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Serway_Physics_6th_Edition_Solutions - 1Physics and...

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Serway_Physics_6th_Edition_Solutions - 1Physics and...

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Unformatted text preview: 1Physics and MeasurementCHAPTER OUTLINE1.1ANSWERS TO QUESTIONSStandards of Length, Mass,and TimeMatter and Model-BuildingDensity and Atomic MassDimensional AnalysisConversion of UnitsEstimates and Order-ofMagnitude CalculationsSignificant FiguresAtomic clocks are based on electromagnetic waves which atomsemit. Also, pulsars are highly regular astronomical clocks.Q1.2Density varies with temperature and pressure. It would benecessary to measure both mass and volume very accurately inorder to use the density of water as a standard.People have different size hands. Defining the unit preciselywould be cumbersome.Q1.4(a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilogramsQ1.5(b) and (d). You cannot add or subtract quantities of differentdimension.Q1.61.7Q1.1Q1.31.21.31.41.51.6A dimensionally correct equation need not be true. Example:1 chimpanzee = 2 chimpanzee is dimensionally correct. If anequation is not dimensionally correct, it cannot be correct.Q1.7If I were a runner, I might walk or run 10 1 miles per day. Since I am a college professor, I walk about10 0 miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day onvacation.Q1.8On February 7, 2001, I am 55 years and 39 days old.55 yrF 365.25 d I + 39 d = 20 128 dFG 86 400 s IJ = 1.74 10GH 1 yr JKH 1d K9s ~ 10 9 s .Many college students are just approaching 1 Gs.Q1.9Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10.Q1.10The mass of the forty-six chapter textbook is on the order of 10 0 kg .Q1.11With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr.12Physics and MeasurementSOLUTIONS TO PROBLEMSSection 1.1Standards of Length, Mass, and TimeNo problems in this sectionSection 1.2P1.1Matter and Model-BuildingFrom the figure, we may see that the spacing between diagonal planes is half the distance betweendiagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from thePythagorean theorem, Ldiag = L2 + L2 . Thus, since the atoms are separated by a distanceL = 0.200 nm , the diagonal planes are separated bySection 1.3*P1.21 2L + L2 = 0.141 nm .2Density and Atomic MassModeling the Earth as a sphere, we find its volume as4 3 4 r = 6.37 10 6 m33ej3= 1.08 10 21 m 3 . Itsm 5.98 10 24 kg== 5.52 10 3 kg m3 . This value is intermediate between theV 1.08 10 21 m 3tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to3 000 kg m3 . The average density of the Earth is significantly higher, so higher-density materialmust be down below the surface.density is then =P1.3afbge jWith V = base area height V = r 2 h and ==afa = 2.15 10 kg m3393.mfor both. Then iron = 9.35 kg V andV19.3 10 3 kg / m3= 23.0 kg .= 9.35 kg7.86 10 3 kg / m3Let V represent the volume of the model, the same in = gold =P1.5F 10 mm If GH 1 m JK1 kgm=2 r h 19.5 mm 2 39.0 mm4*P1.4m, we haveVm goldVV = Vo Vi ==. Next, gold iron43 r2 r133e=m gold9.35 kgand m goldFGHjFG IJ eH Ke34 r2 r13m43, so m = V = r2 r13 =V33jj.IJKChapter 1P1.6344 3 r and the mass is m = V = r 3 . We divide this equation33for the larger sphere by the same equation for the smaller:For either sphere the volume is V = 4 r 3 3 r 3m=== 5.m s 4 rs3 3 rs3a fThen r = rs 3 5 = 4.50 cm 1.71 = 7.69 cm .P1.7Use 1 u = 1.66 10 24 g .F 1.66 10GH 1 uF 1.66 10= 55.9 uGH 1uF 1.66 10= 207 uGH 1u-24I = 6.64 10JKgIJK = 9.29 10gIJK = 3.44 10g24g .23g .22g .(a)(b)For Fe, m 0(c)*P1.8For He, m 0 = 4.00 uFor Pb, m 0(a)The mass of any sample is the number of atoms in the sample times the mass m 0 of oneatom: m = Nm 0 . The first assertion is that the mass of one aluminum atom is-2424m 0 = 27.0 u = 27.0 u 1.66 10 27 kg 1 u = 4.48 10 26 kg .Then the mass of 6.02 10 23 atoms ism = Nm 0 = 6.02 10 23 4.48 10 26 kg = 0.027 0 kg = 27.0 g .Thus the first assertion implies the second. Reasoning in reverse, the second assertion can bewritten m = Nm 0 .0.027 0 kg = 6.02 10 23 m 0 , so m 0 =0.027 kg6.02 10 23= 4.48 10 26 kg ,in agreement with the first assertion.(b)The general equation m = Nm 0 applied to one mole of any substance gives M g = NM u ,where M is the numerical value of the atomic mass. It divides out exactly for all substances,giving 1.000 000 0 10 3 kg = N 1.660 540 2 10 27 kg . With eight-digit data, we can be quitesure of the result to seven digits. For one mole the number of atoms isN=F 1 I 10GH 1.660 540 2 JK3 + 27= 6.022 137 10 23 .(c)The atomic mass of hydrogen is 1.008 0 u and that of oxygen is 15.999 u. The mass of onemolecule of H 2 O is 2 1.008 0 + 15.999 u = 18.0 u. Then the molar mass is 18.0 g .(d)For CO 2 we have 12.011 g + 2 15.999 g = 44.0 g as the mass of one mole.bgbg4Physics and MeasurementP1.91b gFGH 10kgg IJK = 4.5 10kg IJK = 3.27 10 kg .Mass of gold abraded: m = 3.80 g 3.35 g = 0.45 g = 0.45 gF 1.66 10GH 1 u27Each atom has mass m 0 = 197 u = 197 u34kg .25Now, m = N m 0 , and the number of atoms missing isN =m=m04.5 10 4 kg3.27 10 25 kg= 1.38 10 21 atoms .The rate of loss isNtNt=FGH1 yr1.38 10 21 atoms365.25 d50 yr= 8.72 10 11 atoms s .ejejP1.11(a)m = L3 = 7.86 g cm 3 5.00 10 6 cm(b)P1.10IJ FG 1 d IJ FG 1 h IJ FG 1 min IJK H 24 h K H 60 min K H 60 s KN=(a)3= 9.83 10 16 g = 9.83 10 19 kgThe cross-sectional area is9.83 10 19 kgm== 1.06 10 7 atomsm 0 55.9 u 1.66 10 27 kg 1 uejaf afafaA = 2 0.150 m 0.010 m + 0.340 m 0.010 m= 6.40 1032m .f.The volume of the beam isjaefV = AL = 6.40 10 3 m 2 1.50 m = 9.60 10 3 m3 .Thus, its mass iseFIG. P1.11je9.60 10 m j = 72.6 kg .F 1.66 10 kg I = 9.28 10The mass of one typical atom is m = a55.9 ufGH 1 u JK3m = V = 7.56 10 kg / m(b)333270m = Nm 0 and the number of atoms is N =26kg . Now72.6 kgm== 7.82 10 26 atoms .26m 0 9.28 10kgChapter 1P1.12(a)F 1.66 10GH 1 u27The mass of one molecule is m 0 = 18.0 ukgI = 2.99 10JK265kg . The number ofmolecules in the pail isN pail =(b)1.20 kgm== 4.02 10 25 molecules .m 0 2.99 10 26 kgSuppose that enough time has elapsed for thorough mixing of the hydrosphere.N both = N pailF m I = (4.02 10GH M JKpail25F 1.20 kg I ,GH 1.32 10 kg JKmolecules)total21orN both = 3.65 10 4 molecules .Section 1.4P1.13Dimensional AnalysisThe term x has dimensions of L, a has dimensions of LT 2 , and t has dimensions of T. Therefore, theequation x = ka m t n has dimensions ofeL = LT 2j aTfmnor L1 T 0 = Lm T n 2 m .The powers of L and T must be the same on each side of the equation. Therefore,L1 = Lm and m = 1 .Likewise, equating terms in T, we see that n 2m must equal 0. Thus, n = 2 . The value of k, adimensionless constant, cannot be obtained by dimensional analysis .*P1.14(a)Circumference has dimensions of L.(b)Volume has dimensions of L3 .(c)Area has dimensions of L2 .e jExpression (i) has dimension L L21/2= L2 , so this must be area (c).Expression (ii) has dimension L, so it is (a).Expression (iii) has dimension L L2 = L3 , so it is (b). Thus, (a) = ii; (b) = iii, (c) = i .e j6Physics and Measurement*P1.16(a)This is incorrect since the units of ax are m 2 s 2 , while the units of v are m s .(b)P1.15This is correct since the units of y are m, and cos kx is dimensionless if k is in m 1 .(a)a faFor a = kFrepresents the proportionality of acceleration to resultant force andmmthe inverse proportionality of acceleration to mass. If k has no dimensions, we havea = k(b)P1.17In units,M LT2=kg ms2FFLM L, F =, 2 =1.m TMT2, so 1 newton = 1 kg m s 2 .Inserting the proper units for everything except G,LM kg m OP = G kgNs Q m22Multiply both sides by mSection 1.5*P1.18.m32and divide by kg ; the units of G arekg s 2.Conversion of UnitsafafEach of the four walls has area 8.00 ft 12.0 ft = 96.0 ft 2 . Together, they have areae4 96.0 ft 2P1.1922mjFGH 3.128 ft IJK2= 35.7 m 2 .Apply the following conversion factors:1 in = 2.54 cm , 1 d = 86 400 s , 100 cm = 1 m , and 10 9 nm = 1 mFG 1H 32IJ b2.54 cm inge10 m cmje10K86 400 s day2in day9j=nm m9.19 nm s .This means the proteins are assembled at a rate of many layers of atoms each second!*P1.208.50 in 3 = 8.50 in 3FG 0.025 4 m IJH 1 in K3= 1.39 10 4 m 3Chapter 1P1.21Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we shouldexpect the area to be about A 30 m 50 m = 1 500 m 2 .afafCategorize: We model the lot as a perfect rectangle to use Area = Length Width. Use theconversion: 1 m = 3.281 ft .aAnalyze: A = LW = 100 ft1m1mf FGH 3.281 ft IJK a150 ftfFGH 3.281 ft IJK = 1 390 m2= 1.39 10 3 m 2 .Finalize: Our calculated result agrees reasonably well with our initial estimate and has the properunits of m 2 . Unit conversion is a common technique that is applied to many problems.P1.22(a)afafafV = 40.0 m 20.0 m 12.0 m = 9.60 10 3 m 333bgV = 9.60 10 m 3.28 ft 1 m(b)3= 3.39 10 5 ft 3The mass of the air isejejm = air V = 1.20 kg m3 9.60 10 3 m3 = 1.15 10 4 kg .The student must look up weight in the index to findejejFg = mg = 1.15 10 4 kg 9.80 m s 2 = 1.13 10 5 N .Converting to pounds,jbegFg = 1.13 10 5 N 1 lb 4.45 N = 2.54 10 4 lb .P1.23(a)Seven minutes is 420 seconds, so the rate isr=(b)30.0 gal= 7.14 10 2 gal s .420 sConverting gallons first to liters, then to m3 ,er = 7.14 10 2 gal sjFGH 3.1786 L IJK FGH 10 1 Lm IJKgal33r = 2.70 10 4 m3 s .(c)At that rate, to fill a 1-m3 tank would taket=F 1mGH 2.70 1034m3IF 1 h I =s J H 3 600 KKG J1.03 h .78Physics and Measurement*P1.24(a)(b)(c)(d)P1.25FG 1.609 km IJ = 560 km = 5.60 10 m = 5.60 10 cm .H 1 mi KF 0.304 8 m IJ = 491 m = 0.491 km = 4.91 10 cm .Height of Ribbon Falls = 1 612 ftGH 1 ft KF 0.304 8 m IJ = 6.19 km = 6.19 10 m = 6.19 10 cm .Height of Denali = 20 320 ftGH 1 ft KF 0.304 8 m IJ = 2.50 km = 2.50 10 m = 2.50 10 cm .Depth of Kings Canyon = 8 200 ftGH 1 ft K5Length of Mammoth Cave = 348 mi743535From Table 1.5, the density of lead is 1.13 10 4 kg m 3 , so we should expect our calculated value tobe close to this number. This density value tells us that lead is about 11 times denser than water,which agrees with our experience that lead sinks.Density is defined as mass per volume, in ==23.94 g2.10 cm 3F 1 kg I FG 100 cm IJGH 1 000 g JK H 1 m K3m. We must convert to SI units in the calculation.V= 1.14 10 4 kg m3At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Ourresult is indeed close to the expected value. Since the last reported significant digit is not certain, thedifference in the two values is probably due to measurement uncertainty and should not be aconcern. One important common-sense check on density values is that objects which sink in watermust have a density greater than 1 g cm 3 , and objects that float must be less dense than water.P1.26It is often useful to remember that the 1 600-m race at track and field events is approximately 1 milein length. To be precise, there are 1 609 meters in a mile. Thus, 1 acre is equal in area to1a1 acrefFGH 640mi IJK FGH 1 609 m IJKacresmi2*P1.27P1.28The weight flow rate is 1 200FGHton 2 000 lbhton2= 4.05 10 3 m 2 .IJ FG 1 h IJ FG 1 min IJ =K H 60 min K H 60 s K667 lb s .1 mi = 1 609 m = 1.609 km ; thus, to go from mph to km h , multiply by 1.609.