# Answer to Question #21643 in Mechanics | Relativity for daniel jacobs

Question #21643

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.

Expert's answer

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

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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