# Hooke's law when bungee jumping

• gravythis
In summary: 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.
gravythis
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.

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.

## What is Hooke's Law when bungee jumping?

Hooke's Law is a scientific principle that describes the relationship between the force applied to a spring and the resulting deformation or stretch of the spring. In the context of bungee jumping, it refers to the relationship between the force of gravity acting on the bungee cord and the resulting stretch of the cord.

## How is Hooke's Law applied in bungee jumping?

Hooke's Law is applied in bungee jumping by using a bungee cord that is made of a material with known spring constant. The spring constant is a measure of how easily a material can be stretched or compressed. By knowing the spring constant of the bungee cord, the force of gravity acting on the jumper can be calculated based on the amount of stretch in the cord.

## What factors can affect Hooke's Law when bungee jumping?

The main factors that can affect Hooke's Law when bungee jumping are the length and composition of the bungee cord, the weight and body position of the jumper, and external forces such as wind or air resistance. These factors can impact the amount of stretch in the cord and therefore affect the force of gravity acting on the jumper.

## What safety precautions are taken regarding Hooke's Law in bungee jumping?

In bungee jumping, safety precautions are taken by ensuring that the bungee cord is made of a strong and durable material with a known spring constant. The cord is also checked regularly for any signs of wear and tear. The weight and body position of the jumper are also taken into consideration to ensure that the force of gravity does not exceed the maximum weight capacity of the cord. Additionally, the bungee jumping location is carefully chosen to minimize the impact of external forces.

## Are there any limitations to Hooke's Law in bungee jumping?

Yes, there are limitations to Hooke's Law in bungee jumping. This law assumes that the material being stretched is elastic and will return to its original shape once the force is removed. However, bungee cords can experience permanent deformation or breakage if the force is too great. Additionally, Hooke's Law does not take into account factors such as air resistance or the effect of multiple bounces, which can impact the accuracy of calculations in bungee jumping.

• Introductory Physics Homework Help
Replies
44
Views
4K
• Introductory Physics Homework Help
Replies
9
Views
3K
• Introductory Physics Homework Help
Replies
26
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
3K
• Introductory Physics Homework Help
Replies
14
Views
5K
• Introductory Physics Homework Help
Replies
35
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
3K
• Introductory Physics Homework Help
Replies
1
Views
5K
• Introductory Physics Homework Help
Replies
6
Views
3K
• Introductory Physics Homework Help
Replies
6
Views
3K