Question

An engine pulls a train of 20 freight cars, each having a mass of $5.0 \times 10^{4} \mathrm{kg}$ with a constant force. The cars move from rest to a speed of \(4.0 \mathrm{m} / \mathrm{s}\) in \(20.0 \mathrm{s}\) on a straight track. Ignoring friction, what is the force with which the 10th car pulls the 11th one (at the middle of the train)? (school bus)

Short answer

Answer: The force with which the 10th car pulls the 11th car is \(1.0 \times 10^5\) N.

Step-by-step solution

Calculate the acceleration of the train.

First, let's find the acceleration of the train using the given data: Initial velocity (u) = 0 m/s (since the train starts at rest) Final velocity (v) = 4.0 m/s Time (t) = 20.0 s We can use the equation: v = u + a*t We can plug in the values and solve for 'a': 4.0 m/s = 0 + a*(20.0 s) a = 0.2 m/s^2 The acceleration of the train is 0.2 m/s².

Calculate the total mass of the train.

Now, let's calculate the total mass of the train: Number of freight cars = 20 Mass of each freight car = \(5.0 \times 10^4\) kg Total mass of the train (M) = Number of freight cars * Mass of each freight car M = 20 * \(5.0 \times 10^4\) kg = \(1.0 \times 10^6\) kg.

Calculate the total force acting on the train.

Using Newton's second law of motion, we can calculate the total force exerted on the train: Force (F) = Total mass (M) * Acceleration (a) F = \(1.0 \times 10^6\) kg * 0.2 m/s² F = \(2.0 \times 10^5\) N

Calculate the force between the 10th and 11th freight cars.

Since there are 20 freight cars in the train and we are looking for the force between cars 10 and 11, we can calculate the force on one half of the train (10 cars). Since the force is uniformly distributed, we can divide the total force by 2: Force between the 10th and 11th car = Total Force / 2 Force between the 10th and 11th car = \(2.0 \times 10^5\) N / 2 Force between the 10th and 11th car = \(1.0 \times 10^5\) N The force with which the 10th car pulls the 11th one is \(1.0 \times 10^5\) N.