A child is pulling a red wagon full of toys (total mass is 8.35 kg). His pull is 34 N in magnitude and at 23 degrees above the horizontal. He pulls his wagon at a constant acceleration (could be positive or negative). The coefficient of friction between the wagon's wheels and the ground is 0.33. What is the wagon's acceleration?

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**Title: Calculating the Wagon's Acceleration**

**Problem Statement:**

A child is pulling a red wagon full of toys (total mass is 8.35 kg). His pull is 34 N in magnitude and at 23 degrees above the horizontal. He pulls his wagon at a constant acceleration (which could be positive or negative). The coefficient of friction between the wagon's wheels and the ground is 0.33. What is the wagon's acceleration?

**Given Data:**

- Total mass of the wagon: 8.35 kg
- Force exerted by the child: 34 N
- Angle of the pull: 23 degrees above the horizontal
- Coefficient of friction (μ): 0.33

**Objective:**

Determine the acceleration of the wagon.

**Explanation of Solution:**

1. **Resolve the Pulling Force:**
   - The pulling force has two components:
     - Horizontal component, \( F_{\text{horizontal}} = 34 \cos(23^\circ) \)
     - Vertical component, \( F_{\text{vertical}} = 34 \sin(23^\circ) \)

2. **Calculate the Normal Force:**
   - The normal force \( F_N \) can be determined by considering the gravitational force and the vertical component of the pulling force.
   - The gravitational force is \( F_g = mg = 8.35 \times 9.8 \, \text{N} \)
   - Thus, \( F_N = F_g - F_{\text{vertical}} \)

3. **Calculate the Frictional Force:**
   - The frictional force \( F_f \) is given by \( F_f = \mu F_N \)

4. **Determine the Net Force:**
   - The net force acting on the wagon in the horizontal direction, \( F_{\text{net}} \), is given by \( F_{\text{net}} = F_{\text{horizontal}} - F_f \)

5. **Calculate the Acceleration:**
   - Using Newton's second law, the acceleration \( a \) of the wagon can be determined by \( F_{\text{net}} = ma \)
   - Solve for \( a \) to find the wagon's acceleration.

**Note:**

When performing the calculations, ensure to keep the units consistent and use trigonometric values accurately. The final formula for acceleration (\( a \)) can be
Transcribed Image Text:**Title: Calculating the Wagon's Acceleration** **Problem Statement:** A child is pulling a red wagon full of toys (total mass is 8.35 kg). His pull is 34 N in magnitude and at 23 degrees above the horizontal. He pulls his wagon at a constant acceleration (which could be positive or negative). The coefficient of friction between the wagon's wheels and the ground is 0.33. What is the wagon's acceleration? **Given Data:** - Total mass of the wagon: 8.35 kg - Force exerted by the child: 34 N - Angle of the pull: 23 degrees above the horizontal - Coefficient of friction (μ): 0.33 **Objective:** Determine the acceleration of the wagon. **Explanation of Solution:** 1. **Resolve the Pulling Force:** - The pulling force has two components: - Horizontal component, \( F_{\text{horizontal}} = 34 \cos(23^\circ) \) - Vertical component, \( F_{\text{vertical}} = 34 \sin(23^\circ) \) 2. **Calculate the Normal Force:** - The normal force \( F_N \) can be determined by considering the gravitational force and the vertical component of the pulling force. - The gravitational force is \( F_g = mg = 8.35 \times 9.8 \, \text{N} \) - Thus, \( F_N = F_g - F_{\text{vertical}} \) 3. **Calculate the Frictional Force:** - The frictional force \( F_f \) is given by \( F_f = \mu F_N \) 4. **Determine the Net Force:** - The net force acting on the wagon in the horizontal direction, \( F_{\text{net}} \), is given by \( F_{\text{net}} = F_{\text{horizontal}} - F_f \) 5. **Calculate the Acceleration:** - Using Newton's second law, the acceleration \( a \) of the wagon can be determined by \( F_{\text{net}} = ma \) - Solve for \( a \) to find the wagon's acceleration. **Note:** When performing the calculations, ensure to keep the units consistent and use trigonometric values accurately. The final formula for acceleration (\( a \)) can be
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