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.
To calculate these values, we need to use the basic physics equations of motion. Let's start with calculating the acceleration.
1. The acceleration (a) can be calculated using the formula:
v^2 = u^2 + 2as,
where v is the final velocity (0 m/s since the train stops), u is the initial velocity (converted from 140 km/h to m/s), a is the acceleration, and s is the displacement.
First, let's convert the initial velocity from km/h to m/s:
Initial velocity, u = 140 km/h = 140 * (1000 m/3600 s) = 38.89 m/s
Rearranging the formula:
a = (v^2 - u^2) / (2s)
a = (0^2 - (38.89)^2) / (2 * 1000)
a = -1.5 m/s^2 (negative sign indicates deceleration)
Therefore, the acceleration of the train is -1.5 m/s^2.
2. The longitudinal force on the rail is given by Newton's second law of motion: F = ma, where F is the force, m is the mass, and a is the acceleration.
Mass (m) of the train is given as 200 tonnes, which is equivalent to 200,000 kg.
Therefore, the force (F) on the rail can be calculated as:
F = m * a
F = 200,000 * (-1.5)
F = -300,000 N
Therefore, the longitudinal force on the rail is -300,000 N.
3. The coefficient of friction (μ) between the rails and the wheels can be determined using the formula:
F = μN,
where F is the force of friction, μ is the coefficient of friction, and N is the normal force.
The normal force can be calculated as the weight of the train (W) acting vertically on the rails:
N = W = mg,
where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
N = 200,000 kg * 9.8 m/s^2 = 1,960,000 N.
Substituting the values in the friction formula:
-300,000 N = μ * 1,960,000 N,
μ = -300,000 N / 1,960,000 N ≈ -0.15.
Therefore, the coefficient of friction between the rails and the wheels is approximately -0.15. Note that the negative sign indicates that the force is acting in the opposite direction of motion.