We are tasked with finding the work done to raise the bottom of a 35-pound, 10-foot-long rope to the same height as the top of the rope, essentially folding the rope in half.

Step-by-Step Breakdown

1. Work formula: The formula for the work $W$ done to lift an object is given by:

   $W = \int_{a}^{b} F(x) \, dx$

   Where:

   • $F(x)$ is the force applied as a function of height $x$,
   • $a$ and $b$ are the limits of the distance over which the force is applied (in this case, 0 to 10 feet).

2. Uniform rope assumption: The weight of the rope is uniformly distributed along its length. If the total weight of the rope is 35 pounds and the rope is 10 feet long, the weight per unit length is:

   $\text{Weight per foot} = \frac{35 \, \text{lb}}{10 \, \text{ft}} = 3.5 \, \text{lb/ft}$

3. Consider the height:

   • When we lift the rope from its bottom, the bottom part of the rope (near $x = 10$) must be lifted the full distance (10 feet), while parts of the rope closer to the top are lifted less.
   • Each small piece of the rope at height $x$ (from the top of the rope) has to be lifted a distance $10 - x$ feet.

4. Work on a small slice of rope: A small slice of rope at height $x$ has weight $3.5 \, dx$, where $dx$ is a tiny piece of the rope's length. This slice must be lifted $10 - x$ feet. The work done to lift this slice is:

   $dW = 3.5 \cdot (10 - x) \, dx$

5. Total work: To find the total work, integrate this expression from $x = 0$ (top of the rope) to $x = 10$ (bottom of the rope):

   $W = \int_0^{10} 3.5 \cdot (10 - x) \, dx$

   This simplifies to:

   $W = 3.5 \int_0^{10} (10 - x) \, dx$

6. Solve the integral:

   $W = 3.5 \left[ 10x - \frac{x^2}{2} \right]_0^{10}$

   Evaluating at the limits:

   $W = 3.5 \left[ (10 \cdot 10) - \frac{10^2}{2} \right]$ $W = 3.5 \left[ 100 - 50 \right] = 3.5 \cdot 50 = 175 \, \text{ft-lb}$

Final Answer:

The work done to raise the bottom of the rope to the top is 175 ft-lb.

Would you like more details or have any questions?

Related Questions:

1. How would the work change if the rope's length was increased?
2. How is the concept of center of mass related to this problem?
3. Can you explain why we used integration to solve this?
4. What if the rope was not uniformly distributed in weight?
5. How does the force change as the rope is lifted higher?

Tip:

In work problems involving variable forces, integrating the force over the distance is often essential to capture the changing force accurately.