Physics Chapter 6 Part 1 PDF

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Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. CHAPTER 6 Work and Energy du e r.e ov nd On the popular Behemoth roller coaster at Canada’s Wonderland @a amusement park, riders cresting the top of the hill will pick up speed on the way down, as they convert gravitational potential energy into kinetic energy. This conversion is governed by the principle of conservation of energy, a 28 central topic in this chapter. ra LEARNING OBJECTIVES ab After reading this module, you should be able to… hh 6.1 Determine the work done by a constant force. 6.2 Relate work and kinetic energy. pc 6.3 Use gravitational potential energy. 6.4 Identify conservative and nonconservative forces. 6.5 Use the law of conservation of mechanical energy. 6.6 Solve problems involving nonconservative forces. 6.7 Solve problems involving power. Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. 6.8 Employ the law of conservation of energy. 6.9 Determine the work done by a variable force. 6.1 Work Done by a Constant Force Work is a familiar concept. For example, it takes work to push a stalled car. In fact, du more work is done when the pushing force is greater or when the displacement of the car is greater. Force and displacement are, in fact, the two essential elements of work, r.e as Figure 6.1 illustrates. The drawing shows a constant pushing force that points in the same direction as the resulting displacement.* In such a case, the work W is defined as the magnitude F of the force times the magnitude s of the displacement: W e = Fs. The work done to push a car is the same whether the car is moved north to south ov or east to west, provided that the amount of force used and the distance moved are the same. Since work does not convey directional information, it is a scalar quantity. nd @a 28 ra FIGURE 6.1 Work is done when a force pushes a car through a displacement. ab The equation W = Fs indicates that the unit of work is the unit of force times the unit of hh distance, or the newton · meter in SI units. One newton · meter is referred to as a joule (J) (rhymes with “cool”), in honor of James Joule (1818–1889) and his research into the nature of work, energy, and heat. Table 6.1 summarizes the units for work in several pc systems of measurement. TABLE 6.1 Units of Measurement for Work System Force × Distance = Work SI newton (N) meter (m) joule (J) CGS dyne (dyn) centimeter (cm) erg BE pound (lb) foot (ft) foot · pound (ft · 1b) Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. The definition of work as W = Fs does have one surprising feature: If the distance s is zero, the work is zero, even if a force is applied. Pushing on an immovable object, such as a brick wall, may tire your muscles, but there is no work done of the type we are discussing. In physics, the idea of work is intimately tied up with the idea of motion. If the object does not move, the force acting on the object does no work. Often, the force and displacement do not point in the same direction. For instance, Figure 6.2a shows a suitcase-on-wheels being pulled to the right by a force that is du applied along the handle. The force is directed at an angle θ relative to the displacement. In such a case, only the component of the force along the displacement is used in defining work. As Figure 6.2b shows, this component is F cos θ, and it r.e appears in the general definition below: e DEFINITION OF WORK DONE BY A CONSTANT* FORCE ov The work done on an object by a constant force is (6.1) nd where F is the magnitude of the force, s is the magnitude of the displacement, @a and θ is the angle between the force and the displacement. SI Unit of Work: newton · meter = joule (J) 28 When the force points in the same direction as the displacement, then θ = 0°, and ra Equation 6.1 reduces to W = Fs. Example 1 shows how Equation 6.1 is used to calculate work. ab EXAMPLE 1 hh Pulling a Suitcase-on-Wheels pc Find the work done by a 45.0-N force in pulling the suitcase in Figure 6.2a at an angle θ = 50.0° for a distance s = 75.0 m. Reasoning The pulling force causes the suitcase to move a distance of 75.0 m and does work. However, the force makes an angle of 50.0° with the displacement, and we must take this angle into account by using the definition of work given by Equation 6.1. Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. Solution The work done by the 45.0-N force is The answer is expressed in newton · meters or joules (J). du e r.e ov FIGURE 6.2 (a) Work can be done by a force that points at an angle θ nd relative to the displacement. (b) The force component that points along the displacement is F cos θ. @a The definition of work in Equation 6.1 takes into account only the component of the force in the direction of the displacement. The force component perpendicular to the 28 displacement does no work. To do work, there must be a force and a displacement, and since there is no displacement in the perpendicular direction, there is no work ra done by the perpendicular component of the force. If the entire force is perpendicular to the displacement, the angle θ in Equation 6.1 is 90°, and the force does no work at ab all. Work can be either positive or