/*! 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 1 You push your physics book 1.50 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 1

You push your physics book 1.50 \(\mathrm{m}\) along a horizontal table top with a horizontal push of 2.40 \(\mathrm{N}\) while the opposing force of friction is 0.600 \(\mathrm{N} .\) How much work does each of the following forces do on the book: (a) your \(2.40-\mathrm{N}\) push, (b) the friction force, (c) the normal force from the tabletop, and (d) gravity? (e) What is the net work done on the book?

Short Answer

Expert verified
(a) 3.60 J; (b) -0.90 J; (c) 0 J; (d) 0 J; (e) 2.70 J total net work.

Step by step solution

01

Understand the Work Formula

The work done by a force is calculated using the formula \( W = F \times d \times \cos(\theta) \), where \( F \) is the magnitude of the force, \( d \) is the distance over which the force is applied, and \( \theta \) is the angle between the force and the direction of movement.
02

Calculate Work Done by Push Force

For the force of your push, \( F = 2.40 \, \mathrm{N} \), \( d = 1.50 \, \mathrm{m} \), and \( \theta = 0^\circ \) since the push is horizontal and in the direction of movement. The work done is \( W = 2.40 \times 1.50 \times \cos(0) = 3.60 \, \mathrm{J} \).
03

Calculate Work Done by Friction

The friction force opposes movement, so \( F = 0.600 \, \mathrm{N} \), \( d = 1.50 \, \mathrm{m} \), and \( \theta = 180^\circ \). The work done is \( W = 0.600 \times 1.50 \times \cos(180) = -0.90 \, \mathrm{J} \).
04

Calculate Work Done by Normal Force

The normal force acts perpendicular to the displacement, so \( \theta = 90^\circ \), making the work done \( W = F \times d \times \cos(90) = 0 \) because \( \cos(90) = 0 \).
05

Calculate Work Done by Gravity

Similarly, the gravitational force also acts perpendicular to the displacement. Thus, the angle \( \theta = 90^\circ \) and \( W = F \times d \times \cos(90) = 0 \).
06

Determine the Net Work Done

The net work done is the sum of the work done by all forces: Work by push + Work by friction + Work by normal force + Work by gravity = \( 3.60 + (-0.90) + 0 + 0 = 2.70 \, \mathrm{J} \).

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 91影视!

Key Concepts

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

friction in physics
Friction is a crucial force in physics that resists the relative motion between two surfaces in contact. It can be your ally or your adversary, depending on the situation. It's why we can walk without slipping, yet it also slows down objects.

To better understand friction, remember these key points:
  • **Types of Friction:** There are mainly two types: static friction, which prevents motion, and kinetic friction, which opposes ongoing motion.
  • **Direction:** Friction always acts in the opposite direction to the applied force or movement. In our exercise, friction opposes your push on the book.
  • **Magnitude:** The strength of friction depends on two main factors: the nature of the surfaces in contact and the normal force pressing them together.
In our scenario, the frictional force is 0.600 N, acting opposite to your push, and does negative work. This is typical because friction removes energy from the system, slowed down the motion of the book, and causes wear and tear.
net work done
The concept of net work done is about understanding the total energy change in an object as forces act upon it. Consider it as a balance of work contributed by each individual force.

In the given problem, we use the formula for work: \[ W = F \times d \times \cos(\theta) \] where \( F \) is force, \( d \) is distance, and \( \theta \) is the angle between force and direction of motion.
  • **Work by Applied Force:** Here, the push does positive work, calculated as 3.60 J. This adds energy to the system, moving the book forward.
  • **Negative Work of Friction:** Friction does -0.90 J of work, counteracting the push by taking away energy.
  • **Zero Work by Perpendicular Forces:** The work by normal force and gravity is zero because they act perpendicular to the motion, and \( \cos(90^{\circ}) = 0 \).
Adding these contributions gives the net work done: 2.70 J. This shows the book had a net gain of energy enabling further motion after accounting for friction.
force and motion
Understanding the relationship between force and motion is key to mastering physics. An unbalanced force will cause an object to accelerate, which is central to Newton's laws of motion.

