Bungee Jump relating to force question

[zerogoal]

Homework Statement

A bungee cord is 30.0 {\rm m} long and, when stretched a distance x, it exerts a restoring force of magnitude kx. Your father-in-law (mass 100 kg) stands on a platform 45.0 {\rm m} above the ground, and one end of the cord is tied securely to his ankle and the other end to the platform. You have promised him that when he steps off the platform he will fall a maximum distance of only 41.0 {\rm m} before the cord stops him. You had several bungee cords to select from, and you tested them by stretching them out, tying one end to a tree, and pulling on the other end with a force of 420 N.

When you do this, what distance will the bungee cord that you should select have stretched?

Homework Equations

F=kx

The Attempt at a Solution

I started out by using the F=kx formula
100(9.8)=k(41)
k=23.9
Then 420N/23.9=17.6 as the answer.


However this was not the correct answer. May i know how do i go about doing this question?

 

[aranud]

i think that you can calculate the force on the rope when your stepfather has reached the bottom point. that force will have stretched the rope 41-30= 11 meters.
because you know the force, and the distance you can figure out the k (f=kx)
and then he wants to know what the distance stretched will be when you apply a force of 420N . Since you now know k you can easily calculate that distance

edit:Oh i see your fault, you assumed x as being 41 meters. x is the distance stretched, not the total distance :)

 

[LowlyPion]

You need to first determine k.

I think this is best done by examining the potential energy over the distance he will have fallen (41m) and comparing it to the potential energy that must have gone into the bungee over (11m).

In case you don't have the potential energy of a spring, that would be

PE = 1/2*k*x²

Once you have the k, you can figure how much 420N will stretch it.

 

[zerogoal]

PE=mgh=1/2*k*x^2
=100(9.8)(41)=1/2*k*(11)^2
k=664?

Is it like this? Actually i don't understand how to get the PE that have gone into the bungee for the distance of 11m. Do i take the final PE - initial PE?

 

[LowlyPion]

PE=mgh=1/2*k*x^2
=100(9.8)(41)=1/2*k*(11)^2
k=664?

Is it like this? Actually i don't understand how to get the PE that have gone into the bungee for the distance of 11m. Do i take the final PE - initial PE?

The initial PE of the bungee will be 0, until there is tension at 30m down.

664N/m looks right.

So if you pulled 420N then how far should it stretch?

 

[LowlyPion]

If you aren't familiar with a spring PE, just think about the work needed to compress or stretch a spring.

W = ∫ F⋅ds = ∫ kx⋅dx = 1/2*k*x² |_o^x

The work goes into stored energy, or PE.

 

[zerogoal]

so to get x, i just take 420/664=0.63m?

or do i use mgx=1/2kx^2 again?
(420)(x)=1/2(664)(x^2)
x=1.27m?

which method should i use to find the distance?

 

[LowlyPion]

so to get x, i just take 420/664=0.63m?

Well you've already found k, so k = N/m and he's applying N's, so ... that's all you need do.

 

[zerogoal]

Yup got the answer already. Thanks a lot. Appreciate your effort in explaining to me.