Spring Constants and Conservation?

[franfran]

Spring Constants and Conservation??

Homework Statement

In the dangerous “sport” of bungee-jumping, a daring student jumps from a balloon with a specially designed elastic cord attached to his waist. The unstretched length of the cord is 22.2 m, the student weighs 326 N, and the balloon is 33.6 m above the surface of a river below. Calculate the required force constant of the cord if the student is to stop safely 2.58 m above the river.
Answer in units of N/m

Homework Equations

I know k=mg/d
F=k(x-xo)
W=k(x-xo)d

anymore help with equations would make me very happy! :)

The Attempt at a Solution

F=k(x-xo)
326=k(33.6-22.2)
28.6=k (...i have no idea if this is right!)

W=Fx
W=k(x-xo)d
W=28.6(33.6-22.2)31.02

the 31.02 is from 33.6-2.58, because I think that's what I am supposed to do... I am really a big ball of confusion... and stress...

That's all I have! :(

 

[Chestermiller]

What is the jumper's kinetic energy when he first leaves the balloon, and when he just stops short of the surface of the water? What is his change in potential energy? What is the equation for the stored elastic energy of the bungee chord when its length is greater than its unextended length?

 

[franfran]

The first equation for that would be
KE=1/2mv2
So it would be (1/2)(326/9.8)v2, right?

Don't get upset, but how would would you find velocity in this situation?

The second one would be mgh, and That is (326)(33.6-4)
So his potential would be 294 J? Is that right?

The last one would be
F=ma-k(y-l), I think...

Thank you!

 

[Chestermiller]

The first equation for that would be
KE=1/2mv2
So it would be (1/2)(326/9.8)v2, right?

Don't get upset, but how would would you find velocity in this situation?

How fast is he going when he first leaves the balloon?
How fast is he going when he comes to a complete stop just 2.58 m above the water surface?
What is his change in kinetic energy?

The second one would be mgh, and That is (326)(33.6-4)
So his potential would be 294 J? Is that right?

Where did the 4 come from?

The last one would be
F=ma-k(y-l), I think...

The unstretched length of the cord is 22 m. At what distance above the water surface does the bungy cord start to develop tension? What is the equation for the stored elastic energy in a spring or bungy cord? If you don't know, look it up in your textbook.

Chet

 

[franfran]

Am I going about this the right way?
When He leaves the balloon:
KE=PE
.5mv^2=mgh
.5(326/9.8)v^2=326(33.6)
v^2=658.66
v=25.7 m/sThe 4 is actually supposed to be 2.58. Sorry about that.

Equation for stored elastic energy- PEs=(1/2)kx2

The bungee cord develops tension when everything is unraveled, so 22.2 meters right?

PE=326*33.6-2.58
So... 10112.52?? That seems rather large... Argh.

10112.52 = .5kx2

thanks so much. and for being so patient.

 

[Chestermiller]

Am I going about this the right way?
When He leaves the balloon:
KE=PE
.5mv^2=mgh
.5(326/9.8)v^2=326(33.6)
v^2=658.66
v=25.7 m/s

Why are you using an equation to tell me what his velocity is when he leaves the balloon? Picture it in your mind. As he leaves the balloon, if he doesn't push off, what is his downward velocity?

Equation for stored elastic energy- PEs=(1/2)kx2

Please don't use the symbol PE to represent two different things. You are going to get totally confused (as you have already). Let's use SE to represent the stored elastic energy in the cord. So, SE=(1/2)kx2
where x represents the amount that the cord has been extended, over and above its unextended length (22m). So when the guy gets to a distance 2.58 m above the water surface, how much has the cord stretched beyond its unextended length? Also, in this equation, we don't know k, so we are going to express SE algebraically in terms of k. (We are supposed for find out what value of k is required so that he doesn't go splat.) Later, we'll solve for k.

 

[franfran]

Well his acceleration would be 9.8 m/s2 right? I'm not sure about velocity... :(

Would you use a kinematics equation?

SE=1/2kx2
x=22.2-2.58=19.62

So you use the unstretched length instead of the total length!

Thanks again!

 

[Chestermiller]

Well his acceleration would be 9.8 m/s2 right? I'm not sure about velocity... :(

Well, when he leaves the balloon and he doesn't push off, his vertical velocity should be zero. If you drop a ball off your roof, isn't it's downward velocity zero to start with?

SE=1/2kx2
x=22.2-2.58=19.62

You have the right idea here, but you did the geometry incorrectly to find x. Draw a diagram that shows the balloon height above the water surface and the vertical bungee cord attached to the balloon at its unstretched length, just before it begins to stretch. At what altitude is the jumper (and the lower end of the bungee cord) when the cord just begins to stretch? At what altitude is the jumper (and the lower end of the bungee cord) when the jumper stops short of the water surface? How much has the balloon stretched at that point?

 

[franfran]

DUH! I was trying all these equations! I feel like a fool! So vo=0m/s. Got it.

1. When it just begins to stretch, it would be 22.2m, no?
2.When the jumper stops short, it would have to be 2.58m.
3.It has stretched (33.6-2.58)= 31.02m

So you would use that for the spring constant equation right?
So it would be instead
x=33.6-2.58= 31.02m
? does that sound right?

thanks again!

 

[Chestermiller]

DUH! I was trying all these equations! I feel like a fool! So vo=0m/s. Got it.

1. When it just begins to stretch, it would be 22.2m, no?
2.When the jumper stops short, it would have to be 2.58m.
3.It has stretched (33.6-2.58)= 31.02m

So you would use that for the spring constant equation right?
So it would be instead
x=33.6-2.58= 31.02m
? does that sound right?

No. It's not right. Its final length is 31.02 and its unstretched length is 22.2, so the amount that it has stretched is x = (31.02 - 22.2) = 8.82 m. Yes, it has stretched only 8.82 m.