2007 • 885 Pages • 24.24 MB • English

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PEDAGOGICAL USE OF COLOR Displacement and Torque (t) and position vectors angular momentum (L) vectors Velocity vectors (v) ជ Velocity component vectors Linear or rotational motion directions Force vectors (Fជ) Force component vectors Springs Acceleration vectors (ជa) Acceleration component vectors Electric ﬁelds Capacitors Magnetic ﬁelds Inductors (coils) Positive charges + Voltmeters V Negative charges – Ammeters A Resistors Lightbulbs Batteries and other AC sources DC power supplies – + Switches Ground symbol Light rays Objects Lenses and prisms Images Mirrors

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CONVERSION FACTORS Length Speed 1 m = 39.37 in. = 3.281 ft 1 km/h = 0.278 m/s = 0.621 mi/h 1 in. = 2.54 cm 1 m/s = 2.237 mi/h = 3.281 ft/s 1 km = 0.621 mi 1 mi/h = 1.61 km/h = 0.447 m/s = 1.47 ft/s 1 mi = 5 280 ft = 1.609 km 15 Force 1 light year (ly) = 9.461 ⫻ 10 m 1 angstrom (Å) = 10⫺10 m 1 N = 0.224 8 lb = 105 dynes 1 lb = 4.448 N Mass ⫺5 ⫺6 1 dyne = 10 N = 2.248 ⫻ 10 lb 3 ⫺2 1 kg = 10 g = 6.85 ⫻ 10 slug 1 slug = 14.59 kg Work and energy ⫺27 2 1 u = 1.66 ⫻ 10 kg = 931.5 MeV/c 7 1 J = 10 erg = 0.738 ft ⭈ lb = 0.239 cal 1 cal = 4.186 J Time 1 ft ⭈ lb = 1.356 J 1 min = 60 s 1 Btu = 1.054 ⫻ 103 J = 252 cal 1 h = 3 600 s 1 J = 6.24 ⫻ 1018 eV 1 day = 8.64 ⫻ 104 s 1 eV = 1.602 ⫻ 10⫺19 J 1 yr = 365.242 days = 3.156 ⫻ 107 s 1 kWh = 3.60 ⫻ 106 J Volume Pressure 1 L = 1 000 cm3 = 3.531 ⫻ 10⫺2 ft3 1 atm = 1.013 ⫻ 105 N/m2 (or Pa) = 14.70 lb/in.2 1 ft3 = 2.832 ⫻ 10⫺2 m3 1 Pa = 1 N/m2 = 1.45 ⫻ 10⫺4 lb/in.2 1 gal = 3.786 L = 231 in.3 1 lb/in.2 = 6.895 ⫻ 103 N/m2 Angle Power 180⬚ = rad 1 hp = 550 ft ⭈ lb/s = 0.746 kW 1 rad = 5.730⬚ 1 W = 1 J/s = 0.738 ft ⭈ lb/s ⫺2 1⬚ = 60 min = 1.745 ⫻ 10 rad 1 Btu/h = 0.293 W

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Essentials of College Physics Raymond A. Serway Emeritus, James Madison University Chris Vuille Embry-Riddle Aeronautical University Australia · Brazil · Canada · Mexico · Singapore · Spain · United Kingdom · United States

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Physics Acquisitions Editor: Chris Hall Print/Media Buyer: Karen Hunt Publisher: David Harris Permissions Editor: Bob Kauser Vice President, Editor-in-Chief, Sciences: Michelle Julet Production Service: Joan Keyes, Dovetail Publishing Services, Inc. Development Editor: Ed Dodd Text Designer: Patrick Devine Editorial Assistant: Jessica Jacobs Photo Researcher: Jane Sanders Miller Technology Project Manager: Sam Subity Copy Editor: Kathleen Lafferty Marketing Manager: Mark Santee Illustrator: Rollin Graphics, Progressive Information Technologies Marketing Assistant: Michele Colella Cover Designer: Patrick Devine Marketing Communications Manager: Bryan Vann Cover Image: Adrian Weinbrecht/Photolibrary Project Manager, Editorial Production: Teri Hyde Cover Printer: Quebecor World/Dubuque Creative Director: Rob Hugel Compositor: G&S Typesetters, Inc. Art Director: Lee Friedman Printer: Quebecor World/Dubuque ® ® © 2007 by Raymond A. Serway. ExamView and ExamView Pro are registered trademarks of FSCreations, Inc. Windows is a registered trademark of the Microsoft ALL RIGHTS RESERVED. No part of this work covered by the copy- Corporation used herein under license. Macintosh and Power right hereon may be reproduced or used in any form or by any Macintosh are registered trademarks of Apple Computer, Inc. Used means — graphic, electronic, or