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Sagot :
Sure, let's break down this problem step-by-step to find the power developed by the engine.
### Step 1: Convert the Load
First, we need to convert the load from tonnes to kilograms.
- Given load: [tex]\(2 \text{ tonnes}\)[/tex]
- Conversion factor: [tex]\(1 \text{ tonne} = 1000 \text{ kg}\)[/tex]
[tex]\[ \text{Load in kg} = 2 \times 1000 = 2000 \, \text{kg} \][/tex]
### Step 2: Calculate the Gravitational Force
The gravitational force acting on the load can be calculated using the acceleration due to gravity, [tex]\(g = 9.81 \, \text{m/s}^2\)[/tex].
[tex]\[ \text{Weight (Force due to gravity)} = \text{mass} \times g = 2000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 19620 \, \text{N} \][/tex]
### Step 3: Components of the Weight
Next, we resolve this gravitational force into components parallel and perpendicular to the incline.
- Incline angle of the track, [tex]\(\theta = 30^\circ\)[/tex]
The parallel component of the weight:
[tex]\[ \text{Weight (parallel)} = 19620 \, \text{N} \times \sin(30^\circ) = 19620 \, \text{N} \times 0.5 = 9810 \, \text{N} \][/tex]
The perpendicular component of the weight:
[tex]\[ \text{Weight (perpendicular)} = 19620 \, \text{N} \times \cos(30^\circ) = 19620 \, \text{N} \times 0.866 = 16991.42 \, \text{N} \][/tex]
### Step 4: Calculate the Frictional Force
The frictional force can be determined using the coefficient of friction, [tex]\(\mu = 0.3\)[/tex].
[tex]\[ \text{Frictional force} = \mu \times \text{Weight (perpendicular)} = 0.3 \times 16991.42 \, \text{N} = 5097.43 \, \text{N} \][/tex]
### Step 5: Total Force Parallel to the Incline
The total force parallel to the incline is the sum of the parallel component of the weight and the frictional force.
[tex]\[ \text{Total force} = \text{Weight (parallel)} + \text{Frictional force} = 9810 \, \text{N} + 5097.43 \, \text{N} = 14907.43 \, \text{N} \][/tex]
### Step 6: Adjust the Force for the Applied Force Angle
The applied force is inclined at [tex]\(20^\circ\)[/tex] to the incline, thus we need to find the effective component of the force.
[tex]\[ \text{Effective force} = \frac{\text{Total force}}{\cos(20^\circ)} = \frac{14907.43 \, \text{N}}{0.9397} = 15864.15 \, \text{N} \][/tex]
### Step 7: Calculate the Power
Finally, the power developed by the engine can be calculated using the force and the speed.
- Speed, [tex]\(v = 1.2 \, \text{m/s}\)[/tex]
[tex]\[ \text{Power} = \text{Effective force} \times \text{Speed} = 15864.15 \, \text{N} \times 1.2 \, \text{m/s} = 19036.98 \, \text{W} \][/tex]
### Conclusion
The power developed by the engine is:
[tex]\[ 19036.98 \, \text{W} \quad \text{or} \quad 19.04 \, \text{kW} \][/tex]
### Step 1: Convert the Load
First, we need to convert the load from tonnes to kilograms.
- Given load: [tex]\(2 \text{ tonnes}\)[/tex]
- Conversion factor: [tex]\(1 \text{ tonne} = 1000 \text{ kg}\)[/tex]
[tex]\[ \text{Load in kg} = 2 \times 1000 = 2000 \, \text{kg} \][/tex]
### Step 2: Calculate the Gravitational Force
The gravitational force acting on the load can be calculated using the acceleration due to gravity, [tex]\(g = 9.81 \, \text{m/s}^2\)[/tex].
[tex]\[ \text{Weight (Force due to gravity)} = \text{mass} \times g = 2000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 19620 \, \text{N} \][/tex]
### Step 3: Components of the Weight
Next, we resolve this gravitational force into components parallel and perpendicular to the incline.
- Incline angle of the track, [tex]\(\theta = 30^\circ\)[/tex]
The parallel component of the weight:
[tex]\[ \text{Weight (parallel)} = 19620 \, \text{N} \times \sin(30^\circ) = 19620 \, \text{N} \times 0.5 = 9810 \, \text{N} \][/tex]
The perpendicular component of the weight:
[tex]\[ \text{Weight (perpendicular)} = 19620 \, \text{N} \times \cos(30^\circ) = 19620 \, \text{N} \times 0.866 = 16991.42 \, \text{N} \][/tex]
### Step 4: Calculate the Frictional Force
The frictional force can be determined using the coefficient of friction, [tex]\(\mu = 0.3\)[/tex].
[tex]\[ \text{Frictional force} = \mu \times \text{Weight (perpendicular)} = 0.3 \times 16991.42 \, \text{N} = 5097.43 \, \text{N} \][/tex]
### Step 5: Total Force Parallel to the Incline
The total force parallel to the incline is the sum of the parallel component of the weight and the frictional force.
[tex]\[ \text{Total force} = \text{Weight (parallel)} + \text{Frictional force} = 9810 \, \text{N} + 5097.43 \, \text{N} = 14907.43 \, \text{N} \][/tex]
### Step 6: Adjust the Force for the Applied Force Angle
The applied force is inclined at [tex]\(20^\circ\)[/tex] to the incline, thus we need to find the effective component of the force.
[tex]\[ \text{Effective force} = \frac{\text{Total force}}{\cos(20^\circ)} = \frac{14907.43 \, \text{N}}{0.9397} = 15864.15 \, \text{N} \][/tex]
### Step 7: Calculate the Power
Finally, the power developed by the engine can be calculated using the force and the speed.
- Speed, [tex]\(v = 1.2 \, \text{m/s}\)[/tex]
[tex]\[ \text{Power} = \text{Effective force} \times \text{Speed} = 15864.15 \, \text{N} \times 1.2 \, \text{m/s} = 19036.98 \, \text{W} \][/tex]
### Conclusion
The power developed by the engine is:
[tex]\[ 19036.98 \, \text{W} \quad \text{or} \quad 19.04 \, \text{kW} \][/tex]
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