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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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it will be 0.5kx^2 where k=9.17&x=0.133
\[W = F * D\]
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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Other answers:

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

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