# Potential Energy of a crab

1. Nov 12, 2006

### needhelp83

A scallop forces open its shell with an elastic material called abductin, whose elastic modulus is about 2.0 x 106 N/m2. If this piece of abductin is 3.0 mm thick and has a cross-sectional area of .5 cm2, how much potential energy does it store when compressed 1.0 mm?

I have no idea how to solve for the potential energy when compressed. How would i do this?

2. Nov 12, 2006

### rsk

http://dictionary.laborlawtalk.com/prev_wiki/images/math/f264a029703a35c1bd6a9b835613e7b1.png [Broken]

Does this help?

Last edited by a moderator: May 2, 2017
3. Nov 12, 2006

### radou

The potential energy due to compression equals $$U = \frac{F \Delta L}{2}$$, where $$\Delta L$$ is the length of compression. You only have to find the force F now from the relation $$\Delta L = \frac{F L }{EA}$$. If I understood the problem right, 'thick' does represent some kind of length here, so it equals L (?).

Edit: basically, this is the same what rsk wrote, if $$\lambda$$ is the module of elasticity E, and E the potential energy U.

Last edited: Nov 12, 2006
4. Nov 12, 2006

### needhelp83

How about this...

displacement=3-1 mm=2 mm

E=(lamda)Ax^2/2l
E=(2000000 n/m^2)(0.005 m)(0.002 m)^2 / 2(0.003 m)= 6.67

This is a shot!! Does anybody else agree?

5. Nov 12, 2006

### rsk

5cm^2 is 0.0005 m^2 I think. Also why do you have 0.002 for x? Isn't it 0.001?

6. Nov 12, 2006

### needhelp83

I was thinking the displacement was the uncompressed abductin- the compressed abductin

7. Nov 12, 2006

### radou

The displacement x states the amount of compression. Correct your numbers (as rsk suggested), and this should work just fine.

8. Nov 12, 2006

### needhelp83

Ahh.. okay

E=(2000000 n/m^2)(0.0005 m^2)(0.001 m)^2 / 2(0.003 m)= 0.167 N*m

Here we go

9. Nov 12, 2006

### radou

The cross sectional area is 0.5 cm^2 and not 5 cm^2 (according to the text of the problem). So, 0.5 cm^2 = 0.00005 m^2.

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