An ornament of mass 40.0 g is attached to a vertical ideal spring with a force constant (spring constant) of 20.0 N/m. The ornament is then lowered very slowly until the spring stops stretching. How much does the spring stretch?
A) 0.0816 m
B) 0.00200 m
C) 0.0196 m
D) 0.800 m
E) 0.200 m

Question

An ornament of mass  40.0 g is attached to a vertical ideal spring with a force constant (spring constant) of  20.0 N/m. The ornament is then lowered very slowly until the spring stops stretching. How much does the spring stretch?
A)  0.0816 m
B)  0.00200 m
C)  0.0196 m
D)  0.800 m
E)  0.200 m

Quizplus · Accepted Answer

The spring stretch ($x$) can be found using Hooke's Law, which states $F = kx$, where $F$ is the force applied, $k$ is the spring constant, and $x$ is the displacement of the spring from its equilibrium position. The force applied by the ornament due to gravity is its weight, $F = mg$, where $m$ is the mass of the ornament and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). Substituting the weight into Hooke's Law gives $mg = kx$. Rearranging for $x$ gives $x = \frac{mg}{k}$. Substituting $m = 0.04 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, and $k = 20.0 \, \text{N/m}$ gives $x = \frac{0.04 \times 9.8}{20.0} = 0.0196 \, \text{m}$.

An Ornament of Mass 40