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Decart
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how much work is required to stretch a spring 0.133 m if its force constant is 9.17N/m?
 one year ago
 one year ago
Decart Group Title
how much work is required to stretch a spring 0.133 m if its force constant is 9.17N/m?
 one year ago
 one year ago

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HELLSGUARDIAN Group TitleBest ResponseYou've already chosen the best response.1
it will be 0.5kx^2 where k=9.17&x=0.133
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
\[W = F * D\]
 one year ago

valentin68 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Decart Group TitleBest ResponseYou've already chosen the best response.0
This problem is in the Energy Conservation in Oscilatory Motion section of the textbook why is that?
 one year ago

HELLSGUARDIAN Group TitleBest ResponseYou've already chosen the best response.1
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:)
 one year ago

009infinity Group TitleBest ResponseYou've already chosen the best response.0
.16220j
 one year ago

avijit Group TitleBest ResponseYou've already chosen the best response.0
i agree vth 009infi...
 one year ago

HELLSGUARDIAN Group TitleBest ResponseYou've already chosen the best response.1
@Decart you got my point???
 one year ago

Decart Group TitleBest ResponseYou've already chosen the best response.0
Yes the book likes to throw you of course work was chapter 7 so I did not expect to see it here.'
 one year ago

009infinity Group TitleBest ResponseYou've already chosen the best response.0
but how can it be so 1/2mv^2 is for energy@valentin68
 one year ago
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