/*! 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 34 When you push a \(1.80-\mathrm{k... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When you push a \(1.80-\mathrm{kg}\) book that is resting on a tabletop, a force of \(2.25 \mathrm{~N}\) is required to start the book sliding. Once it is sliding, however, a force of only \(1.50 \mathrm{~N}\) keeps the book moving with constant speed. What are the coefficients of static and kinetic friction between the book and the tabletop?

Short Answer

Expert verified
The coefficient of static friction is approximately 0.128, and the coefficient of kinetic friction is approximately 0.085.

Step by step solution

01

Understanding Static Friction

Static friction (\( f_s \)) is the friction force that keeps an object at rest when a force is applied to it, up to a maximum value. The book starts sliding when the applied force exceeds this static friction.
02

Determining Static Friction Force

The force required to start the book sliding is given as \( 2.25 \mathrm{~N} \). This is the force that overcomes the maximum static friction force.
03

Calculating the Coefficient of Static Friction

The coefficient of static friction (\( \mu_s \)) is calculated using the equation:\( f_s = \mu_s \cdot N \), where \( N \) is the normal force. Here, \( N = m \cdot g = 1.80 \cdot 9.8 \approx 17.64 \mathrm{~N} \). Therefore, \( \mu_s = \frac{f_s}{N} = \frac{2.25}{17.64} \approx 0.128 \).
04

Understanding Kinetic Friction

Kinetic friction (\( f_k \)) is the friction force acting on an object in motion. It is less than the maximum static friction and allows the book to move at constant speed.
05

Determining Kinetic Friction Force

The force required to keep the book moving with constant speed is \( 1.50 \mathrm{~N} \). This is the kinetic friction force.
06

Calculating the Coefficient of Kinetic Friction

The coefficient of kinetic friction (\( \mu_k \)) is calculated using the equation:\( f_k = \mu_k \cdot N \). Using the same normal force \( N = 17.64 \mathrm{~N} \), \( \mu_k = \frac{f_k}{N} = \frac{1.50}{17.64} \approx 0.085 \).

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 91Ó°ÊÓ!

Key Concepts

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

Static Friction
Static friction is the force that prevents an object from moving when it is at rest, even when a force is applied. Imagine trying to push a book on a table. Initially, it does not move no matter how much effort you seem to exert. That is the static friction at work. Static friction acts to stop motion until a certain threshold, called the maximum static friction, is surpassed by an applied force. Without static friction, objects would slide off surfaces far too easily. It acts as the adhesive force between the contact surfaces of the book and the table. In our exercise, a force of 2.25 N is needed to overcome this static grip between the stationary book and the table.

When calculating the coefficient of static friction, ( \( \mu_s \)) ), you can use the formula:\[ f_s = \mu_s \cdot N \] where:
  • \( f_s \) is the force of static friction, here 2.25 N.
  • \( \mu_s \) is the coefficient we aim to find.
  • \( N \) is the normal force exerted by the object onto the surface, calculated as mass times gravity (\( N = m \cdot g \)).
In this example, the book has a normal force of approximately 17.64 N, leading us to find a coefficient of static friction of around 0.128.
Kinetic Friction
Kinetic friction comes into play once the object has started moving. Unlike static friction, it is concerned with objects already in motion. It is this force that acts opposite to the direction of motion, slowing it down or keeping it at a steady speed against an externally applied force. Once the book on the table is sliding, it requires less force to keep it moving compared to what was needed to start it moving. This ongoing resistance is due to kinetic friction and acts parallel to the surface.

In the exercise,when the book slides over the tabletop, only a force of 1.50 N is needed to maintain its constant speed. This shows that the maximum static friction has been overcome, and kinetic friction works to ensure the slide does not pick up speed unrestricted by other forces. The formula to find the coefficient of kinetic friction, ( \( \mu_k \)) ), is given by:\[ f_k = \mu_k \cdot N \] where:
  • \( f_k \) is the kinetic friction force, noted here as 1.50 N.
  • \( \mu_k \) is what we calculate.
  • \( N \) is, again, the normal force of about 17.64 N.
The resulting calculation provides a kinetic friction coefficient of about 0.085, indicating how much easier it is to keep moving rather than start moving the book.
Coefficient of Friction
The coefficient of friction is an important parameter in physics that lets us quantify how much friction exists between two surfaces. It is a dimensionless value, indicating the degree of resistance to movement that the surfaces exert against each other. There are separate coefficients for static (\( \mu_s \)) and kinetic (\( \mu_k \)) friction since the resistance encountered by an object in motion differs from when it is stationary. Understanding these coefficients helps in predicting the force needed to move an object or keep it moving, which is vital in many practical and everyday applications.
  • The higher the coefficient, the more grip or friction present.
  • A lower coefficient points to smoother, easier sliding surfaces.
In the exercise, using our calculated values:
  • \( \mu_s \approx 0.128 \) for static friction indicates the level of resistance before movement begins.
  • \( \mu_k \approx 0.085 \) for kinetic friction shows resistance during the sliding motion.
These coefficients help engineers in designing materials and surfaces, to either maximize grip or minimize friction, depending on the intended use.

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

After hitting a long fly ball that goes over the right fielder's head and lands in the outfield, a batter decides to keep going past second base and try for third base. The \(62-\mathrm{kg}\) player begins sliding \(3.4 \mathrm{~m}\) from the base with a speed of \(4.5 \mathrm{~m} / \mathrm{s}\). (a) If the player comes to rest at third base, how much work was done on the player by friction with the ground? (b) What was the coefficient of kinetic friction between the player and the ground?

A particle moves without friction. At point A the particle has a kinetic energy of \(12 \mathrm{~J}\); at point \(\mathrm{B}\) the particle is momentarily at rest, and the potential energy of the system is \(25 \mathrm{~J}\); at point \(\mathrm{C}\) the potential energy of the system is \(5 \mathrm{~J}\). (a) What is the potential energy of the system when the particle is at point \(\mathrm{A}\) ? (b) What is the kinetic energy of the particle at point \(C\) ?

Assess System 1 has a force of \(10 \mathrm{~N}\) and a speed of \(5 \mathrm{~m} / \mathrm{s}\). System 2 has a force of \(20 \mathrm{~N}\) and a speed of \(2 \mathrm{~m} / \mathrm{s}\). Which system has the greater power? Explain.

A 51-kg packing crate is pulled across a rough floor with a rope that is at an angle of \(43^{\circ}\) above the horizontal. If the tension in the rope is \(120 \mathrm{~N}\), how much work is done on the crate to move it \(18 \mathrm{~m}\) ?

An object moves with no friction or air resistance. Initially, its kinetic energy is \(10 \mathrm{~J}\), and its gravitational potential energy is \(20 \mathrm{~J}\). What is its kinetic energy when its potential energy has decreased to 15 J? What is its potential energy when its kinetic energy has decreased to 5 J?

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.