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Chapter 4: Problem 88

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

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

Step by step solution

01

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.
02

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.
03

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
04

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.

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Most popular questions from this chapter

A toy freight train consists of an engine and three identical cars. The train is moving to the right at constant speed along a straight, level track. Three spring scales are used to connect the cars as follows: spring scale \(A\) is located between the engine and the first car; scale \(\mathrm{B}\) is between the first and second cars; scale \(C\) is between the second and third cars. (a) If air resistance and friction are negligible, what are the relative readings on the three spring scales \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C} ?\) (b) Repeat part (a), taking air resistance and friction into consideration this time. [Hint: Draw an FBD for the car in the middle. \(]\) (c) If air resistance and friction together cause a force of magnitude \(5.5 \mathrm{N}\) on each car, directed toward the left, find the readings of scales $\mathrm{A}, \mathrm{B},\( and \)\mathrm{C}$
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