if a force of 80N extends a spring of natural lenght 8m by 0.4m what will be the lenght of the spring when the force applied is 100N.
Hooke's law: F=k*ΔL k - is a constant called the rate or spring constant ΔL - is the displacement of the spring's end from its equilibrium position F1 = k*ΔL1 F2 = k*ΔL2 F1/F2=ΔL1/ΔL2 ΔL2=ΔL1*F2/F1 = 0.4m *100/80 =0.5 m total length: L=L0+ΔL2 = 8 + 0.5 = 8.5 m L0 - natural length Answer: L=8.5 m
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