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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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9e60ae64-4458-4b89-bb0e-7dd9dd5dc5fe%2Facb80133-3acd-4bd6-9955-13cb32ccb5d4%2Ff6pekq9_processed.png&w=3840&q=75)
Transcribed Image Text:### 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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