mechanical, including photocopying, herein under license. recording, taping, web distribution, information storage and retrieval systems, or in any other manner — without the written permission © 2007 Thomson Learning, Inc. All Rights Reserved. Thomson of the publisher. Learning WebTutor™ is a trademark of Thomson Learning, Inc. Printed in the United States of America Thomson Higher Education 1 2 3 4 5 6 7 10 0 9 0 8 0 7 0 6 10 Davis Drive Belmont, CA 94002-3098 USA For more information about our products, contact us at: Thomson Learning Academic Resource Center Library of Congress Control Number: 2005932024 1-800-423-0563 ISBN 0-495-10619-4 For permission to use material from this text or product, submit a request online at http://www.thomsonrights.com. Any additional questions about permissions can be submitted by e-mail to [email protected]

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The Foundation for Success Building the right course for you and your students is easy when you start with Serway and Vuille’s Essentials of College Physics! Because every course is as unique as its instructor—and its students—you need an approach tai- lored to your distinct needs. No matter how you decide to execute your course, Essentials of College Physics, provides the proven foundation for success. This accessible and focused text includes a broad range of engaging and contemporary applications that motivate student understanding of how physics works in the real world. And with its extraordinary range of powerful teaching and learning resources, it’s easy for you to craft a course that ﬁts your exact requirement with: 䊏 Focused homework management system using pedagogy and content from WebAssign the book, including hints and feedback for students 䊏 Access to the PhysicsNow™ student tutorial system—interactive, integrated learning technology that puts concepts in motion 䊏 Premium book-speciﬁc content for audience response systems that lets you interact with your students directly from your own PowerPoint® lectures 䊏 A multimedia presentation tool that lets you incorporate colorful images and clarifying animations into every lecture We know that providing your students with a solid foundation in the basics is the key to student success . . . and that means providing them with proven, time-tested content. As you peruse the following pages of this PREVIEW, be sure to note the adjacent diagram indicating the different com- ponents of our integrated, interrelated program. No matter what kind of course you want to deliver— whether you offer a more traditional text-based course, you’re interested in using or are currently using an online homework management system, or you are ready to turn your lecture into an interactive learning environment through an audi- ence response system—you can be conﬁdent that proven con- tent provides the foundation for each and every component. Whatever your goals are for you and your students, Essentials of College Physics features the content and the courseware to get you there—without the risk. iii Preview

