A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential en

Physics
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A Hooke's law spring is compressed 12.0 cm from equilibrium and the potential energy stored is 72.0 J. What is the spring constant in this case?

Jun 16th, 2015

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The potential energy of a spring is (P.E)= 1/2 kx^2 where x is the distance stretched or compressed from equilibrium 

we have P.E = 72  and x = 12 cm

we have 72J = 1/2 k(0.12m)^2 => k=2*72J/(0.12m)^2
k=10,000N/m

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 16th, 2015

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Jun 16th, 2015
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