A force of 2 pounds is required to hold a spring stretched 0.4 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.8 feet beyond its natural length? 0.8
A force of 2 pounds is required to hold a spring stretched 0.4 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.8 feet beyond its natural length? 0.8
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement: Spring Work Calculation**
A force of 2 pounds is required to hold a spring stretched 0.4 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.8 feet beyond its natural length?
**Input Box Value:** 0.8
---
**Explanation:**
This problem involves calculating the work done in stretching a spring. In physics, the work \( W \) done in stretching or compressing a spring is given by the formula:
\[ W = \int kx \, dx \]
where:
- \( W \) is the work done,
- \( k \) is the spring constant (in pounds per foot),
- \( x \) is the distance stretched (in feet).
**Given:**
- Force required for 0.4 feet: 2 pounds.
First, calculate the spring constant \( k \) using Hooke's Law:
\[ F = kx \]
\[ 2 = k(0.4) \]
\[ k = \frac{2}{0.4} \]
\[ k = 5 \, \text{pounds per foot} \]
**Calculate Work Done:**
- From the natural length to 0.8 feet, the work done is:
\[ W = \int_0^{0.8} 5x \, dx \]
Compute this integral:
\[ W = 5 \left[ \frac{x^2}{2} \right]_0^{0.8} \]
\[ W = 5 \left( \frac{0.8^2}{2} - \frac{0^2}{2} \right) \]
\[ W = 5 \left( \frac{0.64}{2} \right) \]
\[ W = 5 \times 0.32 \]
\[ W = 1.6 \, \text{foot-pounds} \]
Therefore, the work done in stretching the spring from its natural length to 0.8 feet beyond its natural length is 1.6 foot-pounds.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a3afdbf-a9dd-42b8-8403-1d42a5de04b1%2F333d0a42-7cc1-44d7-b0c3-439349f8fe42%2Fufofn8i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement: Spring Work Calculation**
A force of 2 pounds is required to hold a spring stretched 0.4 feet beyond its natural length. How much work (in foot-pounds) is done in stretching the spring from its natural length to 0.8 feet beyond its natural length?
**Input Box Value:** 0.8
---
**Explanation:**
This problem involves calculating the work done in stretching a spring. In physics, the work \( W \) done in stretching or compressing a spring is given by the formula:
\[ W = \int kx \, dx \]
where:
- \( W \) is the work done,
- \( k \) is the spring constant (in pounds per foot),
- \( x \) is the distance stretched (in feet).
**Given:**
- Force required for 0.4 feet: 2 pounds.
First, calculate the spring constant \( k \) using Hooke's Law:
\[ F = kx \]
\[ 2 = k(0.4) \]
\[ k = \frac{2}{0.4} \]
\[ k = 5 \, \text{pounds per foot} \]
**Calculate Work Done:**
- From the natural length to 0.8 feet, the work done is:
\[ W = \int_0^{0.8} 5x \, dx \]
Compute this integral:
\[ W = 5 \left[ \frac{x^2}{2} \right]_0^{0.8} \]
\[ W = 5 \left( \frac{0.8^2}{2} - \frac{0^2}{2} \right) \]
\[ W = 5 \left( \frac{0.64}{2} \right) \]
\[ W = 5 \times 0.32 \]
\[ W = 1.6 \, \text{foot-pounds} \]
Therefore, the work done in stretching the spring from its natural length to 0.8 feet beyond its natural length is 1.6 foot-pounds.
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