(a)1 mi h = 1.609 km h(b)55 mi h = 88.5 km h(c)65 mi h = 104.6 km h . Thus, v = 16.1 km h .Chapter 19(a)F 6 10 $ I F 1 h I FG 1 day IJ F 1 yr I =GH 1 000 $ s JK GH 3 600 s JK H 24 h K GH 365 days JK(b)P1.29The circumference of the Earth at the equator is 2 6.378 10 3 m = 4.01 10 7 m . The length12190 yearsejof one dollar bill is 0.155 m so that the length of 6 trillion bills is 9.30 10 11 m. Thus, the6 trillion dollars would encircle the Earth9.30 10 11 m= 2.32 10 4 times .4.01 0 7 m1.99 10 30 kgmSun== 1.19 10 57 atomsm atom 1.67 10 27 kgP1.30N atoms =P1.31V = At so t =V 3.78 10 3 m 3== 1.51 10 4 m or 151 m2A25.0 mbafe13.0 acres 43 560 ft 2 acre1V = Bh =3373= 9.08 10 ft ,P1.32gj a481 ftfhoreV = 9.08 10 7 ft 3jFGH 2.83 110ft2m33IJKBFIG. P1.32= 2.57 10 6 m3P1.33*P1.34bgejbgFg = 2.50 tons block 2.00 10 6 blocks 2 000 lb ton = 1.00 10 10 lbsThe area covered by water isa fej a fa fej2A w = 0.70 AEarth = 0.70 4 REarth = 0.70 4 6.37 10 6 m2= 3.6 10 14 m 2 .The average depth of the water isafbgd = 2.3 miles 1 609 m l mile = 3.7 10 3 m .The volume of the water isejejV = A w d = 3.6 10 14 m 2 3.7 10 3 m = 1.3 10 18 m 3and the mass isejejm = V = 1 000 kg m3 1.3 10 18 m3 = 1.3 10 21 kg .10P1.35Physics and Measurement(a)d nucleus, scale = d nucleus, realeFdI = 2.40 10 m FG 300 ft IJ = 6.79 10jH 1.06 10 m KGH dJK eft jb304.8 mm 1 ft g = 2.07 mmatom, scale1510atom, reald nucleus, scale = 6.79 10 3(b)Vatom=Vnucleus34 ratom334 rnucleus3=FG rHrIJ = FG dK Hd3atomnucleusIJ = F 1.06 10K GH 2.40 103atomnucleus1015mmIJK3= 8.62 10 13 times as large*P1.36scale distancebetweenP1.37=FG real IJ FG scale IJ = e4.0 10H distanceK H factorK13mjFGH 71..0410 m IJK =103km200 km9The scale factor used in the dinner plate model is0.25 mS=51.0 10 lightyears= 2.5 10 6 m lightyears .The distance to Andromeda in the scale model will beejejDscale = Dactual S = 2.0 10 6 lightyears 2.5 10 6 m lightyears = 5.0 m .FGHF e6.37 10 mjb100 cm mg I=GGH 1.74 10 cm JJK = 13.4IJ = FG e6.37 10 mjb100 cm mg IJ = 49.1K GH 1.74 10 cm JKP1.39(a)(b)P1.382AEarth 4 rEarthr== Earth2A Moon 4 rMoonrMoonVEarth=VMoon34 rEarth334 rMoon3Fr=GHrIJK8Moon363Earth2628To balance, m Fe = m Al or FeVFe = Al VAl FeFG 4IJ rH 3KFe3FG 4 IJ rH 3KFG IJ = a2.00 cmfFG 7.86 IJH 2.70 KH K= AlAl1/3rAl = rFeFeAl31/3= 2.86 cm .3ft , orChapter 1P1.4011The mass of each sphere ism Al = Al VAl =4 Al rAl 33m Fe = FeVFe =4 Fe rFe 3.3andSetting these masses equal,4 Al rAl 3 4 Fe rFe 3=and rAl = rFe 3 Fe .33 AlSection 1.6P1.41Estimates and Order-of-Magnitude CalculationsModel the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ballas a sphere of diameter 0.038 m. The volume of the room is 4 4 3 = 48 m3 , while the volume ofone ball isFGH4 0.038 m32IJK3= 2.87 10 5 m3 .48~ 10 6 ping-pong balls in the room.2.87 10 5As an aside, the actual number is smaller than this because there will be a lot of space in theroom that cannot be covered by balls. In fact, even in the best arrangement, the so-called best1packing fraction is 2 = 0.74 so that at least 26% of the space will be empty. Therefore, the6above estimate reduces to 1.67 10 6 0.740 ~ 10 6 .Therefore, one can fit aboutP1.42A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus,bgbgbgthe tire would make 50 000 mi 5 280 ft mi 1 rev 8 ft = 3 10 7 rev ~ 10 7 rev .P1.43In order to reasonably carry on photosynthesis, we might expect a blade of grass to require at least1in 2 = 43 10 5 ft 2 . Since 1 acre = 43 560 ft 2 , the number of blades of grass to be expected on a16quarter-acre plot of land is aboutn=afej0.25 acre 43 560 ft 2 acretotal area== 2.5 10 7 blades ~ 10 7 blades .area per blade43 10 5 ft 2 blade12P1.44Physics and MeasurementA typical raindrop is spherical and might have a radius of about 0.1 inch. Its volume is thenapproximately 4 10 3 in 3 . Since 1 acre = 43 560 ft 2 , the volume of water required to cover it to adepth of 1 inch isa1 acrefa1 inchf = a1 acre infFGH 431560 ftacre2I F 144 in I 6.3 10JK GH 1 ft JK226in 3 .The number of raindrops required isn=*P1.45volume of water required 6.3 10 6 in 3== 1.6 10 9 ~ 10 9 .volume of a single drop4 10 3 in 3Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is thena fafafafV = 0.5 1.3 m 0.5 m 0.3 m = 0.10 m3 .The mass of this volume of water isejejm water = water V = 1 000 kg m3 0.10 m3 = 100 kg ~ 10 2 kg .Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. Themass of copper required isejejmcopper = copper V = 8 920 kg m3 0.10 m3 = 892 kg ~ 10 3 kg .P1.46The typical person probably drinks 2 to 3 soft drinks daily. Perhaps half of these were in aluminumcans. Thus, we will estimate 1 aluminum can disposal per person per day. In the U.S. there are ~250million people, and 365 days in a year, soe250 106jbgcans day 365 days year 10 11 cansare thrown away or recycled each year. Guessing that each can weighs around 1 10 of an ounce, weestimate this representse10P1.4711jbgbgbgcans 0.1 oz can 1 lb 16 oz 1 ton 2 000 lb 3.1 10 5 tons year . ~ 10 5 tonsAssume: Total population = 10 7 ; one out of every 100 people has a piano; one tuner can serve about1 000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year).Therefore,# tuners ~F 1 tuner I F 1 piano I (10GH 1 000 pianos JK GH 100 people JK7people) = 100 .Chapter 1Section 1.7*P1.4813Significant FiguresMETHOD ONEWe treat the best value with its uncertainty as a binomial 21.3 0.2 cm 9.8 0.1 cm ,af aA = 21.3a9.8f 21.3a0.1f 0.2a9.8 f a0.2 fa0.1f cmf2.The first term gives the best value of the area. The cross terms add together to give the uncertaintyand the fourth term is negligible.A = 209 cm 2 4 cm 2 .METHOD TWOWe add the fractional uncertainties in the data.a00f FGH 21..23 + 9..1 IJK = 209 cm8faA = 21.3 cm 9.8 cm P1.49af r 2 = 10.5 m 0.2 m(a)2 2% = 209 cm 2 4 cm 22= (10.5 m) 2 2(10.5 m)(0.2 m) + ( 0.2 m) 2= 346 m 2 13 m 2af2 r = 2 10.5 m 0.2 m = 66.0 m 1.3 m(b)3P1.50(a)P1.514af am = a1.85 0.02f kg(b)3(c)(d)fr = 6.50 0. 20 cm = 6.50 0.20 10 2 m=mc h r433also, m 3 r=+.mrIn other words, the percentages of uncertainty are cumulative. Therefore,a f 0.02 3 0.20=+= 0.103 ,6.50 1.85=anda1.85c h e6.5 1043f2jm3= 1.61 10 3 kg m 3af = 1.61 0.17 10 3 kg m3 = 1.6 0.2 10 3 kg m3 .214P1.52Physics and Measurement756.??37.2?0.83+ 2.5?796.53 = 797//(b)0.003 2 2 s.f. 356.3 4 s.f. = 1.140 16 = 2 s.f.(c)*P1.53(a)5.620 4 s.f. > 4 s.f. = 17.656 = 4 s.f.afaaf aP1.55afaff1.117.66We work to nine significant digits:1 yr = 1 yrP1.54fF 365.242 199 d I FG 24 h IJ FG 60 min IJ FG 60 s IJ =GH 1 yr JK H 1 d K H 1 h K H 1 min K31 556 926.0 s .The distance around is 38.44 m + 19.5 m + 38.44 m + 19.5 m = 115.88 m , but this answer must berounded to 115.9 m because the distance 19.5 m carries information to only one place past thedecimal. 115.9 mbV = 2V1 + 2V2 = 2 V1 + V2gafa fa fV = a10.0 mfa1.0 mfa0.090 mf = 0.900 mV = 2e1.70 m + 0.900 m j = 5.2 mU0.12 m== 0.0063 |19.0 m| Vw0.01 m| = 0.006 + 0.010 + 0.011 = 0.027 === 0.010 Vw1.0 m|Vt0.1 cm|== 0.011 |t9.0 cmWV1 = 17.0 m + 1.0 m + 1.0 m 1.0 m 0.09 m = 1.70 m 332333FIG. P1.551113%111Additional ProblemsP1.56It is desired to find the distance x such that1 000 mx=100 mx(i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x). Thus, it is seen thatafbgx 2 = 100 m 1 000 m = 1.00 10 5 m 2and thereforex = 1.00 10 5 m 2 = 316 m .Chapter 1*P1.57Consider one cubic meter of gold. Its mass from Table 1.5 is 19 300 kg. One atom of gold has massam 0 = 197 ufFGH 1.66 110u27kgI = 3.27 10JK25kg .So, the number of atoms in the cube isN=19 300 kg3.27 10 25 kg= 5.90 10 28 .The imagined cubical volume of each atom isd3 =1 m3= 1.69 10 29 m 3 .5.90 10 28Sod = 2.57 10 10 m .P1.58A totalP1.59F IVa fe j FG V IJ e A j = GG V JJ e4 r jH KH KF 3V IJ = 3FG 30.0 10 m IJ = 4.50 m=GH r K H 2.00 10 m KAtotal = N A drop =total6total2total4 r 33dropdrop325One month isbgbgbg1 mo = 30 day 24 h day 3 600 s h = 2.592 10 6 s .Applying units to the equation,ej ejV = 1.50 Mft 3 mo t + 0.008 00 Mft 3 mo 2 t 2 .Since 1 Mft 3 = 10 6 ft 3 ,ej ejV = 1.50 10 6 ft 3 mo t + 0.008 00 10 6 ft 3 mo 2 t 2 .Converting months to seconds,V=e1.50 10 6 ft 3 mo2.592 10 6 s moj et+j0.008 00 10 6 ft 3 mo 2e2.592 10Thus, V [ft 3 ] = 0.579 ft 3 s t + 1.19 10 9 ft 3 s 2 t 2 .6s moj2t2.1516Physics and MeasurementafafP1.61 (deg) (rad)tan sin difference15.020.025.024.024.424.524.624.7P1.600.2620.3490.4360.4190.4260.4280.4290.4310.2680.3640.4660.4450.4540.4560.4580.4600.2590.3420.4230.4070.4130.4150.4160.4183.47%6.43%10.2%9.34%9.81%9.87%9.98%10.1%24.62 r = 15.0 mr = 2.39 mh= tan 55.0rh = 2.39 m tan( 55.0 ) = 3.41 mafh55rFIG. P1.61*P1.62Let d represent the diameter of the coin and h its thickness. The mass of the gold ism = V = At = F 2 dGH 42IJK+ dh twhere t is the thickness of the plating.LM a2.41fNM 4m = 19.3 22a fa+ 2.41 0.178= 0.003 64 gramsfOPPe0.18 10 jQ4cost = 0.003 64 grams $10 gram = $0.036 4 = 3.64 centsThis is negligible compared to $4.98.P1.63The actual number of seconds in a year isb86 400 s daygb365.25 day yrg = 31 557 600 s yr .The percent error in the approximation ise 107j bs yr 31 557 600 s yr31 557 600 s yrg 100% =0.449% .Chapter 1P1.64V = L3 , A = L2 , h = L(a)V = A hL3 = L2 L = L3 . Thus, the equation is dimensionally correct.(b)eVrectangular object =P1.65(a)jwh = a w fh = Ah , whereVcylinder = R 2 h = R 2 h = Ah , where A = R 2The speed of rise may be found fromv=(b)aVol rate of flowf = 16.5 cm(Area:(a)a D24 )3 6 .30 cm4fs2= 0.529 cm s .Likewise, at a 1.35 cm diameter,v=P1.66A= w16.5 cm 3 sa 1.35 cm4f2= 11.5 cm s .1 cubic meter of water has a massejejejm = V = 1.00 10 3 kg cm3 1.00 m 3 10 2 cm m(b)3= 1 000 kgAs a rough calculation, we treat each item as if it were 100% water.cell:m = V = FG 4 R IJ = FG 1 D IJ = e1 000 kg m jFG 1 IJ e1.0 10H3 K H6 KH6 K3336jm3= 5.2 10 16 kgkidney: m = V = FG 4 R IJ = e1.00 10H3 K33kg cm34jFGH 3 IJK ( 4.0 cm)3= 0.27 kgfly:m=FG D hIJ = e1 10H4 K23kg cm 3= 1.3 10 5 kgP1.67V20 mpg =(10 8 cars)(10 4 mi yr )= 5.0 10 10 gal yr20 mi galV25 mpg =(10 8 cars)(10 4 mi yr )= 4.0 10 10 gal yr25 mi galFuel saved = V25 mpg V20 mpg = 1.0 10 10 gal yrjFGH IJK a2.0 mmf a4.0 mmfe10421jcm mm31718P1.68P1.69Physics and MeasurementFGHIFJK GHIFJK GHIFJK GHI FGJK HIJ FGKHIJKfurlongs220 yd0.914 4 m 1 fortnight 1 day1 hr= 8.32 10 4 m sfortnight 1 furlong1 yd14 days24 hrs 3 600 sThis speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps a sloth.v = 5.00The volume of the galaxy isej e10 r 2 t = 10 21 m219jm ~ 10 61 m3 .If the distance between stars is 4 10 16 m , then there is one star in a volume on the order ofe4 10The number of stars is aboutP1.7010503m starThe density of each material is =Al:Cu:Brass:Sn:Fe:P1.7110 61 m 3(a)(b)=ab4 51.5 ggf a3.75 cmf4b56.3 g g== a1.23 cmf a5.06 cmf4b94.4 g g== a1.54 cmf a5.69 cmf4b69.1 g g== a1.75 cmf a3.74 cmf4b 216.1 g g== a1.89 cmf a9.77 cmf 2.52 cmj29.3628.9127.682~ 10 11 stars .7.88gcm3gcm3FG g IJ isH cm KF g IJ isThe tabulated value G 8.92H cm KThe tabulated value 2.70This would takee2% smaller.35% smaller.cm3gcm3gcm3FGHThe tabulated value 7.863.16 10 7 s yr34 3 4 r = 5.00 10 7 m = 5.24 10 19 m3331 m3== 1.91 10 18 micrometeorites5.24 10 19 m3Vmm =3gb3 600 s hrgb24 hr daygb365.25 days yrg =VcubeVmm3m ~ 10 50 m 3 .4mmm.==V r 2h D2h= 2.75216j1.91 10 18 micrometeorites3.16 10 7 micrometeorites yr= 6.05 10 10 yr .gcm3IJ isK0.3% smaller.Chapter 1ANSWERS TO EVEN PROBLEMS5.52 10 3 kg m3 , between the densitiesof aluminum and iron, and greater thanthe densities of surface rocks.P1.341.3 10 21 kgP1.36200 kmP1.423.0 kgP1.38(a) 13.4; (b) 49.1P1.67.69 cmP1.8(a) and (b) see the solution,N A = 6.022 137 10 23 ; (c) 18.0 g;(d) 44.0 gP1.2P1.40Al(a)P1.18P1.20534(a) 3.39 10 ft ; (b) 2.54 10 lb5115.9 m316 m4.50 m 2P1.601.39 10 4 m3(a) 797; (b) 1.1; (c) 17.66P1.5835.7 m 2M L; (b) 1 newton = 1 kg m s 2T2(a) 3; (b) 4; (c) 3; (d) 2P1.56P1.16a209 4f cmP1.52(a) ii; (b) iii; (c) i~ 10 11 cans; ~ 10 5 tonsP1.54P1.14~ 10 9 raindropsP1.50(a) 4.02 10 25 molecules;(b) 3.65 10 4 moleculesP1.44P1.48P1.12~ 10 7 revP1.46(a) 9.83 10 16 g ; (b) 1.06 10 7 atomsP1.2413FeP1.42P1.10P1.22rAl = rFeFG IJH Ksee the solution; 24.6P1.623.64 cents ; noP1.64see the solutionP1.66(a) 1 000 kg; (b) 5.2 10 16 kg ; 0. 