negative, depending on whether a component of the hh force points in the same direction as the displacement or in the opposite direction. Example 2 illustrates how positive and negative work arise. pc EXAMPLE 2 The Physics of Weight Lifting The weight lifter in Interactive Figure 6.3a is bench-pressing a barbell whose weight is 710 N. In part b of the figure, he raises the barbell a distance of 0.65 m above his chest, and in part c he lowers it the same distance. The weight is raised and lowered at a Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. constant velocity. Determine the work done on the barbell by the weight lifter during (a) the lifting phase and (b) the lowering phase. Reasoning To calculate the work, it is necessary to know the force exerted by the weight lifter. The barbell is raised and lowered at a constant velocity and, therefore, is in equilibrium. Consequently, the force exerted by the weight lifter must balance the weight of the du barbell, so F = 710 N. During the lifting phase, the force and displacement are in the same direction, as Interactive Figure 6.3b shows. The angle between them is θ = r.e 0°. When the barbell is lowered, however, the force and displacement are in opposite directions, as in Interactive Figure 6.3c. The angle between the force and the displacement is now θ = 180°. With these observations, we can find the work. e ov Solution (a) During the lifting phase, the work done by the forcend is given by Equation 6.1 as @a (b) The work done during the lowering phase is 28 mtd mtd since cos 180° = −1. The work is negative, because the force is opposite to the ra displacement. Weight lifters call each complete up-and-down movement of the barbell a repetition, or “rep.” The lifting of the weight is referred to as the positive part of the ab rep, and the lowering is known as the negative part. hh pc INTERACTIVE FIGURE 6.3 (a) In the bench press, work is done during both the lifting and lowering phases. (b) During the lifting phase, the force does positive work. (c) During the lowering phase, the force does negative work. Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. Example 3 deals with the work done by a static frictional force when it acts on a crate that is resting on the bed of an accelerating truck. EXAMPLE 3 Accelerating a Crate du Figure 6.4a shows a 120-kg crate on the flatbed of a truck that is moving with an acceleration of a = +1.5 m/s2 along the positive x axis. The crate does not slip with respect to the truck as the truck undergoes a displacement whose magnitude is s = 65 r.e m. What is the total work done on the crate by all of the forces acting on it? Reasoning e ov The free-body diagram in Figure 6.4b shows the forces that act on the crate: (1) the weight of the crate, (2) the normal force exerted by the flatbed, and (3) the nd static frictional force , which is exerted by the flatbed in the forward direction and keeps the crate from slipping backward. The weight and the normal force are @a perpendicular to the displacement, so they do no work. Only the static frictional force does work, since it acts in the x direction. To determine the frictional force, we note that the crate does not slip and, therefore, must have the same acceleration of a = +1.5 28 m/s2 as does the truck. The force creating this acceleration is the static frictional force, and, knowing the mass of the crate and its acceleration, we can use Newton’s second ra law to obtain its magnitude. Then, knowing the frictional force and the displacement, we can determine the total work done on the crate. ab Solution From Newton’s second law, we find that the magnitude fs of the static frictional force is hh pc The total work is that done by the static frictional force and is (6.1) The work is positive, because the frictional force is in the same direction as the displacement (θ = 0°). Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. du e r.e ov nd FIGURE 6.4 (a) The truck and crate are accelerating to the right for a distance @a of s = 65 m. (b) The free-body diagram for the crate. 28 Check Your Understanding ra (The answers are given at the end of the book.) ab 1. Two forces and are acting on the box shown in CYU Figure 6.1, causing the box to move across the floor. The two force vectors are drawn to scale. Which one of hh the following statements is correct? (a) does more work than does. (b) does more work than does. (c) Both forces do the same amount of work. (d) Neither pc force does any work. Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. du CYU FIGURE 6.1 r.e 2. A box is being moved with a velocity by a force (in the same direction as ) e along a level horizontal floor. The normal force is , the kinetic frictional force is , ov and the weight is. Which one of the following statements is correct? (a) does positive work, and do zero work, andnd does negative work. (b) does positive work, and do zero work, and does negative work. (c) does positive @a work, and do zero work, and does negative work. (d) does positive work, and do zero work, and does negative work. 28 3. A force does positive work on a particle that has a displacement pointing in the +x direction. This same force does negative work on a particle that has a displacement ra pointing in the +y direction. In which quadrant of the x, y coordinate system does the force lie? (a) First (b) Second (c) Third (d) Fourth ab 4. A suitcase is hanging straight down from your hand as you ride an escalator. Your hand exerts a force on the suitcase, and this force does work. This work is (a) positive hh when you ride up and negative when you ride down, (b) negative when you ride up and positive when you ride down, (c) positive when you ride up or down, (d) negative when you ride up or down. pc 6.2 The Work–Energy Theorem and Kinetic Energy Most people expect that if you do work, you get something as a result. In physics, when a net force performs work on an object, there is always a result from the effort. The result is a change in the kinetic energy of the object. As we will now see, the relationship that relates work to the change in kinetic energy is known as the work– Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. energy theorem. This theorem is obtained by bringing together three basic concepts that we’ve already learned about. First, we’ll apply Newton’s second law of motion, ΣF = ma, which relates the net force ΣF to the acceleration a of an object. Then, we’ll determine the work done by the net force when the object moves through a certain distance. Finally, we’ll use Equation 2.9, one of the equations of kinematics, to relate the distance and acceleration to the initial and final speeds of the object. The result of this approach will be the work–energy theorem. du e r.e ov nd INTERACTIVE FIGURE 6.5 A constant net external force Σ acts over a @a displacement and does work on the plane. As a result of the work done, the plane’s kinetic energy changes. 28 To gain some insight into the idea of kinetic energy and the work–energy theorem, look at Interactive Figure 6.5, where a constant net external force Σ acts on an airplane ra of mass m. This net force is the vector sum of all the external forces acting on the plane, and, for simplicity, it is assumed to have the same direction as the displacement ab. According to Newton’s second law, the net force produces an acceleration a, given by a = ΣF/m. Consequently, the speed of the plane changes from an initial value of υ0 to hh a final value of υf.* Multiplying both sides of ΣF = ma by the distance s gives pc The left side of this equation is the work done by the net external force. The term as on the right side can be related to υ0 and υf by using (Equation 2.9) from the equations of kinematics. Solving this equation gives Substituting this result into (ΣF)s = mas shows that Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. This expression is the work–energy theorem. Its left side is the work W done by the net external force, whereas its right side involves the difference between two terms, each of which has the form (mass)(speed)2. The quantity (mass)(speed)2 is called kinetic energy (KE) and plays a significant role in physics, as we will see in this chapter and later on in other chapters as well. du DEFINITION OF KINETIC ENERGY r.e The kinetic energy KE of an object with mass m and speed υ is given by e (6.2) ov SI Unit of Kinetic Energy: joule (J) nd The SI unit of kinetic energy is the same as the unit for work, the joule. Kinetic energy, like work, is a scalar quantity. These are not surprising observations, because work and @a kinetic energy are closely related, as is clear from the following statement of the work– energy theorem. 28 THE WORK–ENERGY THEOREM ra When a net external force does work W on an object, the kinetic energy of the object changes from its initial value of KE0 to a final value of KEf, the difference ab between the two values being equal to the work: (6.3) hh The work–energy theorem may be derived for any direction of the force relative to the pc displacement, not just the situation in Interactive Figure 6.5. In fact, the force may even vary from point to point along a path that is curved rather than straight, and the theorem remains valid. According to the work–energy theorem, a moving object has kinetic energy, because work was done to accelerate the object from rest to a speed υf.† Conversely, an object with kinetic energy can perform work, if it is allowed to push or pull on another object. Example 4 illustrates the work–energy theorem and considers a single force that does work to change the kinetic energy of a space probe. Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. Analyzing Multiple-Concept Problems EXAMPLE 4 The Physics of an Ion Propulsion Drive The space probe Deep Space 1 was launched October 24, 1998, and it used a type of du engine called an ion propulsion drive. An ion propulsion drive generates only a weak force (or thrust), but can do so for long periods of time using only small amounts of fuel. Suppose the probe, which has a mass of 474 kg, is traveling at an initial speed of r.e 275 m/s. No forces act on it except the 5.60 × 10−2-N thrust of its engine. This external force is directed parallel to the displacement , which has a magnitude of e (see Figure 6.6). Determine the final speed of the probe, assuming that ov its mass remains nearly constant. nd @a 28 ra ab FIGURE 6.6 An ion propulsion drive generates a single force that points in hh the same direction as the