Question

when a force of 20 n is applied to two springs, $s_1$ and $s_2$, the expansion/stretching of the springs is 4 centimeter and 5 cm, respectively. what is the value of the spring constants of the two springs?
$k_1 = 500\frac{n}{m}$ and $k_2 = 400\frac{n}{m}$
$k_1 = 400\frac{n}{m}$ and $k_2 = 500\frac{n}{m}$
$k_1 = 10,000\frac{n}{m}$ and $k_2 = 8000\frac{n}{m}$
$k_1 = 8000\frac{n}{m}$ and $k_2 = 10,000\frac{n}{m}$

Explanation:

Step1: Recall Hooke's Law

Hooke's Law is $F = kx$, where $F$ is the force applied, $k$ is the spring - constant and $x$ is the extension of the spring. We need to solve for $k$, so $k=\frac{F}{x}$.

Step2: Convert the extensions to SI units

For spring $S_1$, $x_1 = 4\ cm=0.04\ m$. For spring $S_2$, $x_2 = 5\ cm = 0.05\ m$. The force $F = 20\ N$.

Step3: Calculate $k_1$

Using $k=\frac{F}{x}$, for spring $S_1$, $k_1=\frac{F}{x_1}=\frac{20\ N}{0.04\ m}=500\ \frac{N}{m}$.

Step4: Calculate $k_2$

Using $k=\frac{F}{x}$, for spring $S_2$, $k_2=\frac{F}{x_2}=\frac{20\ N}{0.05\ m}=400\ \frac{N}{m}$.

Answer:

$k_1 = 500\ \frac{N}{m}$ and $k_2 = 400\ \frac{N}{m}$