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March 2, 2015

March 2, 2015

Posted by **Anonymous** on Wednesday, December 19, 2012 at 9:53am.

How far must it be stretched for its potential energy to be 21 J?

- physics -
**Steve**, Wednesday, December 19, 2012 at 10:35amwell, let's see. You are only given two pieces of data, so the manipulation must be simple.

energy(N-m) = N/m * m^2

P.E. = 1/2 kx^2

21 = 1/2 (70000)x^2

x^2 = .0006

x = .0245m or 2.45 cm

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