27 kg ;27(a) 560 km = 5.60 10 m = 5.60 10 cm ;(b) 491 m = 0.491 km = 4.91 10 4 cm ;(c) 6.19 km = 6.19 10 3 m = 6.19 10 5 cm ;(d) 2.50 km = 2.50 10 3 m = 2.50 10 5 cmP1.264.05 10 3 m 2P1.28(a) 1 mi h = 1.609 km h ; (b) 88.5 km h ;1.3 10 5 kg(c) 16.1 km hP1.688.32 10 4 m s ; a snailP1.301.19 10 57 atomsP1.70see the solutionP1.322.57 10 6 m3192Motion in One DimensionCHAPTER OUTLINE2.12.22.32.42.52.62.7Position, Velocity, andSpeedInstantaneous Velocity andSpeedAccelerationMotion DiagramsOne-Dimensional Motionwith Constant AccelerationFreely Falling ObjectsKinematic EquationsDerived from CalculusANSWERS TO QUESTIONSQ2.1If I count 5.0 s between lightning and thunder, the sound hastraveled 331 m s 5.0 s = 1.7 km . The transit time for the lightis smaller bybga f3.00 10 8 m s= 9.06 10 5 times,331 m sso it is negligible in comparison.Q2.2Yes. Yes, if the particle winds up in the +x region at the end.Q2.3Zero.Q2.4Yes. Yes.Q2.5No. Consider a sprinter running a straight-line race. His average velocity would simply be thelength of the race divided by the time it took for him to complete the race. If he stops along the wayto tie his shoe, then his instantaneous velocity at that point would be zero.Q2.6We assume the object moves along a straight line. If its averagexvelocity is zero, then the displacement must be zero over the timeinterval, according to Equation 2.2. The object might be stationarythroughout the interval. If it is moving to the right at first, it mustlater move to the left to return to its starting point. Its velocity mustbe zero as it turns around. The graph of the motion shown to theright represents such motion, as the initial and final positions arethe same. In an x vs. t graph, the instantaneous velocity at any timet is the slope of the curve at that point. At t 0 in the graph, the slopeof the curve is zero, and thus the instantaneous velocity at that timeis also zero.t0tFIG. Q2.6Q2.7Yes. If the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, thevelocity of the particle is unchanging, or is a constant.2122Motion in One DimensionafQ2.8Yes. If you drop a doughnut from rest v = 0 , then its acceleration is not zero. A commonmisconception is that immediately after the doughnut is released, both the velocity and accelerationare zero. If the acceleration were zero, then the velocity would not change, leaving the doughnutfloating at rest in mid-air.Q2.9No: Car A might have greater acceleration than B, but they might both have zero acceleration, orotherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in therecent past.Q2.10Yes. Consider throwing a ball straight up. As the ball goes up, itsvvelocity is upward v > 0 , and its acceleration is directed downv0a < 0 . A graph of v vs. t for this situation would look like the figureto the right. The acceleration is the slope of a v vs. t graph, and isalways negative in this case, even when the velocity is positive.aa fftFIG. Q2.10Q2.11(a)Accelerating East(b)Braking East(c)Cruising East(d)Braking West(e)Accelerating West(f)Cruising West(g)Stopped but starting to move East(h)Stopped but starting to move WestQ2.12No. Constant acceleration only. Yes. Zero is a constant.Q2.13The position does depend on the origin of the coordinate system. Assume that the cliff is 20 m tall,and that the stone reaches a maximum height of 10 m above the top of the cliff. If the origin is takenas the top of the cliff, then the maximum height reached by the stone would be 10 m. If the origin istaken as the bottom of the cliff, then the maximum height would be 30 m.The velocity is independent of the origin. Since the change in position is used to calculate theinstantaneous velocity in Equation 2.5, the choice of origin is arbitrary.Q2.14Once the objects leave the hand, both are in free fall, and both experience the same downwardacceleration equal to the free-fall acceleration, g.Q2.15They are the same. After the first ball reaches its apex and falls back downward past the student, itwill have a downward velocity equal to vi . This velocity is the same as the velocity of the secondball, so after they fall through equal heights their impact speeds will also be the same.Q2.16With h =1 2gt ,2af12g 0.707t . The time is later than 0.5t.2(a)0.5 h =(b)The distance fallen is 0.25 h =a f12g 0.5t . The elevation is 0.75h, greater than 0.5h.2Chapter 2Q2.17Above. Your ball has zero initial speed and smaller average speed during the time of flight to thepassing point.SOLUTIONS TO PROBLEMSSection 2.1Position, Velocity, and SpeedP2.1v = 2.30 m s(a)(b)x 57.5 m 9.20 m== 16.1 m st3.00 s(c)*P2.2v=v=x 57.5 m 0 m== 11.5 m st5.00 s(a)v=x 20 ft1m=t 1 yr 3.281 ftFGHIJ FG 1 yr IJ = 2 10 m s or in particularly windy timesK H 3.156 10 s Kx 100 ft F 1 m I F=v=GH 3.281 ft JK GH 3.1561 yr s IJK = 1 10 m s .t1 yr10777(b)The time required must have beent =FGH3 000 mi 1 609 mx=v 10 mm yr 1 miv=x 10 m== 5 mst2sv=5m= 1.2 m s4s(c)v=x 2 x1 5 m 10 m== 2.5 m s4 s2 st 2 t1(d)v=x 2 x 1 5 m 5 m== 3.3 m s7 s4 st 2 t1(e)P2.4(a)(b)P2.36v=x 2 x1 0 0== 0 ms80t 2 t1x = 10t 2 : Forafxamfts= 2.02.13.0=44.19040(a)v=x 50 m== 50.0 m st 1.0 s(b)v=x 4.1 m== 41.0 m st0.1 sIJ FG 10 mm IJ =KH 1 m K35 10 8 yr .2324P2.5Motion in One Dimension(a)Let d represent the distance between A and B. Let t1 be the time for which the walker hasdthe higher speed in 5.00 m s = . Let t 2 represent the longer time for the return trip int1ddd3.00 m s = . Then the times are t1 =and t 2 =. The average speedt25.00 m s3.00 m sis:bv=v=(b)Section 2.2P2.6(a)e2 15.0 m 2 s 28.00 m sj=bbd+d=+ 3 .00d m sd5.00 m sg2db8.00 m sgdg e15.0 m s jg b23.75 m sInstantaneous Velocity and SpeedeThus, at t = 3.00 s: x = e3.00 m s ja3.00 sf =jAt any time, t, the position is given by x = 3.00 m s 2 t 2 .22ijae27.0 m .f2At t f = 3.00 s + t : x f = 3.00 m s 2 3.00 s + t , orbja fg ex f = 27.0 m + 18.0 m s t + 3.00 m s 2 t(c)t 0(a)F x x I = lim e18.0 m s + e3.00 m s jtj =GH t JKfi2t 0x f xit f ti=a2.0 8.0f m = 6.0 m =a4 1.5f s 2.5 s18.0 m s .2.4 m sThe slope of the tangent line is found from points C andD. tC = 1.0 s, x C = 9.5 m and t D = 3.5 s, x D = 0 ,bgbgv 3.8 m s .(c).at ti = 1.5 s , x i = 8.0 m (Point A)at t f = 4.0 s , x f = 2.0 m (Point B)v=(b)2The instantaneous velocity at t = 3.00 s is:v = limP2.72She starts and finishes at the same point A. With total displacement = 0, average velocity= 0 .i(b)Total distance=Total timegThe velocity is zero when x is a minimum. This is at t 4 s .FIG. P2.7Chapter 2P2.8(a)(b)58 m2.5 s54 mAt t = 4.0 s, the slope is v 3s49 mAt t = 3.0 s, the slope is v 3.4 s36 mAt t = 2.0 s , the slope is v 4.0 sAt t = 5.0 s, the slope is v 23 m s .18 m s .14 m s .9.0 m s .v 23 m s 4.6 m s 2t5.0 s(c)(d)(a)v=v=(c)v=(d)*P2.10Initial velocity of the car was zero .(b)P2.9a=v=(5 0 ) m(1 0) s= 5 ms(5 10) m(4 2) s= 2.5 m s(5 m 5 m)(5 s 4 s)0 (5 m)(8 s 7 s )= 0= +5 m sFIG. P2.9Once it resumes the race, the hare will run for a time oft=x f xivx=1 000 m 800 m= 25 s .8 msIn this time, the tortoise can crawl a distanceafx f xi = 0.2 m s ( 25 s)= 5.00 m .2526Motion in One DimensionSection 2.3P2.11AccelerationChoose the positive direction to be the outward direction, perpendicular to the wall.v f = vi + at : a =P2.12(a)afv 22.0 m s 25.0 m s== 1.3410 4 m s 2 .t3.50 103 sAcceleration is constant over the first ten seconds, so at the end,chv f = vi + at = 0 + 2.00 m s 2 (10.0 s)= 20.0 m s .Then a = 0 so v is constant from t = 10.0 s to t = 15.0 s. And over the last five seconds thevelocity changes tochv f = vi + at = 20.0 m s + 3.00 m s 2 (5.00 s)= 5.00 m s .(b)In the first ten seconds,1 212at = 0 + 0 + 2.00 m s 2 (10.0 s) = 100 m .22cx f = x i + vi t +hOver the next five seconds the position changes tox f = xi + vi t +af1 2at = 100 m + 20.0 m s (5.00 s)+ 0 = 200 m .2And at t = 20.0 s ,x f = x i + vi t +*P2.13(a)af1 212at = 200 m + 20.0 m s (5.00 s)+ 3.00 m s 2 (5.00 s) = 262 m .22chdistance traveled. During the firsttquarter mile segment, Secretariats average speed wasThe average speed during a time interval t is v =v1 =0.250 mi 1 320 ft== 52.4 ft s25.2 s25.2 sb35.6 mi hg .During the second quarter mile segment,v2 =1 320 ft= 55.0 ft s24.0 sb37.4 mi hg .For the third quarter mile of the race,v3 =1 320 ft= 55.5 ft s23.8 sb37.7 mi hg ,and during the final quarter mile,v4 =continued on next page1 320 ft= 57.4 ft s23.0 sb39.0 mi hg .Chapter 2(b)27Assuming that v f = v 4 and recognizing that vi = 0 , the average acceleration during the racewasv f via=P2.14(a)=total elapsed time57.4 ft s 0= 0.598 ft s 2 .( 25. 2 + 24.0 + 23.8 + 23.0) sAcceleration is the slope of the graph of v vs t.a (m/s2)2.0For 0 < t < 5.00 s, a = 0 .1.6For 15.0 s < t < 20.0 s , a = 0 .For 5.0 s < t < 15.0 s , a =a=v f vit f ti1.0.8.00 (8.00)15.0 5.000.0= 1.60 m s 2t (s)05a=v f vit f ti(i)For 5.00 s < t < 15.0 s , ti = 5.00 s , vi = 8.00 m s ,t f = 15.0 sv f = 8.00 m sa=(ii)x = 2.00 + 3.00t t 2 , v =At t = 3.00 s :v f vit f ti=af8.00 8.00= 1.60 m s 2 .15.0 5.00ti = 0 , vi = 8.00 m s , t f = 20.0 s , v f = 8.00 m sa=P2.1515FIG. P2.14We can plot a(t ) as shown.(b)10v f vit f ti=8.00 (8.00)20.0 0dxdv= 3.00 2.00t , a == 2.00dtdt(a)x = ( 2.00 + 9.00 9.00) m = 2.00 m(b)v = (3.00 6.00) m s = 3.00 m s(c)a = 2.00 m s 2= 0.800 m s 22028P2.16Motion in One Dimension(a)2At t = 2.00 s , x = 3.00( 2.00) 2.00( 2.00)+ 3.00 m = 11.0 m.a fAt t = 3.00 s , x = 3.00 9.002a f 2.00 3.00 + 3.00 m = 24.0 msov=(b)x 24.0 m 11.0 m= 13.0 m s .=3.00 s 2.00 stAt all times the instantaneous velocity isv=d3.00t 2 2.00t + 3.00 = (6.00t 2.00) m sdtchAt t = 2.00 s , v = 6.00( 2.00) 2.00 m s = 10.0 m s .At t = 3.00 s , v = 6.00(3.00) 2.00 m s = 16.0 m s .v 16.0 m s 10.0 m s== 6.00 m s 2t3.00 s 2.00 s(c)(d)At all times a =(a)a=(b)Maximum positive acceleration is at t = 3 s, and is approximately 2 m s 2 .(c)a = 0 , at t = 6 s , and also for t > 10 s .(d)P2.17a=Maximum negative acceleration is at t = 8 s, and is approximately 1.5 m s 2 .Section 2.4P2.18d(6.00 2.00)= 6.00 m s 2 . (This includes both t = 2.00 s and t = 3.00 s ).dtv 8.00 m s== 1.3 m s 2t6.00 sMotion Diagrams(a)(b)(c)(d)(e)continued on next pageChapter 2(f)Section 2.5P2.1929One way of phrasing the answer: The spacing of the successive positions would changewith less regularity.Another way: The object would move with some combination of the kinds of motion shownin (a) through (e). Within one drawing, the accelerations vectors would vary in magnitudeand direction.One-Dimensional Motion with Constant AccelerationcFrom v 2 = vi2 + 2 ax , we have 10.97 10 3 m sfh2= 0 + 2 a( 220 m) , so that a = 2.7410 5 m s 2which is a = 2.79 10 4 times g .P2.20x f xi =(b)P2.21(a)a=v f vitaf11vi + v f t becomes 40 m = vi + 2.80 m s (8.50 s) which yields vi = 6.61 m s .22c=h2.80 m s 6.61 m s= 0.448 m s 28.50 sGiven vi = 12.0 cm s when x i = 3.00 cm(t = 0) , and at t = 2.00 s , x f = 5.00 cm ,11 22at : 5.00 3.00 = 12.0( 2.00)+ a( 2.00)2232.0a == 16.0 cm s 2 .8.00 = 24.0 + 2 a2x f x i = vi t +*P2.22(a)Let i be the state of moving at 60 mi h and f be at restd0 = b60 mi hg22v xf = v xi + 2 a x x f xiax(b)Similarly,mifFGH 5 1280 ft IJK3 600 mi F 5 280 ft I F 1 h I=GJGJ = 21.8 mi h s242 h H 1 mi K H 3 600 s KF 1 609 m IJ FG 1 h IJ = 9.75 m s= 21.8 mi h s GH 1 mi K H 3 600 s K2a+ 2 a x 121 ft 02b0 = 80 mi hax = (c)ibg26 400 5 280ba+ 2 a x 211 ft 0422 3 600gg2.fmi h s = 22.2 mi h s = 9.94 m s 2 .Let i be moving at 80 mi h and f be moving at 60 mi h .d ib60 mi hg = b80 mi hg + 2a a211 ft 121 ftf2 800b5 280 ga =mi h s = 22.8 mi h s = 10.2 m s2a90fb3 600g22v xf = v xi + 2 a x x f x i22xx2.30*P2.23Motion in One Dimension(a)Choose the initial point where the pilot reduces the throttle and the final point where theboat passes the buoy:x i = 0 , x f = 100 m , v xi = 30 m s , v xf = ?, a x = 3.5 m s 2 , t = ?x f = xi + v xi t +1axt 2 :2af100 m = 0 + 30 m s t +13.5 m s 2 t 22chc1.75 m s ht a30 m sft + 100 m = 0 .22We use the quadratic formula:t=t=cb b 2 4ac2ah30 m s 900 m 2 s 2 4 1.75 m s 2 (100 m)c2 1.75 m s2h=30 m s 14.1 m s3.5 m s 2= 12.6 s or 4.53 s .The smaller value is the physical answer. If the boat kept moving with the same acceleration,it would stop and move backward, then gain speed, and pass the buoy again at 12.6 s.ej(b)P2.24v xf = v xi + a x t = 30 m s 3.5 m s 2 4.53 s = 14.1 m s(a)Total displacement = area under the v , t curve from t = 0to 50 s.a fbbga f bga fgaf150 m s 15 s + 50 m s 40 15 s21+ 50 m s 10 s2x = 1 875 mx =(b)From t = 10 s to t = 40 s , displacement is(c)bga f bga f150 m s + 33 m s 5 s + 50 m s 25 s = 1 457 m .2v (50 0) m s0 t 15 s : a1 === 3.3 m s 2t15 s 015 s < t < 40 s : a 2 = 0x =40 s t 50 s : a 3 =continued on next pagev (0 50) m s== 5.0 m s 2t50 s 40 sFIG. P2.24Chapter 2(d)11a1 t 2 = 3.3 m s 2 t 2 or x1 = 1.67 m s 2 t 222chch(i)x1 = 0 +(ii)1x 2 = (15 s) 50 m s 0 + 50 m s (t 15 s) or x 2 = 50 m s t 375 m2(iii)For 40 s t 50 s ,ax3 =fafFG area under v vs t IJ + 1 a (t 40 s) + a50 m sf(t 40 s)H from t = 0 to 40 sK 223orx 3 = 375 m + 1 250 m +ja15.0 m s 2 t 40 s2ef + b50 m sgat 40 sf2which reduces tobg ejx 3 = 250 m s t 2.5 m s 2 t 2 4 375 m .total displacement 1 875 m== 37.5 m stotal elapsed time50 s(e)P2.25v=(a)Compare the position equation x = 2.00 + 3.00t 4.00t 2 to the general formx f = xi + vi t +1 2at2to recognize that x i = 2.00 m, vi = 3.00 m s, and a = 8.00 m s 2 . The velocity equation,v f = vi + at , is thenchv f = 3.00 m s 8.00 m s 2 t .The particle changes direction when v f = 0 , which occurs at t =time is:ax = 2.00 m + 3.00 m s(b)fFGH 3 sIJK c4.00 m s hFGH 3 sIJK8823s . The position at this82= 2.56 m .2v1 2at , observe that when x f = xi , the time is given by t = i . Thus,a2when the particle returns to its initial position, the time isFrom x f = xi + vi t +t=ca2 3.00 m s8.00 m sand the velocity is v f = 3.00 m s 8.00 m s 22f=3 shFGH 3 sIJK =443.00 m s .3132*P2.26Motion in One DimensionThe time for the Ford to slow down we find from1v xi + v xf t22 250 m2 xt=== 6.99 s .v xi + v xf 71.5 m s + 0x f = xi +daifIts time to speed up is similarlyt=2(350 m)0 + 71.5 m s= 9.79 s .The whole time it is moving at less than maximum speed is 6.99 s + 5.00 s + 9.79 s = 21.8 s . TheMercedes travelsafb ga11v xi + v xf t = 0 + 71.5 + 71.5 m s 21.8 s22= 1 558 mdx f = xi +ifwhile the Ford travels 250 + 350 m = 600 m, to fall behind by 1 558 m 600 m = 958 m .P2.27(a)chvi = 100 m s , a = 5.00 m s 2 , v f = vi + at so 0 = 100 5t , v 2 = vi2 + 2 a x f xi sof2ch0 = (100 ) 2(5.00) x f 0 . Thus x f = 1 000 m and t = 20.0 s .