# Answer to Question #16632 in Mechanics | Relativity for dellashantae

Question #16632

If a mass of 12 kg causes a spring to be extended from its equilibrium position by 1.2 mm, how far would the spring elongate if a 175 kg was used?

Expert's answer

According to the Hooke's law:

m*g = k*Δl ==> k = m*g/ΔL = 12[kg]*9.8[m/s²]/0.0012[m] = 98000 [kg/s²].

M*g = k*ΔL ==> ΔL = M*g/k = 175[kg]*9.8[m/s²]/98000[kg/s²] = 0.0175 m.

So, the elongation of the spring will be 1.75 cm.

m*g = k*Δl ==> k = m*g/ΔL = 12[kg]*9.8[m/s²]/0.0012[m] = 98000 [kg/s²].

M*g = k*ΔL ==> ΔL = M*g/k = 175[kg]*9.8[m/s²]/98000[kg/s²] = 0.0175 m.

So, the elongation of the spring will be 1.75 cm.

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