A pen contains a spring with a spring constant of 240 N/m. When the tip of the pen is in its retracted position, the spring is compressed 5.00 mm from its unstrained length. In order to push the tip out and ock it into its writing position, the spring must be compressed an additional 6.60 mm. How much work
A pen contains a spring with a spring constant of 240 N/m. When the tip of the pen is in its retracted position, the spring is compressed 5.00 mm from its unstrained length. In order to push the tip out and ock it into its writing position, the spring must be compressed an additional 6.60 mm. How much work
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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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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![### Physics of Springs: Work Done by a Spring Force in a Pen Mechanism
**Problem Statement:**
A pen contains a spring with a spring constant of 240 N/m. When the tip of the pen is in its retracted position, the spring is compressed 5.00 mm from its unstrained length. In order to push the tip out and lock it into its writing position, the spring must be compressed an additional 6.60 mm. How much work is done by the spring force to ready the pen for writing? Be sure to include the proper algebraic sign with your answer.
**Solution Approach:**
To solve this problem, we need to calculate the work done by the spring force. The work \( W \) done by a spring is given by Hooke's Law:
\[ W = \frac{1}{2} k (x_f^2 - x_i^2) \]
where:
- \( k \) is the spring constant (240 N/m in this case),
- \( x_i \) is the initial compression (5.00 mm or 0.005 m),
- \( x_f \) is the final compression (5.00 mm + 6.60 mm = 11.60 mm or 0.0116 m).
### Step-by-Step Calculation:
1. **Convert all measurements to meters:**
\- Initial compression: \( x_i = 0.005 \) meters.
\- Final compression: \( x_f = 0.0116 \) meters.
2. **Substitute the values into the work formula:**
\[ W = \frac{1}{2} \cdot 240 \, \frac{\text{N}}{\text{m}} \left((0.0116)^2 - (0.005)^2\right) \]
3. **Calculate each squared term:**
\[ (0.0116 \, \text{m})^2 = 0.00013456 \, \text{m}^2 \]
\[ (0.005 \, \text{m})^2 = 0.000025 \, \text{m}^2 \]
4. **Find the difference of the squared terms:**
\[ 0.00013456 \, \text{m}^2 - 0.000025 \, \text{m}^2 = 0.00010956 \, \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70495d9e-d834-4a3c-919e-f23d4c42494e%2F83a2d86c-212c-4bba-b323-6a247c79fbb4%2Ffnax12t_processed.png&w=3840&q=75)
Transcribed Image Text:### Physics of Springs: Work Done by a Spring Force in a Pen Mechanism
**Problem Statement:**
A pen contains a spring with a spring constant of 240 N/m. When the tip of the pen is in its retracted position, the spring is compressed 5.00 mm from its unstrained length. In order to push the tip out and lock it into its writing position, the spring must be compressed an additional 6.60 mm. How much work is done by the spring force to ready the pen for writing? Be sure to include the proper algebraic sign with your answer.
**Solution Approach:**
To solve this problem, we need to calculate the work done by the spring force. The work \( W \) done by a spring is given by Hooke's Law:
\[ W = \frac{1}{2} k (x_f^2 - x_i^2) \]
where:
- \( k \) is the spring constant (240 N/m in this case),
- \( x_i \) is the initial compression (5.00 mm or 0.005 m),
- \( x_f \) is the final compression (5.00 mm + 6.60 mm = 11.60 mm or 0.0116 m).
### Step-by-Step Calculation:
1. **Convert all measurements to meters:**
\- Initial compression: \( x_i = 0.005 \) meters.
\- Final compression: \( x_f = 0.0116 \) meters.
2. **Substitute the values into the work formula:**
\[ W = \frac{1}{2} \cdot 240 \, \frac{\text{N}}{\text{m}} \left((0.0116)^2 - (0.005)^2\right) \]
3. **Calculate each squared term:**
\[ (0.0116 \, \text{m})^2 = 0.00013456 \, \text{m}^2 \]
\[ (0.005 \, \text{m})^2 = 0.000025 \, \text{m}^2 \]
4. **Find the difference of the squared terms:**
\[ 0.00013456 \, \text{m}^2 - 0.000025 \, \text{m}^2 = 0.00010956 \, \
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