(b)P2.28At this acceleration the plane would overshoot the runway: No .(a)Take ti = 0 at the bottom of the hill where x i = 0 , vi = 30.0 m s, a = 2.00 m s 2 . Use thesevalues in the general equationx f = xi + vi t +to finda1 2at2f 1 c2.00 m s ht22x f = 0 + 30.0t m s +when t is in secondsc2hx f = 30.0t t 2 m .ejTo find an equation for the velocity, use v f = vi + at = 30.0 m s + 2.00 m s 2 t ,v f = (30.0 2.00t ) m s .(b)The distance of travel x f becomes a maximum, x max , when v f = 0 (turning point in themotion). Use the expressions found in part (a) for v f to find the value of t when x f has itsmaximum value:From v f = (3.00 2.00t ) m s , v f = 0 when t = 15.0 s. Thench2x max = 30.0t t 2 m = (30.0)(15.0)(15.0) = 225 m .Chapter 2P2.2933In the simultaneous equations:R|Sx|Tv xf = v xi + a x tf xi =U|| we have Rv = v c5.60 m s h(4.20 s)U .S 62.4 m = 1 v + v (4.20 s) |VVt|hW||ch2TWxf1v xi + v xf2cSo substituting for v xi gives 62.4 m =2xixixf1v xf + 56.0 m s 2 ( 4.20 s)+ v xf ( 4.20 s)2ch14.9 m s = v xf +15.60 m s 2 ( 4.20 s).2chThusv xf = 3.10 m s .P2.30Take any two of the standard four equations, such assubstitute into the other: v xi = v xf a x tx f xi =R|Sx|Tv xf = v xi + a x tf xi =1v xi + v xf2cU|. Solve one for vht V|W1v xf a x t + v xf t .2chThus1x f xi = v xf t a x t 2 .2Back in problem 29, 62.4 m = v xf ( 4.20 s)v xf =P2.31v f vi(a)a=(b)x f = vi t +t=632e j=5 2803 6001.40afFGH125.60 m s 2 ( 4. 20 s)2ch62.4 m 49.4 m= 3.10 m s .4.20 s662 ft s 2 = 202 m s 2I a f a fa fJK5 2801 211.40 662 1.40at = 63223 60022= 649 ft = 198 mxi ,and34P2.32Motion in One Dimension(a)The time it takes the truck to reach 20.0 m s is found from v f = vi + at . Solving for t yieldst=v f via=20.0 m s 0 m s2.00 m s 2= 10.0 s .The total time is thus10.0 s + 20.0 s + 5.00 s = 35.0 s .(b)The average velocity is the total distance traveled divided by the total time taken. Thedistance traveled during the first 10.0 s isx 1 = vt =FG 0 + 20.0 IJ(10.0)= 100 m .H 2 KWith a being 0 for this interval, the distance traveled during the next 20.0 s isx 2 = vi t +1 2at = ( 20.0)( 20.0)+ 0 = 400 m.2The distance traveled in the last 5.00 s isx 3 = vt =FG 20.0 + 0 IJ(5.00)= 50.0 m.H 2 KThe total distance x = x1 + x 2 + x 3 = 100 + 400 + 50 = 550 m , and the average velocity isx 550= 15.7 m s .given by v = =t 35.0P2.33We have vi = 2.00 10 4 m s, v f = 6.00 10 6 m s , x f xi = 1.50 102 m .cchh2 1.50 102 m2 x f xi1== 4.98 109 svi + v f t : t =462vi + v f2.00 10 m s + 6.00 10 m sch(a)x f xi =(b)v 2 = vi2 + 2 a x x f xi :fdax =iv 2 vi2f2( x f xi )e6.00 10=6msj e2.00 1022(1.50 10 2 m)4msj2= 1.20 10 15 m s 2Chapter 2*P2.34(a)chc22v xf = v xi + 2 a x x f x i : 0.01 3 10 8 m sax(b)c310=6h2ms80 m= 0 + 2 a x ( 40 m)h2= 1.12 10 11 m s 2We must find separately the time t1 for speeding up and the time t 2 for coasting:x f xi =x f xi =11v xf + v xi t1 : 40 m = 3 10 6 m s + 0 t122t1 = 2.67 10 5 sdiej11v xf + v xi t 2 : 60 m = 3 10 6 m s + 3 10 6 m s t 222t 2 = 2.00 10 5 sdiejtotal time = 4.67 105 s .*P2.35(a)Along the time axis of the graph shown, let i = 0 and f = t m . Then v xf = v xi + a x t givesv c = 0 + am tmam =(b)vc.tmThe displacement between 0 and t m isx f xi = v xi t +1 vc 211axt 2 = 0 +t m = v c tm .2 tm22The displacement between t m and t 0 isx f xi = v xi t +af1a x t 2 = v c t0 tm + 0 .2The total displacement isx =FGH11v c t m + v c t 0 v c t m = v c t 0 tm22IJK.(c)For constant v c and t 0 , x is minimized by maximizing t m to t m = t 0 . Thenv t1x min = v c t 0 t 0 = c 0 .22(e)This is realized by having the servo motor on all the time.(d)We maximize x by letting t m approach zero. In the limit x = v c t 0 0 = v c t 0 .(e)This cannot be attained because the acceleration must be finite.FGHIJKaf3536*P2.36Motion in One DimensionLet the glider enter the photogate with velocity vi and move with constant acceleration a. For itsmotion from entry to exit,1axt 2212= 0 + vi t d + at d = v d t d21v d = vi + at d2x f = xi + v xi t +(a)The speed halfway through the photogate in space is given by2v hs = vi2 + 2 aFG IJ = vH 2K2i+ av d t d .v hs = vi2 + av d t d and this is not equal to v d unless a = 0 .(b)The speed halfway through the photogate in time is given by v ht = vi + aFG t IJ and this isH 2Kdequal to v d as determined above.P2.37(a)Take initial and final points at top and bottom of the incline. If the ball starts from rest,vi = 0 , a = 0.500 m s 2 , x f xi = 9.00 m .Thendijaefv 2 = vi2 + 2 a x f xi = 0 2 + 2 0.500 m s 2 9.00 mfv f = 3.00 m s .(b)x f x i = vi t +1 2at210.500 m s 2 t 22t = 6.00 se9.00 = 0 +(c)Take initial and final points at the bottom of the planes and the top of the second plane,respectively:vi = 3.00 m s, v f = 0 , x f xi = 15.00 m.chv 2 = vi2 + 2 a x f xi givesfa=(d)jv 2 vi2fc2 x f xih=a0 3.00 m s2(15.0 m)f2= 0.300 m s 2 .Take the initial point at the bottom of the planes and the final point 8.00 m along the second:vi = 3.00 m s, x f xi = 8.00 m , a = 0.300 m s 2di bv 2 = vi2 + 2 a x f xi = 3.00 m sfv f = 2.05 m s .g + 2e0.300 m s ja8.00 mf = 4.20 m222s2Chapter 2P2.3837Take the original point to be when Sue notices the van. Choose the origin of the x-axis at Sues car.For her we have x is = 0 , vis = 30.0 m s , a s = 2.00 m s 2 so her position is given byx s (t )= x is + vis t +af11a s t 2 = 30.0 m s t + 2.00 m s 2 t 2 .22chFor the van, x iv = 155 m, viv = 5.00 m s , a v = 0 andx v (t )= xiv + viv t +af1a v t 2 = 155 + 5.00 m s t + 0 .2To test for a collision, we look for an instant t c when both are at the same place:230.0t c t c = 155 + 5.00t c20 = t c 25.0t c + 155 .From the quadratic formula2tc =25.0 ( 25.0) 4(155)2= 13.6 s or 11.4 s .The smaller value is the collision time. (The larger value tells when the van would pull ahead againif the vehicles could move through each other). The wreck happens at positionaf155 m + 5.00 m s (11.4 s)= 212 m .*P2.39As in the algebraic solution to Example 2.8, we let trepresent the time the trooper has been moving. We graphx car = 45 + 45tx (km)1.5car1and2x trooper = 1.5t .They intersect atpoliceofficer0.510t = 31 s .2030FIG. P2.3940t (s)38Motion in One DimensionSection 2.6P2.40Freely Falling ObjectsafChoose the origin y = 0 , t = 0 at the starting point of the ball and take upward as positive. Thenyi = 0 , vi = 0 , and a = g = 9.80 m s 2 . The position and the velocity at time t become:y f yi = vi t +111 2at : y f = gt 2 = 9.80 m s 2 t 2222ejandchv f = vi + at : v f = gt = 9.80 m s 2 t .129.80 m s 2 (1.00 s) = 4.90 m212at t = 2.00 s : y f = 9.80 m s 2 ( 2.00 s) = 19.6 m212at t = 3.00 s : y f = 9.80 m s 2 (3.00 s) = 44.1 m2ccchhh(a)at t = 1.00 s : y f = (b)at t = 1.00 s : v f = 9.80 m s 2 (1.00 s)= 9.80 m sat t = 2.00 s : v fat t = 3.00 s : v fP2.41ch= c9.80 m s h( 2.00 s)= 19.6 m s= c9.80 m s h(3.00 s)= 29.4 m s22Assume that air resistance may be neglected. Then, the acceleration at all times during the flight isthat due to gravity, a = g = 9.80 m s 2 . During the flight, Goff went 1 mile (1 609 m) up and then1 mile back down. Determine his speed just after launch by considering his upward flight:div 2 = vi2 + 2 a y f yi :fjbe0 = vi2 2 9.80 m s 2 1 609 mvi = 178 m s .gHis time in the air may be found by considering his motion from just after launch to just beforeimpact:y f yi = vi t +af11 2at : 0 = 178 m s t 9.80 m s 2 t 2 .22chThe root t = 0 describes launch; the other root, t = 36.2 s , describes his flight time. His rate of paymay then be found frompay rate =bgbg$1.00= 0.027 6 $ s 3 600 s h = $99.3 h .36.2 sWe have assumed that the workmans flight time, a mile, and a dollar, were measured to threedigit precision. We have interpreted up in the sky as referring to the free fall time, not to thelaunch and landing times. Both the takeoff and landing times must be several seconds away fromthe job, in order for Goff to survive to resume work.Chapter 2P2.42391We have y f = gt 2 + vi t + yi2h acf0 = 4.90 m s 2 t 2 8.00 m s t + 30.0 m .Solving for t,t=8.00 64.0 + 588.9.80Using only the positive value for t, we find that t = 1.79 s .P2.431 22at : 4.00 = (1.50)vi (4.90)(1.50) and vi = 10.0 m s upward .2(a)y f yi = vi t +(b)v f = vi + at = 10.0 (9.80)(1.50) = 4.68 m sv f = 4.68 m s downwardP2.44The bill starts from rest vi = 0 and falls with a downward acceleration of 9.80 m s 2 (due to gravity).Thus, in 0.20 s it will fall a distance ofy = vi t 1 22gt = 0 4.90 m s 2 (0. 20 s) = 0.20 m .2chafThis distance is about twice the distance between the center of the bill and its top edge 8 cm .Thus, David will be unsuccessful .*P2.45(a)From y = vi t +1 2at with vi = 0 , we have2t=a f=2 y2(23 m)a9.80 m s 2c= 2.17 s .h(b)The final velocity is v f = 0 + 9.80 m s 2 ( 2.17 s)= 21.2 m s .(c)The time take for the sound of the impact to reach the spectator ist sound =yv sound=23 m= 6.76 102 s ,340 m sso the total elapsed time is t total = 2.17 s + 6.76 10 2 s 2.23 s .40P2.46Motion in One DimensionAt any time t, the position of the ball released from rest is given by y1 = h 1 2gt . At time t, the21 2gt . The time at which the2h1hfirst ball has a position of y1 = is found from the first equation as = h gt 2 , which yields222hh. To require that the second ball have a position of y 2 = at this time, use the secondt=g2position of the ball thrown vertically upward is described by y 2 = vi t equation to obtainF IGH JKhh 1h= vi g. This gives the required initial upward velocity of the second2g 2 gball as vi = gh .P2.47(a)v f = vi gt : v f = 0 when t = 3.00 s , g = 9.80 m s 2 . Therefore,chvi = gt = 9.80 m s 2 (3.00 s)= 29.4 m s .(b)y f yi =1v f + vi t2chy f yi =*P2.48(a)bfga129.4 m s 3.00 s = 44.1 m2Consider the upward flight of the arrow.d i0 = b100 m sg + 2e 9.8 m s jy22v yf = v yi + 2 a y y f yi2y =(b)210 000 m 2 s 219.6 m s 2= 510 mConsider the whole flight of the arrow.y f = yi + v yi t +b1ayt 22g0 = 0 + 100 m s t +19.8 m s 2 t 22ejThe root t = 0 refers to the starting point. The time of flight is given byt=P2.49100 m s4.9 m s 2= 20.4 s .Time to fall 3.00 m is found from Eq. 2.12 with vi = 0 , 3.00 m =19.80 m s 2 t 2 , t = 0.782 s.2ch(a)With the horse galloping at 10.0 m s, the horizontal distance is vt = 7.82 m .(b)t = 0.782 sChapter 2P2.50Take downward as the positive y direction.(a)While the woman was in free fall,y = 144 ft , vi = 0 , and a = g = 32.0 ft s 2 .Thus, y = vi t +before impact is:1 2at 144 ft = 0 + 16.0 ft s 2 t 2 giving t fall = 3.00 s . Her velocity just2chchv f = vi + gt = 0 + 32.0 ft s 2 (3.00 s)= 96.0 ft s .