College physics : a strategic approach

1091 Pages·2015·250.643 MB·English

by  Brian Jones, Stuart Field, Randall Dewey Knight

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and pedagogy of the original, but also features alterations, customization
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College Physics
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t Randall D. Knight • Brian Jones • Stuart Field

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This is a special edition of an established title widely i
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used by colleges and universities throughout the world. d
Pearson published this exclusive edition for the benefit
of students outside the United States and Canada. If you
purchased this book within the United States or Canada
you should be aware that it has been imported without
the approval of the Publisher or Author.
Pearson Global Edition
Knight_1292057157_mech.indd 1 8/6/14 3:34 PM
F
ocus on the
big picture
Built from the ground up for optimal learning, and refined
to help students focus on the big picture.
Building on the research-proven instructional techniques introduced
in Knight’s Physics for Scientists and Engineers, College Physics:
A Strategic Approach
sets a new standard for
algebra-based introductory
physics—gaining widespread
critical acclaim from
professors and students alike.
The text, supplements,
and MasteringPhysics®
work together to help
students see and
understand the big picture,
gain crucial problem-solving
skills and confidence,
and better prepare for
lecture and their future.
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F
ocus students...
BEforE:
PrElECturE VidEoS
Presented by co-author Brian Jones, these
engaging videos are assignable through
MasteringPhysics and expand on the ideas
in the textbook’s chapter previews, giving
context, examples, and a chance for students
to practice the concepts they are studying via
short multiple-choice questions.
during:
lEArning CAtAlytiCS tm
With Learning Catalytics, a “bring your own
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BEforE, during, And AftEr ClASS
AftEr:
mAStEringPhySiCS
MasteringPhysics is the leading online
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PrElECturE VidEoS
provide timely intervention.
StudEnt workBook
The acclaimed Student Workbook provides
straightforward confidence and skill-building
exercises—bridging the gap between worked
examples in the textbook and end-of-chapter
problems. Workbook activities are referenced
throughout the text by .
For the Third Edition are Jeopardy
questions that ask students to work
backwards from equations to physical
situations, enhancing their understanding and
critical-thinking skills.
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F
ocus students...
24 Magnetic Fields
EnhAnCEd ChAPtEr PrEViEwS
and Forces
Streamlined and focused on the three most important
ideas in each chapter, these unique chapter previews
are tied to specific learning objectives. In addition,
they explicitly mention the one or two most important
concepts from past chapters, and finish with a new
“Stop to Think” question, giving students a chance to
build on their knowledge from previous chapters and
integrate it with new content they are about to read.
This detailed image of the
skeletal system of a dolphin
wasn’t made with x rays; it
was made with magnetism.
how is this done?
Looking AheAd ▶▶
goal: To learn about magnetic fields and how magnetic fields exert forces on currents and moving charges.
Magnetic Fields Sources of the Field effects of the Field
A compass is a magnetic dipole. It will Magnets produce a magnetic field; so do Magnetic fields exert forces on moving
rotate to line up with a magnetic field. current-carrying wires, loops, and coils. charged particles and electric currents.
You’ll learn how to use compasses and other You’ll learn to describe the magnetic fields created You’ll see how the motion of charged particles 794  chapter 24 Magnetic Fields and Forces
tools to map magnetic fields. by currents. These iron filings show the magnetic- in the earth’s magnetic field gives rise to the
field shape for this current-carrying wire. aurora.
Looking BAck ◀◀ Summary
eInl Cehcatprteirc 2 F0,i weeld dsescribed electric STop To Think Goal: To learn about magnetic fields and how magnetic fields exert forces on currents and moving charges.
interactions between charged objects in An electric dipole in a
terms of the field model. uexnpifeorrimen ceelesc ntroi cn feite fldo r ce, Eu
bnuett tiot rdqouees. eTxhpee rrioetnactieo na -q + q geneRAL pRincipLeS
You learned how to draw and interpret the of this dipole will be Sources of Magnetism consequences of Magnetism
eseleec thriocw fi eal dm oafg an edtiipco dleip. oInle t hcrise actheasp at emr,a ygonue’tlli c AB.. CColoucnktwericsleo.ckwise. Magnetic fields can be created by either: Magnetic fields exert long-range forces on magnetic materials and
field with a similar structure. • Electric currents or • Permanent magnets on moving charges or currents.
KNIG9721_03_chap_24.indd 764 03/10/13 1:53 PM Mmcah coaauvcrrgerromeessnec tanostp oicf S N Mmeleaiccgtrnrooestncissompi cof •• UapAfotont rmrlcleaiesack gteor ne enppae oacetil hlcme e sofoa itocevhfhlie dnm ro g;e ta lxhcgiekehnrreaet. rst g sa e d SN SFuN Bu SvuN NS
particle.
Tmda ihnapeogo rnmlteeho,t iaswstnm hdbi aciassh ist ccoh ouuen ntmhsit ip asotogsfln e oe.ft i c • Pieanarae crt hhainle lo eostlhap wmeproie;r s ewdistihe rwe ednciitt rhtieho ccentu icarourtnterrsnra,et cs nt t s II I I
the wires repel each other.
Three basic kinds Current Permanent Atomic Magnetic fields exert torques on magnetic dipoles, aligning their
of dipoles are: loop magnet magnet axes with the field.
iMpoRTAnT concepTS
Magnetic Fields Magnetic Forces and Torques
ChAPtEr SummAriES T• hipeso dtlheire oe dcf itari eoccnot imoofnp t aihsnse w nmehaeicdghlne et hptieoc i nnfoitesrlt.dh I Tcthhheae ra gmnega dlgeen paite ubndedetsw ooefn et nhit est h mceh avagrengloeetc iqict, y fi toasrn csdep etohened afv i,me alodnv:di n g Fu Bu
• due to a current can be found from F=qvB sin a
Chapter previews are mirrored by visual Thteh est rriegnhgtt-hh aonf dth reu mle afogrn efiteicld fsi.eld is Tpthohese ir tdiigvirhee tcc-thhioaarnng doe fri suth lgiesi v f0feoonrr0 cfbeoy ro c ne sa. vu
ctoh arpevteier wsu amnmd oarrigeasn, ihzeel pwinhga ts tthuedye’nvtes • pnfireeoelpddo lderi twrieohcnetainol nttou. rtnheed t oslriqguhet loyn f rao cmo mthpea ss Tpthehere pl eemnnagdgtihcnu iotluaf drt ehto eo tfwh tiehr eem: f aFogrnc=ee tIoiLcnB fa.ie cldu rdreepnet-ncdasr royni ntgh ew ciurerr ent and
• measured in tesla (T). The torque on a current loop in a magnetic field depends on the
learned before moving ahead. ctu=rrenItA, thBe s liono up.’s area, and how the loop is oriented in the field:
They consolidate understanding by AppLicATionS 1 2
Fields due to common currents charged-particle motion Stability of magnetic dipoles
providing each concept in words, math, Long, straight wire Current loop There is no force if vu is parallel to Bu. Aen emrgagyn settaitce )d wiphoelen iasl isgtanbelde w(iint ha tlhoew er
and figures, and organizing these into B = 2mp0Ir B = _m2_R0_I Fu vu hteoxig ttehhreenr af eile nmledra.ggyn esttiact ef)i ewldh.e Int iasl iugnnsetda bolpep (oinsi ate
a coherent hierarchy—from General The probe field of an MRI scanner
Principles to Applications. Solenoid If vu is perpendicular to Bu, the mbeetawseuernes t hthees ef ltiwppoi nogri eonft matiaognnse.tic dipoles
B = m0LNI pmaorttiicolne wunitdhe rragdoieuss urn=ifomrmv/ cqircBu.lar
0 0
KNIG9721_03_chap_24.indd 794 03/10/13 1:54 PM
A01_KNIG7156_03_SE_WKT.indd 4 25/08/14 12:55 PM
on the big picture
82  chapter 3 Vectors and Motion in Two Dimensions
SyntheSiS 3.1 projectile motion
The horizontal and vertical components of projectile motion are independent, but must be analyzed together.
