Physics Help Forum Elastic string question

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 Dec 19th 2009, 09:34 AM #1 Junior Member   Join Date: Dec 2009 Posts: 10 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.
 Dec 19th 2009, 10:42 AM #2 Junior Member   Join Date: Jun 2009 Posts: 6 hope you understand my writing Attached Thumbnails
 Dec 20th 2009, 08:22 PM #3 Physics Team     Join Date: Jul 2009 Posts: 310 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. __________________ "Dissent is the highest form of patriotism." - Thomas Jefferson. "Give me control of a nation's money and I care not who makes her laws." -Mayer Amschel Rothschild I study Mathematical Physics at the University of Waterloo. -DC
 Dec 20th 2009, 11:35 PM #4 Physics Team   Join Date: Feb 2009 Posts: 1,425 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!
Dec 21st 2009, 02:13 AM   #5
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 Originally Posted by Deco 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?

 Dec 21st 2009, 08:46 AM #6 Junior Member   Join Date: Dec 2009 Posts: 10 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.
Dec 24th 2009, 03:33 AM   #7
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 Originally Posted by physicsquest 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%

Dec 26th 2009, 08:54 AM   #8
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Join Date: Jun 2009
Posts: 6
 Originally Posted by physicsquest 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)

$\Delta&space;PE_{elastic}=\frac{1}{2}kx{^2}{_{final}}-\frac{1}{2}kx{^2}{_{initial}}=\frac{1}{2}k(x{^2}{_{final}}-x{^2}{_{initial}})$

and this is diferent than:

$\Delta&space;PE_{elastic}=\frac{1}{2}k(x{_{final}}-x{_{initial}})^2$

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

 Tags elastic, question, string

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