/*! 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 54 A book slides across a level, ca... [FREE SOLUTION] | 91Ó°ÊÓ

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A book slides across a level, carpeted floor at an initial speed of \(4 \mathrm{~m} / \mathrm{s}\) and comes to rest after \(3.25 \mathrm{~m}\). Calculate the coefficient of kinetic friction between the book and the carpet. Assume the only forces acting on the book are friction, weight, and the normal force.

Short Answer

Expert verified
To calculate the coefficient of kinetic friction between the book and the carpet, you first calculate acceleration using the given initial velocity and stopping distance. Then, you use the equation for kinetic friction to isolate and solve for the coefficient of kinetic friction.

Step by step solution

01

Identify given parameters

The initial speed \(v_0\) of the book is given as 4 m/s and it comes to rest after sliding a distance \(d\) of 3.25 m. Also, we're told to consider only friction, weight, and the normal force.
02

Apply the equation of motion

The equation of motion to be utilized here is \(v^2 = v_0^2 + 2a d\) , where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(a\) is acceleration and \(d\) is the distance covered. Given the book comes to rest, \(v = 0\). So, we can adapt the equation to solve for the acceleration \(a\) as follows: \(a = -v_0^2 / (2d)\). The negative sign indicates that this is a deceleration.
03

Use Newton’s second law

Concluding that the total force acting on the book is zero (since the book is not accelerating upwards or downwards, the normal force cancels out the weight), Newton's second law (friction = mass * acceleration) can be rearranged as: friction/mass = acceleration. We know that the friction force can be given by \(f_k = \mu_k * m * g\) where \(f_k\) is kinetic friction, \(\mu_k\) is the coefficient of kinetic friction, \(m\) is mass, and \(g\) is gravity. Replacing mass * acceleration with friction in our equation gives us \(\mu_k * g = a\).
04

Solve for the coefficient of kinetic friction

Isolate \(\mu_k\) by dividing both sides of the equation by \(g\), to get \(\mu_k = a / g\). Substituting the previously obtained acceleration \(a\) into this equation will give us the coefficient of kinetic friction. Remember that \(g\) is the gravitational acceleration and on Earth it is approximately 9.8 m/s².

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Key Concepts

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

Newton's Second Law
Newton's Second Law is central to understanding how forces interact with masses to produce motion. The law states that the force exerted on an object is equal to the mass of the object times its acceleration. This can be simplified into the equation: \[ F = m imes a \] Where:
  • \( F \) is the net force acting on the object,
  • \( m \) is the mass, and
  • \( a \) is the acceleration.
In the context of the exercise, Newton's Second Law helps us understand that although the book is moving, the forces acting vertically, like weight and normal force, balance each other out. Thus, the horizontal force (friction) is what slows down the book. The frictional force equals the mass of the book times its deceleration. Hence, by knowing how fast the book slows down, we can pinpoint the strength of the friction.
Equations of Motion
Equations of motion are formulas that relate the five main variables of an object's motion: displacement, initial velocity, final velocity, acceleration, and time. These equations are applicable only when the acceleration is constant. One of the main equations used in this problem is: \[ v^2 = v_0^2 + 2a d \]
  • \( v \) is the final velocity, which is 0 m/s since the book comes to rest.
  • \( v_0 \) is the initial velocity.
  • \( a \) is the acceleration, which we solve for.
  • \( d \) is the distance.
When the book stops, its final velocity is zero, simplifying the formula and allowing us to solve for deceleration. This deceleration is crucial for determining the force of friction acting against the movement.
Kinetic Friction
Kinetic friction occurs when two surfaces are moving relative to each other, such as a book sliding over a carpet. This force opposes the motion of an object, causing it to decelerate and eventually stop. Kinetic friction depends on:
  • The nature of the surfaces in contact (rougher surfaces have higher friction).
  • The normal force, which is the support force exerted by a surface perpendicular to the object. For a book on a flat surface, this equals its weight.
The equation to express kinetic friction is: \[ f_k = \mu_k \times N \] where \( \mu_k \) is the coefficient of kinetic friction and \( N \) represents the normal force. In this exercise, by rearranging Newton's Second Law and using the calculated acceleration, we found the coefficient to understand friction's role in bringing the book to a halt.
Gravitational Acceleration
Gravitational acceleration is the rate at which all objects accelerate towards Earth due to gravity. On Earth, this is approximately \(9.8 \ m/s^2\). This constant affects every falling object and applies to scenarios where gravity acts as a force component. In our problem, gravitational acceleration helps determine the normal force. The book's weight equals the gravitational force pulling it down. Since these forces balance each other on a flat surface, the normal force is equal in magnitude to the weight. By using gravitational acceleration as part of the normal force calculation, we can solve for kinetic friction. Understanding how gravity interacts with objects at rest or in motion helps us understand other forces, like friction, that affect those objects.

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Most popular questions from this chapter

A child slides down a snow-covered slope on a sled. At the top of the hill, her mother gives her a push at a speed of \(1 \mathrm{~m} / \mathrm{s}\) to get her started. The frictional force acting on the sled is one-fifth of the combined weight of the child and the sled. If she travels for a distance of \(25 \mathrm{~m}\) and her speed at the bottom is \(4 \mathrm{~m} / \mathrm{s}\), calculate the angle that the hill makes with the horizontal.

The force that acts on an object is given by \(\vec{F}=2.2 \hat{x}+4.5 \hat{y}\). Calculate the work done on the object when the force results in a displacement given by \(\vec{d}=12 \hat{x}+20 y\). (The multiplicative constants in both expressions carry SI units.)

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