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The strength of Essentials of College Physics starts with the foundation Essentials of College Physics provides students with a clear and logical presentation of the basic concepts and principles of physics. With the text as the foundation, coupled with extraordinary media integration, it’s easy to build a course that gives every student the maximum opportunity for success. Briefer than the average college physics text, Essentials of College Physics compre- 154 Chapter 7 Rotational Motion and the Law of Gravity hensively covers all the standard topics in Exercise 7.3 classical and modern physics. Instructors (a) What are the angular speed and angular displacement of the disc 0.300 s after it begins to rotate? (b) Find the tangential speed at the rim at this time. will notice a clean and clear dialogue with Answers (a) 10.6 rad/s; 1.58 rad (b) 0.472 m/s the student, the book’s uncluttered look and feel, and attention paid to language. The authors’ clear, logical, relaxed, Vectors are denoted in bold- v 7Fig.4ure 7C.5aE sNhoTwRs aI PcaEr TmAovLin gA iCn Ca cEirLcuElaRr ApaTthI OwiNth constant linear speed v. Even satnydle e fnagcialigtaintegs face with arrows over them. Or tshtaonudg hth tihs,e c ocanrs idmeorv tehse adt eaﬁ ncoinngs teaqnut astpioene dfo, ri ta svteirlla ghea sa cacne laecrcaetiloenra:tion. To under- hqeunicskio cno mpre- Ttoh irse cmoagkneizse t.hem easier (a) The numerator represents the dif:afearve⫽nce:vt fb⫺ett:wvi ieen the velocity vectors :vf an[7d. 1:v2i]. 훽 vi 훾 vf Titfh hethe cseeay r v’hse acvvteeol rodsc iimftfyea raye shn iat vmdei rotehvcteiso s niansm, teh eme iacrig rdnciuftfuleadrree ,p ncacoterh rc eiasn pc’ot nedtqiinunuagla tlzoley r tcohh.e a Tsnahgmein edg is,r peaecse tsidho,no bw uontf in Figure 7.5b. For circular motion at constant speed, the acceleration vector r r always points toward the center of the circle. Such an acceleration is called a centripetal (center-seeking) acceleration. Its magnitude is given by O (b) ac ⫽ vr2 [7.13] F l oa( caba ifcr gr)i a pt urAsa nrmvste edhto lhe o7fvrerc.igo n5icotmgaye r wsv 훽( meaiatc )hoct oev Ccer nio훾 srtc ncrah,ius lpatolhaneanetrgga temd l st sihp,or esetoic eoc tdtinhro. cenouf- a(vevssliuoT⫽comi tvdeyf e ⫽t:rvhivaveat) t.E : vTtqiomua acenta idlotc inu:va lfn7a.dtd1e 3i tft,hfh eceron an ocsancitdel yepl eroir niFan tigitdo ui훾nrre,e cw w7tei.ot6 hbnae ;. vgAetihnlnoe wcoiriibt tyjhme :cvEatfgq inaus itaﬁt tuarido stlnea as t7 et a.rp1r o2eti,i mnthte e훽 tfs.wa mWithe acceleration. :aav ⫽ :v tf ⫺ :tvi i ⫽ ⌬⌬:vt [7.14] awlhsoe rvee r⌬y:v sm⫽a:lvlf. ⫺In: vFi iigsu trhee 7c.h6abn, g:vef ins avlemloocsitt yp. aWrahlleenl t⌬ot :ivsi ,v aenryd stmhael lv,e ⌬ctsoarn ⌬d:v ⌬ius arpe- proximately perpendicular to them, pointing toward the center of the circle. In 훽 v ⌬i s 훾 vf tcahececne tleliemrr aiottfiin otghn e c:a ac.s ieFr crwloehm,e a nEn q⌬du ttabhteieoc noavm 7e.er1as4 gv,e a:a naicascnhedinl e⌬grl:avyt ispomonia n:lalt, a iv⌬nb: vtehpceo msinamets e et hxdaeirc eitnlcyts itoaonw ta(airnnde tothhuies r ⌬u r limTith),e sotr itahneg ilnes tiann tFaingeuorue s 7a.c6cae, lewrhaitciohn hpaosi nsitds etos t⌬hse acnendt er,r oisf tshime iclairc lteo. the one formed by the vectors in Figure 7.6b, so the ratios of their sides are equal: O (a) ⌬vv ⫽ ⌬rs vf or ⌬v ⌬u –vi ⌬v ⫽ vr ⌬s [7.15] Figure 7.6 ((ab)) As the particle Substituting the result of Equation 7.15 into aav ⫽ ⌬v/⌬t gives im: vtfso. vv(eeblso) fcTriothyme v ec훽octnotsort r c훾uhc,at tniohgnee sdf ofireo dcmetit o:evnri- toof aav ⫽ vr ⌬ts [7.16] m veilnoicnitgy t⌬h:ve ,d wirheicthio ins tofw tahred cthaenge in But ⌬s is the distance traveled along the arc of the circle in time ⌬t, and in the lim- center of the circle. iting case when ⌬t becomes very small, ⌬s/⌬t approaches the instantaneous value 䉳 Important statements and deﬁnitions are set in boldface type or are highlighted with a background screen for added emphasis and ease of review. Important equations are highlighted with a tan back- ground. The International System of units (SI) is used throughout the book. The U.S. customary system of units is used only to a limited extent in the problem sets of the early chapters on mechanics. iv Foundation

