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

When the mass is in equilibrium on the spring, the weight of the mass (\(F_g = mg\)) is balanced by the spring force (\(F_s = kx\)), where \(m\) is mass, \(g\) is acceleration due to gravity (\(9.8\ m/s^2\)), \(k\) is spring constant, and \(x\) is compression. So \(mg=kx\).

Step2: Solve for \(k\)

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

Answer:

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