anonymous
  • anonymous
how much work is required to stretch a spring 0.133 m if its force constant is 9.17N/m?
Physics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
it will be 0.5kx^2 where k=9.17&x=0.133
anonymous
  • anonymous
\[W = F * D\]
valentin68
  • valentin68
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

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anonymous
  • anonymous
This problem is in the Energy Conservation in Oscilatory Motion section of the textbook why is that?
anonymous
  • anonymous
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:)
anonymous
  • anonymous
.16220j
anonymous
  • anonymous
i agree vth 009infi...
anonymous
  • anonymous
@Decart you got my point???
anonymous
  • anonymous
Yes the book likes to throw you of course work was chapter 7 so I did not expect to see it here.'
anonymous
  • anonymous
but how can it be so 1/2mv^2 is for energy@valentin68

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