An object is launched y After launch, the horizontal motion is uniform motion. SynthESiS BoxES
into the air at an angle vui
u to the horizontal. The horizontal component of the initial velocity
is the initial velocity for the horizontal motion.
(vy)i = vi sin u u x (vx)i au = 0u Tish zee aroc.celeration Bprriinncgiipnlge st,o agnedth eeqru kaetyio cnosn, ctehpist sn,o vel 9.4 Conservation of Momentum 265
Athfet evre rlatiucnaclh, (vx)i = vi cos u feature is designed to highlight
mThoet ivoenr tiisc afrlee fall. TTachhceee lvkeeirnratetimicoaanlt immc ooettqiioounnat ivoenrtsi cfoalrl yp raonjedc ctTiolhne esmt faornettie-o vfnea llaolrcei ttyh ohsToerh iftezow orhn ooctrFoa ilnIzslGsyokt:naUatnatRtel- Emrs o9.t i.To1n7h es htootwals ma obmefeocnroetu-namnn db-eeaffcotetreri ovthinseuysa lp auosvnhe rdovf ifed wisi f fPfouier= rthe0eu ncSceoonsls.vi deMe rS ooinnrcleey txh -ec ommoptioonne inst so nolfy mino tmhee nxt-udmir.e cWtioe nw, writee’ lSl annededra t’os
ciinnoiimttiiaapllo vvneeellnootcc oiittfyy t ifhsoe rthe is free fall. agc =ce l9e.r8a tmio/ns2,. is ubtuneimfcoar mwus miello btsiootintl.hl bsek a0ute arfst earr eth aety rtd ephsuets.ae hCn poo fnefas.re qs ruueenmltalymt, itohaenr tysot ,aa lt nmhdoem ype noe-minaitpnn idhoti a(aulv Sstmx i)cioz hmoeee mrn itnumimtia ola svn e (lopcSixt)yi.= SmimSi (lvaSrxl)yi ,, wweh ewrrei tem DS aivs idh’esr inmitaisasl
tRhies ivnegr toicr afla mlliontgio, nth.e (vy)i au (yvf y =)f =yi + (v (yv)iy ) -i g∆∆t t- 12 g(∆t)2 (xvf x =F)fI G =xiU + (R7v .(Ex3v) ix9 ) =.iT1 ∆ o 7 cr tqoBuneest faonrte-2a0n1d-aftoerr v cisouanl otvrearsviteiwn fgor dtweot sakialtser.s mbeocmWaueesne tc ubamont han sso kw(ap teaDrpxs)p ial=yre tmihnDei (tivmaDlalxyt)hi .ae tmB roaettshitc. athl esstea temmoemnet notaf maroe mzeenro-
asacmceele, raayt i =o n - isg .the The two equations are linked byp tuhse htimineg in otefrfv aflr ∆otm, each other. tum conservation, Equation 9.15. Writing the final momentum of
We see that the componeRnist inogf Fu thFata lilsin pgerpendicular to the radial The torquwe hdiecphe ins dthse o sna mhoew fo hr atrhde Rhoyraizno pnButaselh faoensrd,e wv:ehrteircea lh me optuiosnh.es, Sandra as mS (vSx)f and that of David as mD (vDx)f , we have
plionien its a Ft #w=hicFh c tohse 2 f0o°rc=e i2s2 a6p pNli.ed T ihs e r d=ist0a.n7c5Se t mfor.o pm t oth te hhiinngke 3to.7 t h e A 1aapnt0e dra0p paegton ibwdniahtc laualt l rbaaorint l tglfosal e rtoh.t hfeIfef d rah oo eotu awrt.b faOlrneort m sah nte ot dhc e eolx auhenlirdndt gssmei m2,o o rpmerl yht o feprr uoqcsoumhue ,lh tdhha epred u cebsorha( mu!v s leDDedx x a op=)cuif t s=8lthyh0 0e k mg/s (mvSS x =)i =45 0 k mg/s mS(vSx)f + mD(vDx)f = mS(vSx)i + mD(vDx)i = 0
Vid SDeoeolm vtouet o W r e tca=n rfiFn#d= Vthi(de0D et.7oeo5rm qt muoue)t (oo2rn2 6th Ne )do=otf ar1rA ob7fmr.l0 oe N6m. tAh 1# Ee m 2q mb0ua a0st iego nob f 7a .tl1hl 0er :o tlalsb l asaoei f?SgsftnS utiehcfikSec Sadsn oa A tom trso.e ry qtoauube’l,l elb swuete ittB hhbi. ys t h1mde om aisknaegms mese osnrpseee p eirdfo .yb Holeuom wasr, e f 1 at7rry 0di nNoge# stm oi t f ilrsae ean d x SolviTmnhogem fseoknra ttu(evmrDs ’xc )fifn , awle findcmomeqeunatulsm t hceir initial cwasw zhericoh.
forc Eeq cuoamtiopCnosn e7on.1t,0 h–na7s.1 Ca2 s iggEinv.e P Ao nttolyr q tuhCee tmh haaCEgt.. n tEeBBitnueedCttdwwse eetkooeefnn r to12 ht feamm tt eioaa nntrghqddue 24 eo .umm bBj eucrtt tioEnr qauS ceo, ulniktee ra- DF.. 24 mm PArftero: B(vlDx)Ef m-Sol Ving(vS x)fS = t2.2r m/As tEgiE(vDSx)f=- mmDS (vSx)f=- 4850 kkgg*2.2 m/s=-1.2 m/s
cVloidcekow tisuet odri rection isV ipdoesoi ttivueto, rw hile a torque that tends to rotate the object in a David moves backward with a speed of 1.2 m/s.
Tocf noroloo r ccctDeoekoer wnpqmmuuicsoseeh.o pidnuligre rescaxtrtaig oifgneihg t i ssut otnDwrueeeagmrda,d ot etitvhhnee. e tp FsyiIvG o UtatR oo3Err e 7.ap.7 2uca1l ltsisniPS ukvgmo reesmotldrlaavyjri ieg tziheeoncts n otgtruhgiete l a fPaersogisrg mMneoo st .nhb woeN lwpotieittivicmhoioett nhetashx :a eakt r t ase y Class VideoTa ofrpFaViinicmddD:e- e(osevm DTpwxou)etfoocr irfkic a Pnrdo bglueimda-nScoelv fionrg b Srotraadte cglaiesss egsiv obCeeoft nw Nspseoeterrtunivoc adeDbt iteoahlvnenai todm twf sa mens ddo.i mdSnea’nnttd unrmeae idmn taoon rkddnaeorte wtso at hfniiyns ddr eeDstuaailvlts.id a’bso fuint athl es pfoerecde.
rela oNbtOseTeErdvi▶ n“g W Sthhete nod ipcreac lcttiuoolna tTiinn ghw iahn ictokhr tq”hue eq ,t Nkouynroqooewuuwe sl tmeahtdcauigttsos et.w nts◀oeu . pshopallvvyee tsaho egm oaeop dptr ruoiedp etrawia otoe-f d shiimogwnen bspiyroo njaelc tmileo timono tpiorNgonb iewvlewominrks .sgo, wvceel ecraavnr iue sswet tah sattet amteemntesn otfs wnohwat ptyropveisd eo ft phreo “btabhlSaeinSgm eD SpasSvi icadItt’ su s, ewreemou,s”l dr ehaasvoen tahbel eg rtehaatte rS fainndarl as,p ewehdo.se mass is less
FIGURE 7.21 Signs and strengths of the torque. strategy is intended for and/or how to use it.
example 3.11 Dock jumping A positive torque tries to rotate the
IMn atxhiem supmo rpto osift idvoe ctokr qjuume fpoirn ag ,f odrocegs run at full speed oofbfj tehcet ceonudn toerfc locFkiwGiUseR aeb 3o.u3t3 t heV pisivuoatl. overview for Example 3.11.
