What is the best way to approach solving this bungee jumping problem?

[lily55934]

Homework Statement

A daredevil plans to bungee jump from a balloon 65.0m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at point 10.0m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke❝s law. In a preliminary test he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.50 m. he will drop from rest at point where the top end of a longer section of the cord is attached to the stationary balloon. (a)What length of cord should he use? (b) What maximum acceleration will he experience?

Homework Equations

Potential energy = mgh
Spring potential energy = kx^2/2
Spring force = kx

The Attempt at a Solution

I tried using conservation of energy to find the spring constant k. I said the stretched length of the short cord was h1 and the distance it stretched was x1:

mgh1 = kx1^2/ 2 => k = (2mgh1) / x1^2

I did the same with the longer cord:

mgh2 = kx2^2 / 2 => x2^2 = 2mg(h2) / k

Then I replaced k with what I found:

x2^2 = 2mg(h2) / (2mg(h1)/x1^2) => x2^2 = 2mg(h2) x1^2 / 2mg(h1) = (h2) x1^2 / h1

Is what I did right or not at all? I'm very confused about this problem, thanks for your help!

 

[Student100]

Homework Statement

A daredevil plans to bungee jump from a balloon 65.0m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at point 10.0m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke❝s law. In a preliminary test he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.50 m. he will drop from rest at point where the top end of a longer section of the cord is attached to the stationary balloon. (a)What length of cord should he use? (b) What maximum acceleration will he experience?

Homework Equations

Potential energy = mgh
Spring potential energy = kx^2/2
Spring force = kx

The Attempt at a Solution

I tried using conservation of energy to find the spring constant k. I said the stretched length of the short cord was h1 and the distance it stretched was x1:

mgh1 = kx1^2/ 2 => k = (2mgh1) / x1^2

I did the same with the longer cord:

mgh2 = kx2^2 / 2 => x2^2 = 2mg(h2) / k

Then I replaced k with what I found:

x2^2 = 2mg(h2) / (2mg(h1)/x1^2) => x2^2 = 2mg(h2) x1^2 / 2mg(h1) = (h2) x1^2 / h1

Is what I did right or not at all? I'm very confused about this problem, thanks for your help!

For the 5 meter cord what is the force given by the problem?

 

[haruspex]

I tried using conservation of energy to find the spring constant k. I said the stretched length of the short cord was h1 and the distance it stretched was x1:

mgh1 = kx1^2/ 2 => k = (2mgh1) / x1^2

The information regarding the 5m cord does not involve any transformation of energy. The mass does not fall distance h1, or any distance. This is just hanging at equilibrium.

I did the same with the longer cord:

Be careful here. A spring constant is a property of a specific spring, not of the material of which it is made. A spring of twice the length has a different constant.

 

[lily55934]

Hello, thanks a lot for your help :-)
So if I understand well when the jumper is hanging at equilibrium I should use Newton to find k? Sum of forces = kx-mg => k= mg/x
But I'm still lost about the other part of the problem, because now there are 2 things I don't know (the spring constant for the longer cord and its length, I only know the stretched length). How can I find the spring constant for a rope twice the length for example? Is is k/2? How can I find the ratio between the 2 cords when the lengths I know are completely different (one is at equilibrium (5+1.5m) and the other not (55m))? Please help, it's been several weeks, and I still don't understand anything :(

 

[haruspex]

How can I find the spring constant for a rope twice the length for example? Is is k/2?

Suppose two identical ropes of spring constant k are tied end to end. A tension T is applied to them. How far does each stretch?

 

[lily55934]

Do they each stretch a distance x? So the double spring stretches 2x, and F=2kx, and the k of the 2 ropes is half the k of one shorter rope?

 

[haruspex]

Do they each stretch a distance x? So the double spring stretches 2x, and F=2kx, and the k of the 2 ropes is half the k of one shorter rope?

Yes.