In our scenario, several forces interact to influence the motion of the book:
  • **Applied Force:** The push you provide acts in the direction of the book's motion, causing it to move horizontally.
  • **Frictional Force:** This opposes the applied force and aims to stop the book, representing resistance in the system.
  • **Normal Force and Gravity:** These forces act vertically, balancing each other and not affecting horizontal motion. They ensure the book doesn't float away or fall through the table.
The lesson here is that motion results from unbalanced forces. The book moves because the push is stronger than the frictional force, leading to a positive net work done. This is a vivid demonstration of how forces and motion interplay in practical 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

BIO Chin-Ups, While doing a chin-up, a man lifts his body 0.40 \(\mathrm{m} .\) (a) How much work must the man do per kilogram of body mass? (b) The muscles involved in doing a chin-up can generate about 70 \(\mathrm{J}\) of work per kilogram of muscle mass. If the man can just barely do a \(0.40-\mathrm{m}\) chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical \(70-\mathrm{kg}\) man with 14\(\%\) body fat is about 43\(\%\) . (c) Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 \(\mathrm{J}\) of work per kilogram of muscle mass. (d) Adults and children have about the same percentage of muscle in their bodies. Explain why children can commonly do chin-ups more easily than their fathers.

A boxed 10.0 -kg computer monitor is dragged by friction 5.50 \(\mathrm{m}\) up along the moving surface of a conveyor belt inclined at an angle of \(36.9^{\circ}\) above the horizontal. If the monitor's speed is a constant 2.10 \(\mathrm{cm} / \mathrm{s}\) , how much work is done on the monitor by (a) friction, (b) gravity, and (c) the normal force of the conveyor belt?

Automotive Power II. (a) If 8.00 hp are required to drive a \(1800-\) -kg automobile at 60.0 \(\mathrm{km} / \mathrm{h}\) on a level road, what is the total retarding force due to friction, air resistance, and so on? (b) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) up a 10.0\(\%\) grade (a hill rising 10.0 \(\mathrm{m}\) vertically in 100.0 \(\mathrm{m}\) horizon- tally)? (c) What power is necessary to drive the car at 60.0 \(\mathrm{km} / \mathrm{h}\) down a 1.00\(\%\) grade? (d) Down what percent grade would the car coast at 60.0 \(\mathrm{km} / \mathrm{h}\) ?

CALC 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. On the rough section, the coefficient of friction is not constant, but 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\(?\)

Six diesel units in series can provide 13.4 \(\mathrm{MW}\) of power to the lead car of a freight train. The diesel units have total mass \(1.10 \times 10^{6} \mathrm{kg}\) . The average car in the train has mass \(8.2 \times 10^{4} \mathrm{kg}\) and requires a horizontal pull of 2.8 \(\mathrm{kN}\) to move at a constant 27 \(\mathrm{m} / \mathrm{s}\) on level tracks. (a) How many cars can be in the train under these conditions? (b) This would leave no power for accelerating or climbing hills. Show that the extra force needed to accelerate the train is about the same for a \(0.10-\mathrm{m} / \mathrm{s}^{2}\) acceleration or a 1.0\(\%\) slope (slope angle \(\alpha=\arctan 0.010 )\) . (c) With the 1.0\(\%\) slope, show that an extra 2.9 \(\mathrm{MW}\) of power is needed to maintain the \(27-\mathrm{m} / \mathrm{s}\) speed of the diesel units. (d) With 2.9 \(\mathrm{MW}\) less power available, how many cars can the six diesel units pull up a 1.0\(\%\) slope at a constant 27 \(\mathrm{m} / \mathrm{s} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Particle Model of Matter

Read Explanation

Kinematics Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Force

Read Explanation

Electricity and Magnetism

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.