/*! 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 126 A locomotive pulls a train of 10... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 126

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?

Short Answer

Expert verified
Answer: The force exerted by the last car to the west is \(2.0 \times 10^6 N\) directed west.

Step by step solution

01

Identify given information and what is asked for

The given information is that the locomotive is pulling the train with a force of \(2.0 \times 10^6 N\) directed east. We are asked to find the force with which the last car to the west pulls on the rest of the train.
02

Apply Newton's third law to the situation

According to Newton's third law, if object A exerts a force on object B, then object B must exert an equal and opposite force on object A. In this case, the locomotive (object A) exerts a force on the first car (object B), so the first car exerts an equal and opposite force on the locomotive.
03

Consider the force exerted by the last car to the west

The force exerted by the last car to the west is equal and opposite to the force exerted by the locomotive. This is due to the fact that the train is in equilibrium, with no net acceleration, and the forces acting on it must cancel out.
04

Calculate the force exerted by the last car to the west

Since the force exerted by the last car to the west is equal and opposite to the force exerted by the locomotive, we can calculate its value as follows: Force exerted by the last car = Force exerted by the locomotive This means that the force exerted by the last car to the west is \(2.0 \times 10^6 N\) directed west.

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Most popular questions from this chapter

A crate of oranges weighing \(180 \mathrm{N}\) rests on a flatbed truck $2.0 \mathrm{m}$ from the back of the truck. The coefficients of friction between the crate and the bed are \(\mu_{\mathrm{s}}=0.30\) and $\mu_{\mathrm{k}}=0.20 .\( The truck drives on a straight, level highway at a constant \)8.0 \mathrm{m} / \mathrm{s} .$ (a) What is the force of friction acting on the crate? (b) If the truck speeds up with an acceleration of \(1.0 \mathrm{m} / \mathrm{s}^{2},\) what is the force of the friction on the crate? (c) What is the maximum acceleration the truck can have without the crate starting to slide?
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