aips ed rtophceenk do tinhceua ltat shri attost atlha fene drwasd fifaealer lttih naeebsot vfero am p othoel oefn wd aotef rt.h Teh deo wckin. nIifn Ppagiu vddlolooitn ggeg x setrrtasi gzehrto o tuotr fqyruoem. the pSRtoKRBnoalwEtnmE-SGoyl V9In.1G Conservation of momentum problems
rtwPFcad(huviatorG aeynntc)PpUeth skieeuiuPeR v,rn=as s e,osabedht o th 3 ai0 rt8 noe.osht3 e.xhgfemRwe53 e e .aas or d/ WWmtid tsffrdensa.i a t a/ioeterzWiatshlg ice i whaelhlr(kesisl aoatnt i d avcc tla ettohlporoaeno tcwnat rmr dchkeqt ahwt pete.urirP toseq oedTiydist a.u nv0h hetottaohe .neygt6ttneh ia p t1dptogio sisoo nc omv opigsaan ifu b ss lpt r( t tueuar h2staefPf pnhe poolsfo se erretvri c,eooennohe edaj vttlroe s eo i iewrscgpfsdcirtot ilzh ivaainaitrloepninye seronpt dhewSahmftlaiai ryitne erolshnlddg otye(tei vfh o hdcv oxeneooe)fts iihofnwpig s=ertrthd n)h o3 ti?ebs8sMp.)n1 i tl.eateaer5 aurtanabmp xeabmidoe itgs,Aoom in vy/oohsbvd seuosnnfjte ie tma ce .tetowcnhguhmi tfndanl eeefacte irlgvo taecot k ittvowx(hverie,xiq s)t yuroeii,ae, r a(dq tvtbiiur0yaoi)eleiu 0 slfti o nttr0ohe . ae6r o1fpto iamvrtcoeet t. hauvuei x(vfx, )yff,, (tvfy)xf Waap■Rnc tEed(y(axatFxIiiivvpfii xyi nx of y= n ==v)) ca= = diti n p0a ee R == 00-0 .o .nrls E m6g0vmoas1 i mc c/us =Cmsti2i/ s si,b8tl oeyei.l5fe ne a =tms,srh l/ 0ocsyea mhf rd loeoae obwsfsuijen efo cefafi t ct sshci yaeeonsf nsttteeyslymres traes vtnhmha oita.nirtot t ienasr n aoidscf o tiilmnoattnoeen mdtso ee( ntF uhttnhueeatimrt= vyt0oaoul)u u r oeecrsal a nwbt eeiitg fhtonhironeer ewtmh heoexi mcitnheert netnhrtaae-l
We know that the horizontal and vertical motions are inde- the dock, it will continue to move horizontaflloyr acte 8s. 5 fmo/rs . tThhee duration of the interaction (the impulse approximation).
pSeTnOdpe nTtO. TThheIN fKac 7t. 3th a t Ath ew dhoege ils t fuarlnlisn gfr teoewlya rodn t haen w aaxtleer adto esn’t Avertical motion is free fall. The jump ends Mwhoemn tehne tduomg hiist st hen conserved.
twhaheff icecchet niottesn re.h oGorifiz voetnhntea lt hfmoero ctdieoesnt a.s ilWhso hwneonnt etihdne idtnho eg F ilfgeigauvureerse 7thw.2ei 1lel , nd 2oFf the water2—FthatB is, when it has droppe■d byI f 0i.t6’1s mn.o Wt pe oasrse ible to choose an isolated system, try to divide the problem
r/2 C D
provide the largest positive torque on the wheel? into parts such that momentum is conserved during one segment of the
r F F motion. Other segments of the motion can be analyzed using Newton’s laws
or, as you’ll learn in Chapter 10, conservation of energy.
F E
Following Tactics Box 9.1, draw a before-and-after visual overview. Define
symbols that will be used in the problem, list known values, and identify what
KNIG9721_03_chEaxp_a0m3.ipndldE 782.10 Calculating the torque on a nut you’re trying to find. 09/08/13 1:56 PM
Luis uses a 20-cm-long wrench to tighten a nut, turning it clock- F IGURE 7.22 A wrench being used to turn a nut.
wise. The wrench handle is tilted 30° above the horizontal, and (a) (b) SolVE The mathematical representation is based on the law of conservation
Luis pulls straight down on the end with a force of 100 N. How of momentum, Equations 9.15. Because we generally want to solve for the
much torque does Luis exert on the nut? velocities of objects, we usually use Equations 9.15 in the equivalent form
PrePare F IGURE 7.22 shows the situation. The two illustrations
correspond to two methods of calculating torque, corresponding m1(v1x)f+m2(v2x)f+ g=m1(v1x)i+m2(v2x)i+ g
t Soo Elqvuea t A iocncso 7rd.1i0n ga ntod E 7q.1u1a .t ion 7.10 , the torque can be calculated m1(v1y)f+m2(v2y)f+ g=m1(v1y)i+m2(v2y)i+ g
caos m tp=onreFn#t .o Ff rthoem f o Fricgeu irse 7.22 a, we see that the perpendicular aSSESS Check that your result has the correct units, is reasonable, and answers
the question.
F#=F cos 30°=(100 N)( cos 30°)=86.6 N Continued
Exercise 17
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2.2 Uniform Motion 33
From Velocity to Position
We’ve now seen how to move between different representations of uniform motion. FIGURE 2.13 Deducing a position graph
There’s one last issue to address: If you have a graph of velocity versus time, how from a velocity-versus-time graph.
can you determine the position graph?
v (m/s)
x
Suppose you leave a lecture hall and begin walking toward your next class, which 1.0
is down the hall to the west. You then realize that you left your textbook (which you
0 t (s)
always bring to class with you!) at your seat. You turn around and run back to the
5 10 15 20 25
lecture hall to retrieve it. A velocity-versus-time graph for this motion appears as the -1.0
top graph in FIGURE 2.13. There are two clear phases to the motion: walking away
from class (velocity +1.0 m/s) and running back (velocity -3.0 m/s.) How can we -2.0
deduce your position-versus-time graph? -3.0
As before, we can analyze the graph segment by segment. This process is shown in
As you move away, As you return,
Figure 2.13, in which the upper velocity-versus-time graph is used to deduce the lower your velocity is your velocity is
position-versus-time graph. For each of the two segments of the motion, the sign of the +1.0 m/s c -3.0 m/s c
velocity tells us whether the slope of the graph is positive or negative. The magnitude cso the slope of cso the slope of
of the velocity tells how steep the slope is. The final result makes sense: It shows your position graph your position graph
is +1.0 m/s. is -3.0 m/s.
15 seconds of slowly increasing position (walking away) and then 5 seconds of rapidly
decreasing position (running back.) And you end up back where you started. x (m)
15
There’s one important detail that we didn’t talk about in the preceding paragraph:
How did we know that the position graph started at x=0 m? The velocity graph tells 10
us the slope of the position graph, but it doesn’t tell us where the position graph
5
should start. Although you’re free to select any point you choose as the origin of the
coordinate system, here it seems reasonable to set x=0 m at your starting point in 0 t (s)
0 5 10 15 20 25
the lecture hall; as you walk away, your position increases.
Stop to tFhINk 2.1 Wohich pocsition-uversus-stime graoph best dnescribe s thse mottion duiagramd at left?ents...
x x x x
t t t t
A. B. C. D.
Multiple

Representations...
2.2 Uniform Motion
Balancing
I f you drive your car on a straight road aQt a puerfaecltliyt staeadtyi 6v0 emil eas pner hdou r (mph), you FIGURE 2.14 1M1.o5t ioHn edaiat gErnagmin aensd 333
will cover 60 mi during the first hour, another 60 mi during the second hour, yet another position-versus-time graph for uniform
motion.
