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 relevant formula

When a mass is placed on a vertical spring and is in equilibrium, the force exerted by the spring (Hooke's law, \( F = kx \)) is equal to the weight of the mass (\( F = mg \)). So we have \( kx = mg \), and we can solve for the spring constant \( k \) as \( k=\frac{mg}{x} \).

Step2: Plug in the values

We know the mass \( m = 9.09\space kg \), the acceleration due to gravity \( g = 9.8\space m/s^{2} \), and the compression \( x = 0.0418\space m \).

Substitute these values into the formula:

\( k=\frac{9.09\times9.8}{0.0418} \)

First, calculate the numerator: \( 9.09\times9.8 = 9.09\times(10 - 0.2)=90.9 - 1.818 = 89.082 \)

Then, divide by the denominator: \( k=\frac{89.082}{0.0418}\approx2131.15\space N/m \)

Answer:

The spring constant of the spring is approximately \(\boldsymbol{2131\space N/m}\) (or more precisely \(\boldsymbol{2131.15\space N/m}\)).