When the crampton coal-fired train engine was built in 1852 , its mass was 48.3t ( 1.0 t= 1.0*10^3) and its force capability was rated at 22.4 kN. Assuming it was pulling train cars whose total mass doubled its own was and the total friction on the engine and cars was 10.1 kN, what was the magnitude of the acceleration of the train?
a=Fn/m1+m2
a=8.25X10^-2 m/s^2.
The correct answer.
m1 = 48.3*1000 = 48,300 kg. = Mass of
engine.
m2 = 2 * 48,300 = 96,600 kg = Mass of
load.
a = Fn/m2 = (22400-10100) / 96600 = 0.127 m/s^2.
To find the magnitude of the acceleration of the train, we need to apply Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. We can use this equation:
Net Force = Mass * Acceleration
First, let's convert the given masses and forces to SI units:
Mass of the train engine (m1) = 48.3 t = 48.3 * 10^3 kg
Force capability of the train engine (F1) = 22.4 kN = 22.4 * 10^3 N
Total friction on the engine and cars (f) = 10.1 kN = 10.1 * 10^3 N
Now we can set up the equation:
Net Force = F1 - f
Net Force = (m1 + total mass of cars) * acceleration
Since the total mass of the cars is double the mass of the train engine, we can write:
Total mass of cars = 2 * m1 = 2 * 48.3 * 10^3 kg = 96.6 * 10^3 kg
Substituting these values into the equation, we have:
F1 - f = (m1 + 96.6 * 10^3 kg) * acceleration
Now we can solve for acceleration:
acceleration = (F1 - f) / (m1 + 96.6 * 10^3 kg)
Plugging in the values we have:
acceleration = (22.4 * 10^3 N - 10.1 * 10^3 N) / (48.3 * 10^3 kg + 96.6 * 10^3 kg)
Calculating the numerical result:
acceleration = 12.3 * 10^3 N / 144.9 * 10^3 kg
acceleration = 0.0847 m/s^2
So, the magnitude of the acceleration of the train is approximately 0.0847 m/s^2.