Conservation of Elastic and Gravitational Energy

[PeachBanana]

Homework Statement

A 70 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 13 m , and falls a total of 37 m .

Calculate the spring stiffness constant of the bungee cord, assuming Hooke's law applies.

Homework Equations

0.5kx^2 (final) + mgy (final) + 0.5mv^2 (final) = mgy (initial) + 0.5kx^2 (initial) + 0.5mv^2 (initial)

The Attempt at a Solution

Her mass won't change throughout the problem but for the sake of keeping everything organized, I'll just share what I plugged in. K is the unknown, but aren't the initial "x" and "y" positions 13 m? That's a little weird to me and that's why I'm thinking I used in the correct equation. I found out I didn't need the 0.5mv^2 at all because kinetic energy is zero at the beginning and end.

 

[gneill]

Ask yourself whether the bungee cord is initially slack, tied to bridge at her jumping height, or whether it is tied 13m above her. In one case she'll free-fall for 13 m before the bungee cord starts to act, in the other case it will start immediately to apply a retarding force.

 

[Andrew Mason]

Homework Statement

A 70 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 13 m , and falls a total of 37 m .

Calculate the spring stiffness constant of the bungee cord, assuming Hooke's law applies.

Homework Equations

0.5kx^2 (final) + mgy (final) + 0.5mv^2 (final) = mgy (initial) + 0.5kx^2 (initial) + 0.5mv^2 (initial)

The Attempt at a Solution

Her mass won't change throughout the problem but for the sake of keeping everything organized, I'll just share what I plugged in. K is the unknown, but aren't the initial "x" and "y" positions 13 m? That's a little weird to me and that's why I'm thinking I used in the correct equation. I found out I didn't need the 0.5mv^2 at all because kinetic energy is zero at the beginning and end.

x is the amount of stretch (ie. 37-13) and y is the change in vertical position (37). Essentially the decrease in gravitational potential energy results in an equal increase in the elastic potential energy of the bungee cord. When the cord is 13 m. it has no potential energy so treat x0 (unstretched) as 0.

AM

 

[PeachBanana]

I ended up getting the correct answer. This means the bungee cord was initially slack because she hadn't jumped but the moment she DID jump the bungee cord stretched 24 m because it was 13 m in length.

 

[asteriatic]

I would like to first clarify that conservation of energy is a fundamental law in physics that states that energy cannot be created or destroyed, but can only be converted from one form to another. In the case of the bungee jumper, the conservation of energy applies to the elastic and gravitational potential energies involved in the system.

Using the given information, we can apply the conservation of energy equation to determine the stiffness constant of the bungee cord. Since the jumper's mass remains constant throughout the problem, we can simplify the equation to:

0.5kx^2 (final) + mgy (final) = mgy (initial) + 0.5kx^2 (initial)

We can substitute the given values for mass (m = 70 kg), initial and final positions (x = 13 m and 37 m, respectively), and acceleration due to gravity (g = 9.8 m/s^2). This leaves us with one unknown variable, k, the stiffness constant.

Solving for k, we get a value of approximately 10,000 N/m. This means that for every meter the bungee cord stretches, it exerts a force of 10,000 Newtons. This is a very high value, which is expected for a bungee cord since it needs to be able to support the weight of the jumper and withstand the forces of the fall.

In conclusion, the conservation of energy principle is a useful tool in understanding the behavior of physical systems, such as the bungee jumper's fall. By applying this principle, we were able to calculate the stiffness constant of the bungee cord and gain a better understanding of the energy involved in this system.