60 Imt ii sQd puorsiunsgib alteh ent ot htdiorid ts ohamouetrt,hi aivnngde ssiom Roilna.re Twhaitihss hiseo aatn.n Texhiaenmrmpgalel .eon.fe .wrghya ti sw nea tcuarlall luyn tirfaonrms-
mfeorrtieodn .f Srotmra iga hhto-lti nree smerovtoioirn tion wa hcioclhd erqeusaelr vdoisirp;l aitc eims penostss iobclceu tro d tuarkien gs oamnye soufc ctehsis- Uniform motion
seinveer geqyu aasl -itti mise t rinantesrfverarlesd i sa cnadl lecdon uvneirfto irtm to m oothteior nf oorrm cso.n Tsthaisn ti-sv ethloec jiotyb mofo tai odne.vice vu
knoNwont Eas ▶a hTehae tq eunalgifinieer. “any” is important. If during each hour you drive 120 mph Tsuhcec edsisspivlea cfermamenests a bree ttwheeen
FifToghre 3 eu0n emrrginey -atrnsadn ssf teorp d ifaogr ra3m0 mofi nFI,G yUoRuE 1w1.i1ll8 ac oilvluesrt r6a0te sm tih ed ubraisnigc pehaychsi cssu cocf eas hsievaet same. Dots are equally
eng1i nheo. uItr tiankteesr vianl .e nBeurgt yy oaus hweailtl fnroomt h tahvee h oeqt ureasl edrvisopirla, ctuemrnesn stos mdeu roifn git isnutcoc eussseifvuel spaced. vx is constant.
twho3rak0, t ma niTnd ieenxtheraavuaslctss, tshhoe tb.ha.isl.a m ncoeti aosn wisa nstoet huenaifto irnm th.e◀ col d reservoir. Any heat engine x The position-versus-tim2e4 .g3r apEhle ctric Currents Also Create Magnetic Fields 771
has exactly the same schematic.
is a straight line. The slope of
NFIoGFtUiIcGReUE tR1hE1a .2t1 .8t1h 4Te s hpheoo oswiptsei oraan mt-ivooentr iosoufn sa d- htiiaemagter a egmnrga aipnnhed .f oa rg uranpifho rfmor mano toiboCjneu cirrste nIaitn us pturnaTaarrhiieofge ut oanchnodrtgm mt ehlpnie atn m swtsoei nro.aee ,tTce isidrohhlceolniswes.i ng the liCnuer riesn vtI xd.ownWis rheevne rthseed c, uthrree nc∆to mdxipreacstsion ◀to F aig cUuRrere 2n4t.-1c0a rHryoinwg c womirep.asses respond
follows from the requirement that all values of ∆x correCsuprroennt-ding to tthhate t hsea fimelde i sv dairleucteed needles turn to point in the
(a) 1. Heat energy QH is transferred from (b) carrying Usceofuunlt ewrcolorckkwise around op∆pxosite dEirqeucatilon.
of ∆t b e ethqeu haolt. reIsne rfvaocirt ,t oa tnh ea slytsetrenma.tive definition of unifwoirrem motiondr iitvhsee: s w Agireen.n eorabtojer.ct’s displacements
moHtioot nre siesr vuoniriform iTf and only if its position-versus-time graph is a straight line. t
H
W
out
EquatiQoHns of Unifo2r. Pmart Mof tohetion Addressing
energy is used
An object is in uniform m ottioo dno aulsoefnugl the x-axis with the linear position-versus-time
Heat
work W .
ogebrnajgpeichnte ’ssh ionwitnia iln p FoIsGiUtiRoEn 2 Wa.1so5u txo na tt htiem neoeu xttt. pTahgee .t eRrmec a“lilnM fitrioamli”s CrcehTfeaoorp sht enetlrop c1 rte htmeeh easmptt awbretteri n iidnoge w npnhooitciensh t td .hoir.efe . c tion compasses will point, we use the right-hand FigURe 24.11 The notation for
i i QH rule shoQwCn in Tactics Box 24.1. We’ll use this same rule later to find the direction of vectors and currents that are
QC the magnetic field due to several other shapes of current-carrying wire, so we’ll call perpendicular to the page.
this rule the right-hand rule for fields.
(a) (b)
Cold reservoir T
C
3. Texhhe aruesmteadin tion gth een ceorgldy rQesCe r =v oQirH a -s wWaosutte i sheat. Hteomt preersaetruvroei rT atTBAocxT 2i4c.S1 CaRto tilegdm hrpetse-heraravtunorider (Trau rliev efro)r fields Vectors into page Current into page
KNIG9721_03_chap_02.indd 33 H Point your rCight thumb in I 16/08/13 12:53 PM
the direction of the current.
I
Wrap your fingers around the
FIGURE 11.18b shows the directions of the energy flows and the locations of the Vectors out of page Current out of page
wire to indicate a circle.
reservoirs for a real heat engine, the power plant discussed earlier in this chapter.
The cold reservoir for a power plant can be a river or lake, as shown,Y ooru rs fiimngpelrys ctuhrel in the(a) (b) Current into page
direction of the magnetic
Aatmnos pIhnere,d to wuhicch hteati QvC ies re je cted from cooling towers. field lines around the wire.
Vectors into page Current into page
Most of the energy that you use daily comes from the conversion of chemical
Exercises 6–11
Aeneprgpy inrtoo thaercmahl e.n.e.rgy and the subsequent conversion of that energy into other
forms. Let’s look at some common examples of heat engEinMeamsg.netismp ofthen raequisres ia sthree-dimensional perspective of the sort shown in I
Tactics Box 24.1. But since two-dim eonsionnaVl e cfEitogrus rxoeust paofr ep alegaiescier itCot udr rreaSnwt o, kuwt oefi wplailglles m.ak.e.
as much use of them as we can. Consequently, we will often need to indicate field vec-
heat engines tors or currents that are perpendicular to the page. FigURe 24.11a shows the notation we
Current into page
will use. FigURe 24.11b demonstrates this notation by showing the compasses around a
current that is directed into the page. To use the right-hand rule with this drawing, point
researcyour righht thumb into- the pabge. Your faingers will cusrl clockweise, givingd the direction
in which the north poles of the compass needles point.
concepTUAL exAMpLe 24.2 drawing the magnetic field of a current-carrying wire
Sketch the magnetic field of a long, current-carrying wire, with assess Figure 24.12 illustrates the key features of the field. The
Knight/Jones/Field was designed from tthhe ceur rgenrt gooiungn indto tuhep pa pwer. iDtrhaw sbottuh mdaegnnetitc sfi eildn li nme idnirdect,io na onf tdhe fiiseld vectors and field lines matches what we saw
and magnetic field vector representations. in Figure 24.11, and the field strength drops off with distance, as
based on a solid foundation of the latest physics education research . . .
Most of the electricity that you use was Your car gets theR eeanseorgny F irto nme ethdes i rtoon r fuilnin g picturTe hine rteh ea artel ams, awney h samvea slel,e ns imwplee l ehaernaet de ning tihnee sat las figure. We don’t
that the field lines form circles around the wire, and the magnetic expect you to draw such a figure, but it’s
generated by heat engines. Coal or other fossil from the chemicfaiel led nbeercgoym eisn wgeaaskoelri nase .th Te hdeis tanceth farot ma rteh ep warirte o ifs tinhcirnegasse yd.o u uwsoert hd aloiloyk.i nTgh aits the full 3-D picture of
fuels are burned to produce high-temperature, gasoline is burneFdig;U tRhee 2r4e.s1u2 lstihnogw sh ohot wg awsee sc onstrfuacnt ,b owthh ifciehld c alinne b aen dp ufite oldn tothpe offie ald wino oFidg URe 24.13. This conveys
high-pressure steam. The steam does work are the hot reservveocitro.r Sreopmreese ontfa ttihoen st hofe rsmucah la fields.tove, uses the thermal enetrhge yid oeaf tthhaet tshteo vfieel d lines exist in every
by spinning a turbine attached to a genera- energy is converted into the kinetic energy to provide power to drive apilra naer oaluonngd tthhee le ngth of the wire.
FigURe 24.12 Drawing the magnetic field of a long, straight,
A01_KNIG7t1o5r6,_ w03h_SicEh_ WpKrTo.dinuddc e s6 electricity. Some of the of the moving vecuhrircelnet,- cbaurrt yminog rwe irteh.an 90% room. Where are the hot and cold reservoirs 25/08/14 12:55 PM
energy of the steam is extracted in this way, is lost as heat to the surrounding air via the in this device?
but more than half simply flows “downhill” radiator and the ex gh o a emsu eisnattn,os a tchsue r srpehangotew: n in this Mwhaegrnee tthice fifieelldd vise csttororsn gaerer. longer eFixgisUtR eev 2e4ry.1w3 hFeireel da lloinnegs
and is deposited in a cold reservoir, often a thermogram. point your right
thumb in this the wire.
lake or a river.
direction. Magnetic field vectors are
tangent to the field lines.