(b)While crushing the box, vi = 96.0 ft s , v f = 0 , and y = 18.0 in. = 1.50 ft . Therefore,a=(c)v 2 vi2fa f2 y=a0 96.0 ft sf22(1.50 ft )Time to crush box: t == 3.07 10 3 ft s 2 , or a = 3.07 10 3 ft s 2 upward .2(1.50 ft)yy= v +v =or t = 3.13 102 s .fiv0 + 96.0 ft s2P2.51a fy = 3.00t 3 : At t = 2.00 s , y = 3.00 2.003= 24.0 m andvy =Ady= 9.00t 2 = 36.0 m s .dtIf the helicopter releases a small mailbag at this time, the equation of motion of the mailbag isy b = y bi + vi t 11 2gt = 24.0 + 36.0t (9.80)t 2 .22Setting y b = 0 ,0 = 24.0 + 36.0t 4.90t 2 .Solving for t, (only positive values of t count), t = 7.96 s .*P2.52Consider the last 30 m of fall. We find its speed 30 m above the ground:y f = yi + v yi t +1ayt 22a f 1 e9.8 m s ja1.5 sf220 = 30 m + v yi 1.5 s +v yi =230 m + 11.0 m= 12.6 m s .1.5 sNow consider the portion of its fall above the 30 m point. We assume it starts from restdib12.6 m sg = 0 + 2e9.8 m s jy22v yf = v yi + 2 a y y f yi2y =2160 m 2 s 219.6 m s 2Its original height was then 30 m + 8.16 m = 38.2 m .= 8.16 m .4142Motion in One DimensionSection 2.7P2.53(a)Kinematic Equations Derived from CalculusJ=da= constantdtda = Jdtza = J dt = Jt + c 1but a = ai when t = 0 so c 1 = ai . Therefore, a = Jt + aidvdtdv = adta=z zbv = adt =gJt + ai dt =but v = vi when t = 0, so c 2 = vi and v =1 2Jt + ai t + c 221 2Jt + ai t + vi2dxdtdx = vdtv=z z FGHx = vdt =IJK1 2Jt + ai t + vi dt21 3 1 2Jt + ai t + vi t + c 362x = xix=when t = 0, so c 3 = xi . Therefore, x =(b)aa 2 = Jt + aicf2= J 2 t 2 + ai2 + 2 Jai ta 2 = ai2 + J 2 t 2 + 2 Jai tFGH1 3 1 2Jt + ai t + vi t + xi .62h1 2a 2 = ai2 + 2 JJt + ai t2IJKRecall the expression for v: v =af1 21Jt + ai t + vi . So v vi = Jt 2 + ai t . Therefore,22aa 2 = ai2 + 2 J v vif.Chapter 2P2.54(a)See the graphs at the right.Choose x = 0 at t = 0.At t = 3 s, x =af18 m s (3 s)= 12 m .2afAt t = 5 s, x = 12 m + 8 m s ( 2 s)= 28 m .At t = 7 s, x = 28 m +af18 m s ( 2 s)= 36 m .28 ms= 2.67 m s 2 .3sFor 3 < t < 5 s, a = 0 .(b)(c)For 5 s < t < 9 s , a = (d)At t = 6 s, x = 28 m + 6 m s (1 s)= 34 m .(e)P2.55For 0 < t < 3 s, a =At t = 9 s, x = 36 m +(a)16 m s= 4 m s 2 .4saa=faf18 m s ( 2 s)= 28 m .2FIG. P2.54dv d= 5.00 10 7 t 2 + 3.00 10 5 tdt dtcha = 10.0 10 7 m s 3 t + 3.00 10 5 m s 2Take x i = 0 at t = 0. Then v =dxdtz zett00x 0 = vdt =j5.00 10 7 t 2 + 3.00 10 5 t dtt3t2+ 3.00 10 5323 37x = 1.67 10 m s t + 1.50 10 5 m s 2 t 2 .x = 5.00 10 7e(b)j echThe bullet escapes when a = 0 , at 10.0 10 7 m s 3 t + 3.00 10 5 m s 2 = 0t=(c)jchcNew v = 5.00 10 7 3.00 1033.00 10 5 s= 3.00 103 s .10.0 10 7h + c3.0010 hc3.0010 h253v = 450 m s + 900 m s = 450 m s .(d)chcx = 1.67 10 7 3.00 103h + c1.5010 hc3.0010 h3x = 0.450 m + 1.35 m = 0.900 m53 24344P2.56Motion in One Dimensiona=dv= 3.00 v 2 , vi = 1.50 m sdtSolving for v,dv= 3.00 v 2dtzvztv 2 dv = 3.00 dtv = viWhen v =t =01 11 1+ = 3.00t or 3.00t = .v viv vivi1, t== 0.222 s .23.00 viAdditional Problems*P2.57a fThe distance the car travels at constant velocity, v 0 , during the reaction time is x 1 = v 0 t r . Thetime for the car to come to rest, from initial velocity v 0 , after the brakes are applied ist2 =v f via=0 v0v= 0aaand the distance traveled during this braking period isaxf2= vt 2 =FvGH+ vif2I t = FG 0 + v IJ FG v IJ = v .JK H 2 K H a K 2 a02020Thus, the total distance traveled before coming to a stop isa f + a x fsstop = x*P2.58(a)12= v 0 t r 2v0.2a2v0(See the solution to Problem 2.57) from the2aintersection of length s i when the light turns yellow, the distance the car must travel beforethe light turns red isv2x = sstop + si = v 0 t r 0 + si .2aIf a car is a distance sstop = v 0 t r Assume the driver does not accelerate in an attempt to beat the light (an extremelydangerous practice!). The time the light should remain yellow is then the time required forthe car to travel distance x at constant velocity v 0 . This isv2t light(b)vs x v 0 t r 2 0 + sia=== t r 0 + i .v0v02 a v0With si = 16 m, v = 60 km h , a = 2.0 m s 2 , and t r = 1.1 s ,t light = 1.1 s F 0.278 m s I + 16 m F 1 km h I =GJGJ2e 2.0 m s j H 1 km h K 60 km h H 0.278 m s K60 km h26. 23 s .Chapter 2*P2.59(a)(b)As we see from the graph, from about 50 s to 50 sAcela is cruising at a constant positive velocity inthe +x direction. From 50 s to 200 s, Acelaaccelerates in the +x direction reaching a top speedof about 170 mi/h. Around 200 s, the engineerapplies the brakes, and the train, still traveling inthe +x direction, slows down and then stops at350 s. Just after 350 s, Acela reverses direction (vbecomes negative) and steadily gains speed in thex direction.200v100t100 200 300 400050 0t (s)100FIG. P2.59(a)The peak acceleration between 45 and 170 mi/h is given by the slope of the steepest tangentto the v versus t curve in this interval. From the tangent line shown, we finda = slope =(c)45v (155 45) mi h== 2. 2 mi h s = 0.98 m s 2 .(100 50) staLet us use the fact that the area under the v versust curve equals the displacement. The trainsdisplacement between 0 and 200 s is equal to thearea of the gray shaded region, which we haveapproximated with a series of triangles andrectangles.f200x 0 200 s = area 1 + area 2 + area 3 + area 4 + area 55100041 20bga f bga f+ b160 mi hga100 sf1+ a50 sfb100 mi hg21+ a100 sfb170 mi h 160 mi hg2= 24 000bmi hgasf3100 200 300 400 50 mi h 50 s + 50 mi h 50 sFIG. P2.59(c)Now, at the end of our calculation, we can find the displacement in miles by convertinghours to seconds. As 1 h = 3 600 s ,x 0 200 s F 24 000 mi I asf =GH 3 600 s JK6.7 mi .t (s)46*P2.60Motion in One DimensionAverage speed of every point on the train as the first car passes Liz:x 8.60 m== 5.73 m s.1.50 stThe train has this as its instantaneous speed halfway through the 1.50 s time. Similarly, halfway8.60 mthrough the next 1.10 s, the speed of the train is= 7.82 m s . The time required for the speed1.10 sto change from 5.73 m/s to 7.82 m/s is11(1.50 s)+ (1.10 s)= 1.30 s22so the acceleration is: a x =P2.61v x 7.82 m s 5.73 m s== 1.60 m s 2 .t1.30 sThe rate of hair growth is a velocity and the rate of its increase is an acceleration. Thenmm dv xi = 1.04 mm d and a x = 0.132. The increase in the length of the hair (i.e., displacement)wduring a time of t = 5.00 w = 35.0 d isFGHx = v xi t +bIJK1axt 22f 1 b0.132 mm d wga35.0 dfa5.00 wf2gax = 1.04 mm d 35.0 d +or x = 48.0 mm .P2.62Let point 0 be at ground level and point 1 be at the end of the engine burn. Letpoint 2 be the highest point the rocket reaches and point 3 be just beforeimpact. The data in the table are found for each phase of the rockets motion.a fv 2 80.0f(0 to 1)2a fb= 2 4.00 1 000gsogiving120 = 80.0 +( 4.00)tc2(1 to 2)0 (120) = 2(9.80) x f xihv f = 120 m st = 10.0 sgivingx f xi = 735 m0 120 = 9.80tgivingThis is the time of maximum height of the rocket.afbv 2 0 = 2 9.80 1 735f(2 to 3)gv f = 184 = (9.80)t(a)t total = 10 + 12.2 + 18.8 = 41.0 s(b)cxf xihtotal= 1.73 kmcontinued on next paget = 12.2 sgivingt = 18.8 sFIG. P2.62Chapter 2(c)v final = 184 m s0#1#2#3P2.63t0.010.022.241.0LaunchEnd ThrustRise UpwardsFall to Earthax01 0001 7350fa+4.00+4.009.809.80Distance traveled by motorist = 15.0 m s t1Distance traveled by policeman = 2.00 m s 2 t 22ch(a)intercept occurs when 15.0t = t 2 , or t = 15.0 s(b)v(officer)= 2.00 m s 2 t = 30.0 m s(c)P2.64v801200184x(officer )=ch12.00 m s 2 t 2 = 225 m2chArea A1 is a rectangle. Thus, A1 = hw = v xi t .11Area A 2 is triangular. Therefore A 2 = bh = t v x v xi .22The total area under the curve isbA = A1 + A 2 = v xi t +bvxgvxvxA2vxigA1 v xi t20and since v x v xi = a x ttFIG. P2.64A = v xi t +1axt 2 .2The displacement given by the equation is: x = v xi t +same result as above for the total area.1a x t 2 , the2t4748P2.65Motion in One Dimension(a)Let x be the distance traveled at acceleration a until maximum speed v is reached. If this isachieved in time t1 we can use the following three equations:x=aaff1v + vi t1 , 100 x = v 10.2 t1 and v = vi + at1 .2The first two giveFG 1 t IJ v = FG10.2 1 t IJ atH 2 K H 2 K200a=b20.4 t gt .100 = 10.2 11For Maggie: a =11200a18.4fa2.00f =200For Judy: a =a17.4fa3.00f =(b)15.43 m s 23.83 m s 2v = a1 ta fa fJudy: v = a3.83fa3.00f =Maggie: v = 5.43 2.00 = 10.9 m s(c)11.5 m sAt the six-second markx=fa fa f + a10.9fa4.00f = 54.3 ma fa f + a11.5fa3.00f = 51.7 m15.43 2.0021Judy: x = 3.83 3.002Maggie: x =a1 2at1 + v 6.00 t1222Maggie is ahead by 2.62 m .P2.66a1 = 0.100 m s 21122x = 1 000 m = a1 t1 + v1 t 2 + a 2 t 222atat1121 000 = a1 t1 + a1 t1 1 1 + a 2 1 12a22a2FGHIJKFG IJH Ka 2 = 0.500 m s 2t = t1 + t 2 and v1 = a1 t1 = a 2 t 221 000 =t1 =t2 =a1 t112.9= 26 sa 2 0.500FGHIJKa12a1 1 1 t12a220 000= 129 s1.20Total time = t = 155 sChapter 2P2.67Let the ball fall 1.50 m. It strikes at speed given bych22v xf = v xi + 2 a x f x i :ch2v xf = 0 + 2 9.80 m s 2 (1.50 m)v xf = 5.42 m sand its stopping is described byd22v xf = v xi + 2 a x x f x ib0 = 5.42 m sax =g2e2.00 10j+ 2 a x 10 2 m29.4 m 2 s 22im= +1.47 10 3 m s 2 .Its maximum acceleration will be larger than the average acceleration we estimate by imaginingconstant acceleration, but will still be of order of magnitude ~ 10 3 m s 2 .*P2.68(a)x f = xi + v xi t +1a x t 2 . We assume the package starts from rest.2145 m = 0 + 0 +t=(b)x f = xi + v xi t +19.80 m s 2 t 22c2(145 m)9.80 m s 2h= 5. 44 s112a x t 2 = 0 + 0 + 9.80 m s 2 (5.18 s) = 131 m22chdistance fallen = x f = 131 mej(c)speed = v xf = v xi + a x t = 0 + 9.8 m s 2 5.18 s = 50.8 m s(d)The remaining distance is145 m 131.5 m = 13.5 m .During deceleration,v xi = 50.8 m s, v xf = 0, x f xi = 13.5 mch22v xf = v xi + 2 a x x f x i :a0 = 50.8 m sax =f2+ 2 a x (13.5 m)2 580 m 2 s 2= +95.3 m s 2 = 95.3 m s 2 upward .2 13.5 maf4950P2.69Motion in One Dimension(a)11 2at = 50.0 = 2.00t + (9.80)t 2 ,224.90t 2 + 2.00t 50.0 = 0y f = v i1 t +t=2.00 + 2.00 2 4( 4.90)(50.0)2( 4.90)Only the positive root is physically meaningful:t = 3.00 s after the first stone is thrown.(b)1 2at and t = 3.00 1.00 = 2.00 s212substitute 50.0 = vi 2 ( 2.00)+ (9.80)( 2.00) :2y f = vi 2 t +vi2 = 15.3 m s downward(c)v1 f = vi1 + at = 2.00 +(9.80)(3.00)= 31.4 m s downwardv 2 f = vi 2 + at = 15.3 +(9.80)( 2.00)= 34.8 m s downwardP2.70(a)12d = (9.80)t12t1 + t 2 = 2.4024.90t 2d = 336 t 2a336t 2 = 4.90 2.40 t 2 359.5t 2 + 28.22 = 0t2 =359.5 358.75= 0.076 5 st2 =9.80(b)P2.71(a)(d)359.5 359.5 2 4( 4.90)( 28.22)9.80d = 336 t 2 = 26.4 mIn walking a distance x , in a time t , the lengthof rope is only increased by x sin .xsin . The pack lifts at a ratetxxsin = v boy = v boytx2x + h2FG IJHKd 1dv v boy dx=+ v boy xdtdtdtv boy v boy x dxd= v = v boya = v boy 2, butdtdt222v boyv boy h 2h 2 v boyx2 a==1 2 =23 2x 2 + h2a=FGH(c)212Ignoring the sound travel time, d = (9.80)( 2.40) = 28.2 m , an error of 6.82% .2v=(b)sof2v boyh,0v boy , 0IJKchFIG. P2.71Chapter 2P2.72h = 6.00 m, v boy = 2.00 m s v =However, x = v boy t : v =(a)c2v boy tc2v boy t 2+hc 4t4t2+ 36h12.0.320.631.50.8921.112.51.2831.413.51.5241.604.51.6651.71FIG. P2.72(a)From problem 2.71 above, a =a f aem s j2h 2 v boycx2+hh2 3 2=2h 2 v boyc2v boy t 2+hh2 3 2=c4t1442+ 36h32.2ts00.670.50.6410.571.50.4820.382.50.303.0.243.50.184.0.144.50.115(a)=00.5P2.73h2 12ha f vb m s gts01(b)v boy xxxsin = v boy =.12tx 2 + h20.09FIG. P2.72(b)We require x s = x k when t s = t k + 1.00jb13.50 m s 2 t k + 1.002t k + 1.00 = 1.183t kxs =et k = 5.46 s .ja14.90 m s 2 5.46 s2ef2(b)xk =(c)v k = 4.90 m s 2 5.46 s = 26.7 m svseja f= e3.50 m s ja6.46 sf =2= 73.0 m22.6 m sg2=jb g14.90 m s 2 t k2e2= xk5152P2.74Motion in One Dimension3.000.130.252.600.380.252.160.630.440.251.760.880.340.251.361.130.240.250.961.380.140.250.561.630.030.250.121.880.060.250.242.130.170.250.682.380.280.251.122.630.370.251.482.880.480.251.923.130.570.252.283.380.680.252.723.630.790.253.163.880.880.253.524.130.990.253.964.381.090.254.364.631.190.254.764.886.941.000.256.400.75midpt timet (s)5.750.50v(m/s)0.540.25t(s)0.65Heighth (m)5.00h(m)0.75Timet (s)0.007.381.257.721.507.961.758.102.008.132.258.072.507.902.757.623.007.253.256.773.506.203.755.524.004.734.253.854.502.864.751.775.000.58TABLE P2.74acceleration = slope of line is constant.a =1.63 m s 2 = 1.63 m s 2 downwardFIG. P2.7453Chapter 2P2.75The distance x and y are always related by x 2 + y 2 = L2 .Differentiating this equation with respect to time, we haveyBxdydx2x + 2y=0dtdtLydydx= v .is v B , the unknown velocity of B; anddtdtFrom the equation resulting from differentiation, we haveNowOFG IJH KFGHAxdyx dxx= (v).