Answer:
1.43 cm
Explanation:
Given the force constant from part a, we can use Hooke's Law to find the displacement of the spring.
Solving:
[tex]\subsection*{Given:}\begin{itemize} \item Mass of the object: \( m = 1.00 \, \text{kg} \) \item Force constant of the spring: \( k = 686.17 \, \text{N/m} \) \item Acceleration due to gravity: \( g = 9.8 \, \text{m/s}^2 \)\end{itemize}[/tex]
[tex]\hrulefill[/tex]
[tex]\subsection*{Hooke's Law:}\\\\Use Hooke's Law to determine the displacement \( x \) of the spring:\[F = k \cdot x\]\[x = \frac{F}{k}\]\[x = \frac{9.8 \, \text{N}}{686.17 \, \text{N/m}} \approx \boxed{0.0143 \, \text{m}}\][/tex]
[tex]\[x \approx 0.0143 \, \text{m} \times 100 = \boxed{1.43 \, \text{cm}}\][/tex]
Therefore, the spring stretches approximately 1.43 cm when a 1.00 kg object is suspended from it.