**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 \( k \).

Compression of spring(x)=0.135 menergy stored in spring(U)=2.98 J

Answered: A spring that is compressed 13.5 cm…

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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Concept explainers

Spring Potential Energy

The potential energy is the energy possessed by a body by virtue of its position or the energy held by the object due to its position relative to other objects, its force like stresses, charge, and other factors affecting the particles of the body. The potential energy is the measure of the net work done by or on the body. The spring potential energy is the potential energy stored due to the deformation of an elastic spring and this elasticity is due to the stretched spring. The elasticity is the ability of a solid-state matter to come back to its equilibrium position or original position after any kind of external force is acted on it. It is also known as Elastic potential energy. The potential energy here is caused due to the displacement of the spring from its equilibrium position to another position and this potential energy is equal to the net work done due to the stretching of the spring. It also resembles the same mechanics of the oscillation of a simple pendulum, where due to the external force, the object moves from its equilibrium position to its maximum ends and the periodic motion continues till it comes back to the equilibrium position to rest. 

Question

**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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