A train of mass 200 tonne travelling at 140 km/h applies the brakes fiercely , so that the brakes lock and the wheels slides along the rails.it stops sliding 1000m.
calculate.
1. The acceleration.
2. The longitudinal force on the rail.
3. The coefficient of friction between the rails and the wheels.
Oh, hello there! It seems like you have a mathematical question for me. Well, I'll do my best to provide some humorous answers to your inquiries about the train. Here goes nothing!
1. The acceleration: Well, the train is definitely going from "wheee!" to "stoppp!" in a hurry. To calculate the acceleration, we can use the good ol' physics formula: acceleration = (final velocity - initial velocity) / time. Since the final velocity is 0 km/h (because it stops sliding), the initial velocity is 140 km/h, and the time it takes is... umm... let's just say it's very fast. So, the acceleration is something like "Whoa, that's quick!"
2. The longitudinal force on the rail: When the wheels of the train lock and slide along the rails, they rub against each other like a really intense game of rock-paper-scissors. This rubbing creates a force known as friction. Now, the longitudinal force on the rail is simply a fancy way of saying "how hard the train is pushing against the tracks." In this case, it's as if the train is saying to the rails, "I won't let go!" So, the longitudinal force can be described as a comical tug-of-war between the train and the tracks, with the force being "Grip it and rip it!"
3. The coefficient of friction between the rails and the wheels: Ah, the coefficient of friction, a fancy term for how much the rails and the wheels like each other. In other words, it's a measure of their "stickiness." In this case, since the train is sliding along the rails, I'd say their relationship is more like a roller coaster ride with lots of ups and downs. So, the coefficient of friction could be described as "It's complicated, but they stick together for the thrill!"
Remember, my answers are meant to bring a smile to your face, so take them with a pinch of laughter! If you need any more help, feel free to ask.
To solve this problem, we can use the following formulas:
1. Acceleration (a) = Change in velocity (Δv) / Time taken (Δt)
2. Longitudinal force (F) = Mass (m) × Acceleration (a)
3. Coefficient of friction (μ) = Longitudinal force (F) / Weight of the train (W)
Given:
- Mass of the train (m) = 200 tonnes = 200,000 kg
- Initial velocity (u) = 140 km/h = 140,000 m/3600 s = 38.89 m/s
- Final velocity (v) = 0 m/s
- Distance covered (s) = 1000 m
Let's calculate each step:
1. To calculate the acceleration, we need the change in velocity and the time taken. Here, the initial velocity (u) is known, the final velocity is 0 m/s, and the distance covered is 1000 m.
Δv = v - u = 0 - 38.89 m/s = -38.89 m/s
Δt = s / u = 1000 m / 38.89 m/s ≈ 25.74 s
Acceleration (a) = Δv / Δt = -38.89 m/s / 25.74 s ≈ -1.51 m/s²
2. To calculate the longitudinal force on the rail, we can use the mass of the train and the acceleration.
F = m × a = 200,000 kg × -1.51 m/s² ≈ -302,000 N
The negative sign indicates that the force is acting in the opposite direction to the motion of the train.
3. Finally, we can calculate the coefficient of friction using the longitudinal force and the weight of the train.
Weight of the train (W) = m × g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
W = 200,000 kg × 9.8 m/s² = 1,960,000 N
Coefficient of friction (μ) = F / W = (-302,000 N) / (1,960,000 N) ≈ -0.154
Since the coefficient of friction cannot be negative, we can consider the magnitude of the value, which is approximately 0.154.
To calculate the required values, we can use the following formulas:
1. The acceleration formula is:
```acceleration = (final velocity^2 - initial velocity^2) / (2 * distance)```
2. The longitudinal force formula is:
```force = mass * acceleration```
3. The coefficient of friction formula is:
```coefficient of friction = force / (mass * gravity)```
where:
- Mass is the mass of the train (in this case, 200 tonnes or 200,000 kg)
- Initial velocity is the velocity before the brakes are applied (140 km/h or 38.9 m/s)
- Final velocity is the velocity after the train stops sliding (0 m/s)
- Distance is the distance traveled while sliding (1000 m)
- Acceleration is the rate of change of velocity during sliding
- Force is the longitudinal force on the rail
- Gravity is the gravitational acceleration (9.8 m/s^2)
Now let's calculate each value step by step:
1. The acceleration:
```acceleration = (final velocity^2 - initial velocity^2) / (2 * distance)```
```acceleration = (0 - (38.9)^2) / (2 * 1000)```
```acceleration = (-1509.21) / (2 * 1000)```
```acceleration = -0.7546 m/s^2``` (assuming deceleration is negative)
2. The longitudinal force on the rail:
```force = mass * acceleration```
```force = 200,000 kg * (-0.7546 m/s^2)```
```force = -150,920 N``` (again, negative because it opposes the motion)
3. The coefficient of friction between the rails and the wheels:
```coefficient of friction = force / (mass * gravity)```
```coefficient of friction = -150,920 N / (200,000 kg * 9.8 m/s^2)```
```coefficient of friction = -0.0772``` (again, negative due to the direction of force)
Note: The negative sign indicates that the forces are opposing the motion of the train.