A spring on a horizontal surface can be stretched and held 0.4 m from its equilibrium position with a force of 40 N.

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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}{
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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