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### Author Topic: A Bungee Jump Question!  (Read 9066 times)

#### Seany

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##### A Bungee Jump Question!
« on: 15/04/2011 10:05:48 »
Thanks for the help everyone

A 'bungee jumper' standing on a high platform above the ground is attached to a bungee cord of unstretched length 26m and spring constant k. When she jumps she falls a maximum distance of 96m where she is momentarily stationary at the lowest point of her jump and begin moving upwards. Ignore air resistance.

Then there's a graph of Tension against Extension.

i) Using the graph, calculate the elastic potential energy stored in the bungee cord when the jumper is at the lowest point.

Is that by working out the area under the graph? Which I got to be 56,000J

ii) Initial gravitational potential energy (with respect to ground) possessed by the jumper on the platform is 7.0 x 10^4J and her mass is 60kg. Calculate how high she is above the ground at the bottom of her fall.

Here, I did mgh = 7.0x10^4 and got h = 118.9. I then took away 70m as that was the highest extension as shown on the graph.

iii) After bouncing a few times, the bungee jumper eventually comes to rest hanging a distance, d, below the platform

Now I'm shtuck Have I got all the bits before right?

#### yor_on

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##### A Bungee Jump Question!
« Reply #1 on: 17/04/2011 01:44:46 »
To start you need to define the cord you jump with. The cords weight restrict, as well as its elongation potential. And then you need the length of the jump (cord) and the weight of the jumper. Knowing those, and the air resistance our jumper will meet, you now with confidence can proceed to the calculation. Remember to use Hooke's Law of Elasticity which will describe the tension that cord (or spring) can endure. There is a slight problem with it though, in that rubber fundamentally is a non-linear material where temperature plays a role for the elasticity and breaking-point of the material (tension).

Hooke's law of Elasticity is a linear description primary, made for springs calculating their tension and breaking-points, but it will give you a reasonable estimate of the jump.

"The force necessary to either stretch or compress a spring is represented as F=kx Where F=force, k=spring constant and x=length the spring is is stretched (or compressed). The spring constant is a number used to characterize the rigidity of the spring. This number varies with different springs. Hooke’s law does have its limits though. If a force is too great, the elastic limit will be exceeded. After the elastic limit is an area called the plastic region. The object may be stretched further in this area, however it will not returned to its original shape and will be permanently distorted. Eventually, the object will reach its breaking point. If the force applied is within the elastic limit, the object will return to its original shape.

This principle may be used in regards to the sport of bungee jumping. A bungee cord works in the same manner as a spring. Familiarity with Hooke’s laws has great significance with bungee jumping. For example, cord producer must design cords to not only support specific weights, but to be a certain length at maximum (stretched out) distance.

a) A cord is 20m long un-stretched. If its elasticity is 50% and the spring constant is 20 N/m, what is the maximum force that can be held?

Use the formula F=kx.

X is calculated by taking 50% of the un-stretched cord. Therefore x=10m.

Simply plug the values into the formula.

F=(20N/m)(10m)

F=200N

This is the maximum force that the cord can hold.

b) A cord is 45m long un-stretched. It has an elasticity of 75%. If the maximum force the cord can hold is 350N, what is the spring constant?

X is Calculated as in part a. x=33.75m

F=kx

350N=k(33.75m)

k= [(350N)/(33.75m)]

k=10.370 N/m

This is the spring constant of the particular cord."

Or one could ah, cheat a little and use Mathlab?

#### yor_on

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##### A Bungee Jump Question!
« Reply #2 on: 17/04/2011 02:22:09 »
Hmm missed that one. The elastic potential energy stored in the bungee cord is equal to the work done on the bungee cord. And Work is force times distance, but then you also need to know the spring constant. And the force applied by the bungee changes as the displacement changes

"Since the force in a linear spring is not constant, we need to use the integral form,

W = ∫ F(x) dx

(Note, this is a definite integral, from the x value at the initial position to the x value at the final position.)

In a linear spring, Hooke's law applies, F = k∙x So we can write

W = ∫ k∙x dx
W = k ∫ x dx
W = k x^2 / 2

The limits of integration are from the undisplaced condition, x = 0 m, to the displaced condition, x = 96 m."

W = k (96 m)^2 / 2 - k (0 m)^2 / 2
W = k * 96 m^2

Plugging in k = ?? N/m, we get

W = ?? N/m * 96 m^2
W = ?? N∙m
W = ?? J

That's the work done, and it all goes into the elastic potential energy. I hope that helps!"

I manhandled this second quote somewhat, so if I did it wrong , don't blame the original :) But I think it should be correct, and using the two first examples i cited before.. a) F=kx  Where 'x' is calculated by taking 50% of the un-stretched cord and b.) Then lastly go to this you will solve it.

But it was tougher than I first though :) Using those two first examples you can find the spring constant for your bungee cord and then use this second example to get to the Joule. Don't kill me if I'm wrong though :) But I think this is the correct way to proceed.

#### Geezer

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##### A Bungee Jump Question!
« Reply #3 on: 17/04/2011 05:54:18 »
Seany,

I would have thought the potential energy in the bungee would be equal to the difference in the kinetic energy of the jumper between the point where bungee started to stretch and the point where it was at maximum extension (where the jumper has zero kinetic energy).

So, you should be able to get there if you know the distances and the mass of the jumper. If you don't know the mass of the jumper, you should be able to calculate that from the spring rate of the bungee and the extension (think of the bungee as a sort of weighing device).

#### yor_on

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##### A Bungee Jump Question!
« Reply #4 on: 17/04/2011 06:18:08 »
F=ma seems wrong to use in those situations?  F= ma?

Anyway, if you have a graph you should be able to look at it and see what Newton she has at 70 meters.  Let us assume that it shows 1500 Newton at the 70 meter cord extension.

Then what is the 'Elastic Potential Energy' of the spring at this position?

We know that F= 1500 N when x= 70 meter-  And since F= kx, then you can turn it around and make k= F/x. So (1500 N)/(70 m) = 21,4~ N/m.

So the EPE (Elastic Potential Energy)= (1/2)kx2 = (1/2)(21.4 N/m)(70 m)2= Tam tara tam.. Joules.

Somewhat simpler than my first suggestion :)

It's not many Joules stored, but that has to do with that it measures the potential joules in that bungee cord as I understands it. Hooke's law is a (elastic) conservative force, it stores (potential) energy. And now I'm starting to worry for my sanity, better avoid the rest of it, if I now got it right.

#### JP

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##### A Bungee Jump Question!
« Reply #5 on: 17/04/2011 14:29:56 »
For iii), if you know the mass of the jumper, you know the force they pull down with.  If you know how the tension of the cord varies with stretch, can you tell me at what stretch length these forces cancel each other out?

#### Seany

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##### A Bungee Jump Question!
« Reply #6 on: 18/04/2011 11:17:34 »
Wow - thanks for the answers.. Just a bit of physics past papers, but having no mark scheme doesn't help much!

Must say, the first two replies were a little beyond me

#### The Naked Scientists Forum

##### A Bungee Jump Question!
« Reply #6 on: 18/04/2011 11:17:34 »