Physics Gr.12, Energy and Momentum problem

[Twinbird24]

This is basically what I am given. I need to use all those variables as a function of the minimum height: I need to figure out what the minimum height of the bridge should be for the person to just barely touch the water, taking into account gravity, the mass of the person, etc. (all listed in the image). I need to take into account the Law of Conservation of Energy (LCE).

This is what I came up with, but I'm not sure if it's right: h = (mv²+kx² / 2mg) + (l+L) *the lower case L (l) is supposed to be handwritten, like it appears in the image.

Thanks for any help!

 

Edit: btw, this is how I got the equation:

mgh = mv²/2 + kx²/2 (by LCE)

but h should actually be h minus the height of the person and length of the bungie cord, so:

mg(h-(l+L)) = mv²/2 + kx²/2

then I just rearranged the above equation.

 

mgh is the equation for gravitational potential energy

mv²/2 is for kinetic energy

kx²/2 is for elastic energy (bungie cord)

Am I missing any other energies or variable?

 

Thanks again for any help!!

Edited January 2, 2011 by Twinbird24

[Mr Skeptic]

What's more notable is that you have something extra. Per the problem, you need the person to (briefly) come to a complete stop at the end (he'll bounce back though). What's the kinetic energy for someone who's not moving?

 

Now for the missing things (poorly defined problem-- teacher's fault, not yours):

If you want to annoy your teacher you might want to point out that it is really difficult for a person to have their center of mass at exactly where they're jumping off. For example, the person would be higher up if he walked off than if he rolled off. I suppose the rope could be tied at waist level though. The problem isn't really clear about where the center of mass of the person is when he jumps, but you should consider that if his center of mass starts at where the rope is tied, the length the bungee cord gets stretched will be different than the length for the gravitational potential equation. Also, the center of mass of the person is important for both parts of the problem but isn't defined. For example, you'll need to know how far off the ground the center of mass is at maximum stretch. It also matters where the bungee cord is tied, since if it is tied at the feet it will need to be shorter than if it was tied at the center of mass, and also that's another difference between the height the person falls and the length the cord gets stretched.

[Twinbird24]

Thank you! I forgot, the person has no kinetic energy to start and non at the bottom.

I didn't really think about the center of mass, we never really took that into account when doing other problems either, it was just ignored :/ (we always used boxes in our energy and momentum problems and just put the center of mass in the middle, and that was it)

so I'm not sure how to create the equation to consider the center of mass of the person, the image the teacher gave us is like the one I drew, so the rope is tied around the persons feet.

This is the equation I have now: h = (kx²/2mg)+(l+L)

So this above equation works if the person is dropped from the hanging position (like on the image)?

If the person falls from atop the bridge, would the equation look like this? h = (kx²/2mg)-(l+L)

Thanks!

[Twinbird24]

Actually, I've figured out the equation (not quite the one that I posted above), thanks for the previous reply though. I need some help with my equation now, though, refer to this post (I thought it should be in it's own thread).

Edited January 3, 2011 by Twinbird24