A train engine pulls out of a station along a straight horizontal track with five identical freight cars behind it, each of which weighs 88.0 kN. The train reaches a speed of 13.0 m/s within 6.50 min of starting out. Assuming the engine pulls with a constant force during this interval, and ignore air resistance and friction on the freight cars. Find the tension in the coupling between cars 2 and 3???

Question

A train engine pulls out of a station along a straight horizontal track with five identical freight cars behind it, each of which weighs 88.0 kN. The train reaches a speed of 13.0 m/s within 6.50 min of starting out. Assuming the engine pulls with a constant force during this interval, and ignore air resistance and friction on the freight cars. Find the tension in the coupling between cars 2 and 3???

Answer 1

To find the tension in the coupling between cars 2 and 3, we need to consider the forces acting on the cars and apply Newton's laws of motion.

First, let's calculate the total weight of the freight cars. Since each freight car weighs 88.0 kN, the total weight of the five cars is (88.0 kN) x 5 = 440.0 kN.

Next, we need to calculate the net force acting on the train. The net force is equal to the product of the train's mass and its acceleration (F = ma). To find the mass of the train, we can use the formula m = W/g, where W is the weight of the train and g is the acceleration due to gravity (approximately 9.8 m/s^2).

The weight of the train is the sum of the weight of the engine and the weight of the freight cars. Since the engine and the five cars are identical, the weight of the engine is equal to the weight of each car, which is 88.0 kN. So the total weight of the train is 88.0 kN + 440.0 kN = 528.0 kN.

Using the formula m = W/g, we can calculate the mass of the train: m = (528.0 kN) / (9.8 m/s^2) = 53.87755 metric tons.

Now, let's find the acceleration of the train. We are given that the train reaches a speed of 13.0 m/s within 6.50 min. To convert the time to seconds, we multiply by 60: t = 6.50 min x 60 s/min = 390 s.

The acceleration is given by the formula a = (v - u) / t, where v is the final velocity, u is the initial velocity, and t is the time. In this case, the initial velocity (u) is 0 m/s since the train starts from rest. So the acceleration is a = (13.0 m/s - 0 m/s) / 390 s = 0.033333333 m/s^2.

Now we can calculate the net force acting on the train using F = ma: F = (53.87755 metric tons) x (0.033333333 m/s^2) = 1.79575517 metric tons * m/s^2.

Since the engine pulls with a constant force, this force is the tension in the coupling between cars 2 and 3. Therefore, the tension in the coupling between cars 2 and 3 is approximately 1.80 metric tons * m/s^2.