A train with a mass 5x10^5 kg is moving with a speed 80km/hr. If the train stops in 2min...1)What is the acceleration of the train? 2) What is the braking force applied to the train?
1. Vo = 80km/h = 80000m/3600s=22.2 m/s.
V = Vo + a*t.
a=(V-Vo)/t = (0-22.2)/120s=-0.185 m/s^2.
2. F = m*a = 5*10^5 * (-0.185)=-92500 N.
To find the acceleration of the train, we can use the formula:
acceleration (a) = (final velocity - initial velocity) / time
1) First, we need to convert the speed of the train from km/hr to m/s. We know that 1 km/hr is equal to 0.2778 m/s.
Converting the speed of the train:
Speed = 80 km/hr * (0.2778 m/s / 1 km/hr) = 22.2222 m/s
Next, we need to convert the time from minutes to seconds. We know that there are 60 seconds in 1 minute.
Converting the time:
Time = 2 minutes * (60 seconds / 1 minute) = 120 seconds
Now, we can calculate the acceleration:
Acceleration = (0 m/s - 22.2222 m/s) / 120 seconds = -0.185 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which makes sense as the train is coming to a stop.
2) To find the braking force applied to the train, we need to use Newton's second law of motion:
force (F) = mass (m) * acceleration (a)
From the given information, the mass of the train is 5x10^5 kg, and the acceleration is -0.185 m/s^2.
Braking Force = 5x10^5 kg * (-0.185 m/s^2) = -92,500 N
The negative sign indicates that the force is in the opposite direction of the initial motion of the train, which is also expected when applying brakes.