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
F₁=kx₁
F₂=kx₂
F₁/F₂=x₁/x₂
x₂=x₁F₂/F₁=0.4•100/80 =0.5 m