Hooke's law when bungee jumping

AI Thread Summary
The discussion revolves around calculating the lowest point Astrid reaches during an illegal bungee jump, using Hooke's law and energy conservation principles. The initial calculation mistakenly only considered gravitational force and the spring constant, leading to an incorrect displacement of 72m. The correct approach involves accounting for both kinetic and potential energy at the moment the bungee cord begins to stretch, which includes elastic potential energy. A key insight is that the energy at the maximum stretch of the cord must equal the initial potential energy of Astrid before the jump. The correct answer for the lowest point reached is 132m below the bridge, emphasizing the importance of energy conservation in this scenario.
gravythis
Messages
2
Reaction score
0
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.
 
Physics news on Phys.org
gravythis said:
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...
 
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 displacement 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.
 
Last edited:
There is actually no need to calculate her velocity. Try making her potential energy equal to the spring's kinetic energy.
 
tal444 said:
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...
 
Sorry, my bad. That's what I meant.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top