Two Elastic String

**Simplicity** · June 1st 2008, 07:24 AM

Question:

Answer:

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.

**bobak** · June 1st 2008, 07:31 AM

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

**Isomorphism** · June 1st 2008, 07:34 AM

The strings elongate right? Those strings contain elastic potential energy. For calculation purposes, read Young's modulus - Wikipedia, the free encyclopedia

**Simplicity** · June 1st 2008, 07:38 AM

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 $\frac12 m (v^2 - u^2) + \frac{\lambda x^2}{2l} = mgh$ ?

**bobak** · June 1st 2008, 07:43 AM

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.

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

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