displacement. The force performs positive work, causing the space probe to gain kinetic energy. pc Reasoning If we can determine the final kinetic energy of the space probe, we can determine its final speed, since kinetic energy is related to mass and speed according to Equation 6.2 and the mass of the probe is known. We will use the work–energy theorem (W = KEf − KE0), along with the definition of work, to find the final kinetic energy. Knowns and Unknowns Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. The following list summarizes the data for this problem: Description Symbol Value Comment Explicit Data Mass m 474 kg Initial speed υ0 275 m/s 5.60 × Magnitude of force F du 10−2 N 2.42 × Magnitude of displacement s 109 m r.e Implicit Data e Angle between force and The force is parallel to the θ 0° displacement. ov displacement Unknown Variable Final speed nd υf ? Modeling the Problem @a STEP 1 Kinetic Energy An object of mass m and speed υ has a kinetic 28 energy KE given by Equation 6.2 as. Using the subscript f to denote the final kinetic energy and the final speed of the probe, we have that ra (1) ab Solving for υf gives Equation 1 at the right. The mass m is known but the final kinetic energy KEf is not, so we will turn to Step 2 to evaluate hh it. pc (1) STEP 2 The Work–Energy Theorem The work–energy theorem (2) relates the final kinetic energy KEf of the probe to its initial kinetic energy KE0 and the work W done by the force of the engine. According to Equation 6.3, this theorem is W = KEf − KE0. Solving for KEf shows that Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. The initial kinetic energy can be expressed as , so the expression for the final kinetic energy becomes This result can be substituted into Equation 1, as indicated at the right. Note from the data table that we know the mass m and the initial speed υ0. The work W is not known and will be evaluated in du Step 3. r.e STEP 3 Work The work W is that done by the net external force acting on the space probe. Since there is only the one force acting on the e probe, it is the net force. The work done by this force is given by ov Equation 6.1 as (1) nd (2) where F is the magnitude of the force, θ is the angle between the @a force and the displacement, and s is the magnitude of the displacement. All the variables on the right side of this equation are known, so we can substitute it into Equation 2, as shown in the right column. 28 ra Solution Algebraically combining the results of the three steps, we have ab hh pc The final speed of the space probe is mtd mtd Related Homework: Problem 22 Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. In Example 4 only the force of the engine does work. If several external forces act on an object, they must be added together vectorially to give the net force. The work done by the net force can then be related to the change in the object’s kinetic energy by using the work–energy theorem, as in the next example. Analyzing Multiple-Concept Problems du EXAMPLE 5 r.e Downhill Skiing A 58-kg skier is coasting down a 25° slope, as Figure 6.7a shows. Near the top of the e slope, her speed is 3.6 m/s. She accelerates down the slope because of the ov gravitational force, even though a kinetic frictional force of magnitude 71 N opposes her motion. Ignoring air resistance, determine the speed at a point that is displaced 57 m downhill. nd @a 28 ra ab hh FIGURE 6.7 (a) A skier coasting downhill. (b) The free-body diagram for the pc skier. Reasoning The skier’s speed at a point 57 m downhill (her final speed) depends on her final kinetic energy. According to the work–energy theorem, her final kinetic energy is related to her initial kinetic energy (which we can calculate directly) and the work done by the net external force that acts on her. The work can be evaluated directly from its definition. Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. Knowns and Unknowns The data for this problem are listed below: Description Symbol Value Comment Explicit Data Mass m 58 kg 3.6 υ0 du Initial speed m/s Magnitude of kinetic fk 71 N r.e frictional force Magnitude of skier’s s 57 m displacement e Angle of slope above ov 25° horizontal Implicit Data Angle between net force nd The skier is accelerating down the slope, so the direction of the net acting on skier and her θ 0° force is parallel to her @a displacement displacement. Unknown Variable 28 Final speed υf ? Modeling the Problem ra ab STEP 1 Kinetic Energy The final speed υf of the skier is related to her final kinetic energy KEf and mass m by Equation 6.2: (1) hh Solving for υf gives Equation 1 at the right. Her mass is known, but her pc final kinetic energy is not, so we turn to Step 2 to evaluate it. (1) STEP 2 The Work–Energy Theorem The final kinetic energy KEf of the (2) skier is related to her initial kinetic energy KE0 and the work W done by the net external force acting on her. This is according to the work– energy theorem (Equation 6.3): KEf = KE0 + W. The initial kinetic energy Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. can be written as , so the expression for the final kinetic energy becomes This result can be substituted into Equation 1, as indicated at the right. The mass m and the initial speed υ0 are known. The work W is not known, and will be evaluated in Steps 3 and 4. du STEP 3 Work The work W done by the net external force acting on the r.e skier is given by Equation 6.1 s e (1) ov where ΣF is the magnitude of the net force, θ is the angle between the (2) net force and the displacement, and s is the magnitude of the (3) nd displacement. This result can be substituted into Equation 2, as shown at the right. The variables θ and s are known (see the table of knowns and unknowns), and the net external force will be determined in Step @a 4. 