Two blocks connected by a rope of negligible mass are being dragged by a force at a 29° angle above horizontal (see figure below), so that the blocks are already moving to the right. Suppose F = 67.0 N, m1 = 10.0 kg, m2 = 26.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.136. Use g=9.8 m/s2.
5-p-047
(b) Determine the acceleration of the system.
  
Correct: Your answer is correct.
 m/s2

(c) Determine the tension T in the rope.
  
Incorrect: Your answer is incorrect.
 N

To solve the problem involving two blocks connected by a rope and dragged by a force at an angle, we need to break it down into two main parts:

### Part (b): **Determine the acceleration of the system**

Given:
- Force applied, \( F = 67.0 \, \text{N} \)
- Angle above the horizontal, \( \theta = 29^\circ \)
- Mass of the first block, \( m_1 = 10.0 \, \text{kg} \)
- Mass of the second block, \( m_2 = 26.0 \, \text{kg} \)
- Coefficient of kinetic friction, \( \mu_k = 0.136 \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)

First, let’s calculate the total friction force acting on the two blocks:

1. **Normal force on each block**:  
   Since the blocks are on a horizontal surface, the normal force is equal to the weight minus the vertical component of the applied force. However, the vertical component of the force reduces the normal force on both blocks.

   For both blocks, the combined weight is:
   \[
   (m_1 + m_2)g = (10.0 + 26.0) \times 9.8 = 35.0 \times 9.8 = 343.0 \, \text{N}
   \]

   The vertical component of the applied force is:
   \[
   F_{\text{vertical}} = F \sin \theta = 67.0 \times \sin 29^\circ \approx 67.0 \times 0.4848 = 32.48 \, \text{N}
   \]

   So, the total normal force acting on the blocks is:
   \[
   N = 343.0 \, \text{N} - 32.48 \, \text{N} = 310.52 \, \text{N}
   \]

2. **Frictional force**:  
   The kinetic friction force is:
   \[
   f_k = \mu_k \times N = 0.136 \times 310.52 \approx 42.23 \, \text{N}
   \]

3. **Net horizontal force**:  
   The horizontal component of the applied force is:
   \[
   F_{\text{horizontal}} = F \cos \theta = 67.0 \times \cos 29^\circ \approx 67.0 \times 0.8746 = 58.60 \, \text{N}
   \]

   The net force is the horizontal component of the applied force minus the frictional force:
   \[
   F_{\text{net}} = F_{\text{horizontal}} - f_k = 58.60 \, \text{N} - 42.23 \, \text{N} = 16.37 \, \text{N}
   \]

4. **Acceleration of the system**:  
   The total mass of the system is \( m_1 + m_2 = 10.0 + 26.0 = 36.0 \, \text{kg} \).

   Using Newton's second law:
   \[
   a = \frac{F_{\text{net}}}{m_1 + m_2} = \frac{16.37}{36.0} \approx 0.454 \, \text{m/s}^2
   \]

Thus, the **acceleration of the system** is approximately \( 0.454 \, \text{m/s}^2 \).

---

### Part (c): **Determine the tension in the rope**

Now, we focus on the tension \( T \) in the rope connecting the two blocks. To do this, we analyze the forces acting on the second block, \( m_2 \), which is being pulled by the tension.

1. **Forces on block 2 (m2 = 26.0 kg)**:
   The only forces acting on \( m_2 \) are the tension in the rope pulling it to the right and the friction force opposing its motion.

2. **Friction force on block 2**:
   The friction force on block 2 is:
   \[
   f_{k2} = \mu_k \times N_2 = \mu_k \times m_2 \times g = 0.136 \times 26.0 \times 

Two blocks connected by a rope of negligible mass are being dragged by a force at a 29° angle above horizontal, so that the blocks are already moving to the right. Suppose F = 67.0 N, m1 = 10.0 kg, m2 = 26.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.136. Use g=9.8 m/s^2. (b) Determine the acceleration of the system. (c) Determine the tension in the rope.

Learn how to calculate the acceleration and tension in a system of two blocks connected by a rope with a force applied at a 29° angle. This problem involves concepts from Newton's laws, forces, friction, and trigonometry, ideal for students in Grades 10-12.

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