Asked by Anonymous
                It takes 2.5 J of energy to stretch a helical spring by 15 cm.  Determine the spring 
constant.
            
        constant.
Answers
                    Answered by
            Elena
            
    PE=kx²/2
k=2•PE/ x²=
=2•2.5/0.15²=222 N/m
    
k=2•PE/ x²=
=2•2.5/0.15²=222 N/m
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