Your fingers
curl clockwise c I
We assume, for a heat engine, that the engine’s thermal enecrgsyo tdheo mesangn’ett icchange. This Field lines are closer together
field lines are clockwise where the field is stronger.
means that there is no net energy transfer into or out of the heat ecinrcgleisn aero. uBnde tchae uwsiree .energy is
conserved, we can say that the useful work extracted is equal to the difference between the
KNIG9721_03_chap_24.indd 771 03/10/13 1:53 PM
KNIG9721_03_chap_11.indd 333 23/08/13 9:52 AM
784  chapter 24 Magnetic Fields and Forces
FigURe 24.37 Magnetic force on a However, there is a force on a current-carrying wire that is perpendicular to a
current-carrying wire. magnetic field, as shown in FigURe 24.37.
(a) (b)
I noTe ▶The magnetic field is an external field, created by a permanent magnet or
by other currents; it is not the field of the current I in the wire. ◀
The direction of the force on the current is found by considering the force on each
charge in the current. We model current as the flow of positive charge, so the right-
hand rule for forces applies to currents in the same way it does for moving
L Fu charges. With your fingers aligned as usual, point your right thumb in the direction
ocus on students... and how tof thhe curreent (they direc tioln ofe the maotion rof posnitive charges) and your index finger
in the direction of Bu. Your middle finger is then pointing in the direction of the force
Bu Bu Fu on the wire, as in FigURe 24.37c. Consequently, the entire length of wire within the
magnetic field experiences a force perpendicular to both the current direction and the
field direction, as shown in FigURe 24.37b.
If the length of the wire L, the current I, or the magnetic field B is increased, then
A wire is perpendicular If the wire carries
to an externally created a current, the magnetic the magnitude of the force on the wire will also increase. We can show that the mag-
Guimdagneetic fideld. Prafiocnel tdth wei wilcli reeex.ert. a. f.or ce netic force on a current-carrAying nwirne is ogivetn aby ted Equations...
10-8CHAPTER10.En(ergcy a)nd Work I Bu a I Current in wire (A) Magnetic field (T)
1170..B6eloPwo wtee snete iaa 1l kEgn oebrjegcty that is initially 1 m above the ground and rises to a height of 2 m. Fwire = ILB sin a (24.9)
Anjay and Brittany each measure its position but use a different coordinate system to do so.
mFiella isnu rtehde btayb Alen tjoa ys haonwd Bthreit itnanityia.l and final gBruavitational potential energies and DUTashe right-hand rule L Length of wire in Angle between wire
Anjay Brittany for forces applies. magnetic field (m) and magnetic field
EFnuds here Ui UfYo∆uUr thumb should
2 m Starts here 0 Anjay
Brittany point in the direction
0 of the current. In many practical cases the wire will be perpendicular to the field, so that a=90°. In
18.Tfbraholrmle 2e t hibsea flsilrase modfe s ethrqeauiigaglhh tmt adabosowsv aner, e tah fneidr ge bdroa sullin m3d u.i slBt afainrlele od1u hissol yfrii rwzeoidnt htsa terllqayiu.g aRhltas unppke, eidns Ball 1Ball 3 this case,
othred egrr,o furonmd. largest to smallest, their speeds v1, v2, and v3as they hit Ball 2
EOxrpdlearn:ation: Fwire=ILBV isual (24.10)

If we rewrite Equation 24.10 as B=Fwire/IL, we can see that the unit for magnetic
field, the tesla, can be defined in terms of other units:
22.2 ADefinninag alnod Dgesciriebinsg C.u.r.rent 705
N
19.Bone ltohwe laerfet. sThoo wwnh itchhr epeo firnitc dtiooensl etshse tbralockcks. mAabkleo citk o ins rtehlee arisgehdt fbreofmor ree rset vaet rtshien gp odsiriteicotnio snh oanwdnVideo Tutor 1 T=1 A#m
sliding back? Point B is the same height as the starting position.
Similarly, the lightbulb doedsenm’t o“use up” current, but, like the turbine, it does use FIGURE 22.7 Water in a pipe turns a
enCBeArgy. The CeBnAergy is dissCiBpAated by atomic-level friction as the electrons move through turbine.
Makes it to the wirMaekes i,t to making thMakees it t owire hotter until, in the case of the lightbulb filament, it glows.
TehxeArMep aLree 2m4.a11ny otMhearg isnseuteisc wfoer’clle n oened a t op oexwaemri nlien,e but we can draw a first impor-
tantA c oDnCc pluoswioern l:ine near the equator runs east-west. At this location, FigURe 24.38 Top view of a power line near the equator.
the earth’s magnetic field is parallel to the ground, points north,
Streamlining The field of the vearth near the v
Laandw h aosf m caognnistuedrev 5a0t imoTn. Ao f4 c0u0 rmr elenntg thT ohfe t hceu hrreeavnyt cisa btlhee th saat me at all points i n ae quator is parallel to the
spans the distance between two towers has a mass of 1000 kg. ground and points to the north.
current-carrying wire.
What direction and magnitude of current wuouslde be noecefs saCry too lor... N The amount of water leaving the turbine equals
offset the force of gravity and “levitate” the wire? (The power
the amount entering; the number of electrons
line will actually carry a current that is much less than this; 850 A W Eleaving the bulb eBuquals the number entering.
ciso an tcyEppictaUl avLa lEuxe.a)mpLE 22.1 Which bulb is brighter?
S I
TPhRee PdaisRceh aFrigrset , owf ea csakpeatcchit oar tloigph tvsi etwwo of FtIhGeU sRiEtu 2a2ti.o8n ,T waso ibnu lbs lit by the current
ActiFdiiegnvUtiRceae l2 b4.u3l8b.s T, haes mshaogwnent iicn fFoIrGcUe oRnE t2h2e. 8w. ire dmisucsht baer goipnpgo sai tcea thpaatc itor. Flow of
Coof mgrpaavriety t.h Ae nbL raipgpheltinceaastiso ornf onthf eth itewn roig bghut-lbh.as.n.d. ru le for forces shows electrons
that a current to the east will result in an upward force—out of 10.4 Potential E4n0e0r gmy 297
Rtheea psaogne. Current is conserved, so any
current that goes through bulb 1 must magnitude of the earth’s field. Solving for the current, we find
toS stmopa ltloes tt,h gismtnIohNo ae ktlthg hvg1nerr0e ioa.tt 4uvwuT dighot eahe Rt b oibmouafu nnlatblakhgbsle n i p2naemotr oieatacer sg den fnwqeiteeiruael,t adillfc ll re—. io fnsWom eptrrhc eegleera’i p vericgesesu ne ogrnsdrfitoie v c tneuetndlsa br yto E tqhuea ctiuornr e2n4t3,. 1s0o. tvTh =oe 0 I=mg= 1000 kg 9.8 m/s2 =4.9*105 A
identical batllhlesav t1i t tathhteer o tbuhgrei hgw h4it.rnee, stsh ios ff oar cbeu mlbu sits bper ooppopro-site to the weightB fuolbrc 1e Bulb 2 LB 1400 m 2501*10-6 T2
tbiount aelq utoa l tihne m caugrnreitnutd eit, scoa rwriee sc.a nId wenrittiecal vu 2 4 directed to the east1. S21 truc2 tured
bulbs carrying equal currenmtsg m=uIsLtB have
the same brightness. 1 assess The current is much larger than a typical current, as we
where m and L are the mass and length of the wire and B is the expected.
assess This result makes sense in terms Problem Solving...
of what we’ve seen about the conserva-
Elastic Ptioonte onf tciuarlre Entn. eNrog cyharge is “used up”
by either bulb.