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Essentials of College Physics provides a wealth of outstanding examples and problem sets to help students develop critical prob- lem-solving skills and conceptual understanding. 䉴 All worked Examples include six parts: Goal, Problem, Strategy, Solution, Remarks, 8.4 Examples of Objects in Equilibrium 181 and Exercise/Answer. The “Solution” portion of EXAMPLE 8.4 Locating Your Lab Partner’s Center of Gravity every Example is presented in two-columns to Goal Use torque to ﬁnd a center of gravity. enhance student learning and to help reinforce L Problem In this example, we show how to ﬁnd the loca- L/2 physics concepts. In addition, the authors have tion of a person’s center of gravity. Suppose your lab part- ner has a height L of 173 cm (5 ft, 8 in) and a weight w of n F taken special care to present a graduated level 715 N (160 lb). You can determine the position of his O of difﬁculty within the Examples so students are cbeonatredr osuf pgproarvtiteyd bya t haovnine g ehnidm bsytr eatc hsc oaluet, oans as huonwinfo rimn xcg better prepared to work the end-of-chapter F reigaudrineg8 F.9i.s I3f. 5t0he⫻ b1o0a2rdN’,s ﬁwnedig thte wdbis itsa n4c9e No f aynodu rt lhaeb spcaarlte- w wb Problems. ner’s center of gravity from the left end of the board. cFeignuterre o 8f .g9ravi(tEy.xample 8.4) Determining your lab partner’s Strategy To ﬁnd the position xcg of the center of gravity, compute the torques using an axis through O. Set the sum of the torques equal to zero and solve for xcg. ⌷ Solution Apply the second condition of equilibrium. There is no ⌺ ⫽ 0 t aormrq uise z deuroe. to the normal force :n because its moment ⫺wxcg ⫺ wb(L/2) ⫹ FL ⫽ 0 Solve for xcg and substitute known values: x cg ⫽ FL ⫺ wwb(L/2) ⫽ (350 N)(173 cm7)1⫺5 N(49 N)(86.5 cm) ⫽ 79 cm Remarks The given information is sufﬁcient only to determine the x -coordinate of the center of gravity. The other two coordinates can be estimated, based on the body’s symmetry. Exercise 8.4 Suppose a 416-kg alligator of length 3.5 m is stretched out on a board of the same length weighing 65 N. If the board is supported on the ends as in Figure 8.9, and the scale reads 1 880 N, ﬁnd the x -component of the alligator’s center of gravity. Answer 1.59 m 8.4 EXAMPLES OF OBJECTS IN EQUILIBRIUM Recall from Chapter 4 that when an object is treated as a geometric point, equilib- rium requires only that the net force on the object is zero. In this chapter, we have shown that for extended objects a second condition for equilibrium must also be satisﬁed: the net torque on the object must be zero. The following general proce- dure is recommended for solving problems that involve objects in equilibrium. Math Focus 6.1 One-Dimensional Elastic Collisions The usual notation and subscripts used in the equa- simplify, obtaining tions of one-dimensional collisions often obscure the underlying simplicity of the mathematics. In an elastic ⫺3 ⫽ X ⫹ 2Y Proble(1m) -Solving Strategy Objects in Equilibrium c aonldlis6i.o1n1,atrheeursaetdh,ecroforrremspidoanbdlein-lgootokicnognEseqruvationso6f.10 27 ⫽ X 2 ⫹ 2Y 2 1. Diagram(2 t)he system. Include