F e. A block with a mass of 25.5 kg is moving at constant velocity because a force of 138 N is applied to the right at an upward angle of 29.9 degrees as shown. Use g-10 N/kg. What is the magnitude of the friction force? 119.631 V o units n What is the magnitude of the normal force? units n

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Chapter1: Units, Trigonometry. And Vectors
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**Problem Statement:**

A block with a mass of 25.5 kg is moving at constant velocity because a force of 138 N is applied to the right at an upward angle of 29.9 degrees as shown. Use \( g = 10 \, \text{N/kg} \).

**Questions:**

1. What is the magnitude of the friction force?
   - Answer: \( 119.631 \, \text{N} \)

2. What is the magnitude of the normal force?
   - Answer: [Blank in image]

**Diagram Explanation:**

The diagram depicts a block on a horizontal surface. A force \( F \) is applied at an angle \( \theta \) (29.9 degrees) to the right, above the horizontal. 

**Key Concepts:**

- The block is in equilibrium in the horizontal direction since it moves with constant velocity.
- The horizontal component of the applied force counteracts the frictional force.
- The vertical component of the force affects the normal force acting on the block.

To find the magnitudes of the forces:

- **Friction Force (\( f \)):**
  The horizontal component of force \( F \) is given by \( F \cdot \cos(\theta) \).
  Since the block is moving at constant velocity, friction force \( f = F \cdot \cos(\theta) = 138 \cdot \cos(29.9°) \).

- **Normal Force (\( N \)):**
  The normal force is affected by both the gravitational force and the vertical component of the applied force. It can be calculated using:
  \( N = mg - F \cdot \sin(\theta) \).
  
These calculations lead to the determination of the respective forces.
Transcribed Image Text:**Problem Statement:** A block with a mass of 25.5 kg is moving at constant velocity because a force of 138 N is applied to the right at an upward angle of 29.9 degrees as shown. Use \( g = 10 \, \text{N/kg} \). **Questions:** 1. What is the magnitude of the friction force? - Answer: \( 119.631 \, \text{N} \) 2. What is the magnitude of the normal force? - Answer: [Blank in image] **Diagram Explanation:** The diagram depicts a block on a horizontal surface. A force \( F \) is applied at an angle \( \theta \) (29.9 degrees) to the right, above the horizontal. **Key Concepts:** - The block is in equilibrium in the horizontal direction since it moves with constant velocity. - The horizontal component of the applied force counteracts the frictional force. - The vertical component of the force affects the normal force acting on the block. To find the magnitudes of the forces: - **Friction Force (\( f \)):** The horizontal component of force \( F \) is given by \( F \cdot \cos(\theta) \). Since the block is moving at constant velocity, friction force \( f = F \cdot \cos(\theta) = 138 \cdot \cos(29.9°) \). - **Normal Force (\( N \)):** The normal force is affected by both the gravitational force and the vertical component of the applied force. It can be calculated using: \( N = mg - F \cdot \sin(\theta) \). These calculations lead to the determination of the respective forces.
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