A spring that is compressed 13.5 cm from its equilibrium position stores 2.98 J of potential energy. Determine the spring constant k. k = 22.1 Incorrect

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Problem Statement:**

A spring that is compressed 13.5 cm from its equilibrium position stores 2.98 J of potential energy. Determine the spring constant \( k \).

---

**Answer:**

\[ k = \]

A field is provided for input, with the answer 22.1 displayed. The answer is marked as "Incorrect" in red below the input field.

**Explanation:**

The problem involves calculating the spring constant using the formula for potential energy stored in a spring:

\[
PE = \frac{1}{2} k x^2
\]

Where:
- \( PE \) is the potential energy (2.98 J),
- \( x \) is the compression distance from the equilibrium position (13.5 cm or 0.135 m),
- \( k \) is the spring constant to be determined.

Converting the given distance to meters and rearranging the formula:
\[
k = \frac{2 \times PE}{x^2}
\]

Substitute the given values to solve for \( k \).
Transcribed Image Text:**Problem Statement:** A spring that is compressed 13.5 cm from its equilibrium position stores 2.98 J of potential energy. Determine the spring constant \( k \). --- **Answer:** \[ k = \] A field is provided for input, with the answer 22.1 displayed. The answer is marked as "Incorrect" in red below the input field. **Explanation:** The problem involves calculating the spring constant using the formula for potential energy stored in a spring: \[ PE = \frac{1}{2} k x^2 \] Where: - \( PE \) is the potential energy (2.98 J), - \( x \) is the compression distance from the equilibrium position (13.5 cm or 0.135 m), - \( k \) is the spring constant to be determined. Converting the given distance to meters and rearranging the formula: \[ k = \frac{2 \times PE}{x^2} \] Substitute the given values to solve for \( k \).
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