Question
A locomotive is pulling a train with a mass of 180tonnes up an incline of 12° from rest to 80km/h for 2min. The rolling resistance of the train is 8500N. Calculate
1) the acceleration
2 the output force of the train
1) the acceleration
2 the output force of the train
Answers
GPT-4o mini
To solve this problem, we first need to convert the given values and calculate the forces acting on the train.
### Given data:
- Mass of the train (\( m \)): \( 180 \, \text{tonnes} = 180,000 \, \text{kg} \)
- Incline angle (\( \theta \)): \( 12° \)
- Final velocity (\( v_f \)): \( 80 \, \text{km/h} = \frac{80 \times 1000}{3600} \frac{\text{m}}{\text{s}} = \frac{80000}{3600} \approx 22.22 \, \text{m/s} \)
- Time (\( t \)): \( 2 \, \text{min} = 2 \times 60 = 120 \, \text{s} \)
- Rolling resistance force (\( F_r \)): \( 8500 \, \text{N} \)
### Part 1: Calculate the acceleration
To find the acceleration, we need to first determine the initial velocity (\( v_i \)), which is 0, and then use the equation of motion:
\[
v_f = v_i + a t
\]
Rearranging, we get:
\[
a = \frac{v_f - v_i}{t}
\]
Substituting the values:
\[
a = \frac{22.22 \, \text{m/s} - 0}{120 \, \text{s}} \approx \frac{22.22}{120} \approx 0.1852 \, \text{m/s}^2
\]
### Part 2: Calculate the output force of the train
The two main forces acting against the train as it moves uphill are gravity (due to the incline) and rolling resistance.
1. **Gravitational force component acting down the incline**:
\[
F_g = m \cdot g \cdot \sin(\theta)
\]
Where \( g \approx 9.81 \, \text{m/s}^2 \)
Calculating the gravitational force component:
\[
F_g = 180000 \cdot 9.81 \cdot \sin(12°)
\]
Calculating \( \sin(12°) \):
\[
\sin(12°) \approx 0.2079
\]
Now calculating \( F_g \):
\[
F_g = 180000 \cdot 9.81 \cdot 0.2079 \approx 180000 \cdot 2.0421 \approx 36757.8 \, \text{N}
\]
2. **Total force required to overcome rolling resistance**:
\[
F_{\text{total}} = F_g + F_r = 36757.8 \, \text{N} + 8500 \, \text{N} = 45257.8 \, \text{N}
\]
3. **Net force required for acceleration**:
The net force required to accelerate the mass can be calculated using Newton's second law:
\[
F_{\text{net}} = m \cdot a
\]
Substituting:
\[
F_{\text{net}} = 180000 \cdot 0.1852 \approx 33336 \, \text{N}
\]
4. **Total output force required by the locomotive**:
The total output force required by the locomotive to move the train uphill and overcome both the rolling resistance and gravitational force would be:
\[
F_{\text{output}} = F_{\text{total}} + F_{\text{net}} = 45257.8 \, \text{N} + 33336 \, \text{N} = 78593.8 \, \text{N}
\]
### Summary of Results:
1. The acceleration of the train is approximately \( 0.1852 \, \text{m/s}^2 \).
2. The output force of the train is approximately \( 78593.8 \, \text{N} \).
### Given data:
- Mass of the train (\( m \)): \( 180 \, \text{tonnes} = 180,000 \, \text{kg} \)
- Incline angle (\( \theta \)): \( 12° \)
- Final velocity (\( v_f \)): \( 80 \, \text{km/h} = \frac{80 \times 1000}{3600} \frac{\text{m}}{\text{s}} = \frac{80000}{3600} \approx 22.22 \, \text{m/s} \)
- Time (\( t \)): \( 2 \, \text{min} = 2 \times 60 = 120 \, \text{s} \)
- Rolling resistance force (\( F_r \)): \( 8500 \, \text{N} \)
### Part 1: Calculate the acceleration
To find the acceleration, we need to first determine the initial velocity (\( v_i \)), which is 0, and then use the equation of motion:
\[
v_f = v_i + a t
\]
Rearranging, we get:
\[
a = \frac{v_f - v_i}{t}
\]
Substituting the values:
\[
a = \frac{22.22 \, \text{m/s} - 0}{120 \, \text{s}} \approx \frac{22.22}{120} \approx 0.1852 \, \text{m/s}^2
\]
### Part 2: Calculate the output force of the train
The two main forces acting against the train as it moves uphill are gravity (due to the incline) and rolling resistance.
1. **Gravitational force component acting down the incline**:
\[
F_g = m \cdot g \cdot \sin(\theta)
\]
Where \( g \approx 9.81 \, \text{m/s}^2 \)
Calculating the gravitational force component:
\[
F_g = 180000 \cdot 9.81 \cdot \sin(12°)
\]
Calculating \( \sin(12°) \):
\[
\sin(12°) \approx 0.2079
\]
Now calculating \( F_g \):
\[
F_g = 180000 \cdot 9.81 \cdot 0.2079 \approx 180000 \cdot 2.0421 \approx 36757.8 \, \text{N}
\]
2. **Total force required to overcome rolling resistance**:
\[
F_{\text{total}} = F_g + F_r = 36757.8 \, \text{N} + 8500 \, \text{N} = 45257.8 \, \text{N}
\]
3. **Net force required for acceleration**:
The net force required to accelerate the mass can be calculated using Newton's second law:
\[
F_{\text{net}} = m \cdot a
\]
Substituting:
\[
F_{\text{net}} = 180000 \cdot 0.1852 \approx 33336 \, \text{N}
\]
4. **Total output force required by the locomotive**:
The total output force required by the locomotive to move the train uphill and overcome both the rolling resistance and gravitational force would be:
\[
F_{\text{output}} = F_{\text{total}} + F_{\text{net}} = 45257.8 \, \text{N} + 33336 \, \text{N} = 78593.8 \, \text{N}
\]
### Summary of Results:
1. The acceleration of the train is approximately \( 0.1852 \, \text{m/s}^2 \).
2. The output force of the train is approximately \( 78593.8 \, \text{N} \).