Bungee Jumping and Conservation of Energy

[cse63146]

[SOLVED] Bungee Jumping and Conservation of Energy

Homework Statement

A bored college student decides to try bungee jumping. He attaches an elastic bungee cord to his ankles and happily jumps off a tall bridge across a river. He ends up barely touching the water before the cord jerks him back up.

Find the mass of the student m.
Express the student's mass in terms of variables given in the problem introduction and any appropriate constants.

Homework Equations

mgy = 0.5K$$X^{2}$$

The Attempt at a Solution

so I get m = (0.5K$$X^{2}$$)/gy

but this doesn't seem right. Any ideas?

 

[Doc Al]

What doesn't seem right about it? Looks OK to me. (Not a good model for a real bungee cord, but that's OK.)

 

[cse63146]

would I use $$y_{e}$$ instead of y and X?

 

[Doc Al]

would I use $$y_{e}$$ instead of y?

What is $y_{e}$? If that's how much the cord stretches, then you should use that instead of X.

 

[cse63146]

It's there on the picture, but it looks like it only goes half way (y_e) of the bungee cord

I tried inputing this:$$o.5k(y_e)^2/gy_0$$ and it says it's wrong. Any suggestions?

 

[Doc Al]

It's there on the picture, but it looks like it only goes half way (y_e) of the bungee cord

Without a clear definition, we can only guess. It's not clear from the diagram, but I presume y_e is either the unstretched length of the bungee cord, or the amount that it stretches. There's no additional information? (What text is this?)

 

[cse63146]

It's masteringphysics (online assignment). It's says that y_e is the length the of relaxed cord y_o is the distance between the bridge and the water and y_o>y_e, so once the cord stretches, would it have a length of y_o?

 

[Doc Al]

Yes. Now that you know what y_e and y_0 mean, you can find the amount of stretch in the cord at the lowest point. (Of course, the cord is attached to his ankles, so it doesn't stretch the full distance to y_0. Lacking any data, I would ignore that issue.)

 

[cse63146]

Got it, thanks again.