/*! 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 18 \(\cdot\) \(\cdot\) \(\cdot\) A ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 18

\(\cdot\) \(\cdot\) \(\cdot\) A 1.50 \(\mathrm{kg}\) book is sliding along a rough horizontal sur- face. 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) -6.56 J; (b) 0.75 m/s; (c) 1.60 m/s.

Step by step solution

01

Determine the Work Done (Part a)

To find the work done on the book between points A and B, we use the work-energy principle: \[ W_{AB} = rac{1}{2}m(v_B^2 - v_A^2) \]where \( m = 1.50 \; \text{kg} \), \( v_A = 3.21 \; \text{m/s} \), and \( v_B = 1.25 \; \text{m/s} \).Substitute these values in: \[W_{AB} = \frac{1}{2} \times 1.50 \times (1.25^2 - 3.21^2) \]Solving, we find:\[W_{AB} = \frac{1}{2} \times 1.50 \times (1.5625 - 10.3041) \]\[W_{AB} = \frac{1}{2} \times 1.50 \times (-8.7416) \]\[W_{AB} = -6.5562 \; \text{J} \]
02

Velocity at Point C with -0.750 J Work Done (Part b)

Using the work-energy principle again, the change in kinetic energy from B to C is equal to the work done: \[ W_{BC} = \frac{1}{2}m(v_C^2 - v_B^2) \]Given \( W_{BC} = -0.750 \; \text{J} \) and \( v_B = 1.25 \; \text{m/s} \),\[-0.750 = \frac{1}{2} \times 1.50 \times (v_C^2 - 1.25^2) \]\[-0.750 = 0.75 \times (v_C^2 - 1.5625) \]\[-1 = v_C^2 - 1.5625 \]\[v_C^2 = 0.5625 \]\[v_C = \sqrt{0.5625} = 0.75 \; \text{m/s} \]
03

Velocity at Point C with +0.750 J Work Done (Part c)

If instead, +0.750 J of work is done, set up the equation:\[ W_{BC}' = \frac{1}{2}m(v_C'^2 - v_B^2) \]where \( W_{BC}' = +0.750 \; \text{J} \),\[0.750 = \frac{1}{2} \times 1.50 \times (v_C'^2 - 1.25^2) \]\[0.750 = 0.75 \times (v_C'^2 - 1.5625) \]\[1 = v_C'^2 - 1.5625 \]\[v_C'^2 = 2.5625 \]\[v_C' = \sqrt{2.5625} = 1.60 \; \text{m/s} \]

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 91影视!

Key Concepts

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

Kinetic Energy Change
When objects move, they have what's called kinetic energy, which is the energy of motion. The formula for kinetic energy is \[ KE = \frac{1}{2}mv^2 \]where:
  • \( KE \) is the kinetic energy
  • \( m \) is the mass of the object
  • \( v \) is its velocity
Kinetic energy changes when the object's speed changes. When a book slides on a surface and slows down from a speed of 3.21 m/s to 1.25 m/s, its kinetic energy decreases. This change in kinetic energy can be calculated by using its initial and final velocities. This principle is crucial because it helps us understand how energy is transferred or converted in different systems. In this example, the energy is transferred by the force of friction as it does work against the book's motion.
Velocity Calculations
Velocity tells us how fast an object is moving in a specific direction. To find out how fast the book is moving at different points, we can use the work-energy principle. This principle relates the work done on an object to its change in kinetic energy. For instance, when the book goes from point B to point C, and we know the work done (-0.750 J or +0.750 J), we can calculate the new velocity:1. **With -0.750 J of work:** We can set up our equation: - \(-0.750 = \frac{1}{2} \cdot 1.50 \cdot (v_C^2 - 1.25^2)\) - By solving this equation, we find \(v_C = 0.75 \; \text{m/s}\) at point C.2. **With +0.750 J of work:** If instead the work done was +0.750 J, we use the same method and find: - \(v_C' = \sqrt{2.5625} = 1.60 \; \text{m/s}\)These calculations show how work affects the speed of an object by changing its kinetic energy, helping us gain insights into how different forces influence motion.
Work Done on Objects
Work is one of the main ideas that helps us understand how forces act on objects. When a force is applied over a distance, it does work. The formula for work can be written as:\[ W = F \cdot d \cdot \cos(\theta) \]where:
  • \( W \) is the work done
  • \( F \) is the force applied
  • \( d \) is the distance over which the force is applied
  • \( \theta \) is the angle between the force and the direction of movement
In our exercise, work is done to the book by the force of friction, which is aligned against the movement. We calculated work when the book slowed from 3.21 m/s to 1.25 m/s by using the change in kinetic energy. Negative work indicates that the force is opposing the movement, leading to a decrease in speed. Understanding how work is calculated and how it affects motion is key to solving many problems in physics and helps us design systems like brakes, engines, and many other mechanical tools.

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 loaded 375 kg toboggan is traveling on smooth hori- zontal snow at 4.5 \(\mathrm{m} / \mathrm{s}\) when it suddenly comes to a rough region. The region is 7.0 \(\mathrm{m}\) long and reduces the tobogan's speed by 1.5 \(\mathrm{m} / \mathrm{s}\) . (a) What average friction force did the rough region exert on the toboggan? (b) By what percent did the rough region reduce the toboggan's (i) kinetic energy and (ii) speed?

\(\bullet\) A force of magnitude 800.0 \(\mathrm{N}\) stretches a certain spring by 0.200 \(\mathrm{m}\) from its equilibrium position. (a) What is the force constant of this spring? (b) How much elastic potential energy is stored in the spring when it is: (i) stretched 0.300 \(\mathrm{m}\) from its equilibrium position and (ii) compressed by 0.300 \(\mathrm{m}\) from its equilibrium position? (c) How much work was done in stretch- ing the spring by the original 0.200 \(\mathrm{m} ?\)

\(\bullet\) \(\bullet\) 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?

\(\bullet\) \(\bullet\) \(\bullet\) A ball is thrown upward with an initial velocity of 15 \(\mathrm{m} / \mathrm{s}\) at an angle of \(60.0^{\circ}\) above the horizontal. Use energy conservation to find the ball's greatest height above the ground.

\(\bullet\) A 72.0 -kg swimmer jumps into the old swimming hole from a diving board 3.25 m above the water. Use energy conserva- tion to find his speed just he hits the water (a) if he just holds his nose and drops in, (b) if he bravely jumps straight up (but just beyond the board!) at \(2.50 \mathrm{m} / \mathrm{s},\) and \((\mathrm{c})\) if he manages to jump downward at 2.50 \(\mathrm{m} / \mathrm{s}\) .
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Work Energy and Power

Read Explanation

Radiation

Read Explanation

Physics of Motion

Read Explanation

Electricity and Magnetism

Read Explanation

Thermodynamics

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.