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?



Expert's answer

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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