Hooke's law when bungee jumping


by gravythis
Tags: bungee, hooke, jumping
gravythis
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#1
Mar15-12, 03:58 PM
P: 2
This was on a practice test for an upcoming exam.

You have persuaded your friend Astrid to attempt an illegal bungee jump from a Bridge. You will provide the bungee cord which has a total length of 40 m and a spring constant of k = 16 N/m. During the jump, Astrid will first fall freely for a distance equal to the length of the cord, after which the cord will begin to stretch, obeying Hooke’s law. Astrid’s mass is 52 kg. The lowest point she reaches before rebounding is _____________ below the bridge

I used F=kx, plugging in the gravitational weight for F and 16 for k to get a displacement of 32m, added to 40 would give an answer of 72m.

The actual answer is 132m and I can't figure out what I'm missing.
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berkeman
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#2
Mar15-12, 04:34 PM
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Quote Quote by gravythis View Post
This was on a practice test for an upcoming exam.

You have persuaded your friend Astrid to attempt an illegal bungee jump from a Bridge. You will provide the bungee cord which has a total length of 40 m and a spring constant of k = 16 N/m. During the jump, Astrid will first fall freely for a distance equal to the length of the cord, after which the cord will begin to stretch, obeying Hooke’s law. Astrid’s mass is 52 kg. The lowest point she reaches before rebounding is _____________ below the bridge

I used F=kx, plugging in the gravitational weight for F and 16 for k to get a displacement of 32m, added to 40 would give an answer of 72m.

The actual answer is 132m and I can't figure out what I'm missing.
Welcome to the PF.

You are leaving out the part where Astrid has a downward velocity when the spring starts to stretch. He has both KE and PE when the spring starts to stretch. He has no KE at the bottom, and a different PE there.

Try using energy balance equations to make this easier...
gravythis
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#3
Mar15-12, 05:10 PM
P: 2
So I decided to use K1+U1=U2 since you said to look over energy conservation and as you pointed out there is no KE once it reaches it's max displacment as v=o, unfortunately I don't know what to do with this.

I started out finding K1 by finding v after dropping 40 m,

droptime = √[(2*40)/9.8]=2.86 s
v= -9.8 * 2.86 = -28 m/s
K1=.5*52*282= 20384 J

But from here I can't figure out what y1 should be in U1 as I don't know how high the bridge is above the ground.

I then tried a different approach by rearanging to get
y2-y1=(.5*mv2)/mg in the hopes that this would give me the total displacement and I wouldn't have to figure out where y1 is with repect to the origin but this didn't give me the correct answer either.

Needless to say my head hurts right now.

Edit:

Just realized my mistake in forgetting elastic potential energy.

If I add this to my equation I get K1+U1=UE2+U2
but I still can't figure out what to assign y1 in my equation.

tal444
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#4
Mar15-12, 05:29 PM
P: 157

Hooke's law when bungee jumping


There is actually no need to calculate her velocity. Try making her potential energy equal to the spring's kinetic energy.
berkeman
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#5
Mar15-12, 06:12 PM
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Quote Quote by tal444 View Post
There is actually no need to calculate her velocity. Try making her potential energy equal to the spring's kinetic energy.
Not the spring's kineting energy (that's probably just a typo) -- energy stored in a spring is considered PE.

But the hint does apply. You have an initial situation with no energy in the spring and the jumper standing on the bridge. Then at the moment at the bottom of the jump, she is way lower, and the spring is stretched...
tal444
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#6
Mar15-12, 06:38 PM
P: 157
Sorry, my bad. That's what I meant.


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