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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![**Title: Work Done in Stretching and Compressing a Spring on a Horizontal Surface**
### Problem Statement:
A spring on a horizontal surface can be stretched and held 0.4 m from its equilibrium position with a force of 40 N.
1. **Questions:**
a. How much work is done in stretching the spring 2.5 m from its equilibrium position?
b. How much work is done in compressing the spring 1.5 m from its equilibrium position?
2. **Integral Setup:**
Set up the integral that gives the work done in stretching the spring 2.5 m from its equilibrium position. Use increasing limits of integration.
∫[limits] ( ) dx
(Type exact answers.)
### Explanation:
#### Understanding the Problem:
To solve the problem, we need to calculate the work done in stretching and compressing a spring using the concept of work and Hooke's Law. The force needed to stretch or compress a spring is given by Hooke's Law:
\[ F(x) = kx \]
Where:
- \( F \) is the force.
- \( k \) is the spring constant.
- \( x \) is the displacement from the equilibrium position.
#### Step-by-Step Solution:
**Step 1: Determine the Spring Constant (\( k \))**
Given:
\[ x = 0.4 \, m \]
\[ F = 40 \, N \]
Using Hooke’s Law:
\[ F = kx \]
\[ k = \frac{F}{x} = \frac{40 \, N}{0.4 \, m} = 100 \, N/m \]
**Step 2: Calculate the Work Done in Stretching the Spring 2.5 m**
The work done in stretching or compressing a spring is given by:
\[ W = \int_0^x F(x) \, dx \]
\[ F(x) = kx \]
So,
\[ W = \int_0^x kx \, dx = \int_0^{2.5} 100x \, dx \]
Evaluating the integral:
\[ W = 100 \int_0^{2.5} x \, dx = 100 \left[ \frac{x^2}{2} \right]_0^{2.5} = 100 \left( \frac{(2.5)^2}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0bf917ec-3bd5-4ecc-ad53-7da034b3ca56%2Fe84010fe-7d49-4734-b9c5-7315058ca3ff%2Fprwko6.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Work Done in Stretching and Compressing a Spring on a Horizontal Surface**
### Problem Statement:
A spring on a horizontal surface can be stretched and held 0.4 m from its equilibrium position with a force of 40 N.
1. **Questions:**
a. How much work is done in stretching the spring 2.5 m from its equilibrium position?
b. How much work is done in compressing the spring 1.5 m from its equilibrium position?
2. **Integral Setup:**
Set up the integral that gives the work done in stretching the spring 2.5 m from its equilibrium position. Use increasing limits of integration.
∫[limits] ( ) dx
(Type exact answers.)
### Explanation:
#### Understanding the Problem:
To solve the problem, we need to calculate the work done in stretching and compressing a spring using the concept of work and Hooke's Law. The force needed to stretch or compress a spring is given by Hooke's Law:
\[ F(x) = kx \]
Where:
- \( F \) is the force.
- \( k \) is the spring constant.
- \( x \) is the displacement from the equilibrium position.
#### Step-by-Step Solution:
**Step 1: Determine the Spring Constant (\( k \))**
Given:
\[ x = 0.4 \, m \]
\[ F = 40 \, N \]
Using Hooke’s Law:
\[ F = kx \]
\[ k = \frac{F}{x} = \frac{40 \, N}{0.4 \, m} = 100 \, N/m \]
**Step 2: Calculate the Work Done in Stretching the Spring 2.5 m**
The work done in stretching or compressing a spring is given by:
\[ W = \int_0^x F(x) \, dx \]
\[ F(x) = kx \]
So,
\[ W = \int_0^x kx \, dx = \int_0^{2.5} 100x \, dx \]
Evaluating the integral:
\[ W = 100 \int_0^{2.5} x \, dx = 100 \left[ \frac{x^2}{2} \right]_0^{2.5} = 100 \left( \frac{(2.5)^2}{
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