Lifting an with a motor at the top has a multistrand cable weighing 4.5 lb/ft. When the car is at the first floor, 180 ft of cable are paid out, and effectively 0 ft are out when the car is at the top floor. How much work does the motor do just lifting the cable when it takes the car from the first floor to the top?
Lifting an with a motor at the top has a multistrand cable weighing 4.5 lb/ft. When the car is at the first floor, 180 ft of cable are paid out, and effectively 0 ft are out when the car is at the top floor. How much work does the motor do just lifting the cable when it takes the car from the first floor to the top?
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![### Lifting an Elevator Cable
**Problem Statement:**
An electric elevator with a motor at the top has a multistrand cable weighing 4.5 lb/ft. When the car is at the first floor, 180 ft of cable are paid out, and effectively 0 ft are out when the car is at the top floor. How much work does the motor do just lifting the cable when it takes the car from the first floor to the top?
**Explanation and Solution:**
To understand and solve this problem, let's break it down step by step:
1. **Cable Weight Distribution:**
- Cable weight: 4.5 pounds per foot (lb/ft).
- Total length of the cable: 180 feet.
2. **Initial and Final Conditions:**
- **Initial Condition (First Floor):** The entire cable length of 180 feet is out.
- **Final Condition (Top Floor):** No cable length is out (0 feet).
3. **Concept of Work:**
- Work done (W) against gravity is expressed as \( W = \int F \cdot dh \), where \( F \) is the force applied over a distance \( h \).
4. **Force Due to Gravity:**
- The force exerted by the cable weight at any height \( h \) is proportional to the length of the cable remaining below that height.
- Since the weight per unit length is 4.5 pounds per foot, for a given height \( x \), the remaining length of the cable is \( (180 - x) \).
5. **Setting Up the Integral:**
- Force at height \( x \) is given by \( F(x) = 4.5 \times (180 - x) \).
- The work done by the motor in lifting the cable from 0 to 180 feet is the integral of the force over the distance:
\[
W = \int_0^{180} 4.5 (180 - x) \, dx
\]
6. **Calculating the Integral:**
- Simplifying the integral:
\[
W = 4.5 \int_0^{180} (180 - x) \, dx
\]
- We split the integral:
\[
W = 4.5 \left[ \int_0^{180} 180 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a617c73-16c8-405c-b9c4-72dd4741d892%2Ff67a424d-57fd-4c1e-9480-4418789db230%2Feoaqtye_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Lifting an Elevator Cable
**Problem Statement:**
An electric elevator with a motor at the top has a multistrand cable weighing 4.5 lb/ft. When the car is at the first floor, 180 ft of cable are paid out, and effectively 0 ft are out when the car is at the top floor. How much work does the motor do just lifting the cable when it takes the car from the first floor to the top?
**Explanation and Solution:**
To understand and solve this problem, let's break it down step by step:
1. **Cable Weight Distribution:**
- Cable weight: 4.5 pounds per foot (lb/ft).
- Total length of the cable: 180 feet.
2. **Initial and Final Conditions:**
- **Initial Condition (First Floor):** The entire cable length of 180 feet is out.
- **Final Condition (Top Floor):** No cable length is out (0 feet).
3. **Concept of Work:**
- Work done (W) against gravity is expressed as \( W = \int F \cdot dh \), where \( F \) is the force applied over a distance \( h \).
4. **Force Due to Gravity:**
- The force exerted by the cable weight at any height \( h \) is proportional to the length of the cable remaining below that height.
- Since the weight per unit length is 4.5 pounds per foot, for a given height \( x \), the remaining length of the cable is \( (180 - x) \).
5. **Setting Up the Integral:**
- Force at height \( x \) is given by \( F(x) = 4.5 \times (180 - x) \).
- The work done by the motor in lifting the cable from 0 to 180 feet is the integral of the force over the distance:
\[
W = \int_0^{180} 4.5 (180 - x) \, dx
\]
6. **Calculating the Integral:**
- Simplifying the integral:
\[
W = 4.5 \int_0^{180} (180 - x) \, dx
\]
- We split the integral:
\[
W = 4.5 \left[ \int_0^{180} 180 \
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