Question #312027

Calculate the energy stored in elastic material with K=1500NM and applied force at 750N

1
Expert's answer
2022-03-17T13:55:31-0400

Answer


Stored energy

U=kx22=F22K=7507503000=187.5JU=\frac{kx^2}{2}=\frac{F^2}{2K}\\=\frac{750*750}{3000}\\=187.5J




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