Question

1. when a 9.09-kg mass is placed on top of a vertical spring, the spring compresses 0.0418 m. find the spring constant of the spring.

Explanation:

Step1: Identify the forces

The mass is in equilibrium, so the spring force ($F_s = kx$) equals the weight of the mass ($F_g = mg$). Here, $k$ is the spring constant, $x$ is the compression, $m$ is the mass, and $g = 9.8\ m/s^2$ (acceleration due to gravity).

Step2: Set up the equation

From equilibrium: $kx = mg$. Solve for $k$: $k=\frac{mg}{x}$.

Step3: Substitute the values

Given $m = 9.09\ kg$, $g = 9.8\ m/s^2$, $x = 0.0418\ m$.
Substitute into the formula: $k=\frac{9.09\ kg\times9.8\ m/s^2}{0.0418\ m}$.
First, calculate the numerator: $9.09\times9.8 = 89.082$.
Then divide by the denominator: $k=\frac{89.082}{0.0418}\approx2131\ N/m$.

Answer:

The spring constant is approximately $\boldsymbol{2130\ N/m}$ (or more precisely $\boldsymbol{2131\ N/m}$)