Question #275394

A force of 486N stretches a spring by 0.08m. What is the force constant of the spring? What force will elongate the spring by 0.09m? What is the potential elastic energy when stretched by 0.18m?



1
Expert's answer
2021-12-06T09:44:31-0500

The force constant:


k=F/x=6075 N/m.k=F/x=6075\text { N/m}.

The force for 0.09 m:


F=kx2=547 N.F=kx_2=547\text{ N}.

The energy:


E=12kx32=98.4 J.E=\frac12kx_3^2=98.4\text{ J}.


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