/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 84 A train consists of 50 cars, eac... [FREE SOLUTION] | 91Ó°ÊÓ

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A train consists of 50 cars, each of which has a mass of \(6.8 \times 10^{3} \mathrm{kg} .\) The train has an acceleration of \(+8.0 \times 10^{-2} \mathrm{m} / \mathrm{s}^{2} .\) Ignore friction and determine the tension in the coupling (a) between the 30th and 31st cars and (b) between the 49th and 50th cars.

Short Answer

Expert verified
Part (a): 10,880 N; Part (b): 544 N.

Step by step solution

01

Determine the Mass of the Remaining Cars

First, for part (a), consider only the cars being pulled by the coupling between the 30th and 31st cars. This means the 31st car to the 50th car, which is a total of 20 cars. For part (b), consider only the car being pulled by the coupling between the 49th and 50th cars, which is just the 50th car.
02

Calculate Total Mass for the 31st to 50th Cars

For 20 cars, each with a mass of \(6.8 \times 10^3 \text{ kg}\), the total mass of these cars is:\[20 \times 6.8 \times 10^3 \text{ kg} = 1.36 \times 10^5 \text{ kg}\]
03

Calculate Total Mass for the 50th Car

For the single 50th car, the mass is simply:\[1 \times 6.8 \times 10^3 \text{ kg} = 6.8 \times 10^3 \text{ kg}\]
04

Use Newton's Second Law to Find Tension (Part a)

Using Newton's Second Law, \( F = ma \), the tension in the coupling between the 30th and 31st cars is calculated based on the total mass of the 31st to 50th cars (1.36 \( \times 10^5 \text{ kg} \)) and the given acceleration (\(+8.0 \times 10^{-2} \text{ m/s}^2\)): \[F = (1.36 \times 10^5) \times (8.0 \times 10^{-2}) = 1.088 \times 10^4 \text{ N}\]
05

Use Newton's Second Law to Find Tension (Part b)

For the coupling between the 49th and 50th cars, using the single car's mass of \(6.8 \times 10^3 \text{ kg} \):\[F = (6.8 \times 10^3) \times (8.0 \times 10^{-2}) = 544 \text{ N}\]
06

Conclusion

The tension in the coupling between the 30th and 31st cars is \(1.088 \times 10^4 \text{ N}\) and between the 49th and 50th cars is \(544 \text{ N}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Tension
Tension is a force that transmits energy through strings, ropes, cables or any other form of medium when forces are applied at opposite ends. In the context of this exercise, tension is transmitted through the couplings between train cars as they are pulled forward by the locomotive. When evaluating tension, it’s about considering the forces acting to extend and hold the medium. In this specific train scenario, tension helps us understand the pulling forces required to accelerate the cars. Each coupling between the train cars experiences tension from the mass of the cars it pulls and the train's acceleration. By calculating the tension, we determine how much force is needed in the couplings to move these specific sections of the train.
Defining Mass
Mass is a measure of the amount of matter in an object, and it's a fundamental property that affects how an object responds to forces. In our train problem, knowing the mass of each car is crucial because it enables us to compute the force needed to accelerate the cars. Each car in the train has a mass of \(6.8 \times 10^3\) kg, which we use to find the total mass of segmented parts of the train. For instance, to find the tension in the coupling between the 30th and 31st cars, we need to calculate the combined mass of the 20 cars being pulled behind this point, resulting in a total mass of \(1.36 \times 10^5\) kg. This total mass directly influences the calculation of tension as it determines the gravitational force, which impacts the acceleration.
Comprehending Acceleration
Acceleration is the rate of change of velocity of an object with respect to time, indicating how quickly it speeds up or slows down. In this exercise, the train has a given acceleration of \(+8.0 \times 10^{-2}\) m/s². This value is important because it combines with the mass of the train segment being considered to calculate the required force. The acceleration tells us that for every second, the velocity of the train's cars increases by \(+8.0 \times 10^{-2}\) m/s. To find the force or the tension, we must apply this acceleration to the total mass of the cars being pulled by any given coupling. By using Newton's Second Law, we can translate the acceleration's effect on these mass formations into tangible tension results.
Mastering Force Calculation with Newton's Second Law
Newton's Second Law of Motion states that the force acting on an object is equal to the mass of that object multiplied by its acceleration, formulated as \( F = ma \). This fundamental principle enables us to calculate the forces acting within dynamic systems, such as the tension in the train's couplings. For the situation where we need to find the tension between the 30th and 31st cars, we multiply the combined mass of these cars, \(1.36 \times 10^5\) kg, by the train’s acceleration, \(+8.0 \times 10^{-2}\) m/s², resulting in a force of \(1.088 \times 10^4\) N, or Newtons. This calculation method efficiently allows us to quantify the force necessary to maintain the train's motion, taking into account both the mass being acted upon and the rate of change in its velocity. Through understanding Newton’s Second Law, we gain deeper insights into how forces interact with mass and acceleration to produce such tension.

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Most popular questions from this chapter

As part \(a\) of the drawing shows, two blocks are connected by a rope that passes over a set of pulleys. One block has a weight of 412 \(\mathrm{N}\) , and the other has a weight of 908 \(\mathrm{N}\) . The rope and the pulleys are mass-less and there is no friction. (a) What is the acceleration of the lighter block? (b) Suppose that the heavier block is removed, and a downward force of 908 \(\mathrm{N}\) is provided by someone pulling on the rope, as part \(b\) of the drawing shows. Find the acceleration of the remaining block. (c) Explain why the answers in (a) and (b) are different.

The central ideas in this problem are reviewed in Multiple-Concept Example 9. One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and the magnitude of the force is slowly increased. When the force reaches 47.0 \(\mathrm{N}\), the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block so that it just begins to slide out from under the upper block?

A block is pressed against a vertical wall by a force \(\overrightarrow{\mathbf{P}},\) as the drawing shows. This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The magnitude of the force is different in the two cases, while the directional angle \(\theta\) is the same. Kinetic friction exists between the block and the wall, and the coefficient of kinetic friction is 0.250 . The weight of the block is \(39.0 \mathrm{N},\) and the directional angle for the force \(\overrightarrow{\mathbf{P}}\) is \(\theta=30.0^{\circ} .\) Determine the magnitude of \(\overrightarrow{\mathbf{P}}\) when the block slides \((\mathbf{a})\) up the wall and (b) down the wall.

A billiard ball strikes and rebounds from the cushion of a pool table perpendicularly. The mass of the ball is \(0.38 k g\). The ball approaches the cushion with a velocity of \(+2.1 \mathrm{m} / \mathrm{s}\) and rebounds with a velocity of \(-2.0 \mathrm{m} / \mathrm{s}\) . The ball remains in contact with the cushion for a time of \(3.3 \times 10^{-3} \mathrm{s}\) What is the average net force (magnitude and direction) exerted on the ball by the cushion?

ssm mm A rocket of mass \(4.50 \times 10^{5} \mathrm{kg}\) is in flight. Its thrust is directed at an angle of \(55.0^{\circ}\) above the horizontal and has a magnitude of \(7.50 \times 10^{6} \mathrm{N}\) . Find the magnitude and direction of the rocket's acceleration. Give the direction as an angle above the horizontal.

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