Answer: [tex]200 \frac{N}{m}[/tex]
Step-by-step explanation:
Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force [tex]F[/tex] applied to it, as long as the spring is not permanently deformed:
[tex]F=k \Delta x[/tex]
Where:
[tex]F=36 N[/tex]
[tex]k[/tex] is the elastic constant of the spring
[tex]\Delta x[/tex] is the displacement of the spring after applying the force
In this case we have two springs with constants [tex]k_{1}[/tex] and [tex]k_{2}[/tex], displacement [tex]\Delta x_{1}=6 cm \frac{1 m}{100 cm}=0.06 m[/tex] and [tex]\Delta x_{2}=9 cm \frac{1 m}{100 cm}=0.09 m[/tex], and the same force is applied to both.
For spring 1:
[tex]k_{1}=\frac{F}{\Delta x_{1}}[/tex]
[tex]k_{1}=\frac{36 N}{0.06 m}[/tex]
[tex]k_{1}=600\frac{N}{m}[/tex]
For spring 2:
[tex]k_{2}=\frac{F}{\Delta x_{2}}[/tex]
[tex]k_{2}=\frac{36 N}{0.09 m}[/tex]
[tex]k_{2}=400\frac{N}{m}[/tex]
Calculating the difference between them:
[tex]k_{1}-k_{2}=600\frac{N}{m}-400\frac{N}{m}[/tex]
Finally:
[tex]k_{1}-k_{2}=200\frac{N}{m}[/tex]
