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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on either side of the canal. Each horse pulls with a force of \(560 \mathrm{N}\) at an angle of \(15^{\circ}\) with the centerline of the canal. Find the sum of the two forces exerted by the horses on the barge.
Find the altitudes above the Earth's surface where Earth's gravitational field strength would be (a) two thirds and (b) one third of its value at the surface. [Hint: First find the radius for each situation; then recall that the altitude is the distance from the surface to a point above the surface. Use proportional reasoning.]
A woman of mass \(51 \mathrm{kg}\) is standing in an elevator. (a) If the elevator floor pushes up on her feet with a force of $408 \mathrm{N},$ what is the acceleration of the elevator? (b) If the elevator is moving at \(1.5 \mathrm{m} / \mathrm{s}\) as it passes the fourth floor on its way down, what is its speed 4.0 s later?
Oliver has a mass of \(76.2 \mathrm{kg} .\) He is riding in an elevator that has a downward acceleration of \(1.37 \mathrm{m} / \mathrm{s}^{2} .\) With what magnitude force does the elevator floor push upward on Oliver?
A bike is hanging from a hook in a garage. Consider the following forces: (a) the force of the Earth pulling down on the bike, (b) the force of the bike pulling up on the Earth, (c) the force of the hook pulling up on the bike, and (d) the force of the hook pulling down on the ceiling. Which two forces are equal and opposite because of Newton's third law? Which two forces are equal and opposite because of Newton's first law?
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free