coordinates and choose a convenient rotation axis momentum and conservation of energy, respectively. Equation (1) is that of a straight line, whereasfEorq cuoamtiopunting the net torque on the object. t lIoinec sai t aiteryesp oiacflal t lgh pievr eocnob,l leliedmain,v itgnh goe b tmwjeoacs tuse.n sSk uannbodswt itnthuse,t i tnihngei t tﬁihanela vmle vloeor-cei- (a2⫺n)d3ds⫺eusbc2srYtiibt)ue2ts⫹eainn2etYoll2iE,pwqsehu.aicSthioolcnvae(nE2b)q,euosabitmitoapinnli(iﬁn1eg)2d f.2ot7oDiornb⫽rXgjae wocn ta. i(ftM.r eFoeos-rbt ospydrsoyte bdmlieasmg wrsai twmhi lmol fho tarhvee t oah basijnen cgotln eoe fo oibnbjetjecrcte to,s tfd, irsnahtweo rwae isnetp.g)a arlalt exdtieargnralm fo frocre es aaccht- c goemthmero wni-tlho othkien kgn voawrinab qluesa,n Xtit⫽iesv 1yifealdnsd e Yqu⫽atvio2nf,s t foo-r Y 2 ⫹ 2Y ⫺ 3 ⫽ 0 a straight line (the momentum equation) and an el- In general, the quadratic formula must now be ap- l tiposne t(htehne reendeurcgeys e tqou ﬁantidoinn)g. Tthhee imntaetrhsecmtiaotnic aolf saolu- 1plmie/ds, obru t⫺ t3hmis /eqs.u Oatniolyn t hfaec tﬁorrsst, agnivsiwnegr ,Y 1⫽m/vs2, fm⫽akes straight line and an ellipse. sense. Substituting it into Equation (1) yields Example: In a one-dimensional collision, suppose the X ⫽ v1f ⫽ ⫺5m/s ﬁrst object has mass m1 ⫽ 1kg and initial velocity v1i ⫽ 3m/s, whereas the second object has mass It is also possible to use Equation 6.10 together with m2 ⫽ 2kg and initial velocity v2i ⫽ ⫺3m/s. Find the the derived Equation 6.14. This situation, illustrated ﬁnal velocities for the two objects. (For clarity, signiﬁ- in Example 6.5, is equivalent to ﬁnding the intersec- cant ﬁgure conventions are not observed here.) tion of two straight lines. It’s easier to remember the equation for the conservation of energy than the spe- Solution: Substitute the given values and X ⫽ v1f and cial Equation 6.14, so it’s a good idea to be able to Y ⫽ v2f into Equations 6.10 and 6.11, respectively, and solve such problems both ways. 䉱 New! Just-In-Time Math Tutorials! An emphasis on quantitative problem-solving is provided in the Math Focus boxes. These boxes develop mathematical methods important to a particular area of physics, or point out a technique that is often overlooked. Each Math Focus box has been placed within the applicable section of the text, giving students just-in-time support. A complete Appendix provides students with additional math help applied to speciﬁc physics concepts. v Foundation

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Building critical-thinking skills and conceptual understanding Essentials of College Physics includes time-tested as well as new peda- gogy that adheres to the ﬁndings of physics education research to help students improve their conceptual understanding. 䉳 A wealth of interesting and relevant 288 Chapter 11 Energy in Thermal Processes Applications reveals the role physics plays The same process occurs when a radiator raises the temperature of a room. in our lives and in other disciplines. These The hot radiator warms the air in the lower regions of the room. The warm air Applications are woven throughout the text expands and, because of its lower density, rises to the ceiling. The denser cooler air from above sinks, setting up the continuous air current pattern shown in narrative and are indicated with a margin note. Figure 11.9. An automobile engine is maintained at a safe operating temperature by a For biology and pre-med students, icons point combination of conduction and forced