Let's break down the problem step by step: ### **Problem Restatement** A cable weighing **6 lb/ft** is used to lift **800 ft** of cable up a mine shaft that is **800 ft deep**. Find the **total work done**. --- ### **Step 1: Understanding the Problem** - **Cable length:** 800 ft - **Weight per foot:** 6 lb/ft - **Depth:** 800 ft (so, entire cable hangs vertically down the shaft) When you start lifting, you are lifting the entire 800 ft of cable. As you pull up, less and less cable remains hanging, so the weight you are lifting decreases. --- ### **Step 2: Riemann Sum Approximation** Suppose we break the cable into \( n \) small pieces, each of length \( \Delta x \). At a depth \( x \) **from the top** (i.e., \( x = 0 \) at the top, \( x = 800 \) at the bottom): - The amount of cable **still hanging** is \( 800 - x \) feet. - The weight hanging at depth \( x \) is: \[ \text{Weight} = 6 \text{ lb/ft} \times (800 - x) \text{ ft} \] The small amount of work to lift a piece \( \Delta x \) at depth \( x \) to the top is: \[ \text{Work for small piece} = \text{weight of hanging cable} \times \text{distance to lift} = 6(800 - x) \Delta x \] So, the Riemann sum for the total work is: \[ \sum_{i=1}^n 6[800 - x_i] \Delta x \] where \( x_i \) is the distance from the top to the \(i\)-th piece. --- ### **Step 3: Express as an Integral** As \( n \to \infty \), the Riemann sum becomes the integral: \[ \text{Total Work} = \int_{x=0}^{x=800} 6(800 - x) \, dx \] --- ### **Step 4: Evaluate the Integral** Calculate: \[ \int_{0}^{800} 6(800 - x) \, dx \] Expand: \[ = 6 \int_{0}^{800} (800 - x) dx = 6 \left[ 800x - \frac{1}{2}x^2 \right]_{0}^{800} \] Evaluate at the bounds: - At \( x = 800 \): \( 800 \times 800 - \frac{1}{2}(800)^2 = 640,000 - 320,000 = 320,000 \) - At \( x = 0 \): \( 0 - 0 = 0 \) So, \[ 6 \times (320,000 - 0) = 1,920,000 \text{ ft-lb} \] --- ### **Final Answers** **Step-by-step answers:** 1. **Riemann sum:** \[ \sum_{i=1}^n 6[800 - x_i] \Delta x \] 2. **Integral form:** \[ \int_0^{800} 6(800 - x) \, dx \] 3. **Evaluate the integral:** \[ 1,920,000 \text{ ft-lb} \] --- **Boxed Final Answer:** \[ \boxed{1,920,000} \text{ ft-lb} \]

Let's carefully work through this problem with detailed calculations. --- ### **Problem Restatement** - A cable weighing **6 lb/ft** is used to lift **800 ft** of cable from a mine shaft that is **800 ft deep**. - Find the **total work done** to lift the entire cable to the top. --- ### **Understanding the problem** - **Cable length:** 800 ft - **Weight per foot:** 6 lb/ft - **Depth of shaft:** 800 ft As we lift the cable, the length of cable remaining hanging decreases from 800 ft to 0 ft. The key point: **The work to lift a small segment of cable depends on how much cable is still hanging at that position**, and the distance it needs to be lifted. --- ### **Step 1: Set up variables** Let: - \( x \) = the depth from the top (so \( x = 0 \) at the top, \( x = 800 \) at the bottom). - \( dx \) = a small segment of the cable at depth \( x \). At depth \( x \): - The length of cable still hanging is \( 800 - x \) ft. - The weight of this segment is: \[ \text{Weight} = 6 \times (800 - x) \quad \text{(lb)} \] The effort to lift this segment from depth \( x \) to the top involves lifting it a vertical distance of \( x \) ft. --- ### **Step 2: Express differential work** The differential work \( dW \) to lift the segment at depth \( x \): \[ dW = \text{(weight of segment)} \times \text{distance to lift} = 6(800 - x) \times x \] This is because the segment at depth \( x \) must be lifted **to the top**, a distance of \( x \) ft. --- ### **Step 3: Set up the integral for total work** Total work \( W \): \[ W = \int_{x=0}^{800} 6(800 - x) \times x\, dx \] --- ### **Step 4: Simplify the integrand** Multiply out: \[ W = 6 \int_0^{800} x(800 - x) dx \] Distribute: \[ W = 6 \int_0^{800} (800x - x^2) dx \] --- ### **Step 5: Integrate** Compute each integral separately: \[ W = 6 \left[ \int_0^{800} 800x dx - \int_0^{800} x^2 dx \right] \] Calculate the first integral: \[ \int_0^{800} 800x dx = 800 \times \frac{x^2}{2} \bigg|_0^{800} = 800 \times \frac{(800)^2}{2} \] Calculate the second integral: \[ \int_0^{800} x^2 dx = \frac{x^3}{3} \bigg|_0^{800} = \frac{(800)^3}{3} \] --- ### **Step 6: Plug in the limits** Compute: - \( (800)^2 = 640,000 \) - \( (800)^3 = 640,000 \times 800 = 512,000,000 \) Now: \[ W = 6 \left[ 800 \times \frac{640,000}{2} - \frac{512,000,000}{3} \right] \] Simplify each term: \[ 800 \times \frac{640,000}{2} = 800 \times 320,000 = 256,000,000 \] So, \[ W = 6 \left[ 256,000,000 - \frac{512,000,000}{3} \right] \] Express both terms with denominator 3: \[ 256,000,000 = \frac{768,000,000}{3} \] Thus, \[ W = 6 \left[ \frac{768,000,000}{3} - \frac{512,000,000}{3} \right] = 6 \times \frac{(768,000,000 - 512,000,000)}{3} \] Subtract numerator: \[ 768,000,000 - 512,000,000 = 256,000,000 \] Now, \[ W = 6 \times \frac{256,000,000}{3} = 6 \times 85,333,333.\overline{3} \] Finally, \[ W = 6 \times 85,333,333.\overline{3} = 512,000,000 \] --- ### **### **Final Answer:** \[ \boxed{ \textbf{Total work} = 512,000,000 \text{ ft-lb} } \] **Therefore, the total work done to lift the 800 ft of cable is \(\boxed{512,000,000}\) foot-pounds.**