Asked by Amarech

2. Two blocks connected by a rope of negligible mass are being dragged by a horizontal force F (Fig.(1)).Suppose that F=68.0N, M1=12.0kg, M2=18.0kg, and the coefficient of kinetic friction between each block and and the surface is 0.20.
A, Draw a free body diagram for each block.
B, Determine the tension T and the magnitude of the acceleration of the system .
M1-----T-----M2----F
Figure(1)

Answers

There are no human answers yet.
Answered by GPT-4o mini
To solve this problem, we'll go through the steps systematically. We'll start with part A (the free body diagrams) and then move on to part B (determining the tension and acceleration).

### A. Free Body Diagrams

1. **Block M1 (mass = 12.0 kg)**:
- **Forces acting on M1**:
- Weight (W1) acting downward: \( W_1 = M_1 \cdot g = 12.0 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 117.72 \, \text{N} \)
- Normal force (N1) acting upward: \( N_1 = W_1 = 117.72 \, \text{N} \) (since the surface is horizontal)
- Tension (T) acting to the right
- Friction force (f1) acting to the left:
\[ f_1 = \mu_k \cdot N_1 = 0.20 \cdot 117.72 \, \text{N} = 23.544 \, \text{N} \]

- **Free Body Diagram for M1**:
```
↑ N1
|
|
---|--- T →
| ← f1
↓ W1
```

2. **Block M2 (mass = 18.0 kg)**:
- **Forces acting on M2**:
- Weight (W2) acting downward: \( W_2 = M_2 \cdot g = 18.0 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 176.58 \, \text{N} \)
- Normal force (N2) acting upward: \( N_2 = W_2 = 176.58 \, \text{N} \)
- tension (T) acting to the left
- Friction force (f2) acting to the left:
\[ f_2 = \mu_k \cdot N_2 = 0.20 \cdot 176.58 \, \text{N} = 35.316 \, \text{N} \]

- **Free Body Diagram for M2**:
```
↑ N2
|
|
← T |
|
| ← f2
↓ W2
```

### B. Determine Tension T and Magnitude of the Acceleration

1. **Net force on the system**:
- The total horizontal force applied is \( F = 68.0 \, \text{N} \).
- The total frictional force opposing the motion is the sum of frictional forces acting on both blocks:
\[
f_{\text{total}} = f_1 + f_2 = 23.544 \, \text{N} + 35.316 \, \text{N} = 58.86 \, \text{N}
\]

2. **Net force exerted on the system**:
\[
F_{\text{net}} = F - f_{\text{total}} = 68.0 \, \text{N} - 58.86 \, \text{N} = 9.14 \, \text{N}
\]

3. **Total mass of the system**:
\[
M_{\text{total}} = M_1 + M_2 = 12.0 \, \text{kg} + 18.0 \, \text{kg} = 30.0 \, \text{kg}
\]

4. **Calculate the acceleration (a)**:
Using Newton's second law, \( F = m \cdot a \):
\[
a = \frac{F_{\text{net}}}{M_{\text{total}}} = \frac{9.14 \, \text{N}}{30.0 \, \text{kg}} \approx 0.3047 \, \text{m/s}^2
\]

5. **Determine the tension (T)**:
We can analyze either block to find the tension. Let’s consider Block M2:
\[
F_{\text{net on M2}} = T - f_2 = M_2 \cdot a
\]
\[
T - 35.316 \, \text{N} = 18.0 \, \text{kg} \cdot 0.3047 \, \text{m/s}^2
\]
\[
T - 35.316 \, \text{N} = 5.48 \, \text{N}
\]
\[
T = 5.48 \, \text{N} + 35.316 \, \text{N} = 40.796 \, \text{N} \approx 40.8 \, \text{N}
\]

**Final Results**:
- The magnitude of the acceleration of the system is approximately \( 0.30 \, \text{m/s}^2 \).
- The tension in the rope (T) is approximately \( 40.8 \, \text{N} \).

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