### Problem Statement: A child is pulling a red wagon full of toys (total mass is 8.75 kg). His pull is 36 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 0.35. What is the wagon's acceleration?### Solution Outline:To find the acceleration of the wagon, follow these steps:1. **Force Components:** - Calculate the horizontal ($F_{x}$) and vertical ($F_{y}$) components of the pulling force: \[ F_{x} = F \cos(\theta) \] \[ F_{y} = F \sin(\theta) \] - Here, $ F = 36 \, \text{N} $ and $ \theta = 23^\circ $.2. **Normal Force (N):** - Determine the normal force using the weight of the wagon and the vertical component of the pulling force: \[ N = mg - F_{y} \] - Where $ m = 8.75 \, \text{kg} $ and $ g = 9.81 \, \text{m/s}^2 $.3. **Frictional Force (F_f):** - Calculate the frictional force using the normal force and the coefficient of friction ($ \mu $): \[ F_{f} = \mu N \] - Given $ \mu = 0.35 $.4. **Net Force (F_{net}):** - Compute the net force acting on the wagon along the horizontal direction: \[ F_{net} = F_{x} - F_{f} \]5. **Acceleration (a):** - Finally, find the acceleration using Newton's second law: \[ F_{net} = ma \] \[ a = \frac{F_{net}}{m} \]### Detailed Calculation:1. **Force Components:** \[ F_{x} = 36 \cos(23^\circ) \approx 33.17 \, \text{N} \] \[ F_{y} = 36 \sin(23^\circ) \approx 14.07 \, \text

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A child is pulling a red wagon full of toys (total mass is 8.75 kg). His pull is 36 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 0.35. What is the wagon's acceleration?

A child is pulling a red wagon full of toys (total mass is 8.75 kg). His pull is 36 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 0.35. What is the wagon's acceleration?

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Author:Raymond A. Serway, John W. Jewett

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Chapter5: The Laws Of Motion

Section: Chapter Questions

Problem 5.19CQ: Give reasons for the answers to each of the following questions: (a) Clan a normal force be...

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Concept explainers

Newton's Second Law

Newton’s laws of motion describe the relationship between the motion of an object and forces acting on it. Newton’s law of motion has two types as follows:

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### Problem Statement:

A child is pulling a red wagon full of toys (total mass is 8.75 kg). His pull is 36 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 0.35. What is the wagon's acceleration?

### Solution Outline:

To find the acceleration of the wagon, follow these steps:

1. **Force Components:**
- Calculate the horizontal ($F_{x}$) and vertical ($F_{y}$) components of the pulling force:
\[ F_{x} = F \cos(\theta) \]
\[ F_{y} = F \sin(\theta) \]
- Here, $ F = 36 \, \text{N} $ and $ \theta = 23^\circ $.

2. **Normal Force (N):**
- Determine the normal force using the weight of the wagon and the vertical component of the pulling force:
\[ N = mg - F_{y} \]
- Where $ m = 8.75 \, \text{kg} $ and $ g = 9.81 \, \text{m/s}^2 $.

3. **Frictional Force (F_f):**
- Calculate the frictional force using the normal force and the coefficient of friction ($ \mu $):
\[ F_{f} = \mu N \]
- Given $ \mu = 0.35 $.

4. **Net Force (F_{net}):**
- Compute the net force acting on the wagon along the horizontal direction:
\[ F_{net} = F_{x} - F_{f} \]

5. **Acceleration (a):**
- Finally, find the acceleration using Newton's second law:
\[ F_{net} = ma \]
\[ a = \frac{F_{net}}{m} \]

### Detailed Calculation:

1. **Force Components:**
\[ F_{x} = 36 \cos(23^\circ) \approx 33.17 \, \text{N} \]
\[ F_{y} = 36 \sin(23^\circ) \approx 14.07 \, \text

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