convection. Water (actually, a mixture of water and antifreeze) circulates in the interior of the engine. As the metal of the way to various practical and interesting P stheoamto garnadp htu orbf ua ltenatk ceotnlev,e schtiowni naigr tchoeolenrgwinaetebrlobcyk tinhceremaasel scionntdeumcptieorna.tuTreh,eenweartgeyr pausmseps frfomrcetshewhatoetrmoeuttalotof the Applications of physical principles to biology currents. engine and into the radiator, carrying energy along with it (by forced convection). and medicine. With this edition the authors In the radiator, the hot water passes through metal pipes that are in contact with the cooler outside air, and energy passes into the air by conduction. The cooled wa- have increased the number of life science- A P P L I C AT I O N ter is then returned to the engine by the water pump to absorb more energy. The Cooling Automobile Engines process of air being pulled past the radiator by the fan is also forced convection. oriented applications and end-of-chapter The algal blooms often seen in temperate lakes and ponds during the spring or fall are caused by convection currents in the water. To understand this process, Problems to help motivate students to master A P P L I C AT I O N cgorandsiednerts ,F wigiuthr ea n1 1u.1p0p.e Dr, uwrainrmg tlhayee rsu omf mwaetre, rb soedpiaersa toefd w farotemr dae lvoewloepr, cteomldp learyaetru brey the content. Algal Blooms in Ponds a buffer zone called a thermocline. In the spring or fall, temperature changes in and Lakes the water break down this thermocline, setting up convection currents that mix the water. The mixing process transports nutrients from the bottom to the surface. The nutrient-rich water forming at the surface can cause a rapid, tempo- rary increase in the algae population. Figure 11.9 Convection currents Warm Layer 25°–22°C Tip notations address common student 9.3 Density and Pressure 211 are set up in a room warmed by a Thermocline 20°–10°C radiator. Cool layer 5°–4°C misconTche ppretsisuoren ast a aspnecdiﬁc spoiitnut ian tai ﬂounid scanin be wmehasiucrhed wsithu t-he device pic- dents tsuporefintdge i ntnh Fa itgf houarlsel ob9e.w7ebn: paurne nveivopaucrsuloya tcdeadlu icbycrlaitnteidvde ewr ie tnhpc klaonsotinhwgns aw. leiigghhtt ps.i sAtos nth ceo dnenveicete ids stuob a- TEqIPu a9ti.o1n 9.F7o mrcaeke as na dcl eParre dsissutirnec- (a) Summer layering of wateAr ppromtxheirm gsperdai nitnge aul nyﬂtu ili1 dth,0 eth 0ien wﬂTauriidp pforsrecseese ces xdteoirowtend osbny tathhere e tﬂo upfi odo fui sth nbea dlpain sticonend atbnhyd et hcoem opurtwesasreds tAfoionrcnoe tibhse eatr wv iemcetpnoro farotnradcne tp ardenissdstu ipnrerc eitsis osaun sr cieas. ltahr.at marginfosrc,e epxerrotevd ibdy itnheg sp srintgu. dLeet Fnbtes t hwe mitahgn ituhde o fh theel pfo rcteh oen ythe piston and There is no direction associated with need tAhoeth seap ravirneoga iiodsf s tpchreeo atdom po usmut rofoavcener to hmf et heiens tpitriaset okanree. aNs, om tioctei vtahtaint gt hoeu fro frocrem thala td ecoﬁmniptiroens soesf pfporerercspeseu anrsdes,oi cbcuiualtat ert hdtoe w dtihtihree tschutiero fpnacr oefs o stufhree is pressure: interest. and misunderstandings. If F is the magnitude of a force exerted perpendicular to a given surface of 䊴 Pressure area A, then the pressure P is the force divided by the area: P ⬅ AF [9.7] SI unit: pascal (Pa) Because pressure is deﬁned as force per unit area, it has units of pascals (newtons per square meter). The English customary unit for