Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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**Problem Statement:**

A 4-meter chain with linear mass density \(\rho(x) = 2x(5 - x)\) kg/m lies on the ground. Calculate the work required to lift the chain from its front end so that its bottom is 3 meters above the ground.

*Note: Round your answer to one decimal place.*

**Solution:**

To solve this problem, we need to calculate the work done in lifting the chain. The chain's linear mass density is a function of position, \(x\), and is given by \(\rho(x) = 2x(5 - x)\).

### Explanation:

1. **Linear Mass Density:** 
   - \(\rho(x) = 2x(5 - x)\) kg/m, where \(x\) is the position along the chain.

2. **Total Length of the Chain:**
   - Length = 4 m

3. **Objective:**
   - Lift the chain so that its bottom is 3 meters above the ground.

4. **Work Calculation:**
   - The work done on a small segment of the chain can be calculated using the integral of force over distance:
   - \(\text{Work} = \int \text{Force} \cdot \text{Distance}\)

5. **Integration and Calculation:**
   - As the chain is lifted, each small segment \(dx\) moves varying distances depending on its initial position \(x\).
   - The limits of integration will need to be from 0 to 4 (the entire length of the chain).
   - For each segment, the force is given by the weight \(\rho(x) \cdot g \cdot dx\), where \(g\) is the gravitational acceleration.
   - We integrate this force over the distance moved by each segment to find the total work done.

6. **Final Result:**
   - Calculate the work and round it to one decimal place to find the required energy in Joules.

**Answer Field:**

\[ W = \_\_\_\_\_\_ \ \text{J} \]
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Transcribed Image Text:**Problem Statement:** A 4-meter chain with linear mass density \(\rho(x) = 2x(5 - x)\) kg/m lies on the ground. Calculate the work required to lift the chain from its front end so that its bottom is 3 meters above the ground. *Note: Round your answer to one decimal place.* **Solution:** To solve this problem, we need to calculate the work done in lifting the chain. The chain's linear mass density is a function of position, \(x\), and is given by \(\rho(x) = 2x(5 - x)\). ### Explanation: 1. **Linear Mass Density:** - \(\rho(x) = 2x(5 - x)\) kg/m, where \(x\) is the position along the chain. 2. **Total Length of the Chain:** - Length = 4 m 3. **Objective:** - Lift the chain so that its bottom is 3 meters above the ground. 4. **Work Calculation:** - The work done on a small segment of the chain can be calculated using the integral of force over distance: - \(\text{Work} = \int \text{Force} \cdot \text{Distance}\) 5. **Integration and Calculation:** - As the chain is lifted, each small segment \(dx\) moves varying distances depending on its initial position \(x\). - The limits of integration will need to be from 0 to 4 (the entire length of the chain). - For each segment, the force is given by the weight \(\rho(x) \cdot g \cdot dx\), where \(g\) is the gravitational acceleration. - We integrate this force over the distance moved by each segment to find the total work done. 6. **Final Result:** - Calculate the work and round it to one decimal place to find the required energy in Joules. **Answer Field:** \[ W = \_\_\_\_\_\_ \ \text{J} \]
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