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: Recall Hooke's Law and Equilibrium

When the mass is placed on the spring and in equilibrium, the force exerted by the spring (Hooke's Law: \( F = kx \), where \( k \) is the spring constant and \( x \) is the compression) balances the weight of the mass (\( F = mg \), where \( m \) is mass and \( g = 9.8\ m/s^2 \) is acceleration due to gravity). So, \( kx = mg \).

Step2: Solve for Spring Constant \( k \)

Rearrange the formula to solve for \( k \): \( k=\frac{mg}{x} \).
Substitute \( m = 9.09\ kg \), \( g = 9.8\ m/s^2 \), and \( x = 0.0418\ m \):
\( k=\frac{9.09\times9.8}{0.0418} \)
First, calculate the numerator: \( 9.09\times9.8 = 89.082 \)
Then divide by the denominator: \( k=\frac{89.082}{0.0418}\approx2131.15\ N/m \)

Answer:

The spring constant is approximately \( 2130\ N/m \) (or \( 2.13\times10^3\ N/m \), depending on significant figures).