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(Figure 1) shows two 1.0 kgkg blocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 gg. The entire assembly is accelerated upward at 3.0 m/s2m/s2 by force F⃗ →. What is the tension at the top end of rope 2?

We are tasked with finding the tension at the top end of "Rope 2" in the system shown in the image.

### Step-by-Step Solution

#### Given:
- Mass of block A, \( m_A = 1.0 \, \text{kg} \)
- Mass of block B, \( m_B = 1.0 \, \text{kg} \)
- Mass of each rope = \( 250 \, \text{g} = 0.25 \, \text{kg} \)
- Acceleration \( a = 3.0 \, \text{m/s}^2 \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)

#### Step 1: Analyze the forces on Block B
We will begin by calculating the forces acting on Block B and "Rope 2."

For Block B and the mass of "Rope 2," we need to account for:
- Weight of Block B: \( W_B = m_B \cdot g = 1.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 9.8 \, \text{N} \)
- Weight of Rope 2: \( W_{\text{rope2}} = 0.25 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 2.45 \, \text{N} \)

Thus, the total weight of the system below "Rope 2" is the combined weight of Block B and Rope 2:
\[
W_{\text{total}} = W_B + W_{\text{rope2}} = 9.8 \, \text{N} + 2.45 \, \text{N} = 12.25 \, \text{N}
\]

#### Step 2: Apply Newton's Second Law
Since the entire system is accelerating upward at \( 3.0 \, \text{m/s}^2 \), we use Newton's Second Law:
\[
F_{\text{net}} = m_{\text{total}} \cdot a
\]
where \( m_{\text{total}} \) is the combined mass of Block B and Rope 2:
\[
m_{\text{total}} = 1.0 \, \text{kg} + 0.25 \, \text{kg} = 1.25 \, \text{kg}
\]
Thus, the net force required to accelerate this system is:
\[
F_{\text{net}} = 1.25 \, \text{kg} \cdot 3.0 \, \text{m/s}^2 = 3.75 \, \text{N}
\]

#### Step 3: Calculate the tension in Rope 2
The tension at the top of Rope 2 must overcome both the weight of the system and provide the net upward force for the acceleration. Therefore, the total tension \( T_2 \) at the top of Rope 2 is the sum of the weight and the net force:
\[
T_2 = W_{\text{total}} + F_{\text{net}} = 12.25 \, \text{N} + 3.75 \, \text{N} = 16.0 \, \text{N}
\]

Thus, the tension at the top of Rope 2 is **16.0 N**.

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Do you want more details or have any questions?

Here are five related questions:
1. How would the tension change if the acceleration were downward instead of upward?
2. What would happen to the tension in Rope 1, and how does it compare to Rope 2?
3. How does the mass of the ropes affect the tension calculations?
4. What if Block B had a different mass? How would that affect the system?
5. Could this system be solved using energy conservation principles?

**Tip**: Always remember to account for the mass of ropes when dealing with tension problems in systems with multiple objects.

Two 1.0 kg blocks are connected by a rope, and a second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at 3.0 m/s² by force F⃗. What is the tension at the top end of rope 2?

Calculate the tension at the top of Rope 2 in a system where two 1.0 kg blocks and two 250 g ropes are accelerated upward by a force of 3.0 m/s². Explore key concepts of Newton's Second Law and tension in ropes. Suitable for high school students in Grades 10-12.

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