Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 8

A horizontal force equal to the object's weight is applied to an object resting on a table. What is the acceleration of the moving object when the coefficient of kinetic friction between the object and floor is 1 (assuming the object is moving in the direction of the applied force). a) zero c) Not enough information is b) \(1 \mathrm{~m} / \mathrm{s}^{2}\) given to find the acceleration.

Short Answer

Expert verified
Answer: The acceleration of the object is 0 m/s².

Step by step solution

01

Determine the Applied Force

To determine the acceleration of the object, we first need to find out the applied force. The weight of the object can be calculated using the formula \(W=mg\), where \(W\) is the weight, \(m\) is the mass, and \(g\) is the gravitational acceleration (\(9.81 \mathrm{~m} / \mathrm{s}^{2}\)). The problem states that the applied force is equal to the object's weight, so we have: \(F_{applied} = W = mg\)
02

Calculate the Friction Force

Now, we need to calculate the friction force experienced by the object when it is moving. The friction force (\(F_{friction}\)) can be calculated using the formula: \(F_{friction} = \mu F_N\), where \(\mu\) is the coefficient of friction and \(F_N\) is the normal force. In this case, since the object is resting on a horizontal surface, the normal force is equal to the object's weight. Therefore, we have: \(F_{friction} = \mu mg\) Given that the coefficient of kinetic friction (\(\mu\)) between the object and the floor is 1, the friction force becomes: \(F_{friction} = 1 \cdot mg = mg\)
03

Apply Newton's Second Law

Now, we can apply Newton's second law to find the acceleration of the moving object. According to Newton's second law, the net force acting on an object is equal to the product of its mass and acceleration: \(F_{net} = ma\). The net force acting on the object is the difference between the applied force and the friction force. Since the applied force and friction force are equal, we have: \(F_{net} = F_{applied} - F_{friction} = mg - mg = 0\) Using Newton's second law, we can now calculate the acceleration: \(0 = ma\)
04

Determine the Acceleration

Finally, to determine the acceleration of the moving object, we need to solve the equation obtained in the last step: \(0 = ma\) Dividing both sides by \(m\), we get the acceleration: \(a = 0 \mathrm{~m} / \mathrm{s}^{2}\) Therefore, the correct answer is (a) zero.

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

The gravitational acceleration on the Moon is a sixth of that on Earth. The weight of an apple is \(1.00 \mathrm{~N}\) on Earth. a) What is the weight of the apple on the Moon? b) What is the mass of the apple?

A load of bricks of mass \(M=200.0 \mathrm{~kg}\) is attached to a crane by a cable of negligible mass and length \(L=3.00 \mathrm{~m}\). Initially, when the cable hangs vertically downward, the bricks are a horizontal distance \(D=1.50 \mathrm{~m}\) from the wall where the bricks are to be placed. What is the magnitude of the horizontal force that must be applied to the load of bricks (without moving the crane) so that the bricks will rest directly above the wall?

A car without ABS (antilock brake system) was moving at \(15.0 \mathrm{~m} / \mathrm{s}\) when the driver hit the brake to make a sudden stop. The coefficients of static and kinetic friction between the tires and the road are 0.550 and 0.430 , respectively. a) What was the acceleration of the car during the interval between braking and stopping? b) How far did the car travel before it stopped?

What coefficient of friction is required to stop a hockey puck sliding at \(12.5 \mathrm{~m} / \mathrm{s}\) initially over a distance of \(60.5 \mathrm{~m} ?\)

A car pulls a trailer down the highway. Let \(F_{\mathrm{t}}\) be the be the magnitude of the force on car due to the trailer. If the car and trailer are moving at a constant velocity across level ground, then \(F_{\mathrm{t}}=F_{c}\). If the car and trailer are accelerating up a hill, what is the relationship between the two forces?
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Kinematics Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Radiation

Read Explanation

Dynamics

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