a 5 meter rope is lying on the floor and has a weight of 40 N in total. the left end of the rope is lifted. the pulley makes the lifted part of the rope perpendicular to the floor and there is no friction between the rope and floor. how much work is required to lift up one end of the rope to a height of 3 meters?

To solve this problem, we need to calculate the work required to lift one end of the rope to a height of 3 meters. The key idea is that as the rope is lifted, the work done is equivalent to the potential energy gained by the rope.

### Step 1: Determine the mass per unit length
The weight of the rope is 40 N, and its length is 5 meters. The weight per unit length is given by:
\[
\text{Weight per unit length} = \frac{40 \text{ N}}{5 \text{ m}} = 8 \text{ N/m}
\]

### Step 2: Consider the work required to lift the rope
When lifting one end of the rope to a height of 3 meters, different parts of the rope will be lifted to different heights, ranging from 0 meters to 3 meters.

Let \( y \) be the height at a particular point on the rope. The length of the rope at height \( y \) is \( \frac{y}{3} \times 5 = \frac{5y}{3} \).

The weight of the small segment of the rope at height \( y \) is:
\[
\text{Weight of segment} = \text{(Weight per unit length)} \times \text{(Length of segment)} = 8 \text{ N/m} \times \frac{5y}{3}
\]
This small segment's contribution to the work done is:
\[
\text{Work for small segment} = \text{Weight of segment} \times y = 8 \times \frac{5y}{3} \times y = \frac{40y^2}{3} \text{ J}
\]

### Step 3: Integrate to find total work
To find the total work required to lift the rope, we integrate the work done over the height of the rope from 0 to 3 meters:
\[
W = \int_0^3 \frac{40y^2}{3} \, dy
\]
Evaluating this integral:
\[
W = \frac{40}{3} \int_0^3 y^2 \, dy
\]
The integral of \( y^2 \) is \( \frac{y^3}{3} \), so:
\[
W = \frac{40}{3} \times \left[ \frac{y^3}{3} \right]_0^3 = \frac{40}{9} \times \left[ 27 - 0 \right] = \frac{40 \times 27}{9} = 120 \text{ J}
\]

### Final Answer:
The total work required to lift one end of the rope to a height of 3 meters is **120 Joules**.

Would you like further details on this solution or have any questions?

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**Related Questions:**
1. How does the work change if the rope is 10 meters long?
2. What if the rope had a varying weight per unit length?
3. How would friction between the rope and the floor affect the work done?
4. What happens to the work done if the rope is lifted to a height of 5 meters?
5. How is the center of mass of the rope relevant to this problem?

**Tip:** When dealing with problems involving variable forces or varying mass distribution, integration is often the key to finding the work done.

A 5 meter rope is lying on the floor and has a weight of 40 N in total. The left end of the rope is lifted. The pulley makes the lifted part of the rope perpendicular to the floor and there is no friction between the rope and floor. How much work is required to lift up one end of the rope to a height of 3 meters?

Discover how to calculate the work required to lift one end of a 5-meter rope, weighing 40 N, to a height of 3 meters. This solution involves key concepts from physics and calculus, such as work-energy theorem and integration, making it suitable for students in Grades 11-12.

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