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To solve for the coefficient of kinetic friction (\( \mu_k \)), let's go through the information provided:

1. **Mass of the box, \( m \) = 10.4 kg**
2. **Applied force, \( F \) = 17.4 N**
3. **Angle of applied force, \( \theta = 23.5^\circ \)**
4. **Constant speed** (indicating that the net force along the direction of motion is zero, so the frictional force balances the horizontal component of the applied force).

### Step-by-Step Solution:

#### 1. Find the gravitational force (weight) of the box:
\[
F_{\text{gravity}} = m \cdot g = 10.4 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 101.92 \, \text{N}
\]

#### 2. Resolve the applied force into horizontal and vertical components:
   - **Horizontal component**, \( F_x = F \cos \theta \)
   \[
   F_x = 17.4 \, \text{N} \times \cos(23.5^\circ) \approx 17.4 \times 0.917 = 15.9658 \, \text{N}
   \]

   - **Vertical component**, \( F_y = F \sin \theta \)
   \[
   F_y = 17.4 \, \text{N} \times \sin(23.5^\circ) \approx 17.4 \times 0.398 = 6.9252 \, \text{N}
   \]

#### 3. Calculate the normal force (\( F_N \)):
Since there is a vertical component from the applied force that partially offsets the gravitational force, the normal force \( F_N \) is given by:
\[
F_N = F_{\text{gravity}} - F_y = 101.92 \, \text{N} - 6.9252 \, \text{N} = 94.9948 \, \text{N}
\]

#### 4. Calculate the frictional force:
Since the box is moving at a constant speed, the horizontal component of the applied force \( F_x \) is equal to the frictional force \( F_{\text{friction}} \):
\[
F_{\text{friction}} = F_x = 15.9658 \, \text{N}
\]

#### 5. Determine the coefficient of kinetic friction (\( \mu_k \)):
The frictional force is related to the normal force by \( F_{\text{friction}} = \mu_k \cdot F_N \). Solving for \( \mu_k \):
\[
\mu_k = \frac{F_{\text{friction}}}{F_N} = \frac{15.9658 \, \text{N}}{94.9948 \, \text{N}} \approx 0.168
\]

### Answer:
\[
\mu_k \approx 0.168
\]

Would you like more details on any part of this solution, or have any questions?

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Here are some related questions you might find interesting:
1. How would the coefficient of friction change if the angle of applied force increased?
2. What would happen if the box's mass was doubled?
3. How would you calculate the frictional force if the box was accelerating?
4. How does the normal force change with different angles of applied force?
5. What role does the coefficient of friction play in real-life scenarios like vehicle braking?

**Tip:** Remember, the angle of applied force affects both the horizontal force (which moves the object) and the vertical force (which alters the normal force).

A woman drags a 10.4 kg box of books at a constant speed, applying a 17.4 N force at 23.5°. What is the coefficient of friction?

Determine the coefficient of friction for a 10.4 kg box dragged with a 17.4 N force at an angle of 23.5°. This physics problem covers key concepts such as friction, normal force, and components of forces, with a step-by-step solution suitable for students in Grades 11-12.

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