https://file.aiquickdraw.com/mathbot/file/1727055764987ocax7dmm.jpg

This problem asks you to find the work done to lift both a cable and coal up a 700 ft deep mine shaft, using both a Riemann sum and integral approach. Let's break it down step by step.

### Step 1: Work done using a Riemann sum

We need to approximate the work using a Riemann sum. Let \( x_i^* \) be the distance in feet below the top of the shaft. The sum is:

\[
\lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x
\]

where \( f(x_i^*) \) represents the force required to lift the cable and coal at height \( x_i^* \).

### Step 2: Express the work as an integral

There are two parts to the work: 
1. **Lifting the coal**: This is simply the weight of the coal (900 lb) multiplied by the depth (700 ft). 
   
   The force to lift the coal is constant: \( 900 \, \text{lb} \), and it needs to be lifted 700 feet, so:
   \[
   \text{Work}_{\text{coal}} = 900 \times 700 \, \text{ft-lb}
   \]

2. **Lifting the cable**: The cable weighs \( 8 \, \text{lb/ft} \), and the force required to lift it depends on how much of the cable is still below the top of the shaft. If \( x \) is the distance from the top, the weight of the cable at depth \( x \) is \( 8 \times (700 - x) \, \text{lb} \).

   Therefore, the total work for lifting the cable is expressed as:
   \[
   \text{Work}_{\text{cable}} = \int_0^{700} 8x \, dx
   \]

### Step 3: Evaluate the integral

We need to evaluate the integral for the work done lifting the cable:

\[
\int_0^{700} 8x \, dx
\]

Let’s compute this now.The work required to lift the cable is \( 1,960,000 \, \text{ft-lb} \).

### Step 4: Total work done

Now, let's add the work done to lift the coal and the cable:
\[
\text{Total work} = \text{Work}_{\text{coal}} + \text{Work}_{\text{cable}} = 900 \times 700 + 1,960,000
\]

Let's compute the total work now.The total work done to lift both the coal and the cable is \( 2,590,000 \, \text{ft-lb} \).

Would you like more details or have any questions about this problem?

Here are some related questions you might explore:
1. What is the general formula for work when force varies with distance?
2. How does the work change if the weight of the cable increases to 10 lb/ft?
3. How would the integral change if the depth of the mine was different?
4. Can this method be used to calculate work done in lifting objects in a circular well instead of a straight shaft?
5. What is the importance of using Riemann sums in approximating work?

**Tip:** When solving work problems, always identify if the force is constant or variable along the distance to decide between using simple multiplication or integration!

A cable that weighs 8 lb/ft is used to lift 900 lb of coal up a mine shaft 700 ft deep. Find the work done. Show how to approximate the required work by a Riemann sum. Express the work as an integral and evaluate the integral.

Learn how to calculate the work required to lift a 900 lb load of coal and a cable weighing 8 lb/ft up a 700 ft mine shaft. This problem uses Riemann sums and integration, covering key calculus concepts such as variable force and work. Ideal for undergraduate calculus and physics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation