A 4-m chain with linear mass density p(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 m above ground. (Round your answer to one decimal place.) W =
A 4-m chain with linear mass density p(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 m above ground. (Round your answer to one decimal place.) W =
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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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:**
\[ W = \_\_\_\_\_\_ \ \text{J} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc9b629bd-7b7c-4b52-85cc-46a95271ddc8%2F3963d6e1-40b2-480b-8ca3-f8568518deea%2Fypio86_processed.png&w=3840&q=75)
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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