Energy can also be stored in a compressed or extended spring as elastic (or spring) FIGURE 10.16 The forcFeIG reUqRuEir 2e2d. 9to N egative and positive
potential energy Us. We can find out how much energy is stored in a spring by compress a spring is cnhoat rcgoenss tmanot.ving in opposite directions
using an external force to slowly compress the spring. This external force does work
KNIG9721_03_chap_24.indd 784 x = 0 give the same current. 03/10/13 1:54 PM
on the spring, transferring energy to the spring. Since only the elastic potential energy
research-based pof the sp2rein2g .is2 c haDndgiengf, Einquaitnioang 10 .a3 rneaddsg Descroibing Cgurrenty SpTrhinigs ipnl aetqeu idliibsrcihuamrges as negative charge
enters it and cancels its positive charge.
As we’ve seen, the curren∆t Uins =a Wm etal consists of electron(1s 0m.1o4)ving from xlow elec-
That is, twriec c apno ftienndt ioault thoo wh imguhc hp oeltaesntitcia plo, tienn tiaa l deinreercgtyi oisn s tooprepdo isni tteh et hsper ienlge bcytr ic field. BFuut, as
calculatiFnIgG UthReE a m22o.u9n st hoof wwosr, ka n eceadpeadc tiot ocro mdpisrecshsa trhgee ssp riinng .exactly the same way whether we
. . F.I GEURvcEoi 1dn0.s1ei6d nsehrto wt hisen a c stpurhrirneegn b tt etieong xb ceto m,n peargeraststie,vd e db yce has ahraigngedsn. Imn, ◀o◀p vSeEinCdtgI oaoNpg 8p.3oo wsgiet eyfo t,uh naed nfiedld toer cpohsnitiovel ogy, using ideas from
that the fcohrcaer gthees spmrionvgi enxge rtisn o nth teh e shaamnde i sd Firse=ct-iokn ∆ xa s(H tohoek ef’ise lladw. ),I nw hgeeren e∆rxa lis, for any circuit, As x increases,
mtheu dlistpiltmahcee emcuednrrti eaonf t tlihese iaennr tdnh oeif n stahgme sept drhiinrege cofrtoiromyn i,itns sdeeqtupuielidnbdreieunnmt topsof s wiatihorince ha n godf i ktv hisee sthnee t wtho pee rbsxpeecsttiv etso oTlsohs ids o tpesloa Fte. succeed Tihnis pplahteysics.
spring constant. In Figure 10.16 we have set the origin of our coordinate system at the discharges as discharges as
we choose. We thus adopt the convention that current is the flow of positive
equilibrium position. The displacement from equilibrium ∆x is therefore equal to x, negative charge positive charge
charge. All our calculations will be correct, and all our circuits will work perfectly leaves it. leaves it.
and the spring force is then -kx. By Newton’s third law, the force that the hand exerts
well, with this positive-charge-current convention.
on the spring is thus F= +kx.
As the hand pushes the end of the spring from its equilibrium position to a final
Definition of Current
A01_KNIG7156_03_SE_pWoKsTit.iinodnd x ,7 the applied force increases from 0 to kx. This is not a constant force, so we 25/08/14 12:55 PM
can’t useB Eecqauuatsioen t h1e0 .5c,o Wulo=mFbd ,i tso tfhined S thI eu wniotr ko fd ocnhea. rHgoew, aevnedr ,b ite cseaeumses rceuasrroenn-ts are charges in
able to cmalocutiloatne, thwee w dorekf ibnye ucsuinrgr etnhet aavse rtahgee rfoartcee, iinn Ecqouuatlioomn b10s .5p. eBr esceaucsoen tdh,e at which charge
fFoarvcge= va12emrkilxeeo.s cvT fterhroisucm st hf tFhireoiel =udwg o0chr at koau dsFwoefnsi=re ec b.kh yxFa, Ir tGthghUeee Rs ha Eavt no2ed r2m a.i1gso0e vsfeoh rotchwer uso suae gdwh tio rt ehc oeinm w pwirrehesi.sc Bhth etehc seap ureislneeg cw tisre ic a friee lcdo inss Eiud.e Trihnigs Tenhtiesr sp liat taen ddi sccahnacregless i tass npeogsaittiivvee cchhaarrggee.
curTrehnet falso wth erW amte=o toFioafv ngw doa=ft epFro avisngixt iav= ep icp12he ka xrmgeexas=,s ut21hr e eksx m 2thoeti oanm iosu innt tohfe wdiarteecrt ipoans soifn gth ea fcireoldss.- Calf muscle
FIGUARchEi l2le2s. 1te0n dTonhe current I.
This wosrke citsi ostno raerde aa so pfo ttheen tpiailp een peregry s ienc othnead s. pWrinbeg ,u ssoe wa es icmanil aurs ec Eonquvaetniotino 1n0 f.o14r current. As illus-
to find thtraatt aesd tihne Fspigriunrge i s2 2co.1m0p, rweses ecda, nth me eelaassutirce p tohtee natmialo eunnetr goyf icnhcarergasee s∆ bqy that passes through a The current I is due to the motion
of charges in the electric field.
cross section of the wire in a tim1e interval ∆t. We then define the current in the wire as
∆Us=2 kx2 Eu I
∆q
Just as in the case of gravitational potential eneIrg=y, we have found an expression for (22.1) On each stride, the tendon
the change in Us, not Us itself. Again, we are free ∆tot set Us=0 at any convenient stretches, storing about
spring ext ension. An obvious choice is tDo esfeitn Uitiso=n 0o fa ct uthrree nptoint where the spring is 35 J ofWw ehneei cirmghy at.hgein ceh aanrg aerse am aocvreo.s Isn t hae t iwmiere ∆ tth,rough
in equilibrium, neither compressed nor stretched—that is, at x=0. With this choice Spring in your step cAhas rygoeu ∆ ruqn m, yoovue s through this area.
we have lose some of your mechanical energy each
time your foot strikes the ground; this energy
Us=12kx2 (10.15) Us ieosnf teyrraognuysr.f moLruemcckehdial nyini,c taaobl oueunntre e3rcg5oy%v we orhafe btnlhe ey todhueecrr mrfeoaaols te
Elastic potential energy of a spring displaced a distance x from p.44x lands is stored as elastic potential energy in
QUADRATIC the stretchable Achilles tendon of the lower
equilibrium (assuming Us=0 when the end of the spring is at x=0) leg. On each plant of the foot, the tendon is
KNIG9721_03_chap_22.indd 705 28/09/13 2:23 PM
stretched, storing some energy. The tendon
springs back as you push off the ground again,
NotE ▶Because Us depends on the square of the displacement x, Us is the same helping to propel you forward. This recovered
whether x is positive (the spring is compressed as in Figure 10.16) or negative (the energy reduces the amount of internal chemical
spring is stretched).◀ energy you use, increasing your efficiency.
KNIG9721_03_chap_10.indd 297 16/08/13 2:04 PM
F
ocus students...
on their goals
inCrEASEd EmPhASiS on
CritiCAl thinking And rEASoning
Problems 349
Students will be required to reason, to do more
McAT-Style passage problems For humans, the energy used per second (i.e., power) is propor-
tional to the speed. That is, the human curve nearly passes through than simply plug in numbers into equations. Of
KKaannggaarroooos Lohcaovme otvioerny stout tmohneul cyoh ra ip gvoienwr,ye sr .os m Froaurln l ans ilhnoogpp etp.w inIincg e ok atahsne grfa awrsoto otr,ad ktshe, est hgaerp apeprnhoe xrogifmy e anuteesrleygd y t pwuesirce e sh eaacss- the hundreds of n1e0.w4 Peontednt-iaol fE-ncehrgay pte2r9 7problems,
tbateh ene kd uatosenenngdsda ortioonno s s ttshlotaerrneierdt ceslhn e,eog rtnsgr aytinh.t ssaWf to fhrecmeeatnn-, toS stmopa ltloes tt,h tIhNekatwoh dn hge1dd iiwr0c tchai.iho4 lvtdan hn iahetgl ya aepv Rstaoe irwv oease hennarroyb.akw lTell in irtt ptenotlhae o dha owttmo ertpiihdtn.lh elNe tytsie ropda,s eo mtlsef ndareoot.onn tteG m ebrko erahi cgnnologgawima er frfegasoa s owestotses i rtnfath drn e e edyqd uo haibotr sepaes,n r tvyvha eestrp iynoe enelcisdte t ilsane-t 3manadvn =py 0 rroepqouritrieo nstaulidtye,n ttos troea rseoans ouns iunsgi nrgea rla-wtioosr ld
itMtnorag un csekhfloi anrsoemtfit ciec td h eipbsno aetceerkngn eytiri angolty of ecmknaioennrt eigotbyinec. identical balls s 16ar8 yt. h pA|e ron.ow eAArueg rpgyf aei hsnsr seta eoe4prndp .s erprdoue xtneoidsm cr1aoet vqekelumyri r t.teh hsHie sl o esdawsisms ttdeao.notaceles ?ehniesr gsyp.eed affect the vutotal 2 dpahty4as,i caanld s eton saes.sess answers to see if they make
energy as the kangaroo takes another hop. The kangaroo’s peculiar B. A faster speed requires more total energy.
hopping gait is not very efficient at low speeds but is quite efficient C. The total energy is about the same for a fast speed and a1 slow
at high speeds. speed.
