Calculate Force Constant for Safe Bungee Jumping: F=700N, x=9m, k=77.77 N/m

  • Thread starter Thread starter Marc Briancon
  • Start date Start date
  • Tags Tags
    Bungee
AI Thread Summary
To calculate the force constant for safe bungee jumping, the formula F=kx is used, where F is the force (700 N) and x is the stretch of the cord (9 m). This results in a force constant k of 77.77 N/m. The discussion emphasizes the importance of ensuring the jumper stops safely 4.00 m above the river, indicating that the velocity should be zero at that point. Clarifications are made regarding the conditions under which the calculations apply, particularly focusing on achieving a safe landing. Proper understanding of these calculations is crucial for bungee jumping safety.
Marc Briancon
Messages
3
Reaction score
0
Bungee jumping- The unstretched length of the cord is 25.0m, the student weighs about 700N, and the balloon you start at is 36.0 m above the surface of the river below. Calculate the required force constant of the cord if the student is to stop safely 4.00m above the river.

So would I want to use F=kx, then set k=(F/x)...?

F=700 N
x=9 m
k=77.77 N/m

Is that right?
 
Physics news on Phys.org
Welcome to PF!

Marc Briancon said:
… Calculate the required force constant of the cord if the student is to stop safely 4.00m above the river.

So would I want to use F=kx, then set k=(F/x)...?

F=700 N
x=9 m
k=77.77 N/m

Is that right?

Hi Marc! Welcome to PF! :smile:

Isn't that when acceleration is zero?

You want the velocity to be zero. :wink:
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top