### Problem 5: Calculating Acceleration of a Box on a Horizontal FloorA 33.0 kg box is being pulled by a person across a horizontal floor with a force \( F = 165 \) N. The force makes an angle \( \theta = 24.0° \) with the horizontal, as illustrated below. Determine the magnitude of the acceleration of the box if the coefficient of kinetic friction between the box and the floor is 0.16.Calculate the answer in m/s².#### Diagram Explanation:- The diagram depicts a box on a horizontal floor.- A force \( \vec{F} \) is applied to the box at an angle \( \theta = 24.0° \) above the horizontal.- The force vector \( \vec{F} \) is shown as an arrow pointing above and to the left, representing the direction of the applied force.- The horizontal line represents the floor.**Note: Insert your calculated value in the box provided in the question.**### Steps to Solve:1. **Resolve the applied force (\( F \)) into horizontal (\( F_x \)) and vertical components (\( F_y \)).** - \( F_x = F \cos(\theta) \) - \( F_y = F \sin(\theta) \)2. **Compute the normal force (\( N \)).** - The normal force is influenced by both the weight of the box and the vertical component of the applied force. - \( N = mg - F_y \) - Where \( m \) is the mass of the box, and \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)).3. **Calculate the kinetic friction force (\( f_k \)).** - \( f_k = \mu_k N \) - Where \( \mu_k \) is the coefficient of kinetic friction.4. **Determine the net force (\( F_{\text{net}} \)) in the horizontal direction.** - \( F_{\text{net}} = F_x - f_k \)5. **Use Newton’s Second Law to find the acceleration (\( a \)) of the box.** - \( F_{\text{net}} = ma \) - \( a = \frac{F_{\text{net}}}{m} \

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Description: ### Problem 5: Calculating Acceleration of a Box on a Horizontal FloorA 33.0 kg box is being pulled by a person across a horizontal floor with a force \( F = 165 \) N. The force makes an angle \( \theta = 24.0° \) with the horizontal, as illustrated

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5. A 33.0 kg box is being pulled by a person across a horizontal floor with a force F = 165 N that makes an angle 0 = 24.0° with the horizontal as shown in the figure. What is the magnitude of the acceleration of the box if the coefficient of kinetic friction between the box and the floor is 0.16? m/s² 8

5. A 33.0 kg box is being pulled by a person across a horizontal floor with a force F = 165 N that makes an angle 0 = 24.0° with the horizontal as shown in the figure. What is the magnitude of the acceleration of the box if the coefficient of kinetic friction between the box and the floor is 0.16? m/s² 8

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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 5: Calculating Acceleration of a Box on a Horizontal Floor

A 33.0 kg box is being pulled by a person across a horizontal floor with a force \( F = 165 \) N. The force makes an angle \( \theta = 24.0° \) with the horizontal, as illustrated below. Determine the magnitude of the acceleration of the box if the coefficient of kinetic friction between the box and the floor is 0.16.

Calculate the answer in m/s².

#### Diagram Explanation:
- The diagram depicts a box on a horizontal floor.
- A force \( \vec{F} \) is applied to the box at an angle \( \theta = 24.0° \) above the horizontal.
- The force vector \( \vec{F} \) is shown as an arrow pointing above and to the left, representing the direction of the applied force.
- The horizontal line represents the floor.

**Note: Insert your calculated value in the box provided in the question.**

### Steps to Solve:
1. **Resolve the applied force (\( F \)) into horizontal (\( F_x \)) and vertical components (\( F_y \)).**
- \( F_x = F \cos(\theta) \)
- \( F_y = F \sin(\theta) \)

2. **Compute the normal force (\( N \)).**
- The normal force is influenced by both the weight of the box and the vertical component of the applied force.
- \( N = mg - F_y \)
- Where \( m \) is the mass of the box, and \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)).

3. **Calculate the kinetic friction force (\( f_k \)).**
- \( f_k = \mu_k N \)
- Where \( \mu_k \) is the coefficient of kinetic friction.

4. **Determine the net force (\( F_{\text{net}} \)) in the horizontal direction.**
- \( F_{\text{net}} = F_x - f_k \)

5. **Use Newton’s Second Law to find the acceleration (\( a \)) of the box.**
- \( F_{\text{net}} = ma \)
- \( a = \frac{F_{\text{net}}}{m} \

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