=dty dtyButvFIG. P2.75IJKyvv 31== 0.577 v .= tan so v B =v . When = 60.0 , v B =tan 60.03xtan ANSWERS TO EVEN PROBLEMS(a) 2 10 7 m s ; 1 10 6 m s ;P2.24(b) 5 10 yr(a) 1.88 km; (b) 1.46 km;(c) see the solution;(d) (i) x 1 = 1.67 m s 2 t 2 ;P2.4(a) 50.0 m s ; (b) 41.0 m s(ii) x 2 = 50 m s t 375 m ;P2.6(a) 27.0 m ;2(b) 27.0 m + 18.0 m s t + 3.00 m s 2 t ;(iii) x 3P2.28bg eeja f(c) 18.0 m sP2.8(a), (b), (c) see the solution; 4.6 m s 2 ; (d) 0P2.10(a) 20.0 m s ; 5.00 m s ; (b) 262 mP2.14P2.16bg= b 250 m sgt e 2.5 m s jt22 4 375 m ;(e) 37.5 m sP2.26958 mP2.28(a) x f = 30.0t t 2 m; v f = 30.0 2t m s ;5.00 mP2.12jejaf(b) 225 m1a x t 2 ; 3.10 m s2P2.30x f xi = v xf t (a) see the solution;(b) 1.60 m s 2 ; 0.800 m s 2P2.32(a) 35.0 s; (b) 15.7 m s(a) 13.0 m s; (b) 10.0 m s; 16.0 m s;P2.34(a) 1.12 10 11 m s 2 ; (b) 4.67 10 5 sP2.36(a) False unless the acceleration is zero;see the solution; (b) True(c) 6.00 m s 2 ; (d) 6.00 m s 2P2.18see the solutionP2.20(a) 6.61 m s; (b) 0. 448 m s 2P2.38Yes; 212 m; 11.4 sP2.22(a) 21.8 mi h s = 9.75 m s 2 ;P2.40(a) 4.90 m ; 19.6 m; 44.1 m;(b) 9.80 m s; 19.6 m s; 29.4 m sP2.421.79 s(b) 22.2 mi h s = 9.94 m s 2 ;(c) 22.8 mi h s = 10.2 m s 254Motion in One DimensionP2.44No; see the solutionP2.601.60 m s 2P2.46The second ball is thrown at speedvi = ghP2.62(a) 41.0 s; (b) 1.73 km; (c) 184 m sP2.48(a) 510 m; (b) 20.4 sP2.64v xi t +P2.50(a) 96.0 ft s ;P2.66155 s; 129 sP2.68(a) 5.44 s; (b) 131 m; (c) 50.8 m s ;32(b) a = 3.07 10 ft s upward ;(c) t = 3.13 10 2 sP2.521a x t 2 ; displacements agree2(d) 95.3 m s 2 upward38.2 mP2.70(a) 26.4 m; (b) 6.82%2P2.54(a) and (b) see the solution; (c) 4 m s ;(d) 34 m; (e) 28 mP2.72see the solutionP2.560.222 sP2.74see the solution; a x = 1.63 m s 2P2.58(a) see the solution; (b) 6.23 s3VectorsCHAPTER OUTLINE3.13.2Coordinate SystemsVector and ScalarQuantitiesSome Properties of VectorsComponents of a Vectorand Unit VectorsANSWERS TO QUESTIONSNo. The sum of two vectors can only be zero if they are inopposite directions and have the same magnitude. If you walk10 meters north and then 6 meters south, you wont end upwhere you started.Q3.23.33.4Q3.1No, the magnitude of the displacement is always less than orequal to the distance traveled. If two displacements in the samedirection are added, then the magnitude of their sum will beequal to the distance traveled. Two vectors in any otherorientation will give a displacement less than the distancetraveled. If you first walk 3 meters east, and then 4 meterssouth, you will have walked a total distance of 7 meters, butyou will only be 5 meters from your starting point.Q3.3The largest possible magnitude of R = A + B is 7 units, found when A and B point in the samedirection. The smallest magnitude of R = A + B is 3 units, found when A and B have oppositedirections.Q3.4Only force and velocity are vectors. None of the other quantities requires a direction to be described.Q3.5If the direction-angle of A is between 180 degrees and 270 degrees, its components are bothnegative. If a vector is in the second quadrant or the fourth quadrant, its components have oppositesigns.Q3.6The books displacement is zero, as it ends up at the point from which it started. The distancetraveled is 6.0 meters.Q3.785 miles. The magnitude of the displacement is the distance from the starting point, the 260-milemark, to the ending point, the 175-mile mark.Q3.8Vectors A and B are perpendicular to each other.Q3.9No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction,not magnitude.5556VectorsQ3.10Any vector that points along a line at 45 to the x and y axes has components equal in magnitude.Q3.11A x = B x and A y = B y .Q3.12Addition of a vector to a scalar is not defined. Think of apples and oranges.Q3.13One difficulty arises in determining the individual components. The relationships between a vectorand its components such as A x = A cos , are based on right-triangle trigonometry. Another problemwould be in determining the magnitude or the direction of a vector from its components. Again,22A = A x + A y only holds true if the two component vectors, A x and A y , are perpendicular.Q3.14If the direction of a vector is specified by giving the angle of the vector measured clockwise from thepositive y-axis, then the x-component of the vector is equal to the sine of the angle multiplied by themagnitude of the vector.SOLUTIONS TO PROBLEMSSection 3.1P3.1P3.2Coordinate Systemsa fa fa fy = r sin = a5.50 mf sin 240 = a5.50 mfa 0.866f =x = r cos = 5.50 m cos 240 = 5.50 m 0.5 = 2.75 m(a)4.76 mx = r cos and y = r sin , thereforex1 = 2.50 m cos 30.0 , y1 = 2.50 m sin 30.0 , andafafbx , y g = a2.17 , 1.25f mx = a3.80 mf cos 120 , y = a3.80 mf sin 120 , andbx , y g = a1.90, 3.29f m .1122(b)P3.322d = ( x) 2 + ( y) 2 = 16.6 + 4.16 = 4.55 mThe x distance out to the fly is 2.00 m and the y distance up to the fly is 1.00 m.(a)We can use the Pythagorean theorem to find the distance from the origin to the fly.distance = x 2 + y 2 =(b) = tan 1FG 1 IJ = 26.6 ; r =H 2Ka2.00 mf + a1.00 mf2.24 m, 26.622= 5.00 m 2 = 2.24 mChapter 3P3.4(a)d=bx2 x1g + by22 y1g2=c2.00 3.00 h + a4.00 3.00f22d = 25.0 + 49.0 = 8.60 m(b)a2.00f + a4.00f = 20.0 =F 4.00 IJ = 63.4= tan G H 2.00 K2r1 =1r2 =24.47 m1a3.00f + a3.00f22= 18.0 = 4.24 m 2 = 135 measured from the +x axis.P3.5We have 2.00 = r cos 30.0r=2.00= 2.31cos 30.0and y = r sin 30.0 = 2.31 sin 30.0 = 1.15 .P3.6We have r = x 2 + y 2 and = tan 1(a)FG y IJ .H xKThe radius for this new point isa x f2+ y2 = x2 + y2 = rand its angle istan 1(b)FG y IJ =H x K180 .b g( 2 x) 2 + ( 2 y) 2 = 2r . This point is in the third quadrant if x , y is in the first quadrantb gor in the fourth quadrant if x , y is in the second quadrant. It is at an angle of 180+ .(c)b g( 3 x) 2 + ( 3 y) 2 = 3r . This point is in the fourth quadrant if x , y is in the first quadrantb gor in the third quadrant if x , y is in the second quadrant. It is at an angle of .5758VectorsSection 3.2Vector and Scalar QuantitiesSection 3.3Some Properties of VectorsP3.7x100 mx = 100 m tan 35.0 = 70.0 mtan 35.0 =afFIG. P3.7P3.8R = 14 km = 65 N of ER13 km6 km1 kmFIG. P3.8P3.9 R = 310 km at 57 S of W(Scale: 1 unit = 20 km )FIG. P3.9P3.10(a)Using graphical methods, place the tail ofvector B at the head of vector A. The newvector A + B has a magnitude of6.1 at 112 from the x-axis.yAA+B(b)The vector difference A B is found byplacing the negative of vector B at thehead of vector A. The resultant vectorA B has magnitude 14.8 units at anangle of 22 from the + x-axis.BBABxOFIG. P3.10Chapter 3P3.11(a)d = 10.0 i = 10.0 m since the displacement is in aCstraight line from point A to point B.(b)The actual distance skated is not equal to the straight-linedisplacement. The distance follows the curved path of thesemi-circle (ACB).s=(c)P3.12595.00 mdBAFIG. P3.11b g12 r = 5 = 15.7 m2If the circle is complete, d begins and ends at point A. Hence, d = 0 .Find the resultant F1 + F2 graphically by placing the tail of F2 at the head of F1 . The resultant forcevector F1 + F2 is of magnitude 9.5 N and at an angle of 57 above the x -axis .yF1 + F2F2F1x0 1 2 3 NFIG. P3.12P3.13(a)The large majority of people are standing or sitting at this hour. Their instantaneous foot-tohead vectors have upward vertical components on the order of 1 m and randomly orientedhorizontal components. The citywide sum will be ~ 10 5 m upward .(b)Most people are lying in bed early Saturday morning. We suppose their beds are orientednorth, south, east, west quite at random. Then the horizontal component of their total vectorheight is very nearly zero. If their compressed pillows give their height vectors verticalcomponents averaging 3 cm, and if one-tenth of one percent of the population are on-dutynurses or police officers, we estimate the total vector height as ~ 10 5 0.03 m + 10 2 1 ma3~ 10 m upward .fa f60P3.14VectorsNYour sketch should be drawn to scale, andshould look somewhat like that pictured tothe right. The angle from the westwarddirection, , can be measured to be4 N of W , and the distance R from the1mW15.0 metersRsketch can be converted according to thescale to be 7.9 m .3.50meters30.08.20metersESFIG. P3.14P3.15To find these vector expressions graphically, wedraw each set of vectors. Measurements of theresults are taken using a ruler and protractor.(Scale: 1 unit = 0.5 m )(a)A + B = 5.2 m at 60(b)A B = 3.0 m at 330(c)B A = 3.0 m at 150(d)A 2B = 5.2 m at 300.FIG. P3.15*P3.16The three diagrams shown below represent the graphical solutions for the three vector sums:R 1 = A + B + C , R 2 = B + C + A , and R 3 = C + B + A . You should observe that R 1 = R 2 = R 3 ,illustrating that the sum of a set of vectors is not affected by the order in which the vectors areadded.100 mCBAABR1AR2CBFIG. P3.16R3CChapter 3P3.17The scale drawing for the graphical solutionshould be similar to the figure to the right. Themagnitude and direction of the final displacementfrom the starting point are obtained by measuringd and on the drawing and applying the scalefactor used in making the drawing. The resultsshould be(Scale: 1 unit = 20 ft )d = 420 ft and = 3FIG. P3.17Section 3.4P3.18Components of a Vector and Unit VectorsCoordinates of the super-hero are:a f a fy = a100 mf sina 30.0f =x = 100 m cos 30.0 = 86.6 m50.0 mFIG. P3.18P3.19A x = 25.0A y = 40.022A = Ax + Ay =a25.0f + a40.0f22= 47.2 unitsWe observe thatAytan =.AxFIG. P3.19So = tan 1F A I = tan 40.0 = tan a1.60f = 58.0 .GH A JK 25.0y1xThe diagram shows that the angle from the +x axis can be found by subtracting from 180: = 180 58 = 122 .P3.20afThe person would have to walk 3.10 sin 25.0 = 1.31 km north , andaf3.10 cos 25.0 = 2.81 km east .6162P3.21Vectorsx = r cos and y = r sin , therefore:b g ejb g ejb g ej(a)(b)x = 3.30 cos 60.0 , y = 3.30 sin 60.0 , and x , y = 1.65 i + 2.86 j cm(c)P3.22x = 12.8 cos 150 , y = 12.8 sin 150 , and x , y = 11.1i + 6.40 j mx = 22.0 cos 215 , y = 22.0 sin 215 , and x , y = 18.0 i 12.6 j inaad=*P3.23f a ff a fa25.0 mfi + a43.3 mf jx = d cos = 50.0 m cos 120 = 25.0 my = d sin = 50.0 m sin 120 = 43.3 m(a)Her net x (east-west) displacement is 3.00 + 0 + 6.00 = +3.00 blocks, while her net y (northsouth) displacement is 0 + 4.00 + 0 = +4.00 blocks. The magnitude of the resultantdisplacement isR=b x g + by gnet2net2=a3.00f + a4.00f22= 5.00 blocksand the angle the resultant makes with the x-axis (eastward direction) is = tan 1FG 4.00 IJ = tan a1.33f = 53.1 .H 3.00 K1The resultant displacement is then 5.00 blocks at 53.1 N of E .(b)*P3.24The total distance traveled is 3.00 + 4.00 + 6.00 = 13.0 blocks .NLet i = east and j = north. The unicyclists displacement is, in meters280 j + 220 i + 360 j 300 i 120 j + 60 i 40 j 90 i + 70 j .RR = 110 i + 550 j=a110 mf + a550 mf22at tan 1= 561 m at 11.3 west of north .110 mwest of north550 mThe crows velocity isv= x 561 m at 11.3 W of N=t40 s= 14.0 m s at 11.3 west of north .EFIG. P3.24Chapter 3P3.2563+x East, +y North x = 250 +125 cos 30 = 358 m y = 75 +125 sin 30150 = 12.5 mc xh + c yh = a358f + a12.5fc yh = 12.5 = 0.0349tan =c xh 3582d=222= 358 m = 2.00d = 358 m at 2.00 S of EP3.26The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums ofthe east and north components of the displacements from Dallas to Atlanta (A) and from Atlanta toChicago. In equation form:d DC east = d DA east + d AC east = 730 cos 5.00560 sin 21.0 = 527 miles.d DC north = d DA north + d AC north = 730 sin 5.00+560 cos 21.0 = 586 miles.By the Pythagorean theorem, d = ( d DC east ) 2 + ( d DC north ) 2 = 788 mi .Then tan =d DC north= 1.11 and = 48.0 .d DC eastThus, Chicago is 788 miles at 48.0 northeast of Dallas .P3.27(a)See figure to the right.