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To solve this problem, we need to determine the total force acting upon the sled by accounting for the applied force, friction, and other forces along the horizontal direction.

### Given Data:
- Mass of sled (\( m \)) = 14.0 kg
- Applied force (\( F_{\text{applied}} \)) = 28.2 N
- Angle of applied force (\( \theta \)) = 51.0°
- Coefficient of kinetic friction (\( \mu_k \)) = 0.124
- Gravitational acceleration (\( g \)) ≈ 9.8 m/s²

### Solution Steps:
1. **Calculate the weight of the sled**:
   \[
   W = m \cdot g = 14.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 137.2 \, \text{N}
   \]

2. **Resolve the applied force into horizontal and vertical components**:
   - Horizontal component of \( F_{\text{applied}} \):
     \[
     F_{\text{applied, x}} = F_{\text{applied}} \cdot \cos(\theta) = 28.2 \cdot \cos(51.0^\circ)
     \]
   - Vertical component of \( F_{\text{applied}} \):
     \[
     F_{\text{applied, y}} = F_{\text{applied}} \cdot \sin(\theta) = 28.2 \cdot \sin(51.0^\circ)
     \]

3. **Calculate the normal force**:
   Since there is a vertical component of the applied force, the normal force will be adjusted as follows:
   \[
   F_{\text{normal}} = W - F_{\text{applied, y}}
   \]

4. **Calculate the frictional force**:
   Frictional force (\( F_{\text{friction}} \)) is given by:
   \[
   F_{\text{friction}} = \mu_k \cdot F_{\text{normal}}
   \]

5. **Calculate the net force in the horizontal direction**:
   The net force \( F \) is the difference between the horizontal component of the applied force and the frictional force:
   \[
   F = F_{\text{applied, x}} - F_{\text{friction}}
   \]

### Detailed Calculations:

#### Step 2: Horizontal and Vertical Components
Using trigonometric values:
   \[
   F_{\text{applied, x}} = 28.2 \cdot \cos(51.0^\circ) \approx 28.2 \cdot 0.6293 = 17.74 \, \text{N}
   \]
   \[
   F_{\text{applied, y}} = 28.2 \cdot \sin(51.0^\circ) \approx 28.2 \cdot 0.7771 = 21.92 \, \text{N}
   \]

#### Step 3: Normal Force
   \[
   F_{\text{normal}} = 137.2 - 21.92 = 115.28 \, \text{N}
   \]

#### Step 4: Frictional Force
   \[
   F_{\text{friction}} = 0.124 \cdot 115.28 \approx 14.29 \, \text{N}
   \]

#### Step 5: Net Force
   \[
   F = F_{\text{applied, x}} - F_{\text{friction}} = 17.74 - 14.29 = 3.45 \, \text{N}
   \]

### Answer:
The total force acting upon the sled is:
\[
\mathbf{F = +3.45 \, N}
\]
(indicating the force is positive in the direction of the applied force).

Would you like further details or explanations on any part of this solution?

### Follow-up Questions:
1. How does the angle of the applied force affect the total force?
2. What would happen to the total force if the coefficient of friction increased?
3. How would the solution change if the sled was on an inclined plane?
4. What if the force was applied at a different angle, say 30° instead of 51°?
5. How does the sled's mass influence the frictional force?

### Tip:
Remember that friction depends on the normal force, which can change with the vertical component of an applied force.

A 14.0 kg sled is pulled with a 28.2 N force at a 51.0° angle, across ground where the coefficient of friction (μₖ) is equal to 0.124. What is the total force acting upon the sled?

Learn how to calculate the total force on a 14.0 kg sled pulled with a 28.2 N force at a 51.0° angle across a frictional surface with μₖ = 0.124. This example uses Newton's Second Law, trigonometric components of force, and friction calculations. Ideal for high school physics students.

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