Bungee cord communications problem

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Homework Statement

Most bungee cord problems have a massless cord. How would the equations in terms of force and energy be different if the cord has mass?

Homework Equations

Ee = 1/2 kx^2
Eg = mgh
Ek = 1/2 mv^2

Fe = -kx

The Attempt at a Solution

if the bungee cord has mass, it would need to support its own mass as well as the mass of the person. at the early stage of the jump with the cord unstretched, the net force on the cord is Fg(cord) because the cord and the person are moving at the same speed. The kinetic energy of the system is 1/2 (m(person) + m(cord))v^2. When the person's displacement matches the length of the cord and it begins to stretch, the elastic force must support both the mass of the cord and the mass of the person. This is where I start to get confused. If anyone could explain the concept to to me, I would really appreciate it. Thanks!

 

[anathema]

Hello! You are correct in your understanding that if the bungee cord has mass, it would need to support its own mass as well as the mass of the person. This means that the equations for force and energy would be different.

For the force equation, Fe = -kx would still apply, but now the x value would represent the displacement of both the person and the cord, rather than just the person. This is because the cord is also stretching and contributing to the overall displacement.

For the energy equations, there would be an added term for the elastic potential energy of the cord. So the total elastic potential energy equation would be Ee = 1/2 kx^2 + 1/2 kx(cord)^2, where x(cord) is the displacement of the cord.

Similarly, the total kinetic energy equation would be Ek = 1/2 (m(person) + m(cord))v^2, taking into account the mass of both the person and the cord.

I hope this helps to clarify the concept for you. Let me know if you have any further questions.