Question

when a force of 36 newtons is applied to springs (s_1) and (s_2), the displacement of the springs is 6 centimeters and 9 cm, respectively. what is the difference between the spring - constants of the two springs? enter your answer as a whole number, like this: 425.

Explanation:

Step1: Recall Hooke's Law

Hooke's Law is $F = kx$, where $F$ is the force, $k$ is the spring - constant and $x$ is the displacement. We can rewrite it as $k=\frac{F}{x}$.

Step2: Calculate spring - constant of $S_1$

Given $F = 36\ N$ and $x_1=6\ cm = 0.06\ m$. Using $k_1=\frac{F}{x_1}$, we have $k_1=\frac{36}{0.06}=600\ N/m$.

Step3: Calculate spring - constant of $S_2$

Given $F = 36\ N$ and $x_2 = 9\ cm=0.09\ m$. Using $k_2=\frac{F}{x_2}$, we get $k_2=\frac{36}{0.09}=400\ N/m$.

Step4: Calculate the difference

The difference $\Delta k=k_1 - k_2$. Substituting the values of $k_1$ and $k_2$, we have $\Delta k=600 - 400 = 200$.

Answer:

200