/*! 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 9 Your physics book is resting in ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Your physics book is resting in front of you on a horizontal table in the campus library. You push the book over to your friend, who is seated at the other side of the table, 0.400 m north and 0.300 m east of you. If you push the book in a straight line to your friend, friction does \(-4.8 \mathrm{~J}\) of work on the book. If instead you push the book \(0.400 \mathrm{~m}\) due north and then \(0.300 \mathrm{~m}\) due east, how much work is done by friction?

Short Answer

Expert verified
The work done by friction when the book is pushed 0.400 m north and then 0.300 m east is also -4.8 J.

Step by step solution

01

Understand the force of friction

Understand that friction is a force that resists the motion of an object. In this case, it opposes the movement of the book on the table. The work made by friction on the book is negative, pointing out that it is opposite to the direction of motion.
02

Calculate the Work

Work is done by the force of friction in opposing the motion of the book. In this exercise, the work done by the force of friction when the book is pushed straight to the friend is -4.8 J. Since friction is the same everywhere on this table, and it's realized that the distance does not affect the friction, the work done by friction will be the same when the book is pushed north and then east. Therefore, the work done by friction in this case will also be -4.8 J.

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.

Work and Energy
In the realm of physics, the concept of work highlights how forces cause objects to move. Work is defined as the product of the force applied to an object and the distance over which that force is applied. It is given by the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and the direction of displacement.
In this exercise, the work done by friction is negative because it opposes the book's motion. When you push the book, friction resists by doing negative work, which decreases the book's mechanical energy as it moves.
This example illustrates how energy is transferred from the book's kinetic energy to thermal energy due to friction. The amount of work done depends on the force of friction and the distance, both independent of the path taken. Whether you push the book in a straight line or take a two-part path, the work done by friction remains constant, because the total displacement is the same in both cases.
Forces and Motion
Forces are what cause changes in the motion of objects, and one of the most common is friction. It acts opposite to the direction of movement, slowing down or completely stopping the motion of objects. Understanding how to balance these forces and predict the resulting motion is a fundamental part of physics.
In this particular exercise, friction is the force of interest. It arises because the surfaces in contact—the table and the book—are not perfectly smooth, creating resistance to motion. Even though it's sometimes regarded as a 'limiting' force, friction is crucial in daily life as it allows us to walk without slipping and cars to stop safely.
By calculating the work done by this force, we understand how much motion is resisted by the friction at play. This understanding helps not only in solving physics problems but also in applications like designing safer transportation systems or creating better sports equipment.
Physics Problem Solving
Solving physics problems often involves breaking them into simpler parts and understanding the underlying principles. For this exercise involving friction, the solution requires knowing how friction works and assessing the work it performs.
Start by identifying what information is provided. Here, the work done by friction when pushing the book straight is given and needs to be assessed for a different path. Understanding that friction's resistance is independent of path helps simplify the problem, avoiding unnecessary calculations.
Next, apply known formulas and principles. In this case, because the displacement and frictional force stay the same, the work remains unchanged. This kind of problem-solving is about logical thinking and understanding when certain factors remain constant, aiding in efficient and accurate solutions in physics.

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

CALC An object has several forces acting on it. One of these forces is \(\overrightarrow{\boldsymbol{F}}=\alpha x y \hat{\imath},\) a force in the \(x\) -direction whose magnitude depends on the position of the object, with \(\alpha=2.50 \mathrm{~N} / \mathrm{m}^{2}\). Calculate the work done on the object by this force for the following displacements of the object: (a) The object starts at the point \((x=0, y=3.00 \mathrm{~m})\) and moves parallel to the \(x\) -axis to the point \((x=2.00 \mathrm{~m}, y=3.00 \mathrm{~m})\). (b) The object starts at the point \((x=2.00 \mathrm{~m}, y=0)\) and moves in the \(y\) -direction to the point \((x=2.00 \mathrm{~m}, y=3.00 \mathrm{~m}) .\) (c) The object starts at the origin and moves on the line \(y=1.5 x\) to the point \((x=2.00 \mathrm{~m}, y=3.00 \mathrm{~m})\).

Varying Coefficient of Friction. A box is sliding with a speed of \(4.50 \mathrm{~m} / \mathrm{s}\) on a horizontal surface when, at point \(P,\) it encounters a rough section. The coefficient of friction there is not constant; it starts at 0.100 at \(P\) and increases linearly with distance past \(P\), reaching a value of 0.600 at \(12.5 \mathrm{~m}\) past point \(P .\) (a) Use the work-energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid if the friction coefficient didn't increase but instead had the constant value of \(0.100 ?\)

Stopping Distance. A car is traveling on a level road with speed \(v_{0}\) at the instant when the brakes lock, so that the tires slide rather than roll. (a) Use the work-energy theorem to calculate the minimum stopping distance of the car in terms of \(v_{0}, g,\) and the coefficient of kinetic friction \(\mu_{\mathrm{k}}\) between the tires and the road. ( b) By what factor would the minimum stopping distance change if (i) the coefficient of kinetic friction were doubled, or (ii) the initial speed were doubled, or (iii) both the coefficient of kinetic friction and the initial speed were doubled?

A \(4.00 \mathrm{~kg}\) block of ice is placed against one end of a horizontal spring that is fixed at the other end, has force constant \(k=200 \mathrm{~N} / \mathrm{m}\) and is compressed \(0.025 \mathrm{~m}\). The spring is released and accelerates the block along a horizontal surface. Ignore friction and the mass of the spring. (a) Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length. (b) What is the speed of the block after it leaves the spring?

While hovering, a typical flying insect applies an average force equal to twice its weight during each downward stroke. Take the mass of the insect to be \(10 \mathrm{~g}\), and assume the wings move an average downward distance of \(1.0 \mathrm{~cm}\) during each stroke. Assuming 100 downward strokes per second, estimate the average power output of the insect.

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.