(b)C = A + B = 2.00 i + 6.00 j + 3.00 i 2.00 j = 5.00 i + 4.00 jC = 25.0 + 16.0 at tan 1FG 4 IJ =H 5K6.40 at 38.7D = A B = 2.00 i + 6.00 j 3.00 i + 2.00 j = 1.00 i + 8.00 ja1.00f + a8.00f at tan FGH 81.00 IJK.00D = 8.06 at b180 82.9g = 8.06 at 97.22D=P3.2821bx + x + x g + by + y + y g= a3.00 5.00 + 6.00f + a 2.00 + 3.00 + 1.00 fF 6.00 IJ = 56.3 = tan GH 4.00 Kd=12322121232= 52.0 = 7.21 mFIG. P3.2764P3.29VectorsWe have B = R A :A x = 150 cos120 = 75.0 cmA y = 150 sin 120 = 130 cmR x = 140 cos 35.0 = 115 cmR y = 140 sin 35.0 = 80.3 cmFIG. P3.29Therefore,a fejB = 115 75 i + 80.3 130 j = 190 i 49.7 j cmB = 190 2 + 49.7 2 = 196 cmFGH = tan 1 P3.30IJK49.7= 14.7 .190A = 8.70 i + 15.0 j and B = 13. 2 i 6.60 jA B + 3C = 0 :3C = B A = 21.9 i 21.6 jC = 7.30 i 7.20 jorC x = 7.30 cm ; C y = 7.20 cm(a)aA + Bf = e3 i 2 jj + e i 4 jj =2i 6 j(b)aA Bf = e3i 2 jj e i 4 jj =4i + 2 j(c)A + B = 2 2 + 6 2 = 6.32(d)A B = 4 2 + 2 2 = 4.47(e)P3.31 A+B = tan1 AB = tan1P3.32(a)FG 6 IJ = 71.6=H 2KFG 2 IJ = 26.6H 4K288D = A + B + C = 2i + 4 jD = 2 2 + 4 2 = 4. 47 m at = 63.4(b)E = A B + C = 6 i + 6 jE = 6 2 + 6 2 = 8.49 m at = 135Chapter 3P3.33ejd1 = 3.50 j mejd 2 = 8.20 cos 45.0 i + 8.20 sin 45.0 j = 5.80 i + 5.80 j mejd 3 = 15.0 i maf af e9.20i + 2.30 jj mR = d1 + d 2 + d 3 = 15.0 + 5.80 i + 5.80 3.50 j =(or 9.20 m west and 2.30 m north)The magnitude of the resultant displacement is22R = Rx + R y =FG 2.30 IJ =H 9.20 KThe direction is = arctanP3.3422= 9.48 m .166 .A = 10.0Refer to the sketchR = A + B + C = 10.0 i 15.0 j + 50.0 i= 40.0 i 15.0 ja f + a15.0fR = 40.0a9.20f + a2.30f22 12RB = 15.0C = 50.0= 42.7 yardsFIG. P3.34P3.35F = F1 + F2(a)aaa fF = 60.0 i + 104 j 20.7 i + 77.3 j = e39.3 i + 181 jj NffafF = 120 cos 60.0 i + 120 sin 60.0 j 80.0 cos 75.0 i + 80.0 sin 75.0 jF = 39.3 2 + 181 2 = 185 N = tan 1(b)P3.36F3 = F =Eastx0m1.410.500+0.914R=FG 181 IJ =H 39.3 K77.8e39.3 i 181 jj NWesty4.00 m1.410.8664.5522x + y = 4.64 m at 78.6 N of E6566VectorsA = 3.00 m, A = 30.0B = 3.00 m , B = 90.0A x = A cos A = 3.00 cos 30.0 = 2.60 mP3.37A y = A sin A = 3.00 sin 30.0= 1.50 mejA = A x i + A y j = 2.60 i + 1.50 j mBx = 0 , By = 3.00 mejA + B = 2.60 i + 1.50 j + 3.00 j =P3.38B = 3.00 j msoe2.60i + 4.50 jj mLet the positive x-direction be eastward, the positive y-direction be vertically upward, and thepositive z-direction be southward. The total displacement is thenejejejd = 4.80 i + 4.80 j cm + 3.70 j 3.70k cm = 4.80 i + 8.50 j 3.70k cm .222(a)(b)P3.39The magnitude is d = ( 4.80) +(8.50) + (3.70) cm = 10.4 cm .Its angle with the y-axis follows from cos =8.50, giving = 35.5 .10. 4B = Bx i + By j + Bz k = 4.00 i + 6.00 j + 3.00kB = 4.00 2 + 6.00 2 + 3.00 2 = 7.81 = cos 1 = cos 1 = cos 1P3.40FG 4.00 IJ =H 7.81 KFG 6.00 IJ =H 7.81 KFG 3.00 IJ =H 7.81 K59.239.867.4The y coordinate of the airplane is constant and equal to 7.60 10 3 m whereas the x coordinate isgiven by x = vi t where vi is the constant speed in the horizontal direction.isAt t = 30.0 s we have x = 8.0410 3 , so vi = 268 m s. The position vector as a function of timebg ejP = 268 m s t i + 7.60 10 3 m j .At t = 45.0 s , P = 1. 21 10 4 i + 7.60 10 3 j m. The magnitude isP=c1.2110 h + c7.6010 h4 23 2m = 1.43 10 4 mand the direction is = arctanF 7.6010 I =GH 1.2110 JK3432.2 above the horizontal .Chapter 3A = 8.00 i + 12.0 j 4.00kB=(c)P3.42(a)(b)P3.41C = 3A = 24.0 i 36.0 j + 12.0kA= 2.00 i + 3.00 j 1.00k4R = 75.0 cos 240 i + 75.0 sin 240 j + 125 cos 135 i + 125 sin 135 j + 100 cos 160 i + 100 sin 160 jR = 37.5 i 65.0 j 88.4i + 88.4 j 94.0 i + 34.2 jR = 220 i + 57.6 j2R = (220 ) + 57.6 2 at arctanFG 57.6 IJ above the x-axisH 220 KR = 227 paces at 165P3.43(a)e5.00 i 1.00 j 3.00kj mC=A+B=222C = (5.00) +(1.00) +(3.00) m = 5.92 m(b)D = 2A B =e4.00i 11.0 j + 15.0kj m222D = ( 4.00) +(11.0) +(15.0) m = 19.0 mP3.44The position vector from radar station to ship isejejS = 17.3 sin 136 i + 17.3 cos 136 j km = 12.0 i 12.4 j km.From station to plane, the position vector isejP = 19.6 sin 153 i + 19.6 cos 153 j + 2.20k km,orejP = 8.90 i 17.5 j + 2.20k km.(a)To fly to the ship, the plane must undergo displacementD = S P =(b)e3.12 i + 5.02 j 2.20kj km .The distance the plane must travel is222D = D = (3.12) +(5.02) +( 2.20) km = 6.31 km .6768P3.45VectorsThe hurricanes first displacement isisFG 41.0 km IJ(3.00 h) at 60.0 N of W, and its second displacementH h KFG 25.0 km IJ(1.50 h) due North. With iH h Krepresenting east and j representing north, its totaldisplacement is:FG 41.0 km cos 60.0IJ a3.00 hfe ij + FG 41.0 km sin 60.0IJ a3.00 hf j + FG 25.0 km IJ a1.50 hf j = 61.5 kme ijH hKH hKH hK+144 km j22with magnitude (61.5 km) +(144 km) = 157 km .P3.46(a)aE=(b)ye15.1i + 7.72 jj cmaafe7.72i + 15.1 jj cmfaafe+7.72 i + 15.1 jj cm27.0 27.0fFFIG. P3.46A x = 3.00 , A y = 2.00(a)A = A x i + A y j = 3.00 i + 2.00 j(b)22A = A x + A y = (3.00) +( 2.00) = 3.612tan =AyAx=22.00= 0.667 , tan1 (0.667)= 33.7(3.00)af is in the 2 nd quadrant, so = 180+ 33.7 = 146 .(c)GEG = + 17.0 cm sin 27.0 i + 17.0 cm cos 27.0 jG=P3.47fF = 17.0 cm sin 27.0 i + 17.0 cm cos 27.0 jF=(c)afE = 17.0 cm cos 27.0 i + 17.0 cm sin 27.0 jR x = 0 , R y = 4.00 , R = A + B thus B = R A andBx = R x A x = 0 (3.00)= 3.00 , By = R y A y = 4.00 2.00 = 6.00 .Therefore, B = 3.00 i 6.00 j .27.0xChapter 3P3.48Let +x = East, +y = North,x3001750125y0303150453y= 74.6 N of Ex(a)(b)P3.49 = tan1R = x 2 + y 2 = 470 km(a)R x = 40.0 cos 45.0+30.0 cos 45.0 = 49.5yR y = 40.0 sin 45.030.0 sin 45.0+20.0 = 27.1AR = 49.5 i + 27.1 j(b)a49.5f + a27.1f = 56.4F 27.1 IJ = 28.7 = tan GH 49.5 K2R=2O1B45x45CFIG. P3.49P3.50Taking components along i and j , we get two equations:6.00 a 8.00b + 26.0 = 0and8.00 a + 3.00b + 19.0 = 0 .Solving simultaneously,a = 5.00 , b = 7.00 .Therefore,5.00A + 7.00B + C = 0 .6970VectorsAdditional ProblemsP3.51Let represent the angle between the directions of A and B. SinceA and B have the same magnitudes, A, B, and R = A + B form anisosceles triangle in which the angles are 180 ,magnitude of R is then R = 2 A cosR /2, and . The22FG IJ . [Hint: apply the law ofH 2Kcosines to the isosceles triangle and use the fact that B = A .]Again, A, B, and D = A B form an isosceles triangle with apexangle . Applying the law of cosines and the identityBAADBFIG. P3.51a1 cos f = 2 sin FGH IJK22FG IJ .H 2Kgives the magnitude of D as D = 2 A sinThe problem requires that R = 100D .Thus, 2 A cosFG IJ = 200 A sinFG IJ . This gives tanFG IJ = 0.010 andH 2KH 2KH 2K = 1.15 .P3.52Let represent the angle between the directions of A and B. SinceA and B have the same magnitudes, A, B, and R = A + B form anisosceles triangle in which the angles are 180 ,magnitude of R is then R = 2 A cos, and . The22FG IJ . [Hint: apply the law ofH 2Kcosines to the isosceles triangle and use the fact that B = A . ]Again, A, B, and D = A B form an isosceles triangle with apexangle . Applying the law of cosines and the identitya1 cos f = 2 sin FGH IJK22FG IJ .H 2KF IF IThe problem requires that R = nD or cosG J = n sinG J givingH 2KH 2KF 1I = 2 tan G J .H nKgives the magnitude of D as D = 2 A sin1FIG. P3.52Chapter 3P3.53(a)R x = 2.00 , R y = 1.00 , R z = 3.00(b)222R = R x + R y + R z = 4.00 + 1.00 + 9.00 = 14.0 = 3.74(c)cos x =cos y =cos z =*P3.54F IGH JKF R I = 74.5 from + yGH R JKF R I = 36.7 from + zGH R JKRxR x = cos1 x = 57.7 from + xRRRyRy y = cos1Rz z = cos1RzTake the x-axis along the tail section of the snake. The displacement from tail to head isaaff240 m i + 420 240 m cos 180105 i 180 m sin 75 j = 287 m i 174 m j .22Its magnitude is (287) +(174) m = 335 m . From v =distance, the time for each childs run ista faa fbfbgInge: t =distance 335 m h 1 km 3 600 s== 101 sv12 km 1 000 m 1 hOlaf: t =420 m s= 126 s .3.33 mga fInge wins by 126 101 = 25.4 s .*P3.55The position vector from the ground under the controller of the first airplane isafa f a= e17.4i + 8.11 j + 0.8k j km .faf afafa f a= e16.5 i + 6.02 j + 1.1k j km .faf afr1 = 19.2 km cos 25 i + 19.2 km sin 25 j + 0.8 km kThe second is atr2 = 17.6 km cos 20 i + 17.6 km sin 20 j + 1.1 km kNow the displacement from the first plane to the second isejr2 r1 = 0.863 i 2.09 j + 0.3k kmwith magnitude222(0.863 ) +( 2.09) +(0.3) = 2.29 km .7172*P3.56Vectors3Let A represent the distance from island 2 to island 3. Thedisplacement is A = A at 159 . Represent the displacement from 3to 1 as B = B at 298 . We have 4.76 km at 37 +A + B = 0 .A28 BFor x-componentsa4.76 kmf cos 37+ A cos 159+B cos 298 = 037169C2NE3.80 km 0.934 A + 0.469B = 0FIG. P3.56B = 8.10 km + 1.99 AFor y-components,a4.76 kmf sin 37+ A sin 159+B sin 298 = 02.86 km + 0.358 A 0.883B = 0(a)We solve by eliminating B by substitution:af2.86 km + 0.358 A 0.883 8.10 km + 1.99 A = 02.86 km + 0.358 A + 7.15 km 1.76 A = 010.0 km = 1.40 AA = 7.17 km(b)*P3.57B = 8.10 km + 1.99(7.17 km)= 6.15 km(a)We first express the corners position vectors as sets of componentsa fa fB = a12 mf cos 30 i + a12 mf sin 30 j = 10.4 m i +6.00 mj .A = 10 m cos 50 i + 10 m sin 50 j = 6.43 m i +7.66 m jThe horizontal width of the rectangle is10.4 m 6.43 m = 3.96 m .Its vertical height is7.66 m 6.00 m = 1.66 m .Its perimeter is2(3.96 + 1.66) m = 11.2 m .(b)The position vector of the distant corner is Bx i + A y j = 10.4 mi +7.66 mj = 10.4 2 + 7.66 2 m at7.66 m= 12.9 m at 36.4 .tan110.4 mChapter 3P3.58Choose the +x-axis in the direction of the first force. The total force, yin newtons, is then31 N12.0 i + 31.0 j 8.40 i 24.0 j =e3.60 ij + e7.00 jj N8.4 NThe magnitude of the total force is2xR.12 N35.0horizontal24 N2(3.60) +(7.00) N = 7.87 Nand the angle it makes with our +x-axis is given by tan =(7.00)(3.60) = 62.8 . Thus, its angle counterclockwise from the horizontal is35.0+62.8 = 97.8 .P3.59FIG. P3.58,d 1 = 100 id 2 = 300 ja fa fd = 200 cosa60.0fi + 200 sina60.0f j = 100 i + 173 jR = d + d + d + d = e 130 i 202 jj mR = a 130f + a202f = 240 mF 202 IJ = 57.2 = tan GH 130 Kd 3 = 150 cos 30.0 i 150 sin 30.0 j = 130 i 75.0 j41234221FIG. P3.59 = 180 + = 237P3.60ejdr d 4 i + 3 j 2 t j== 0 + 0 2 j = 2.00 m s jdtdtbgThe position vector at t = 0 is 4i + 3 j . At t = 1 s , the position is 4i + 1 j , and so on. The object ismoving straight downward at 2 m/s, sodrrepresents its velocity vector .dtP3.61af afv = v x i + v y j = 300 + 100 cos 30.0 i + 100 sin 30.0 jejv = 387 i + 50.0 j mi hv = 390 mi h at 7.37 N of E7374P3.62Vectors(a)ejYou start at point A: r1 = rA = 30.0 i 20.0 j m.The displacement to B isrB rA = 60.0 i + 80.0 j 30.0 i + 20.0 j = 30.0 i + 100 j .ejYou cover half of this, 15.0 i + 50.0 j to move to r2 = 30.0 i 20.0 j + 15.0 i + 50.0 j = 45.0 i + 30.0 j .Now the displacement from your current position to C isrC r2 = 10.0 i 10.0 j 45.0 i 30.0 j = 55.0 i 40.0 j .You cover one-third, moving tor3 = r2 + r23 = 45.0 i + 30.0 j +155.0 i 40.0 j = 26.7 i + 16.7 j .3ejThe displacement from where you are to D isrD r3 = 40.0 i 30.0 j 26.7 i 16.7 j = 13.3 i 46.7 j .You traverse one-quarter of it, moving tor4 = r3 +bg11rD r3 = 26.7 i + 16.7 j + 13.3 i 46.7 j = 30.0 i + 5.00 j .44ejThe displacement from your new location to E isrE r4 = 70.0 i + 60.0 j 30.0 i 5.00 j = 100 i + 55.0 jof which you cover one-fifth the distance, 20.0 i + 11.0 j, moving tor4 + r45 = 30.0 i + 5.00 j 20.0 i + 11.0 j = 10.0 i + 16.0 j .The treasure is at (10.0 m, 16.0 m) .(b)Following the directions brings you to the average position of the trees. The steps we tooknumerically in part (a) bring you torA +arthen toarthen toA+ rBf+r2+ rB + rCACarA +rB23f+rf=araf FGHr + rB1rB rA = A22IJKrA + rB + rC3A + rB + rCfr + rB + rC + rD= A344arA +rB +rC +rD f r + r + r + r + rrA + rB + rC + rDr BCDE4+ E= Aand last to.455aD3fThis center of mass of the tree distribution is the same location whatever order we take thetrees in.Chapter 3*P3.63(a)75Let T represent the force exerted by each child. The xcomponent of the resultant force isaf a f a fT cos 0 + T cos 120+T cos 240 = T 1 + T 0.5 + T 0.5 = 0 .The y-component isT sin 0 + T sin 120 + T sin 240 = 0 + 0.866T 0.866T = 0 .FIG. P3.63Thus, F = 0.