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posted by ikye on Tuesday, July 2, 2013 at 1:52pm.
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
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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