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if a force of 80N extends a spring of natural length 8m by 0.4m what will be the length of the spring when the applied force 100N

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Hook’s Law
F₁=kx₁
F₂=kx₂
F₁/F₂=x₁/x₂
x₂=x₁F₂/F₁=0.4•100/80 =0.5 m

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