Solving for Unstretched Length in Jump Problem: ##l_0 = 97.5m##

  • Thread starter Thread starter STEMucator
  • Start date Start date
  • Tags Tags
    Jump Length
Click For Summary
SUMMARY

The problem involves calculating the unstretched length of a bungee cord, denoted as ##l_0##, for a 75 kg man jumping from a height with an initial velocity of 1.5 m/s. Using the conservation of energy principle, the equation derived is ##\frac{1}{2} (75) (1.5)^2 + 0 = 0 - (75)(9.81)(150) + \frac{1}{2} (80) (150 - l_0)^2##. The two solutions obtained are ##l_{0_1} = 203m## and ##l_{0_2} = 97.5m##, with the latter being the physically valid solution as it does not exceed the jump height. The analysis emphasizes the importance of considering the physical constraints of the bungee cord's potential energy.

PREREQUISITES
  • Understanding of conservation of energy principles in physics
  • Familiarity with quadratic equations and their solutions
  • Knowledge of forces acting on a body in free fall
  • Basic concepts of potential and kinetic energy
NEXT STEPS
  • Study the principles of conservation of energy in mechanical systems
  • Learn how to solve quadratic equations effectively
  • Explore the dynamics of bungee jumping and the forces involved
  • Investigate the differences between bungee cords and springs in terms of potential energy
USEFUL FOR

Physics students, engineers, and anyone interested in mechanics and energy conservation principles, particularly in applications involving bungee jumping scenarios.

STEMucator
Homework Helper
Messages
2,076
Reaction score
140

Homework Statement



The ##m = 75kg## man jumps off the bridge with ##v_1 = 1.5 m/s##. Determine the unstretched length ##l_0## of the cord in order that he stops momentarily above the surface of the water. The stiffness is ##k = 80 N/m##.

Picture: http://gyazo.com/80220984e620ed5fc0c58c8207e79ba9

Homework Equations





The Attempt at a Solution



This is one of those problems they don't give the answer for and I'm wondering if this is okay.

I drew my FBD stick figure with the Datum plane located through position A. I then used conservation:

##T_1 + V_1 = T_2 + V_2##

Since I stuck the datum plane at the top where the man is, he currently posses no potential energy at all, only kinetic.

In the final phase, the man has lost all of his kinetic energy and it is now in the form of potential. So we have:

##\frac{1}{2} (75) (1.5)^2 + 0 = 0 - (75)(9.81)(150) + \frac{1}{2} (80) (150 - l_0)^2##

Which is quadratic in ##l_0##. Solving I obtain ##l_{0_1} = 203m##, which makes no physical sense at all since it exceeds the jumping height. ##l_{0_2} = 97.5m##, which seems reasonable.

Does this look okay?
 
Physics news on Phys.org
I believe your answers are correct. He falls for ##l_0## meters at which point the rope becomes taut. Over the next ##150 - l_0## meters, two forces act on him, tension and gravity. Each force does work but ##W_{tension} - W_{gravity} = E_k##. This gives a pretty complicated formula that can be solved for ##l_0##. With some computer help I got the same two answers: 202.55m and 97.45m.
 
  • Like
Likes   Reactions: 1 person
The reason you get the extra root, is because (150-l_0)^2 is positive if l0>150, so the potential energy is positive. This is true for a spring with length l0 , but not for a bungee cord which has 0 potential energy if the cord is longer than the height of the fall.

Your computation is rather dangerous, because it will produce the right result if the bungee jumper comes to a stop at the bottom, but if the bungee jumper does not come to a stop, you might still get a result.
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
6K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
12K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 1 ·
Replies
1
Views
3K