pressure is the pound per inch Figure 11.10 (a) During the summ(ebr,) a F waallr amn du pspreirn lga yueprw oefl lwinagter is separa1stqe.0du 1afrr⫻oemd1 a.0 cA5ootPmlear.o lsopwhereric pressure at sea level is 14.7 lb/in2, which in SI units is layer by a thermocline. (b) Convection currents during the spring or fall mix the Awast ewr ean sde cea nf rcoaumse Equation 9.7, the effect of a given force depends critically on algal blooms. the area to which it’s applied. A 700-N man can stand on a vinyl-covered ﬂoor in regular street shoes without damaging the surface, but if he wears golf shoes, the metal cleats protruding from the soles can do considerable damage to the ﬂoor. With the cleats, the same force is concentrated into a smaller area, greatly elevat- ing the pressure in those areas, resulting in a greater likelihood of exceeding the 䉴 Applying Physics sections allow stu- ultimate strength of the ﬂoor material. dents to review concepts presented in a maSl nfowrcseh ooens uthse tshheo esas mtoe spurpinpcoiprtl et h(eF igp.e 9rs.o8)n.’ sT whe igsnhot.w A ecxceorrtds inang utop wNaerdw tnonor’s- dseecmtioonns. tSraotme eth Ae pcpolyninnegc Ptiohnys bicestweexeanm pthlees toshvioerred st hloaewn v, etthrhyei s l asunrgpoeww a.a rIdef a tf horef c eepa ecisrh s aoscnco oiwsm swhpoeaaenr,i iesnodg t bhsyna ota w thdseoh wopnersew,s astruhdra etf oafrto craecn eye xigsei vrdeteinsdt rp ibobyiun tthe ides Fpsnpiegroreswuoa rdnbe e of c9rvaoe.u8mrs ae s ilStnahnrkegoi newforgs rh aicnroeeet aos, n ptrh retehd veuseo csnfnitno tgwhe is relatively low and the person doesn’t penetrate very deeply into the snow. the pressure on the snow’s surface. concepts presented in that chapter and other scientiﬁc disciplines. Applying Physics 9.1 Bed of Nails Trick After an exciting but exhausting lecture, a physics pro- would be more uncomfortable yet to stand on a bed 䊏 Quick Quiz questions throughout fessor stretches out for a nap on a bed of nails, as in of nails without shoes.) Figure 9.9, suffering no injury and only moderate dis- the book provide students ample comfort. How is this possible? opportunity to assess their conceptual Explanation If you try to support your entire weight understanding. on a single nail, the pressure on your body is your weight divided by the very small area of the end of the nail. The resulting pressure is large enough to pene- 䊏 Checkpoints ask simple questions trate the skin. If you distribute your weight over several hundred nails, however, as demonstrated by based on the text to further reinforce the professor, the pressure is considerably reduced because the area that supports your weight is the total key ideas. area of all nails in contact with your body. (Why is lying on a bed of nails more comfortable than sitting on the same bed? Extend the logic to show that it Figure 9.9 (Applying Physics 9.1) Does anyone have a pillow? vi Foundation GReasrye aSrecthtelers,/ SInccie. nce Source/Photo © Royalty-Free/Corbis

