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Potential Energy of a crab

  1. Nov 12, 2006 #1
    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. jcsd
  3. Nov 12, 2006 #2

    rsk

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    [​IMG]

    Does this help?
     
  4. Nov 12, 2006 #3

    radou

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    The potential energy due to compression equals [tex]U = \frac{F \Delta L}{2}[/tex], where [tex]\Delta L[/tex] is the length of compression. You only have to find the force F now from the relation [tex]\Delta L = \frac{F L }{EA}[/tex]. 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 [tex]\lambda[/tex] is the module of elasticity E, and E the potential energy U. :smile:
     
    Last edited: Nov 12, 2006
  5. Nov 12, 2006 #4
    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?
     
  6. Nov 12, 2006 #5

    rsk

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    5cm^2 is 0.0005 m^2 I think. Also why do you have 0.002 for x? Isn't it 0.001?
     
  7. Nov 12, 2006 #6
    I was thinking the displacement was the uncompressed abductin- the compressed abductin
     
  8. Nov 12, 2006 #7

    radou

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    The displacement x states the amount of compression. Correct your numbers (as rsk suggested), and this should work just fine.
     
  9. Nov 12, 2006 #8
    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
     
  10. Nov 12, 2006 #9

    radou

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    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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