Peyton is using a rope to lift a bucket from the bottom of a 25-foot-deep well. The rope weighs 0.6 lbs/ft. The bucket (by itself) weighs 3 lbs and is *initially* holding 3 lbs of water. Unfortunately, Peyton is lifting this bucket in the middle of a torrential downpour. By the time the bucket reaches the top of the well, it contains 8 lbs of water.

Assuming she lifted the bucket at a constant rate and the rain is filling the bucket at a constant rate, set up, **but do not evaluate**, an integral to represent the work Peyton did in lifting this bucket.


To set up the integral representing the work Peyton did in lifting the bucket, we need to consider the varying forces she had to overcome as the bucket was lifted.

### Key components:
- **Depth of the well**: 25 feet.
- **Weight of the rope**: 0.6 lbs/ft.
- **Weight of the bucket**: 3 lbs.
- **Initial weight of water in the bucket**: 3 lbs (increases to 8 lbs by the time it reaches the top).
- **The rate at which water fills the bucket**: \(\frac{8 \text{ lbs} - 3 \text{ lbs}}{25 \text{ feet}} = \frac{5}{25} = 0.2 \text{ lbs/ft}\).

The work done in lifting the bucket can be found by integrating the force needed to lift the bucket over the distance.

### Step-by-step formulation:

1. **Weight of the rope**: 
   The length of the rope at any point \(x\), where \(x\) is the distance from the bottom of the well, is \(x\) feet, so the weight of the rope at any point \(x\) is \(0.6x\) lbs.

2. **Weight of the bucket**: 
   The weight of the bucket itself is constant at 3 lbs.

3. **Weight of the water**: 
   The weight of the water is increasing linearly as the bucket is lifted. At height \(x\) (where \(x = 0\) is the bottom and \(x = 25\) is the top), the weight of the water is:
   \[
   3 + 0.2x \, \text{lbs}.
   \]
   
4. **Total force at height \(x\)**:
   The total force Peyton is lifting at any point \(x\) (the sum of the weight of the rope, the bucket, and the water) is:
   \[
   0.6x + 3 + (3 + 0.2x) = 0.6x + 6 + 0.2x = 0.8x + 6.
   \]
   
5. **Work done**: 
   The work done to lift the bucket from the bottom of the well to the top is the integral of the force over the distance:
   \[
   W = \int_0^{25} (0.8x + 6) \, dx.
   \]

This integral represents the total work Peyton did in lifting the bucket from the bottom of the well to the top, accounting for the weight of the rope, the bucket, and the water.

---

Would you like further details on the components or help with solving the integral? Here are five related questions:

1. How does the varying weight of the water affect the work Peyton does?
2. What would the work integral look like if there were no water filling the bucket?
3. How would the problem change if the rope weighed more per foot?
4. Could we use a different method to calculate the work besides using an integral?
5. What is the significance of the integral limits in this problem?

**Tip**: When dealing with work problems in physics, always break down the forces involved and how they change over time or distance to set up the integral correctly.

Peyton is using a rope to lift a bucket from the bottom of a 25-foot-deep well. The rope weighs 0.6 lbs/ft. The bucket (by itself) weighs 3 lbs and is initially holding 3 lbs of water. Unfortunately, Peyton is lifting this bucket in the middle of a torrential downpour. By the time the bucket reaches the top of the well, it contains 8 lbs of water. Assuming she lifted the bucket at a constant rate and the rain is filling the bucket at a constant rate, set up, but do not evaluate, an integral to represent the work Peyton did in lifting this bucket.

Learn how to set up an integral to calculate the work done in lifting a bucket from a 25-foot-deep well while accounting for varying forces, including the weight of the rope, the bucket, and the increasing weight of water during a rainstorm. This problem involves calculus concepts and is suitable for Grades 11-12 or early college students.

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