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March 1, 2015

March 1, 2015

Posted by **Kelly** on Saturday, June 2, 2012 at 11:52am.

(a) How much more energy is needed to further stretch the spring to 2 cm beyond its equilibrium length?

(b) From this new position, how much energy is needed to compress the spring to 2 cm shorter than its equilibrium position?

- Physics -
**Elena**, Saturday, June 2, 2012 at 12:19pmPE1 =k•x1²/2

PE2 = k•x2²/2

W = ΔPE =PE2 –PE1 =k(x2² - x1²)/2 = ….

where k =54 N/m, x1 =0.01 m,

x2 =0.01+0.02 = 0.03 m.

this compression will be done by the work of elastic force of the spring. When we release the spring which is stretched by 0.03 m, the released energy is k•(0.003)²/2 which is large than k•(0.02)²/2.

- Physics -
**Kelly**, Saturday, June 2, 2012 at 8:28pmI'm still a little confused. When I use the equation:

W = k(x2^2 - x1^2)/2, I get 0.0216 J. This doesn't seem correct.

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