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Old Dec 19th 2009, 09:34 AM   #1
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Elastic string question

A bungee-jumping athlete of mass 72 kg leaps from a 40-meters tall bridge, held by an elastic string with elastic constant 100 N/m. What is the maximum string's length so that the athlete arrives at the floor with null speed? Consider that Hooke's law is valid for the string and g = 10 m/s^2.


I tried this way:

F = k.x
720 = 100.x
x = 7,2 m

So, the string lenght would be L + 7,2 = 40 -> L = 32,8 meters

But the correct answer is 16 meters. How to obtain it?

Thanks in advance.
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Old Dec 19th 2009, 10:42 AM   #2
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hope you understand my writing
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Old Dec 20th 2009, 08:22 PM   #3
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Rachel, you seemed to have solved for when the tension in the string is equal to the jumper's weight.

Apply Conservation of Energy. Ask yourself how the speed, height and length of the string are connected.

Follow-up if you are still having trouble.
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Old Dec 20th 2009, 11:35 PM   #4
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Rachel, your approach seems in general to be correct (the writing is hard to follow; maybe you should have keyed it in).

However you can solve it in a simpler fashion by not bringing in K.E.

Initially the system has only grav P.E. and finally at full stretch just at ground level, it is again only elastic P.E.

Apply conservation of energy.

If L is the unstretched length of the string, we have

m g 40 = 0.5 k (40 - L)^2

Solving, we get L = 16.

I am sure r.samanta will agree with this method!
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Old Dec 21st 2009, 02:13 AM   #5
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Originally Posted by Deco View Post
Rachel, you seemed to have solved for when the tension in the string is equal to the jumper's weight.

Apply Conservation of Energy. Ask yourself how the speed, height and length of the string are connected.

Follow-up if you are still having trouble.
Yes I did it. I've considered a simple harmonic moviment...is it wrong?
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Old Dec 21st 2009, 08:46 AM   #6
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I couldn't solve this question at first because I thought it was an issue with Newton's second law. It turns out to be obviously simple.

Thanks.
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Old Dec 24th 2009, 03:33 AM   #7
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Originally Posted by physicsquest View Post
Rachel, your approach seems in general to be correct (the writing is hard to follow; maybe you should have keyed it in).

However you can solve it in a simpler fashion by not bringing in K.E.

Initially the system has only grav P.E. and finally at full stretch just at ground level, it is again only elastic P.E.

Apply conservation of energy.

If L is the unstretched length of the string, we have

m g 40 = 0.5 k (40 - L)^2

Solving, we get L = 16.

I am sure r.samanta will agree with this method!
agreed 100%
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Old Dec 26th 2009, 08:54 AM   #8
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Originally Posted by physicsquest View Post
Rachel, your approach seems in general to be correct (the writing is hard to follow; maybe you should have keyed it in).

However you can solve it in a simpler fashion by not bringing in K.E.

Initially the system has only grav P.E. and finally at full stretch just at ground level, it is again only elastic P.E.

Apply conservation of energy.

If L is the unstretched length of the string, we have

m g 40 = 0.5 k (40 - L)^2

Solving, we get L = 16.

I am sure r.samanta will agree with this method!
I think that method is incorrect (or not)



and this is diferent than:




and you have to consider the beggining of SHM which is not at the begining of the jump

I am sorry if I am incorrect
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