FF 4. Find the acceleration of the blocks and the tension in the rope. 8 kg u=0.2 Fn T T mg =8 Ff 4 kg μ=0.2 Fn 80 N 80W mg. 54. .cd-1102

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pls help. acceleration = 4.71m/s^2 and tension = 53.3N but i cant figure it out 

**Problem Statement:**

4. Find the acceleration of the blocks and the tension in the rope.

**Diagram Explanation:**

The diagram illustrates a system with two blocks connected by a rope. The first block has a mass of 8 kg, and the second block has a mass of 4 kg. Both blocks are on a horizontal surface with a coefficient of friction (μ) of 0.2. An external force of 80 N is applied to the 4 kg block horizontally.

**Free Body Diagram Analysis:**

1. **8 kg Block:**

   - **Forces:**
     - Weight (mg) = 8 kg × 9.8 m/s² = 78.4 N downward
     - Normal force (Fn): Upward, equal to the weight
     - Frictional Force (Ff): μ × Fn = 0.2 × 78.4 N = 15.68 N, opposing motion
     - Tension (T): Acts horizontally towards the 4 kg block

2. **4 kg Block:**

   - **Forces:**
     - Weight (mg) = 4 kg × 9.8 m/s² = 39.2 N downward
     - Normal force (Fn): Upward, equal to the weight
     - Frictional Force (Ff): μ × Fn = 0.2 × 39.2 N = 7.84 N, opposing motion
     - Tension (T): Acts horizontally towards the 8 kg block
     - Applied Force: 80 N acting horizontally

These components are essential for understanding the physics of the system, including calculating the net force, acceleration, and tension.
Transcribed Image Text:**Problem Statement:** 4. Find the acceleration of the blocks and the tension in the rope. **Diagram Explanation:** The diagram illustrates a system with two blocks connected by a rope. The first block has a mass of 8 kg, and the second block has a mass of 4 kg. Both blocks are on a horizontal surface with a coefficient of friction (μ) of 0.2. An external force of 80 N is applied to the 4 kg block horizontally. **Free Body Diagram Analysis:** 1. **8 kg Block:** - **Forces:** - Weight (mg) = 8 kg × 9.8 m/s² = 78.4 N downward - Normal force (Fn): Upward, equal to the weight - Frictional Force (Ff): μ × Fn = 0.2 × 78.4 N = 15.68 N, opposing motion - Tension (T): Acts horizontally towards the 4 kg block 2. **4 kg Block:** - **Forces:** - Weight (mg) = 4 kg × 9.8 m/s² = 39.2 N downward - Normal force (Fn): Upward, equal to the weight - Frictional Force (Ff): μ × Fn = 0.2 × 39.2 N = 7.84 N, opposing motion - Tension (T): Acts horizontally towards the 8 kg block - Applied Force: 80 N acting horizontally These components are essential for understanding the physics of the system, including calculating the net force, acceleration, and tension.
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