Calculate the energy stored in elastic material with K=1500NM and applied force at 750N
Answer
Stored energy
U=kx22=F22K=750∗7503000=187.5JU=\frac{kx^2}{2}=\frac{F^2}{2K}\\=\frac{750*750}{3000}\\=187.5JU=2kx2=2KF2=3000750∗750=187.5J
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