28 STEP 4 The Net External Force Figure 6.7b is a free-body diagram for the skier and shows the three external forces acting on her: the ra gravitational force , the kinetic frictional force and the normal force. The net external force along the y axis is zero, because ab there is no acceleration in that direction (the normal force balances the component mg cos 25° of the weight perpendicular to the slope). (1) hh Using the data from the table of knowns and unknowns, we find that (2) the net external force along the x axis is (3) pc mtd mtd All the variables on the right side of this equation are known, so we substitute this expression into Equation 3 for the net external force (see the right column). Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. Math Skills In Figure 6.7b the angle between the gravitational force and the −y axis equals the 25° angle of the slope. Figure 6.8 uses β for this angle and helps to explain why β = 25°. It shows the vertical directional line along which the gravitational force acts, as well as the slope and the x, y axes. Since this directional line is perpendicular to the ground, the triangle ABC is a right triangle. It follows, then, that α + 25° = 90° or α = 65°. Since the y axis is perpendicular to the slope, it is also true that α + β = 90°. Solving this du equation for the angle β and substituting α = 65° gives e r.e ov nd @a 28 ra ab FIGURE 6.8 Math Skills drawing. hh Solution Algebraically combining the four steps, we arrive at the following expression for the pc final speed of the skier: Referring to the table of knowns and unknowns for the values of the symbols in this result, we find that the final speed of the skier is Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. mtd mtd Related Homework: Problem 22 du Problem-Solving Insight Example 5 emphasizes that the work–energy theorem deals with the work done by r.e the net external force. The work–energy theorem does not apply to the work done by an individual force, unless that force happens to be the only one present, in which case it is the net force. e ov If the work done by the net force is positive, as in Example 5, the kinetic energy of the object increases. If the work done is negative, the kinetic energy decreases. If the work is nd zero, the kinetic energy remains the same. Conceptual Example 6 explores these ideas further in a situation where the component of the net force along the direction of the @a displacement changes during the motion. CONCEPTUAL EXAMPLE 6 28 Work and Kinetic Energy ra Figure 6.9 illustrates a satellite moving about the earth in a circular orbit and in an ab elliptical orbit. The only external force that acts on the satellite is the gravitational force. In which orbit does the kinetic energy of the satellite change, (a) the circular orbit or (b) the elliptical orbit? hh pc Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. du r.e FIGURE 6.9 (a) In a circular orbit, the gravitational force is always e perpendicular to the displacement of the satellite and does no work. (b) In ov an elliptical orbit, there can be a component of the force along the displacement, and, consequently, work can be done. nd Reasoning @a The gravitational force is the only force acting on the satellite, so it is the net force. With this fact in mind, we will apply the work–energy theorem, which states that the work done by the net force equals the change in the kinetic energy. 28 Answer (a) is incorrect. ra For the circular orbit in Figure 6.9a, the gravitational force does no work on the ab satellite since the force is perpendicular to the instantaneous displacement at all times. Thus, the work done by the net force is zero, and according to the work–energy theorem (Equation 6.3), the kinetic energy of the satellite (and, hence, its speed) hh remains the same everywhere on the orbit. Answer (b) is correct. pc For the elliptical orbit in Figure 6.9b, the gravitational force does do work. For example, as the satellite moves toward the earth in the top part of Figure 6.9b, there is a component of that points in the same direction as the displacement. Consequently, does positive work during this part of the orbit, and the kinetic energy of the satellite increases. When the satellite moves away from the earth, as in the lower part of Figure 6.9b, has a component that points opposite to the Printed by: [email protected]. Printing is for personal use only. No part of this book may be reproduced or transmitted without publisher's prior permission. Violators will be prosecuted. displacement and, therefore, does negative work. As a result, the kinetic energy of the satellite decreases. Related Homework: Problem 18 du e r.e ov nd @a 28 ra ab hh pc