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Posted by on Wednesday, December 12, 2012 at 8:53am.

It takes 2.5 J of energy to stretch a helical spring by 15 cm. Determine the spring
constant.

  • physics - , Wednesday, December 12, 2012 at 5:11pm

    PE=kx²/2
    k=2•PE/ x²=
    =2•2.5/0.15²=222 N/m

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