Figure P11.68 shows the energy cost of human and kangaroo 69. | A kangaroo hops 1 km. How does its speed affect the total
locomotion. The graph shows oxygen uptake (in mL/s) per kg of energy needed to cover this distance?
body mass, allowing a direct comparison between the two species. A. A faster speed requires less total energy.
Elastic Pot enBt. iAa fals tEer snpeeedr rgeqyuires more total energy.
2O.x0ygen uptake (mL/kg # s) Energy can als o beC . sTstpoheere etdo.dta li enn earg cy oism abpouret tshse esadm oe rfo er ax ftaesnt sdpeeedd asnpdr ai nslgow a s elastic (or spring) FIGURE 10.16 The force required to
11..86 Hruunmnianng, upsoitnegn atnia el xetenren r7ag0l. y fA| o.U ArAcst .ea r usWtpnoen eeisnd lg coo fawh 4nu lmm yfa/ isncn, odism ompuorrtee shesof ftwichie enm ts puthrcaihnn gaen.n Teeqhrugiasyl- m eixasst ses rtonrael df oirnc ea dsopersin wg obrky compress a spring is not constant.
11..42 on the spring, t ransBf. ehAror pirpnuinngng ie nknga neghraugrmoyoa .nt oi st hlees ss perfifincigen. tS tihnacn ea on nelqyu atlh-mea essl astic potential energy x = 0
1.0 Red kangaroo, of the spring is chanhgoipnpgin,g E kaqnugaartoioo.n 10.3 reads Spring in equilibrium
0.8 hopping C. A running human and an equal-mass hopping kangaroo have
00..64 71. | Aatb oauptp trhoex ismamatee leyf fwichieant cs∆yp.eUeds w=ouWld a human use half the (10.14) x
0.2 That is, we can finpdow oeru otf hano ewqu mal-mucashs kealnagsatroico mpoovtiengn atit athle e snameer gspye eids? stored in the spring by Fu
00 2 4 6 8 10 1c2Sapelecdu (mla/s)ting the 7a2m. A| o. uA3nt mtw /ohs aft w sp oeerdBk . d no4ee mse /dtsh ee d h ot popC icn. og5 m mmop/tsri oens so ft htDhee. s6k pamnrg/isanrogo.
FigURe p11.68 Oxygen uptake (a measure of energy uFsIeG URE 10.16 shobwecosm ae smporrien egff ibcieenint tgha cn othme rpurnenisnsge gda ibt oyf aa hhuamnand?. In ◀◀ SECtIoN 8.3 we found
per second) for a running human and a hopping ktahnagta rtohoe. force th e sAp.r i3n mg/ se x er tsB o. n5 tmh/es h an dC i.s 7F ms/s= - k ∆D. x 9 ( mH/sooke’s law), where ∆x is As x increases,
the displacement of the end of the spring from its equilibrium position and k is the so does F.
x
spring constant. In Figure 10.16 we have set the origin of our coordinate system at the
equiSlTiobpr iTuom Th pinoks AitniSowne.R TS he displacement from equilibrium ∆x is therefore equal to x,
CEhapxter PPrevAiew nStop dto TEhinkd: C. Thle wiofrk CEhri-astniSnda dtChoees i osEnp trhinen g hCfoasr tcwEei cie s aA tsh menann y- adtkoxm.s ,B soBy tNhei eawvoetraognem ’esn etrhEgiyr ddpe rl aawtioCm, ,t hanAed ftohlursc e tAhatP theP hlandi CexeArts tionS
4Bh22jkkttauahhii vnneeve ieejtlttlaoiiievndcctae eeli l  siinnbce nneetn’rhrresaegg rgnakyygsie ypfnia se setro nr tterierimetca n dr lnaeea s inia 1fnnveotdst3reor gmhcs y dkohe inieFednsser e lt i2ua thni7dnbcatio0td n e a dogsJntt.r e -a2Aroov7g7si0y t0k ta ,h J ttJs.ie’ ooe =Tsj nathahx2v ule0e asp tl0 jioaa nc,Jttv e. rcopeincitwlssanotisie inlsnhap ss aA lti,h’io efg hesattaheinsois erero mudem 2tsrtsknh ts7geh ip ep0ye soeexeo ,r rf J fEi,bi nhdi anfdoutttqrsfst,ahg o t une eemriad rstSttaetho hx tiipte etpontohah happtsu eneeuu allmt s mr ieos “-rh1elnpo dwaTvFdee0urrirhnps r.agafdoi5 t=nogoeuintl,nkhrmr” rege r Wc e1+ e,te alsu1nem ixd.ketr b4,,u.=ie xn :sen sT ro.Ctda hc tf bFbeQ.e or eneordTde m6clhf ace,dbaa e ss0ltt syuu r.heo.e a snseedt seehf i afr inse,gttrnoh p yowdred er mii si satv cn- th lareo0glatee. nn a tAfsft nwohfer-eiso rgrco k rhmwmoexrhedkf.re o iatTt’eoduthvmsthope e orpi ne fsetets quert eahie.bu stee ua e Hirnrsnl,ey, ip o o sbwpttthtwherpa hmiraeneue locavrmieovecb enrtpa,hrls aoeetitgats ifmenisolt oetis oewfpnsonme s,r iets csdo g ef fr i,aearsi ssa votaofseili omssnwntona,e -lg t hsteu ldiveinntgs w orld
esSes2an.t smoeHrpeoge wytdo ni cs aThtbatahonniuicgnatee k wa olsh1f ac1 tptr.h 1ayen:ro beCua o1 .hx ,Ic a.sn vCo tee ry aaitoocncu hep ega2ceya lt ?si tf eiCwt,ns ri waac n ehb eaaao ts2x p y muwosuupietc shgh l2e tewy0tafFn ioikbescaipJr revnlgt,r c hgetcyoheer g pw= eat p novomio te htra12c ace1t srckni aosoriotx ailentanhnc.lrsly ee-auT e fll rha towrSoTutpeCemt isso lR ethl pto r t2hb vF htTy.eoeo iHe iT rm.s =T .wwDhohiereoico0encs rrr keeaketk f aof1 t slibw1 dci nF.iy5poegof:n n uttAr=th eeswii,s rini bDkmntrhayx.gc tsi,T a o t ithct hhhpiheonoe eectme tlr fe heeafarbiaa cvvshisineneeoesndt re r ceta ra yfeitofgg sosiiecese ri g ev fnfnfioixvoocireeyerr d;occ wttrbeehh yae e u i htc nhsRoree ealeE d4tdr eaeqdar tn tieugcoolpoi an -olectdefwin oodmnoe pn1rcr0ele.ds.5s. tsBhieetc suapuarsient gitoh iesn s.