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Several components from the text are enhanced in the PhysicsNow student tutorial program to reinforce material, including the dynamic Active Figures, which are animated diagrams from the text. Labeled with the PhysicsNow icon, these ﬁgures come to life and allow stu- dents to visualize phenomena and processes that can’t be repre- sented on the printed page. INTERACTIVE EXAMPLE 4.7 Atwood’s Machine 䉳 Over 40 of the text’s worked Examples are identi- Goal Use the second law to solve a two-body problem. ﬁed as Interactive Examples and labeled with the Problem Two objects of mass m1 and m2, with m2 ⬎ m1, PhysicsNow icon. As part of the PhysicsNow web- are connected by a light, inextensible cord and hung T over a frictionless pulley, as in Active Figure 4.15a. Both T based tutorial system, students can engage in an inter- cord and pulley have negligible mass. Find the magni- t thude ec orfd t.he acceleration of the system and the tension in m 1 m 2 dacintgiv ew eoxrkteends iEoxna omf ptlheef rpormob tlheme tesoxtlv. eTdh isn othfte nc oinrcrelusdpeosn- iSnt rathteg ynegTathivee h eya-dvierer cmtiaosns., mS2in, accec etlheer atceosr do wcannw’at rdbse, a1 m1 m1g elements of both visualization and calculation, and may otisnitnvr e mb tacyanh gaden fd ioat,ur 2tcdhiese o,n afbe cutgecaten tolisevpieorpa,on tsa i:Tnotendins na ot2dhf⫽ iert ehu⫺cept aiwto1wan.o rE,d samsoc d ahtsih rsmeacsta tsaiso1r iensis aea pnqcotudesa dial- (a) m2 a2 (b) m2g alrseo g iunivdoelvde t hproeudgicht iothne a sntde pinst unietieodne bdu tilod isnogl.v eS tau dperonbtslem Nfuorerewc e4to .1on5f’ sbg rseahcvoiotwnys d inf rl eatewh- ebf odro yw ednaicwahga rmdamsa sds i,fr oetroc tgtiheotenh .et wAr ocw timivteha sFstehigse.- An(EbeC)xc TtaFeImrVdep Eebl- eybF oaI4G d.l7iyUg) d hRAitEat swg t4roai.on1mdg5’s st fhmoarat tcphaeisn soebs. j(oeacv)tes rT. aw ofr ihcatinogninlegs so pbujelcetys. con- toyp dei ffaenrden at rsec tehneanr iaosk. ed to apply what they have learned equation relating the accelerations, constitutes a set of three equations for the three unknowns—a1, a2, and T. Log into to PhysicsNow at http://physics.brookscole.com/ecp and go to Active Figure 4.15 to adjust the masses of objects on Atwood’s ma- chine and observe the resulting motion. Solution Apply the second law to each of the two masses m1a1 ⫽ T ⫺ m1g (1) m2a2 ⫽ T ⫺ m2g (2) individually: Substitute a2 ⫽ ⫺a1 into the second equation, and m2a1 ⫽ ⫺T ⫹ m2g multiply both sides by ⫺1: Add the stacked equations, and solve for a1: (m1 ⫹ m2)a1 ⫽ m2g ⫺ m1g (3) a1 ⫽ 冢 m 12 ⫹⫺ m 12 冣g Substitute this result into Equation (1) to ﬁnd T: T ⫽ 冢 m2 1m⫹1mm22 冣g Remarks The acceleration of the second block is the same as that of the ﬁrst, but negative. When m2 gets very large compared with m1, the acceleration of the system approaches g, as expected, because m2 is falling nearly freely under the inﬂuence of gravity. Indeed, m2 is only slightly restrained by the much lighter m1. The acceleration of the system can also be found by the system approach, as illustrated in Example 4.10. 䉴 Also in the PhysicsNow student tutorial pro- gram are Coached Problems. These engaging problems reinforce the lessons in the text by taking the same step-by-step approach to problem solving as found in the text. Each Coached Problem gives students the option of breaking down a problem from the text into steps with feedback to ‘coach’ them toward the solution. There are approximately three Coached Problems per chapter. Once the stu- dent has worked through the problem, he or she can click “Try Another” to change the variables in the problem for more practice. An MCAT Test Preparation Guide is contained in the preface to help students prepare for the exam and reach their career goals. The Guide outlines key test concepts and directed review activities from the text and the PhysicsNow student tutorial program to help students get up to speed. Foundation

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