Question

fall 2025 physics a wwva simple harmonic motion simulation with the spring constant set to 200 n/m, slowly drag the applied force 1 slider from 0.0 n to 50.0 n, observing what happens to the values of displacement and applied force. choose the correct answer from each drop-down menu to complete the statements. an applied force of dropdown results in a displacement of 0.100 m. doubling the force results in a displacement of dropdown. therefore, the relationship between applied force and displacement is dropdown.

Explanation:

To solve this, we use Hooke's Law, \( F = kx \), where \( F \) is the applied force, \( k \) is the spring constant, and \( x \) is the displacement.

Step 1: Find the applied force for \( x = 0.100 \, \text{m} \)

Given \( k = 200 \, \text{N/m} \) and \( x = 0.100 \, \text{m} \), substitute into Hooke's Law:
\( F = kx = 200 \, \text{N/m} \times 0.100 \, \text{m} = 20.0 \, \text{N} \).

Step 2: Find displacement when force is doubled

Doubling the force: \( F' = 2 \times 20.0 \, \text{N} = 40.0 \, \text{N} \).
Use \( x' = \frac{F'}{k} \):
\( x' = \frac{40.0 \, \text{N}}{200 \, \text{N/m}} = 0.200 \, \text{m} \).

Step 3: Determine the relationship

From Hooke's Law (\( F = kx \)), \( F \) is directly proportional to \( x \) (since \( k \) is constant).

Answer:

s:

• An applied force of \( \boldsymbol{20.0 \, \text{N}} \) results in a displacement of \( 0.100 \, \text{m} \).
• Doubling the force results in a displacement of \( \boldsymbol{0.200 \, \text{m}} \).
• The relationship between applied force and displacement is \( \boldsymbol{\text{directly proportional}} \).