A sled with a mass of 8 kg is pulled at a 50-degree angle with a force of 20 N. The force of friction acting on the sled is 2.4 N.

The free-body diagram shows the forces acting on the sled.

What is the acceleration of the sled and the normal force acting on it, to the nearest tenth?

A. [tex]a = 1.3 \, \text{m/s}^2, \, F_N = 63.1 \, \text{N}[/tex]
B. [tex]a = 1.6 \, \text{m/s}^2, \, F_N = 65.6 \, \text{N}[/tex]
C. [tex]a = 1.9 \, \text{m/s}^2, \, F_N = 93.7 \, \text{N}[/tex]
D. [tex]a = 2.2 \, \text{m/s}^2, \, F_N = 78.4 \, \text{N}[/tex]



Answer :

To find the acceleration of the sled and the normal force acting on it, we need to analyze the forces acting on the sled. Let's break down the steps:

1. Identifying Forces:
- Mass of the sled, [tex]\( m \)[/tex]: 8 kg
- Pulling force, [tex]\( F_{\text{pull}} \)[/tex]: 20 N
- Angle of pulling force, [tex]\( \theta \)[/tex]: 50 degrees
- Force of friction, [tex]\( F_{\text{friction}} \)[/tex]: 2.4 N

2. Horizontal and Vertical Components of the Pulling Force:
- The horizontal component ([tex]\( F_{\text{horizontal}} \)[/tex]) of the pulling force can be calculated using the cosine of the angle:
[tex]\[ F_{\text{horizontal}} = F_{\text{pull}} \cdot \cos(\theta) \][/tex]
- The vertical component ([tex]\( F_{\text{vertical}} \)[/tex]) of the pulling force can be calculated using the sine of the angle:
[tex]\[ F_{\text{vertical}} = F_{\text{pull}} \cdot \sin(\theta) \][/tex]

3. Calculating the Normal Force ( [tex]\( F_{\text{normal}} \)[/tex]):
- The weight of the sled ([tex]\( F_{\text{gravity}} \)[/tex]) is:
[tex]\[ F_{\text{gravity}} = m \cdot g \][/tex]
where [tex]\( g \)[/tex] (acceleration due to gravity) is approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex].
- The normal force is affected by the vertical component of the pulling force. Thus, the normal force ([tex]\( F_{\text{normal}} \)[/tex]) is:
[tex]\[ F_{\text{normal}} = F_{\text{gravity}} - F_{\text{vertical}} \][/tex]

4. Calculating the Total Horizontal Force:
- The total horizontal force is the horizontal component of the pulling force minus the force of friction:
[tex]\[ F_{\text{total-horizontal}} = F_{\text{horizontal}} - F_{\text{friction}} \][/tex]

5. Calculating the Acceleration:
- According to Newton’s second law ([tex]\( F = ma \)[/tex]), the acceleration [tex]\( a \)[/tex] can be found by:
[tex]\[ a = \frac{F_{\text{total-horizontal}}}{m} \][/tex]

Let's put these steps together to determine the values:

- Step 2: Components of Pulling Force
[tex]\[ F_{\text{horizontal}} = 20 \cdot \cos(50^\circ) \approx 12.86 \, \text{N} \][/tex]
[tex]\[ F_{\text{vertical}} = 20 \cdot \sin(50^\circ) \approx 15.32 \, \text{N} \][/tex]

- Step 3: Normal Force
[tex]\[ F_{\text{gravity}} = 8 \cdot 9.8 = 78.4 \, \text{N} \][/tex]
[tex]\[ F_{\text{normal}} = 78.4 - 15.32 = 63.08 \, \text{N} \][/tex]

- Step 4: Total Horizontal Force
[tex]\[ F_{\text{total-horizontal}} = 12.86 - 2.4 = 10.46 \, \text{N} \][/tex]

- Step 5: Acceleration
[tex]\[ a = \frac{10.46}{8} \approx 1.31 \, \text{m/s}^2 \][/tex]

Rounding these values to the nearest tenth, we get:
- The acceleration [tex]\( a \approx 1.3 \, \text{m/s}^2 \)[/tex]
- The normal force [tex]\( F_{\text{normal}} \approx 63.1 \, \text{N} \)[/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{a = 1.3 \, \text{m/s}^2, \, F_{\text{normal}} = 63.1 \, \text{N}} \][/tex]