/*! 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 4 \(\cdot\) A constant horizontal ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 4

\(\cdot\) A constant horizontal pull of 8.50 \(\mathrm{N}\) drags a box along a hor- izontal floor through a distance of 17.4 \(\mathrm{m} .\) (a) How much work does the pull do on the box? (b) Suppose that the same pull is exerted at an angle above the horizontal. If this pull now does 65.0 \(\mathrm{J}\) of work on the box while pulling it through the same dis- tance, what angle does the force make with the horizontal?

Short Answer

Expert verified
(a) 147.9 J, (b) 64.0 degrees.

Step by step solution

01

Understand the Work Done by a Force

Work done by a force can be calculated using the formula \[ W = F \times d \times \cos(\theta) \]where \( W \) is the work done, \( F \) is the magnitude of the force, \( d \) is the distance over which the force is exerted, and \( \theta \) is the angle between the force and the direction of the movement. When the force is parallel to the direction of movement, \( \theta = 0 \), and \( \cos(0) = 1 \).
02

Calculate the Work Done by the Horizontal Pull

Given that the force is horizontal, we have \( \theta = 0 \). So,\[ W = 8.50 \ \text{N} \times 17.4 \ \text{m} \times \cos(0) = 8.50 \ \text{N} \times 17.4 \ \text{m} \times 1 \]Calculating the work gives:\[ W = 147.9 \ \text{J} \]
03

Set Up the Equation for the Angled Pull

For the work done by the angled pull, the work is given to be 65.0 J. The equation is:\[ 65.0 \ \text{J} = 8.50 \ \text{N} \times 17.4 \ \text{m} \times \cos(\theta) \]
04

Solve for the Angle \(\theta\)

Rearranging the equation to find \( \cos(\theta) \):\[ \cos(\theta) = \frac{65.0 \ \text{J}}{8.50 \ \text{N} \times 17.4 \ \text{m}} \]Calculate this value:\[ \cos(\theta) = \frac{65.0}{147.9} \approx 0.4396 \]Now find \( \theta \) by taking the inverse cosine:\[ \theta = \cos^{-1}(0.4396) \approx 64.0^{\circ} \]
05

Conclusion

For part (a), the work done by the horizontal force is 147.9 J. For part (b), if the work done is 65.0 J, the angle of the force with respect to the horizontal is approximately 64.0 degrees.

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.

Force
Force is a fundamental concept in physics, representing a push or pull that can cause an object to move, stop, or change direction. It is measured in newtons (N) and is a vector quantity, which means it has both a magnitude and a direction. One of the easiest ways to understand force is by observing how it acts on everyday objects. For example, pushing a shopping cart applies a force; pulling a door open involves another force. These forces can act at different angles, affecting how they influence the motion of objects.
  • Magnitude: Indicates how strong or weak a force is.
  • Direction: Shows where the force is applied.
Understanding force helps predict how objects will react when in contact with each other, making it crucial for solving problems involving motion.
Angle of Force
The angle of force is a key factor when analyzing situations where force is applied at any inclination other than directly along the path of motion. This is important because the angle affects how effective the force is in doing its job. In mathematical terms, when force is at an angle to the direction of movement, not all of the force contributes to the motion. Only the component of the force in the direction of motion does work.
  • Calculating Components: Use trigonometric functions to break down the force into horizontal and vertical components.
  • Cosine Function: Specifically for the horizontal component, the cosine angle is used in work calculations.
By understanding how angles affect force, you can evaluate how different angles of application will impact the movement and work done by the force.
Horizontal Motion
Horizontal motion refers to the movement of objects along a straight line on a horizontal plane. In most scenarios, horizontal motion simplifies to forces acting parallel or at an angle to the ground. The importance of understanding horizontal motion lies in how it relates to the application of forces and energy transfer. In physics problems, knowing whether a force acts horizontally helps in determining the work done and the effect on the moving object's speed.
  • Direction of Motion: Straight path along the horizontal axis.
  • Parallel Forces: Forces that align with the direction of motion require simple calculations.
With horizontal motion, the path of movement is clear, making it easier to apply formulas and predict outcomes.
Work Done by a Force
Work done by a force involves transferring energy to an object as it moves it along a path. The formula for calculating work is \(W = F \times d \times \cos(\theta)\), meaning the force multiplied by the distance over which it acts, and the cosine of the angle of the force in relation to the direction of motion.This formula highlights some important features:
  • Direction Dependency: Only the component of force that is in the same direction as the movement does work, thus requiring the cosine to account for angles.
  • Units of Work: Measured in joules (J), where one joule equals the work done by a one Newton force moving an object one meter.
  • Physical Meaning: Work is essentially about how much energy has been transferred to move an object a certain distance.
By understanding work done by a force, students can link the practical effects of force on motion to theoretical calculations, offering a complete view of movement and energy dynamics.

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

\(\bullet\) \(\bullet\) A good workout. You overindulged on a delicious dessert, so you plan to work off the extra calories at the gym. To accomplish this, you decide to do a series of arm raises hold- ing a 5.0 kg weight in one hand. The distance from your elbow to the weight is \(35 \mathrm{cm},\) and in each arm raise you start with your arm horizontal and pivot it until it is vertical. Assume that the weight of your arm is small enough compared with the weight you are lifting that you can ignore it. As is typical, your muscles are 20\(\%\) efficient in converting the food energy they use up into mechanical energy, with the rest going into heat. If your dessert contained 350 food calories, how many arm raises must you do to work off these calories? Is it realistic to do them all in one session?

\(\bullet\) Volcanoes on Io. Io, a satellite of Jupiter, is the most volcanically active moon or planet in the solar system. It has volcanoes that send plumes of matter over 500 \(\mathrm{km}\) high (see the accompanying fig- ure). Due to the satellite's small mass, the acceleration due to gravity on Io is only \(1.81 \mathrm{m} / \mathrm{s}^{2},\) and \(\mathrm{Io}\) has no appreciable atmosphere. As- sume that there is no varia- tion in gravity over the distance traveled. (a) What must be the speed of material just as it leaves the volcano to reach an altitude of 500 \(\mathrm{km} ?\) (b) If the gravitational potential energy is zero at the surface, what is the potential energy for a 25 kg fragment at its maximum height on Io? How much would this gravitational potential energy be if it were at the same height above earth?

\(\cdot\) \(\cdot\) A racing dog is initially running at \(10.0 \mathrm{m} / \mathrm{s},\) but is slow- ing down. (a) How fast is the dog moving when its kinetic energy has been reduced by half? (b) By what fraction has its kinetic energy been reduced when its speed has been reduced by half?

\(\bullet\) Tendons. Tendons are strong elastic fibers that attach mus- cles to bones. To a reasonable approximation, they obey Hooke's law. In laboratory tests on a particular tendon, it was found that, when a 250 g object was hung from it, the tendon stretched 1.23 \(\mathrm{cm}\) . (a) Find the force constant of this tendon in \(\mathrm{N} / \mathrm{m}\) . (b) Because of its thickness, the maximum tension this tendon can support without rupturing is 138 \(\mathrm{N}\) . By how much can the tendon stretch without rupturing, and how much energy is stored in it at that point?

\(\bullet\) Rescue. Your friend (mass 65.0 \(\mathrm{kg}\) ) is standing on the ice in the middle of a frozen pond. There is very little friction between her feet and the ice, so she is unable to walk. Fortu- nately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for 3.00 s and accelerate your friend from rest to a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) while you remain at rest. What is the average power supplied by the force you applied?
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Turning Points in Physics

Read Explanation

Physics of Motion

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Fields in Physics

Read Explanation

Waves Physics

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.