# Two Elastic String

• June 1st 2008, 07:24 AM
Simplicity
Two Elastic String
Question:
http://i148.photobucket.com/albums/s...sticString.jpg

http://i148.photobucket.com/albums/s...ticStringA.jpg

My Problem:
I don't understand what they have done in the mark scheme where I have put a red box around the part that I don't understand. Please help. Thanks in advance.
• June 1st 2008, 07:31 AM
bobak
they are using

Gain in Kinetic energy + gain in elastic potential energy = Loss in Gravitational potential energy.

note $E.P.E = \frac{\lambda x^2}{2l}$ for each string.

Bobak
• June 1st 2008, 07:34 AM
Isomorphism
The strings elongate right? Those strings contain elastic potential energy. For calculation purposes, read Young's modulus - Wikipedia, the free encyclopedia
• June 1st 2008, 07:38 AM
Simplicity
Quote:

Originally Posted by bobak
they are using

Gain in Kinetic energy + gain in elastic potential energy = Loss in Gravitational potential energy.

note $E.P.E = \frac{\lambda x^2}{2l}$ for each string.

Bobak

Oh, I see. Thanks.

I thought they were using $\mathrm{KE}_{\mathrm{GAIN}} = \frac12 m (v^2 - u^2)$ which got me confused. I believe $u$ is zero hence $u^2$ is zero? If it was a value, would it be $\frac12 m (v^2 - u^2) + \frac{\lambda x^2}{2l} = mgh$?
• June 1st 2008, 07:43 AM
bobak
Quote:

Originally Posted by Air
Oh, I see. Thanks.

I thought they were using $\mathrm{KE}_{\mathrm{GAIN}} = \frac12 m (v^2 - u^2)$ which got me confused. I believe $u$ is zero hence $u^2$ is zero?

Yes your told that it is released form rest so the initial velocity is zero.

Quote:

would it be $\frac12 m (v^2 - u^2) + \frac{\lambda x^2}{2l} = mgh$?
With u = 0 and remember you have 2 elastic strings so you need to double the E.P.E term.

Bobak