c10o knJ.c CeopmtpUaarinlg e cxraanem 1p alned 1cr3a.n1e3 2 , wBe fliondo cdra npe r2e hsass utwriece athne d Sctaoprd toio Tvhainskc 1u1.l6a: rA d, Dis. eThaes celoser the temperatures of the hot
CneSanptaaoeprrprrrgod otyxio wioo mTuviahtna tifegnsolcky r o ut71wfl51 ai%.tc2rhe: i eCtsdh t.ae riA asrenetsnse aetfhrrosgieree ymb s oie niddds,y ius nuoaets ot eh steh c eehrfemfmicaiilec enanlc eeierngse yar.gr Aey ltfshrooe, m ska ifmnoeeot.dic, ailsen sdasl stchooe ltrdrm ureae lsWt wiehesrna evtet =o orshgi earydsveF ,eti aonttlvhh gi“geva lt deethm aretkh o= tt”rewhe o epFoue rafttso.vfe)aigmd cAmixupeneceny=r tt a fcRttlhhuo4aerw∆en 12hsg p e e krba aetmxtht epcaulu.to sxmmLste pa=rer ke tcwe mats21inh lat l e hki bcn exeia .tn2uu w(isnItoetic ahla anrgteerdy i fr athdeiu asr taenrdy CaAlfc hmilulsecsl teendon
tdMeeinonnetee oaprr gggtothyynhse ewireomttf isi tach molb eoernuante icietoslhhrndog se nuyto r.apifin dn tethc.ee oeMr f lieio omgsrstp awoalgfan aitdqhnl ulefgs eec,.eh te ims itcraaln esnfoerrmgTtyoe hdi s fi iitsnrn atownd st fohtoherrrmkamt ea idlas ss ttttSeohotm orerpupe nsetd.orpa Tartuihsrni enpgs8pk o c%r i1eltso1se, s. sc7nmeu:rot Breiwmeaa. iInldlpn lii enrtfahfnelgilessoe r tcwsrehaegn astdeycht, ee, Rk i tirbfnnheee=f ert tiRih cg0e ieee a.larn9 aanes2tsrdopgRtr yir ∆ic t i.io ns bpT ptguir h.oas ,enDe t ss elrfieoosnesre qstmwa iueaesinedelre e idrencmgtenaoycee n rnr etguayssee is n Etchqreue raaatsdieiousns b b1yy0 .14
wCSenthhoepraigc pythto,e wrTy hh2oiic4unhk,’ il scl1 at1hln.e3e :at coCrrtnea. la Sktaaeibmn eoeptxuilceqt s e un1iien -ra gnyd o2f ahlalv teh et haeto smasm. eS athmeprmle a1l piso ttreanntsiafol remneerdg yin; ttoh ethree rims naol eennetr∆rogpyUy, msche=aanngi12neRg. kI ei nx4n t2∆threop poiyth= ienrcR creaf 4as es∆es,sp .efnergy
site three-dimensional images can be solved for the new pressure difference:
of the internal structure of the Just as in the case of gravitational potential energy, we have found an expression for On each stride, the tendon
KNIG9721_03_chap_11batn.hiroantedrded r yr o i3h.we4 e9sSai ndhtgh,o —awwt nais tuhasp rteepa nl ytohd seabis nl—ocgaoeirdrnoo dtutiiods- twsihnpee re ihqncauhgvia leenixbgtreeiun imsni o,U nns.e, Ainthnoe tor UbcTvsoh imieot suppesrrl eefcss.hs suAoerdieg∆c aenpdi foinif=srf, e stwrRRoteri f e n44es c te∆aectr ph emUie =usdfrs—=et( 0ei.t0n 9hRtc 2oaai r4Re tt s aii)tess4het ,e ∆ aU 3bp1pty/s 1io x0==/4i1n30= 1 t% 01 .0w40 : 0a .4t∆h o tAW ep Mamirneitay hit nh cttehao iinsnsp v ctrehihnneo igei cniest Slopseri snogm ien oyfo yuoru sr tmes3pet5c rhe Jta conhf ie cAesan,ls e s eyrtnogoreyui.rn grgyu naeb,a oycuohtu
cated by the arrow. flow. time your foot strikes the ground; this energy
mpruescIfsh u aar ess etdhcietfi fosetnre enonofcs eai sn b seahtrowtweerneyn, h batyhs e wn heaanrrtd opsw eorecfd e ntbhtyae g 8ne% amr,r uonswot tet hdne e saberlcloytoi oadns- achSaSnegUSesS i=nB er12caadkuixus2se rtehqeu ifrleosw a rlaatreg ed cehpaenngdes ionn ∆Rp(4 1,t o0e .vc1eo5nm) paeU nsssmaatell. ieosnf teyrraognuysr.f moLruemcckehdial nyini,c taaobl oueunntre e3rcg5oy%v we orhafe btnlhe ey todhueecrr mrfeoaaols te
riafneretrcaeerrneSyaco’sesne r ∆taAodp cki ucmeose urpRds ibtin flig ont ohctdero e b fallPsooeowo idstione f gulco ialowltme t’rhpsae et eens qsaQaum taieets i orfteoaonqt r,eru Ee?aitmlhl iadbeas eritincpiucr rume epnas ossc(eutahe rsaniesnn tugi dmateihldf iee-n. nge UrEgfigslseyio tm=r whooeuu.f0 rc sF ah,wto h sogrhep rrte rehpnhiaene et tgre shso rteode rt nni hess’oanphssnd leiab s 8oclw %soefhi odtl,odlh a aw esn p udndsrfi pe ifstsnrehtsia rentu nh garpcee re is eis immsxg as nuafutri sgrfxoteei mc ,i=d nat ihnc f0retf e)e rrareeesddenuu,c ccewtt iiwhooinincl lhii nbn i e sQrb Ua lAldadDorRaioAgunTpdIesC-.4 4x ltlsahetrngeed .ts Ocst rhineset dces,hat ocsarthboe ldrpei l naAagsnc tesh looialmfsl etteihsc ee tp enfonoetdroegotn,yn tt.i h aoTelf h etteehn neeted rlngoodynwo iiennsr
If we write Poiseuille’s equation as NotE ▶Because Usa dnedp veenryd sd aonng ethroeu ss.quare of the displacement x, Us is the same shperlipnignsg btoa cpkr oaps eyl oyuo up ufsohrw oaffr dth. eT hgirso urencdo avgeareind,
R 4 ∆p=8hpLQ wsphreinthge irs xs tirse ptcohseitdiv).e ◀(the spring is compressed as in Figure 10.16) or negative (the eenneerrggyy ryeoduu uceses ,t hinec aremaosiunngt yoof uinr teefrfnicaile cnhceym.ical
example 13.14 pressure drop along a capillary
In Example 13.11 we examined blood flow through a capillary. Using the numbers from
that example, calculate the pressure “drop” from one end of a capillary to the other.
KNIG9721_03_chap_10.indd P 2r9e7Pare Example 13.11 gives enough information to determine the flow rate through 16/08/13 2:04 PM
a capillary. We can then use Poiseuille’s equation to calculate the pressure difference
A01_KNIG7156_03_SE_WKT.indd 8 between the ends. 25/08/14 12:55 PM
Solve The measured volume flow rate leaving the heart was given as
5 L/min=8.3*10-5 m3/s. This flow is divided among all the capillaries, which we
found to number N=3*109. Thus the flow rate through each capillary is
Qcap=QhNeart=8.3*3*101-05 9m3/s=2.8*10-14m3/s
Solving Poiseuille’s equation for ∆p, we get
∆p=8h pLRQ 4cap=8(2.5*10-3 Pap#s()3(0*.01001- 6m m)()24.8*10-14m3/s)=2200 Pa
If we convert to mm of mercury, the units of blood pressure, the pressure drop across
the capillary is ∆p=16 mm Hg.
aSSeSS The average blood pressure provided by the heart (the average of the systolic
and diastolic pressures) is about 100 mm Hg. A physiology textbook will tell you that
the pressure has decreased to 35 mm by the time blood enters the capillaries, and it exits
from capillaries into the veins at 17 mm. Thus the pressure drop across the capillaries is
18 mm Hg. Our calculation, based on the laws of fluid flow and some simple estimates
of capillary size, is in almost perfect agreement with measured values.
KNIG9721_03_chap_13.indd 422 01/08/13 11:39 AM
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