Question

When a force of 36 Newtons is applied to springs S1 and S2, the displacement of the springs is 6 centimeters and 9 cm, respectively.
What is the difference between the spring constants of the two springs?

Answer 1

Answer: $200 \frac{N}{m}$

Step-by-step explanation:

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force $F$ applied to it, as long as the spring is not permanently deformed:  

$F=k \Delta x$

Where:  

$F=36 N$

$k$ is the elastic constant of the spring

$\Delta x$ is the displacement of the spring after applying the force

In this case we have two springs with constants $k_{1}$ and $k_{2}$, displacement $\Delta x_{1}=6 cm \frac{1 m}{100 cm}=0.06 m$ and $\Delta x_{2}=9 cm \frac{1 m}{100 cm}=0.09 m$, and the same force is applied to both.

For spring 1:

$k_{1}=\frac{F}{\Delta x_{1}}$

$k_{1}=\frac{36 N}{0.06 m}$

$k_{1}=600\frac{N}{m}$

For spring 2:

$k_{2}=\frac{F}{\Delta x_{2}}$

$k_{2}=\frac{36 N}{0.09 m}$

$k_{2}=400\frac{N}{m}$

Calculating the difference between them:

$k_{1}-k_{2}=600\frac{N}{m}-400\frac{N}{m}$

Finally:

$k_{1}-k_{2}=200\frac{N}{m}$