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Decart

  • 3 years ago

how much work is required to stretch a spring 0.133 m if its force constant is 9.17N/m?

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  1. HELLSGUARDIAN
    • 3 years ago
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    it will be 0.5kx^2 where k=9.17&x=0.133

  2. Yahoo!
    • 3 years ago
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    \[W = F * D\]

  3. valentin68
    • 3 years ago
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    By definition the work needed to stretch a sping having elastic constant k with the dispăplacement x is equal to the elastic potential energy variation W =k*x^2/2 = 9.17*0.133*0.133/2 =0.081 J

  4. Decart
    • 3 years ago
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    This problem is in the Energy Conservation in Oscilatory Motion section of the textbook why is that?

  5. HELLSGUARDIAN
    • 3 years ago
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    this is because the work we do on the spring gets converted to the potential energy of the spring & gets conserved..and can be used when needed:)

  6. 009infinity
    • 3 years ago
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    .16220j

  7. avijit
    • 3 years ago
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    i agree vth 009infi...

  8. HELLSGUARDIAN
    • 3 years ago
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    @Decart you got my point???

  9. Decart
    • 3 years ago
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    Yes the book likes to throw you of course work was chapter 7 so I did not expect to see it here.'

  10. 009infinity
    • 3 years ago
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    but how can it be so 1/2mv^2 is for energy@valentin68

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