the interpretation of the jump height and the point where the cord becomes taut. The maximum acceleration can be calculated using these parameters without ambiguity.

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with spring constants and Hooke's Law
  • Basic knowledge of gravitational acceleration
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study Hooke's Law and its application in real-world scenarios
  • Learn about the dynamics of oscillatory motion in bungee jumping
  • Explore the effects of varying mass on acceleration in spring systems
  • Investigate the physics of free fall and its interaction with elastic forces
USEFUL FOR

Physics students, engineers, and anyone interested in the mechanics of bungee jumping and elastic systems will benefit from this discussion.

dhtml12345
Messages
2
Reaction score
0

Homework Statement



A bungee jumper of mass m kg jumps from 20 meters. The bungie cord's spring constant is k N/m. Let y = 0 be the point where the bungee cord begins to become taught, and let y = 20 be the jump point.

It is known that the point of maximum acceleration is y = -17.

Question: What is the value of the maximum acceleration?

Hint: The acceleration is not equal to gravity.

2. The attempt at a solution

F = ky - mg = ma
a = (ky - mg)/m [FAILS!]
 
Last edited:
Physics news on Phys.org
Thanks, but that didn't work.
 
The problem is a bit unclear about the altitudes used.
It says a jump from 20 m, but then they say that the cord only starts to become taught
at 20 m below the jump point, and the point of maximum acceleration (hence maximum length of the cord) is 17m below that. I hope the jump is from an altitude of at least 37m.

The method of you and anti-meson is correct. a = F/m = (kx - mg)/m, with
F = net force on person
k = spring constant = 100N/m
x = distance cord is stretched = 17m.
m = mass of person = 60 kg.
g = 9.81 m/s^2

only the distance should be 17m and not 37m

I don't see how you got 9.81, -9.81, 6, or -6 as none of them is the answer comes
from the equation with either 17 or 37 m.
 

Similar threads

Replies
44
Views
7K
Replies
9
Views
4K
Replies
26
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 14 ·
Replies
14
Views
6K
  • · Replies 13 ·
Replies
13
Views
16K
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 11 ·
Replies
11
Views
4K