/*! 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 11 \(\cdot\) \(\cdot\) A boxed 10.0... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 11

\(\cdot\) \(\cdot\) A boxed 10.0 kg computer monitor is dragged by friction 5.50 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?

Short Answer

Expert verified
Friction does negative work equal to gravity's work; normal force does zero work.

Step by step solution

01

Understand Work and Its Formula

The work done on an object is given by the formula: \( W = F \cdot d \cdot \cos(\theta) \), where \(F\) is the force applied, \(d\) is the distance moved by the object in the direction of the force, and \(\theta\) is the angle between the force and the direction of movement.
02

Calculate Work Done by Gravity

The gravitational force can be found using \(F_g = m \cdot g\), where \(m = 10.0\, \text{kg}\) and \(g = 9.81\, \text{m/s}^2\). The component of the gravitational force along the incline is \(F_{g\parallel} = F_g \cdot \sin(36.9^\circ)\). The work done by gravity is \(W_g = F_{g\parallel} \cdot d\). Substitute the values to compute \(W_g\).
03

Compute Work Done by Friction

Assuming the monitor moves at a constant speed, the net work done by all forces is zero (due to no change in kinetic energy). As a result, the work done by friction must be equal to and opposite in sign to the work done by both gravity and the conveyor. Calculate the work done by friction, knowing that it should balance the positive work done by gravity along the incline.
04

Determine Work Done by Normal Force

The normal force acts perpendicular to the direction of movement on the incline. Since work is only done by a force component in the direction of motion, the work done by the normal force is zero \(W_n = 0\).
05

Summarize the Calculations

Recall that work done by the friction and by the normal force are both related to the scenario of constant speed and incline. Summarize the work done by each force using the equations and solutions from previous steps.

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
When pulling an object like a computer monitor up an inclined conveyor belt, friction plays a crucial role. Friction is the resistive force that acts opposite to the direction of motion, preventing slipping. In this case, since the monitor moves at a constant speed, friction should perfectly balance the other forces acting on it. Here are a few important features of friction:
  • Direction: Friction always opposes motion. If the motion is upwards along the incline, friction acts downwards.
  • Source: It originates from the contact between the monitor and the surface.
  • Energy Loss: Friction converts some of the mechanical energy into heat.
The work done by friction equals the work done by other forces (like gravity) but with the opposite sign. This ensures constant speed with no net gain in kinetic energy.
Gravitational Force
Gravitational force is a fundamental concept that attracts objects with mass towards each other. For a monitor on an incline, gravity acts downwards. The force is defined by the equation: \( F_g = m \cdot g \), where \( m \) is mass (here 10 kg) and \( g \) is the acceleration due to gravity (approximately 9.81 m/s虏). The work done by gravity along an inclined plane requires calculating the component of this force parallel to the incline, which we find using the sine of the angle:
  • Constant and Predictable: Gravity remains constant and acts downwards.
  • Works Parallel to Incline: By considering the angle of the incline (36.9掳 in this exercise), we find the effective force performing work parallel to the slope.
This parallel component determines the contribution of gravity's work along the slope.
Normal Force
The normal force is the one exerted perpendicular to a surface when an object is in contact with it. In the scenario of the monitor on the conveyor belt, the normal force acts perpendicular to the plane of the conveyor.
  • Perpendicular Action: Normal forces act at a right angle to the surface.
  • No Work Done: Because work is only accomplished by the force component in the direction of motion, the normal force performs no work here. The work done is zero, since there is no displacement along this force鈥檚 direction.
  • Supports the Object: It balances the object against gravitational force pushing it into the surface.
The normal force is crucial to maintaining the object's position on the conveyor while not contributing to the work done.
Inclined Plane
An inclined plane simplifies the understanding of various force components acting on an object. When pulling a monitor up such a plane:
  • Slope Influence: The slope angle affects how much of gravitational force contributes to pulling the object down the incline.
  • Force Breakdown: Forces like gravity can be divided into components parallel and perpendicular to the plane.
  • Reduced Force Needs: Inclines reduce the force necessary to lift an object compared to a vertical lift, making it easier to manage with less energy.
This setup on a conveyor belt helps visualize how different forces might work together and balance in practical scenarios, such as transporting goods or machinery moving uphill.
Constant Speed
Achieving constant speed when moving up an inclined plane means that all forces contributing to the motion are balanced. Constant speed indicates a state of equilibrium, where no net force is acting to accelerate or decelerate the object.
  • Zero Net Work: The sum of work done by all forces equals zero, as there is no net change in kinetic energy.
  • Friction Balance: Friction perfectly counters the forces propelling the monitor upwards, such as the engine force of the conveyor belt, and part of gravity parallel to the plane.
  • Practical Applications: Maintaining constant speed is essential in industrial applications for safe and efficient operation.
Understanding constant speed on an incline provides insight into how forces must be equally balanced to maintain a steady state of motion.

