Answer :
To calculate the acceleration of the sled and the normal force acting on it, let's break down the given forces and the situation into manageable steps.
### Given Information:
- Mass of the sled, [tex]\( m = 8 \, \text{kg} \)[/tex]
- Angle of the pull, [tex]\( \theta = 50^\circ \)[/tex]
- Applied force, [tex]\( F = 20 \, \text{N} \)[/tex]
- Force of friction, [tex]\( F_{\text{friction}} = 2.4 \, \text{N} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
### Step-by-Step Solution:
1. Calculate the Components of the Applied Force:
- The parallel component of the force (along the direction of motion):
[tex]\[ F_{\text{parallel}} = F \cos(\theta) \][/tex]
- The perpendicular component of the force (along the direction perpendicular to the motion):
[tex]\[ F_{\text{perpendicular}} = F \sin(\theta) \][/tex]
2. Net Force in the Direction of Motion:
- The force acting along the direction of motion has to overcome the frictional force:
[tex]\[ F_{\text{net, parallel}} = F_{\text{parallel}} - F_{\text{friction}} \][/tex]
3. Acceleration:
- Using Newton's second law, the net force divided by the mass gives the acceleration:
[tex]\[ a = \frac{F_{\text{net, parallel}}}{m} \][/tex]
4. Normal Force:
- The normal force is the force perpendicular to the surface. It is affected by the perpendicular component of the applied force and the weight of the sled:
[tex]\[ F_{\text{normal}} = mg - F_{\text{perpendicular}} \][/tex]
### Specific Calculation Steps:
1. Convert the Angle to Radians:
- Since trigonometric functions in some contexts might use radians, we convert degrees to radians:
[tex]\[ \theta_{\text{rad}} = \theta \times \left(\frac{\pi}{180}\right) \approx 50 \times 0.01745 \approx 0.873 \text{ radians} \][/tex]
2. Calculate the Parallel and Perpendicular Components:
[tex]\[ F_{\text{parallel}} = 20 \cos(50^\circ) \approx 20 \times 0.6428 \approx 12.856 \, \text{N} \][/tex]
[tex]\[ F_{\text{perpendicular}} = 20 \sin(50^\circ) \approx 20 \times 0.7660 \approx 15.32 \, \text{N} \][/tex]
3. Calculate the Net Force in the Parallel Direction:
[tex]\[ F_{\text{net, parallel}} = 12.856 \, \text{N} - 2.4 \, \text{N} \approx 10.456 \, \text{N} \][/tex]
4. Calculate the Acceleration:
[tex]\[ a = \frac{10.456 \, \text{N}}{8 \, \text{kg}} \approx 1.307 \, \text{m/s}^2 \][/tex]
- Rounded to the nearest tenth: [tex]\( a \approx 1.3 \, \text{m/s}^2 \)[/tex]
5. Calculate the Normal Force:
[tex]\[ F_{\text{normal}} = mg - F_{\text{perpendicular}} \approx (8 \times 9.8) \, \text{N} - 15.32 \, \text{N} = 78.4 \, \text{N} - 15.32 \, \text{N} \approx 63.08 \, \text{N} \][/tex]
- Rounded to the nearest tenth: [tex]\( F_{\text{normal}} \approx 63.1 \, \text{N} \)[/tex]
### Conclusion:
The acceleration of the sled is [tex]\( \boxed{1.3 \, \text{m/s}^2} \)[/tex], and the normal force acting on it is [tex]\( \boxed{63.1 \, \text{N}} \)[/tex].
Thus, the correct answer is:
[tex]\[ a=1.3 \, \text{m/s}^2 ; F_{N}=63.1 \, \text{N} \][/tex]
### Given Information:
- Mass of the sled, [tex]\( m = 8 \, \text{kg} \)[/tex]
- Angle of the pull, [tex]\( \theta = 50^\circ \)[/tex]
- Applied force, [tex]\( F = 20 \, \text{N} \)[/tex]
- Force of friction, [tex]\( F_{\text{friction}} = 2.4 \, \text{N} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
### Step-by-Step Solution:
1. Calculate the Components of the Applied Force:
- The parallel component of the force (along the direction of motion):
[tex]\[ F_{\text{parallel}} = F \cos(\theta) \][/tex]
- The perpendicular component of the force (along the direction perpendicular to the motion):
[tex]\[ F_{\text{perpendicular}} = F \sin(\theta) \][/tex]
2. Net Force in the Direction of Motion:
- The force acting along the direction of motion has to overcome the frictional force:
[tex]\[ F_{\text{net, parallel}} = F_{\text{parallel}} - F_{\text{friction}} \][/tex]
3. Acceleration:
- Using Newton's second law, the net force divided by the mass gives the acceleration:
[tex]\[ a = \frac{F_{\text{net, parallel}}}{m} \][/tex]
4. Normal Force:
- The normal force is the force perpendicular to the surface. It is affected by the perpendicular component of the applied force and the weight of the sled:
[tex]\[ F_{\text{normal}} = mg - F_{\text{perpendicular}} \][/tex]
### Specific Calculation Steps:
1. Convert the Angle to Radians:
- Since trigonometric functions in some contexts might use radians, we convert degrees to radians:
[tex]\[ \theta_{\text{rad}} = \theta \times \left(\frac{\pi}{180}\right) \approx 50 \times 0.01745 \approx 0.873 \text{ radians} \][/tex]
2. Calculate the Parallel and Perpendicular Components:
[tex]\[ F_{\text{parallel}} = 20 \cos(50^\circ) \approx 20 \times 0.6428 \approx 12.856 \, \text{N} \][/tex]
[tex]\[ F_{\text{perpendicular}} = 20 \sin(50^\circ) \approx 20 \times 0.7660 \approx 15.32 \, \text{N} \][/tex]
3. Calculate the Net Force in the Parallel Direction:
[tex]\[ F_{\text{net, parallel}} = 12.856 \, \text{N} - 2.4 \, \text{N} \approx 10.456 \, \text{N} \][/tex]
4. Calculate the Acceleration:
[tex]\[ a = \frac{10.456 \, \text{N}}{8 \, \text{kg}} \approx 1.307 \, \text{m/s}^2 \][/tex]
- Rounded to the nearest tenth: [tex]\( a \approx 1.3 \, \text{m/s}^2 \)[/tex]
5. Calculate the Normal Force:
[tex]\[ F_{\text{normal}} = mg - F_{\text{perpendicular}} \approx (8 \times 9.8) \, \text{N} - 15.32 \, \text{N} = 78.4 \, \text{N} - 15.32 \, \text{N} \approx 63.08 \, \text{N} \][/tex]
- Rounded to the nearest tenth: [tex]\( F_{\text{normal}} \approx 63.1 \, \text{N} \)[/tex]
### Conclusion:
The acceleration of the sled is [tex]\( \boxed{1.3 \, \text{m/s}^2} \)[/tex], and the normal force acting on it is [tex]\( \boxed{63.1 \, \text{N}} \)[/tex].
Thus, the correct answer is:
[tex]\[ a=1.3 \, \text{m/s}^2 ; F_{N}=63.1 \, \text{N} \][/tex]
