/*! 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 14 A \(1.50 \mathrm{~kg}\) book is ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(1.50 \mathrm{~kg}\) book is sliding along a rough horizontal surface. At point \(A\) it is moving at \(3.21 \mathrm{~m} / \mathrm{s},\) and at point \(B\) it has slowed to \(1.25 \mathrm{~m} / \mathrm{s}\). (a) How much work was done on the book between \(A\) and \(B ?\) (b) If \(-0.750 \mathrm{~J}\) of work is done on the book from \(B\) to \(C\) how fast is it moving at point \(C ?\) (c) How fast would it be moving at \(C\) if \(+0.750 \mathrm{~J}\) of work were done on it from \(B\) to \(C ?\)

Short Answer

Expert verified
(a) Work done from A to B: -6.567 J (b) Speed at C with -0.750 J work: 0.750 m/s (c) Speed at C with +0.750 J work: 1.597 m/s

Step by step solution

01

Calculate Kinetic Energy at Point A

The kinetic energy at point A can be calculated using the formula: \( KE_A = \frac{1}{2}mv^2 \). Given that the mass \( m = 1.50 \text{ kg} \) and the velocity \( v = 3.21 \text{ m/s} \), we find \( KE_A = \frac{1}{2} \times 1.50 \times (3.21)^2 = 7.739025 \text{ J} \).
02

Calculate Kinetic Energy at Point B

The kinetic energy at point B is calculated similarly: \( KE_B = \frac{1}{2}mv^2 \). Given the velocity \( v = 1.25 \text{ m/s} \), we find \( KE_B = \frac{1}{2} \times 1.50 \times (1.25)^2 = 1.171875 \text{ J} \).
03

Calculate Work Done from A to B

The work done from A to B is the change in kinetic energy: \( W_{AB} = KE_B - KE_A \). Thus, \( W_{AB} = 1.171875 - 7.739025 = -6.56715 \text{ J} \).
04

Calculate Speed at Point C for Work Done \(-0.750 \text{ J}\)

From B to C, \(-0.750 \text{ J}\) of work is done, so the kinetic energy at C is \( KE_C = KE_B + (-0.750) = 1.171875 - 0.750 = 0.421875 \text{ J}\). Use the kinetic energy formula to find the speed at C: \( 0.421875 = \frac{1}{2} \times 1.50 \times v_C^2 \). Solving for \( v_C \), we find \( v_C = \sqrt{\frac{0.421875}{0.75}} = 0.7496 \text{ m/s} \).
05

Calculate Speed at Point C for Work Done \(+0.750 \text{ J}\)

For \(+0.750 \text{ J}\) of work done from B to C, the kinetic energy at C is \( KE_C = KE_B + 0.750 = 1.171875 + 0.750 = 1.921875 \text{ J}\). Solving for \( v_C \), we get \( v_C = \sqrt{\frac{1.921875}{0.75}} = 1.597 \text{ m/s} \).

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.

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It plays a crucial role in understanding how objects like our book move along surfaces. The general formula for kinetic energy is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) stands for mass and \( v \) represents velocity. When dealing with objects in motion, like the book sliding on a surface, you must calculate its kinetic energy at different points to understand how much energy it has. For instance, at point A, using the equation and the known values of mass (1.50 kg) and velocity (3.21 m/s), the kinetic energy is calculated to be approximately 7.74 Joules.This calculation shows how the speed of an object directly affects its kinetic energy. A higher velocity results in higher kinetic energy, showcasing the direct relationship between motion and energy.
Work-Energy Principle
The work-energy principle is a fundamental concept in physics that connects the work done on an object to its change in kinetic energy. Whenever work is done on an object, it results in a change in the object's energy. In the exercise, as the book moves from point A to point B, work is done due to the forces acting on it, like friction. The work-energy principle is expressed mathematically as:\( \text{Work Done} = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \).For the book, the work done from A to B is calculated by the change in kinetic energy: \( \Delta KE = 1.17 \text{ J} - 7.74 \text{ J} = -6.57 \text{ J} \).This negative value means that energy is being taken away from the book, slowing it down due to resistance forces. The principle helps in understanding how forces affect motion and can be applied in various physical situations to predict the movement of objects.
Conservation of Energy
The conservation of energy is a principle stating that energy cannot be created or destroyed, only transformed from one form to another. This concept helps explain the behavior of a system when forces are involved.In our problem, energy is transformed from one type like mechanical (kinetic) to other forms due to work done by forces such as friction. Between points B and C, different work values illustrate this transformation:- With \(-0.750 \text{ J}\), the energy is further decreased, leading to a lower speed at C.- With \(+0.750 \text{ J}\), energy increases, meaning more energy is available for motion, resulting in a higher speed at C.Energy conservation enables predictions about how an object's speed will change when different amounts of work are applied. Understanding energy conservation helps in exploring how systems behave under various forces and conditions.
Frictional Force Impact
Frictional forces are forces that resist the motion of objects sliding against one another. They play a significant part in how energy is transformed in physical systems. For our sliding book, friction between the book and the surface did work to slow the book down, essentially converting some motion energy into other forms like heat. This is evident in the work done from A to B, resulting in a decrease of kinetic energy by 6.57 J. Frictional forces often act against the direction of movement, causing energy loss in the form of work done to overcome this force. This influences the amount of kinetic energy available and impacts how quickly an object moves. Understanding friction is crucial, as it is a common force that affects nearly all motion. Recognizing its impact in energy calculations helps reveal why objects slow and describes the energy transformations happening in everyday movements.

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

A wooden rod of negligible mass and length \(80.0 \mathrm{~cm}\) is pivoted about a horizontal axis through its center. A white rat with mass \(0.500 \mathrm{~kg}\) clings to one end of the stick, and a mouse with mass \(0.200 \mathrm{~kg}\) clings to the other end. The system is released from rest with the rod horizontal. If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position?

A spring with spring constant \(k\) is anchored to the wall on one side of a hockey rink. A hockey puck is pressed against the spring and then released to slide across the ice. In the process the hockey puck gains a kinetic energy \(K .\) Derive an expression for the initial compression of the spring \(x\) in terms of \(k\) and \(K\).

A \(61 \mathrm{~kg}\) skier on level snow coasts \(184 \mathrm{~m}\) to a stop from a speed of \(12.0 \mathrm{~m} / \mathrm{s} .\) (a) Use the work-energy theorem to find the coefficient of kinetic friction between the skis and the snow. (b) Suppose a \(75 \mathrm{~kg}\) skier with twice the starting speed coasted the same distance before stopping. Find the coefficient of kinetic friction between that skier's skis and the snow.

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?

During the calibration process, the cantilever is observed to deflect by \(0.10 \mathrm{nm}\) when a force of \(3.0 \mathrm{pN}\) is applied to it. What deflection of the cantilever corresponds to a force of \(6.0 \mathrm{pN} ?\) A. \(0.07 \mathrm{nm}\) C. \(0.20 \mathrm{nm}\) B. \(0.14 \mathrm{nm}\) D. \(0.40 \mathrm{nm}\)

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.