Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 86

A 2.0 -kg toy locomotive is pulling a 1.0 -kg caboose. The frictional force of the track on the caboose is \(0.50 \mathrm{N}\) backward along the track. If the train's acceleration forward is \(3.0 \mathrm{m} / \mathrm{s}^{2},\) what is the magnitude of the force exerted by the locomotive on the caboose?

Short Answer

Expert verified
Answer: 3.50 N

Step by step solution

01

Identify the forces acting on the caboose

There are two forces acting on the caboose: the force exerted by the locomotive, which we'll represent as \(F_{L}\), and the frictional force acting in the opposite direction, which is 0.50 N. We will denote the frictional force as \(F_{f}\).
02

Write the equation using Newton's second law

Newton's second law states that the net force acting on an object is equal to the product of its mass and acceleration: \(F_{net} = m \times a\). We can write the net force acting on the caboose as the difference between the force exerted by the locomotive and the frictional force: \(F_{net} = F_{L} - F_{f}\).
03

Substitute the given values and solve for the unknown force

We are given the mass of the caboose (m = 1.0 kg) and the train's acceleration (a = 3.0 m/s²). We also know that the frictional force is \(F_{f} = 0.50\,\mathrm{N}\). Substituting these values into the equation from Step 2, we get: \(F_{net} = F_{L} - F_{f}\) \(m \times a = F_{L} - 0.50\,\mathrm{N}\) \(1.0\,\mathrm{kg} \times 3.0\,\mathrm{m/s^2} = F_{L} - 0.50\,\mathrm{N}\) \(3.0\,\mathrm{N} = F_{L} - 0.50\,\mathrm{N}\) Now, we will solve for \(F_{L}\): \(F_{L} = 3.0\,\mathrm{N} + 0.50\,\mathrm{N}\) \(F_{L} = 3.50\,\mathrm{N}\)
04

State the final answer

The magnitude of the force exerted by the locomotive on the caboose is \(3.50\,\mathrm{N}\).

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.

Start your free trial

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

An astronaut stands at a position on the Moon such that Earth is directly over head and releases a Moon rock that was in her hand. (a) Which way will it fall? (b) What is the gravitational force exerted by the Moon on a 1.0 -kg rock resting on the Moon's surface? (c) What is the gravitational force exerted by the Earth on the same 1.0 -kg rock resting on the surface of the Moon? (d) What is the net gravitational force on the rock?
The mass of the Moon is 0.0123 times that of the Earth. A spaceship is traveling along a line connecting the centers of the Earth and the Moon. At what distance from the Earth does the spaceship find the gravitational pull of the Earth equal in magnitude to that of the Moon? Express your answer as a percentage of the distance between the centers of the two bodies.
The coefficient of static friction between a block and a horizontal floor is \(0.40,\) while the coefficient of kinetic friction is \(0.15 .\) The mass of the block is \(5.0 \mathrm{kg} .\) A horizontal force is applied to the block and slowly increased. (a) What is the value of the applied horizontal force at the instant that the block starts to slide? (b) What is the net force on the block after it starts to slide?
A crate of potatoes of mass \(18.0 \mathrm{kg}\) is on a ramp with angle of incline \(30^{\circ}\) to the horizontal. The coefficients of friction are \(\mu_{\mathrm{s}}=0.75\) and \(\mu_{\mathrm{k}}=0.40 .\) Find the frictional force (magnitude and direction) on the crate if the crate is sliding up the ramp.
Refer to Example 4.19. What is the apparent weight of the same passenger (weighing \(598 \mathrm{N}\) ) in the following situations? In each case, the magnitude of the elevator's acceleration is $0.50 \mathrm{m} / \mathrm{s}^{2} .$ (a) After having stopped at the 15 th floor, the passenger pushes the 8 th floor button; the elevator is beginning to move downward. (b) The elevator is moving downward and is slowing down as it nears the 8 th floor.
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Mechanics and Materials

Read Explanation

Engineering Physics

Read Explanation

Thermodynamics

Read Explanation

Work Energy and Power

Read Explanation

Modern Physics

Read Explanation
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