What is the work done in raising a 18 m rope to the top of the building? The rope weighs 2.3 kg/m and is hanging over the edge of a 37 m tall building so that 9 m of the rope is already on the top of the building?  Take g=9.81m/sec^2 and give an answer that is correctly rounded to 3 decimal digits without the units.

The total work done in raising the rope involves lifting the remaining \( 18 - 9 = 9 \) meters of rope that is still hanging over the edge of the building.

### Step-by-step solution:

1. **Weight of rope per unit length:**
   - The rope weighs \( 2.3 \, \text{kg/m} \).
   - Gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \).

   Therefore, the weight per meter of the rope is:
   \[
   \text{Weight per meter} = 2.3 \times 9.81 = 22.563 \, \text{N/m}.
   \]

2. **Remaining length of rope to be lifted:**
   - The length of the rope to be lifted is \( 9 \, \text{m} \).

3. **Work done in lifting a small section of the rope:**
   Consider an infinitesimal segment of the rope at a height \( x \) meters below the top of the building. The force required to lift this segment is the weight per unit length times its length, and the distance this segment must be lifted is \( x \) meters.

   The work done to lift a small piece of the rope is:
   \[
   dW = \text{Weight per meter} \times x \, dx = 22.563 \times x \, dx.
   \]

4. **Total work done:**
   To find the total work, integrate from 0 to 9 (since we are lifting the bottom part of the remaining 9 meters of rope):
   \[
   W = \int_0^9 22.563 \, x \, dx.
   \]

   This is a simple integral:
   \[
   W = 22.563 \times \frac{x^2}{2} \Big|_0^9.
   \]

   Evaluating the integral:
   \[
   W = 22.563 \times \frac{9^2}{2} = 22.563 \times 40.5 = 913.79265 \, \text{J}.
   \]

5. **Final Answer:**
   Rounded to three decimal places, the work done is:
   \[
   \boxed{913.793}.
   \]

Let me know if you would like any further details or explanations!

### Related questions:
1. How does the total work change if the rope weighs 3 kg/m instead of 2.3 kg/m?
2. What would be the work done if the entire 18 m of rope needed to be lifted?
3. How does gravitational acceleration affect the total work done?
4. If only 5 meters of the rope needed to be lifted, how much work would be done?
5. What is the work done if the rope is lifted halfway (4.5 meters) instead of fully 9 meters?

### Tip:
When dealing with integrals for physical systems, always ensure that your limits and the variable in the integrand accurately represent the system's setup!

What is the work done in raising an 18 m rope to the top of the building? The rope weighs 2.3 kg/m and is hanging over the edge of a 37 m tall building so that 9 m of the rope is already on the top of the building? Take g=9.81m/sec^2 and give an answer that is correctly rounded to 3 decimal digits without the units.

Learn how to calculate the work done in raising an 18-meter rope over a building with a detailed solution involving calculus and the work-energy principle. The rope weighs 2.3 kg/m, and 9 meters of it are already on top. This problem covers definite integration and is suitable for Grades 11-12 physics and calculus students.

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