A locomotive that has a mass of 6.4 x 105 kg is used to pull 2 railway cars. Railway car 1 has a mass of 5.0 x 105 kg, is attached to railway car to that has a mass of 3.6 x 105 kg by a locking mechanism. A railway engineer tests the mechanism and estimates that it can only withstand 2.0 x 105 N of force. Determine the maximum acceleration of the train that does not break the locking mechanism. Explain your reasoning. Assume that friction is negligible.
Well, well, well! Looks like we have ourselves a locomotive conundrum! Don't worry, my friend, Clown Bot is here to put a smile on your face and some knowledge in your brain.
To find the maximum acceleration that the train can have without breaking the locking mechanism, we need to consider the forces at play. The force applied by the locomotive will be used to accelerate the entire mass of the train (locomotive + railway cars). If this force exceeds the maximum force the locking mechanism can withstand, trouble could be on the tracks!
Let's start by calculating the total mass of the train. We have a locomotive with a mass of 6.4 x 105 kg, and railway car 1 with a mass of 5.0 x 105 kg, and railway car 2 with a mass of 3.6 x 105 kg. Adding these up, we get a total mass of:
6.4 x 105 kg + 5.0 x 105 kg + 3.6 x 105 kg = 15.0 x 105 kg
Now, we need to convert the maximum force the locking mechanism can withstand (2.0 x 105 N) into an acceleration by dividing it by the total mass of the train:
Acceleration = Force / Mass
Acceleration = 2.0 x 105 N / 15.0 x 105 kg
Simplifying, we find:
Acceleration = 0.133 m/s²
So, the maximum acceleration that the train can have without breaking the locking mechanism is approximately 0.133 m/s². This means that the locomotive needs to apply a force that keeps the acceleration below this value to keep the locking mechanism intact.
Just remember, in the world of physics, it's all about keeping things in balance. So, make sure that acceleration doesn't go overboard, or you might end up with railway cars flying in all directions!
To determine the maximum acceleration of the train that does not break the locking mechanism, we need to consider the force exerted on the locking mechanism.
The force exerted on the locomotive is given by Newton's second law of motion:
Force = mass × acceleration
The force exerted on the locking mechanism will be the total force exerted on the system (locomotive and railway cars) minus the force exerted by railway car 1 on railway car 2.
The total force exerted on the system is given by:
Force_total = (mass_locomotive + mass_railway_car1 + mass_railway_car2) × acceleration
To calculate the maximum acceleration, we need to find the maximum force that the locking mechanism can withstand. This is given in the question as 2.0 × 10^5 N.
Set up the equation to find the maximum acceleration:
2.0 × 10^5 N = (6.4 × 10^5 kg + 5.0 × 10^5 kg + 3.6 × 10^5 kg) × acceleration
Simplifying the equation:
2.0 × 10^5 N = (15 × 10^5 kg) × acceleration
Dividing both sides by 15 × 10^5 kg:
acceleration = (2.0 × 10^5 N) / (15 × 10^5 kg)
Simplifying:
acceleration = 0.133 m/s^2
Therefore, the maximum acceleration of the train that does not break the locking mechanism is 0.133 m/s^2.
To determine the maximum acceleration of the train that does not break the locking mechanism, we need to consider the total force exerted on the locking mechanism.
First, let's calculate the total mass of the train by adding the masses of the locomotive and the two railway cars:
Total mass = mass of locomotive + mass of railway car 1 + mass of railway car 2
Substituting the given values:
Total mass = 6.4 x 10^5 kg + 5.0 x 10^5 kg + 3.6 x 10^5 kg
Next, we need to calculate the force required to accelerate the train:
Force = mass × acceleration
Now, the maximum force the locking mechanism can withstand is given as 2.0 x 10^5 N. So, we can set up the following equation:
Force <= maximum force (2.0 x 10^5 N)
Substituting the expression for force:
mass × acceleration <= 2.0 x 10^5 N
Since the lock mechanism holds the two railway cars together, the mass involved in the equation would be the sum of the mass of the locomotive and railway car 1:
(mass of locomotive + mass of railway car 1) × acceleration <= 2.0 x 10^5 N
Substituting the values:
(6.4 x 10^5 kg + 5.0 x 10^5 kg) × acceleration <= 2.0 x 10^5 N
Now we can solve for the acceleration:
11.4 x 10^5 kg × acceleration <= 2.0 x 10^5 N
Dividing both sides of the equation by 11.4 x 10^5 kg:
acceleration <= 2.0 x 10^5 N / 11.4 x 10^5 kg
Simplifying:
acceleration <= 0.175 m/s^2
Therefore, the maximum acceleration of the train that does not break the locking mechanism is 0.175 m/s^2.