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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 computer weighing \(87 \mathrm{N}\) rests on the horizontal surface of your desk. The coefficient of friction between the computer and the desk is $0.60 .$ (a) Draw an FBD for the computer. (b) What is the magnitude of the frictional force acting on the computer? (c) How hard would you have to push on it to get it to start to slide across the desk?
Before hanging new William Morris wallpaper in her bedroom, Brenda sanded the walls lightly to smooth out some irregularities on the surface. The sanding block weighs \(2.0 \mathrm{N}\) and Brenda pushes on it with a force of $3.0 \mathrm{N}\( at an angle of \)30.0^{\circ}$ with respect to the vertical, and angled toward the wall. Draw an FBD for the sanding block as it moves straight up the wall at a constant speed. What is the coefficient of kinetic friction between the wall and the block?
Two canal workers pull a barge along the narrow waterway at a constant speed. One worker pulls with a force of \(105 \mathrm{N}\) at an angle of \(28^{\circ}\) with respect to the forward motion of the barge and the other worker, on the opposite tow path, pulls at an angle of \(38^{\circ}\) relative to the barge motion. Both ropes are parallel to the ground. (a) With what magnitude force should the second worker pull to make the sum of the two forces be in the forward direction? (b) What is the magnitude of the force on the barge from the two tow ropes?
During a balloon ascension, wearing an oxygen mask, you measure the weight of a calibrated \(5.00-\mathrm{kg}\) mass and find that the value of the gravitational field strength at your location is 9.792 N/kg. How high above sea level, where the gravitational field strength was measured to be $9.803 \mathrm{N} / \mathrm{kg},$ are you located?
A locomotive pulls a train of 10 identical cars, on a track that runs east- west, with a force of \(2.0 \times 10^{6} \mathrm{N}\) directed east. What is the force with which the last car to the west pulls on the rest of the train?
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