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

\(\bullet\) \(\bullet\) Pebbles of weight \(w\) are launched from the edge of a verti- cal cliff of height \(h\) at speed \(v_{0} .\) How fast (in terms of the quan- tities just given) will these pebbles be moving when they reach the ground if they are launched (a) straight up, (b) straight down, (c) horizontally away from the cliff, and (d) at an angle \(\theta\) above the horizontal? (e) How would the answers to the pre- vious parts change if the pebbles weighed twice as much?

\(\bullet\) \(\bullet\) You and three friends stand at the corners of a square whose sides are 8.0 \(\mathrm{m}\) long in the mid- dle of the gym floor, as shown in the accompanying figure. You take your physics book and You take your physics book and push it from one person to the other. The book has a mass of \(1.5 \mathrm{kg},\) and the coefficient of kinetic friction between the book and the floor is \(\mu_{k}=0.25\) . (a) The book slides from you to Beth and then from Beth to Carlos, along the lines connect- ing these people. What is the work done by friction during this displacement? (b) You slide the book from you to Carlos along the diagonal of the square. What is the work done by friction during this displacement? (c) You slide the book to Kim who then slides it back to you. What is the total work done by fric- tion during this motion of the book? (d) Is the friction force on the book conservative or nonconservative? Explain.

\(\bullet\) \(\bullet\) A pump is required to lift 750 liters of water per minute from a well 14.0 \(\mathrm{m}\) deep and eject it with a speed of 18.0 \(\mathrm{m} / \mathrm{s}\) . How much work per minute does the pump do?

\(\bullet\) The total height of Yosemite Falls is 2425 ft. (a) How many more joules of gravitational potential energy are there for each kilogram of water at the top of this waterfall com- pared with each kilogram of water at the foot of the falls? (b) Find the kinetic energy and speed of each kilogram of water as it reaches the base of the waterfall, assuming that there are no losses due to friction with the air or rocks and that the mass of water had negligible vertical speed at the top. How fast (in \(\mathrm{m} / \mathrm{s}\) and mph) would a 70 \(\mathrm{kg}\) person have to run to have that much kinetic energy? (c) How high would Yosemite Falls have to be so that each kilogram of water at the base had twice the kinetic energy you found in part (b); twice the speed you found in part (b)?

\(\bullet\) \(\bullet\) The spring of a spring gun has force constant \(k=\) 400 \(\mathrm{N} / \mathrm{m}\) and negligible mass. The spring is compressed 6.00 \(\mathrm{cm},\) and a ball with mass 0.0300 \(\mathrm{kg}\) is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 \(\mathrm{cm}\) long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal. (a) Calculate the speed with which the ball leaves the barrel if you can ignore friction. (b) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 \(\mathrm{N}\) acts on the ball as it moves along the barrel. (c) For the situation in part (b), at what posi- tion along the barrel does the ball have the greatest speed, and what is that speed? (In this case, the maximum speed does not occur at the end of the barrel.)
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Geometrical and Physical Optics

Read Explanation

Torque and Rotational Motion

Read Explanation

Atoms and Radioactivity

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.