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Decart Group Title

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

  • 2 years ago
  • 2 years ago

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  1. HELLSGUARDIAN Group Title
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    it will be 0.5kx^2 where k=9.17&x=0.133

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

    • 2 years ago
  3. valentin68 Group Title
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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

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

    • 2 years ago
  5. HELLSGUARDIAN Group Title
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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:)

    • 2 years ago
  6. 009infinity Group Title
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    .16220j

    • 2 years ago
  7. avijit Group Title
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    i agree vth 009infi...

    • 2 years ago
  8. HELLSGUARDIAN Group Title
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    @Decart you got my point???

    • 2 years ago
  9. Decart Group Title
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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.'

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

    • 2 years ago
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