(b)P3.64If the total force is not zero, it must point in some direction. When each child moves one360space clockwise, the total must turn clockwise by that angle,. Since each child exertsNthe same force, the new situation is identical to the old and the net force on the tire must stillpoint in the original direction. The contradiction indicates that we were wrong in supposingthat the total force is not zero. The total force must be zero.(a)From the picture, R 1 = a i + b j and R 1 = a 2 + b 2 .(b)R 2 = ai + b j + ck ; its magnitude is2R1 + c 2 = a 2 + b 2 + c 2 .FIG. P3.6476P3.65VectorsSinceA + B = 6.00 j ,we havebAxg ej+ Bx i + A y + B y j = 0 i + 6.00 jFIG. P3.65givingA x + B x = 0 or A x = Bx[1]A y + B y = 6.00 .[2]andSince both vectors have a magnitude of 5.00, we also have2222A x + A y = Bx + By = 5.00 2 .From A x = Bx , it is seen that22A x = Bx .2222Therefore, A x + A y = Bx + By gives22A y = By .Then, A y = By and Eq. [2] givesA y = By = 3.00 .Defining as the angle between either A or B and the y axis, it is seen thatcos =AyA=ByB=3.00= 0.600 and = 53.1 .5.00The angle between A and B is then = 2 = 106 .Chapter 3*P3.66Let represent the angle the x-axis makes with the horizontal. Sinceangles are equal if their sides are perpendicular right side to rightside and left side to left side, is also the angle between the weightand our y axis. The x-components of the forces must add to zero:xy0.127 NTy0.150 N sin + 0.127 N = 0 .0.150 N = 57.9(b)(a)FIG. P3.66The y-components for the forces must add to zero:af+Ty 0.150 N cos 57.9 = 0 , Ty = 0.079 8 N .(c)P3.67The angle between the y axis and the horizontal is 90.057.9= 32.1 .The displacement of point P is invariant under rotation of2the coordinates. Therefore, r = r and r 2 = r or,bgyb g + by g . Also, from the figure, = F y IF yItan G J = tan G J H xKH x Ky e j tan =x 1 + e j tan x 2 + y2 = x2Py21r1yxxtOyxFIG. P3.67Which we simplify by multiplying top and bottom by x cos . Then,x = x cos + y sin , y = x sin + y cos .ANSWERS TO EVEN PROBLEMSP3.2afaf(a) 2.17 m, 1.25 m ; 1.90 m, 3.29 m ;(b) 4.55 mP3.16see the solutionP3.1886.6 m and 50.0 mP3.4(a) 8.60 m;(b) 4.47 m at 63.4 ; 4.24 m at 135P3.201.31 km north; 2.81 km eastP3.6(a) r at 180 ; (b) 2r at 180+ ; (c) 3r at P3.2225.0 m i + 43.3 m jP3.814 km at 65 north of eastP3.2414.0 m s at 11.3 west of northP3.10(a) 6.1 at 112; (b) 14.8 at 22P3.26788 mi at 48.0 north of eastP3.129.5 N at 57P3.287.21 m at 56.3P3.147.9 m at 4 north of westP3.30C = 7.30 cm i 7.20 cm jx7778VectorsP3.50a = 5.00 , b = 7.00P3.522 tan 1P3.5425.4 sP3.56(a) 7.17 km; (b) 6.15 kmP3.587.87 N at 97.8 counterclockwise from ahorizontal line to the rightP3.32(a) 4.47 m at 63.4; (b) 8.49 m at 135P3.3442.7 yardsP3.364.64 m at 78.6P3.38(a) 10.4 cm; (b) 35.5P3.401.43 10 4 m at 32.2 above the horizontalP3.42220 i + 57.6 j = 227 paces at 165P3.44(a) 3.12 i + 5.02 j 2.20k km; (b) 6.31 kmP3.60ej(b) e 7.72 i + 15.1 jj cm;(c) e +7.72 i + 15.1 jj cmP3.62b2.00 m sg j ; its velocity vector(a) a10.0 m, 16.0 mf ; (b) see the solutionP3.64(a) R 1 = a i + b j ; R 1 = a 2 + b 2 ;(a) 74.6 north of east; (b) 470 kmP3.66P3.46P3.48ej(a) 15.1i + 7.72 j cm;FG 1 IJH nK(b) R 2 = ai + b j + ck ; R 2 = a 2 + b 2 + c 2(a) 0.079 8N; (b) 57.9; (c) 32.14Motion in Two DimensionsCHAPTER OUTLINE4.14.24.34.44.54.6ANSWERS TO QUESTIONSThe Position, Velocity, andAcceleration VectorsTwo-Dimensional Motionwith Constant AccelerationProjectile MotionUniform Circular MotionTangential and RadialAccelerationRelative Velocity andRelative AccelerationQ4.1Yes. An object moving in uniform circular motion moves at aconstant speed, but changes its direction of motion. An objectcannot accelerate if its velocity is constant.Q4.2No, you cannot determine the instantaneous velocity. Yes, youcan determine the average velocity. The points could be widelyseparated. In this case, you can only determine the averagevelocity, which isv=Q4.3(a)ava(b)vavavx.tava vavava v10 i m s9.80 j m s 2Q4.4(a)Q4.5The easiest way to approach this problem is to determine acceleration first, velocity second andfinally position.Vertical: In free flight, a y = g . At the top of a projectiles trajectory, v y = 0. Using this, the(b)di2maximum height can be found using v 2 = viy + 2 a y y f yi .fyHorizontal: a x = 0 , so v x is always the same. To find the horizontal position at maximumheight, one needs the flight time, t. Using the vertical information found previously, the flight timecan be found using v fy = viy + a y t . The horizontal position is x f = vix t .If air resistance is taken into account, then the acceleration in both the x and y-directionswould have an additional term due to the drag.Q4.6A parabola.7980Motion in Two DimensionsQ4.7The balls will be closest together as the second ball is thrown. Yes, the first ball will always bemoving faster, since its flight time is larger, and thus the vertical component of the velocity is larger.The time interval will be one second. No, since the vertical component of the motion determines theflight time.Q4.8The ball will have the greater speed. Both the rock and the ball will have the same verticalcomponent of the velocity, but the ball will have the additional horizontal component.Q4.9(a)Q4.10Straight up. Throwing the ball any other direction than straight up will give a nonzero speed at thetop of the trajectory.Q4.11No. The projectile with the larger vertical component of the initial velocity will be in the air longer.Q4.12The projectile is in free fall. Its vertical component of acceleration is the downward acceleration ofgravity. Its horizontal component of acceleration is zero.Q4.13(a)Q4.1460. The projection angle appears in the expression for horizontal range in the function sin 2 . Thisfunction is the same for 30 and 60.Q4.15The optimal angle would be less than 45. The longer the projectile is in the air, the more that airresistance will change the components of the velocity. Since the vertical component of the motiondetermines the flight time, an angle less than 45 would increase range.Q4.16The projectile on the moon would have both the larger range and the greater altitude. Apolloastronauts performed the experiment with golf balls.Q4.17Gravity only changes the vertical component of motion. Since both the coin and the ball are fallingfrom the same height with the same vertical component of the initial velocity, they must hit the floorat the same time.Q4.18(a)yesno(b)(b)nonoyes(c)(c)noyes(b)(d)(d)yes(e)nonoyesIn the second case, the particle is continuously changing the direction of its velocity vector.Q4.19The racing car rounds the turn at a constant speed of 90 miles per hour.Q4.20The acceleration cannot be zero because the pendulum does not remain at rest at the end of the arc.Q4.21(a)The velocity is not constant because the object is constantly changing the direction of itsmotion.(b)The acceleration is not constant because the acceleration always points towards the center ofthe circle. The magnitude of the acceleration is constant, but not the direction.(a)straight aheadQ4.22(b)in a circle or straight aheadChapter 4Q4.23vvaavQ4.24aa vavQ4.25vaa81vrarra aavvavrrvaThe unit vectors r and are in different directions at different points in the xy plane. At a locationalong the x-axis, for example, r = i and = j, but at a point on the y-axis, r = j and = i . The unitvector i is equal everywhere, and j is also uniform.Q4.26The wrench will hit at the base of the mast. If air resistance is a factor, it will hit slightly leeward ofthe base of the mast, displaced in the direction in which air is moving relative to the deck. If the boatis scudding before the wind, for example, the wrenchs impact point can be in front of the mast.Q4.27(a)The ball would move straight up and down as observed by the passenger. The ball wouldmove in a parabolic trajectory as seen by the ground observer.(b)Both the passenger and the ground observer would see the ball move in a parabolictrajectory, although the two observed paths would not be the same.(a)g downwardQ4.28(b)g downwardThe horizontal component of the motion does not affect the vertical acceleration.SOLUTIONS TO PROBLEMSSection 4.1The Position, Velocity, and Acceleration Vectorsa fa fxm0P4.1ym3 6003 00001 2704 270 m(a)1 2702 330 mNet displacement = x 2 + y 2= 4.87 km at 28.6 S of WFIG. P4.1b20.0 m sga180 sf + b25.0 m sga120 sf + b30.0 m sga60.0 sf =(b)Average speed =(c)Average velocity =180 s + 120 s + 60.0 s4.87 10 3 m= 13.5 m s along R360 s23.3 m s82Motion in Two Dimensionsej(a)r = 18.0t i + 4.00t 4.90t 2 j(b)v=(c)P4.2a=b18.0 m sgi + 4.00 m s e9.80 m s jt j2(d)(e)v 3.00 s = 18.0 m s i 25.4 m s j(f)*P4.3e9.80 m s j jra3.00 sf = a54.0 mf i a32.1 mf ja 3.00 s =2af baf e9.80 m s j jg bg2The sun projects onto the ground the x-component of her velocity:af5.00 m s cos 60.0 = 2.50 m s .P4.4(a)From x = 5.00 sin t , the x-component of velocity isvx =FG IJ bH Kgdxd=5.00 sin t = 5.00 cos tdtdtand a x =dv x= +5.00 2 sin tdtsimilarly, v y =and a y =FG d IJ b4.00 5.00 cos tg = 0 + 5.00 sin tH dt KFG d IJ b5.00 sin tg = 5.00H dt K2cos t .e5.00 i + 0 jj m se0 i + 5.00 jj m s .At t = 0 , v = 5.00 cos 0 i + 5.00 sin 0 j =and a = 5.00 2 sin 0 i + 5.00 2 cos 0 j =(b)2a4.00 mf j + a5.00 mfe sin t i cos t jja5.00 mf cos t i + sin t ja5.00 mf sin t i + cos t jr = xi + yj =v=a=(c)22afThe object moves in a circle of radius 5.00 m centered at 0 , 4.00 m .Chapter 4Section 4.2P4.5Two-Dimensional Motion with Constant Accelerationv f = vi + a t(a)v f via==tr f = ri + v i t +(b)e9.00i + 7.00 jj e3.00i 2.00 jj =3.00e2.00i + 3.00 jj m s21 21a t = 3.00 i 2.00 j t + 2.00 i + 3.00 j t 222eejjejejx = 3.00t + t 2 m and y = 1.50t 2 2.00t mP4.6FG IJ ejH Kdv F d Ia== G J e 12.0t jj = 12.0 j m sdt H dt Kv=(a)e2jr = 3.00 i 6.00 j m; v = 12.0 j m s(b)P4.7drd3.00 i 6.00t 2 j = 12.0t j m s=dtdtea f ejjv i = 4.00 i + 1.00 j m s and v 20.0 = 20.0 i 5.00 j m s v x 20.0 4.00=m s 2 = 0.800 m s 2t20.0 v y 5.00 1.00ay ==m s 2 = 0.300 m s 2t20.0(a)ax =(b) = tan 1(c)At t = 25.0 sFG 0.300 IJ = 20.6 =H 0.800 K339 from + x axisa f a fa fa f a fa f112a x t 2 = 10.0 + 4.00 25.0 + 0.800 25.0 = 360 m22112y f = yi + v yi t + a y t 2 = 4.00 + 1.00 25.0 + 0.300 25.0 = 72.7 m22v xf = v xi + a x t = 4 + 0.8 25 = 24 m sx f = xi + v xi t +a fv = v + a t = 1 0.3a 25f = 6.5 m sFv IF 6.50 IJ = 15.2 = tan G J = tan GH 24.0 KHv Kyfyiy1yx18384P4.8Motion in Two Dimensionsa = 3.00 j m s 2 ; v i = 5.00 i m s ; ri = 0 i + 0 j(a)r f = ri + v i t +v f = vi + a t =(b)LM5.00ti + 1 3.00t jOP m2NQe5.00i + 3.00tjj m s1 2at =22a fa fa f j = e10.0 i + 6.00 jj m13.00 2.002so x f = 10.0 m , y f = 6.00 mt = 2.00 s , r f = 5.00 2.00 i +2a f ej+ v = a5.00f + a6.00fv f = 5.00 i + 3.00 2.00 j = 5.00 i + 6.00 j m s2v f = v f = v xf*P4.9(a)22yf2For the x-component of the motion we have x f = xi + v xi t +ej0.01 m = 0 + 1.80 10 7 m s t +e4 10t==14j e= 7.81 m s1axt 2 .218 10 14 m s 2 t 22ejjm s 2 t 2 + 1.80 10 7 m s t 10 2 m = 0e1.8 10 m sj 4e4 102e 4 10 m s j1.80 10 7 m s 271414jejm s 2 10 2 m21.8 10 7 1.84 10 7 m s8 10 14 m s 2We choose the + sign to represent the physical situationt=Herey f = yi + v yi t +4.39 10 5 m s8 1014ms2= 5.49 10 10 s .11a y t 2 = 0 + 0 + 1.6 10 15 m s 2 5.49 10 10 s22eejej2= 2.41 10 4 m .jSo, r f = 10.0 i + 0.241 j mm .(b)ejev f = v i + at = 1.80 10 7 m s i + 8 10 14 m s 2 i + 1.6 10 15 m s 2 j 5.49 10 10 sej ej em sj i + e8.78 10 m sj jj= 1.80 10 7 m s i + 4.39 10 5 m s i + 8.78 10 5 m s j=(c)(d)e1.84 1075e1.84 10 m sj + e8.78 10 m sj =F 8.78 10 I = 2.73Fv I = tan G J = tan Gv KHH 1.84 10 JK27vf =1yx155721.85 10 7 m sjChapter 4Section 4.3P4.10Projectile Motionx = v xi t = vi cos i tbgafax = 300 m s cos 55.0 42.0 sf3x = 7. 23 10 my = v yi t b1 21gt = vi sin i t gt 222gafaf 1 e9.80 m s ja42.0 sf22y = 300 m s sin 55.0 42.0 s P4.11(a)2= 1.68 10 3 mThe mug leaves the counter horizontally with a velocity v xi(say). If time t elapses before it hits the ground, then since thereis no horizontal acceleration, x f = v xi t , i.e.,t=xfv xi=a1.40 mfv xiFIG. P4.11In the same time it falls a distance of 0.860 m with accelerationdownward of 9.80 m s 2 . Theny f = yi + v yi t +11a y t 2 : 0 = 0.860 m + 9.80 m s 222ejFGH 1.40 m IJKvxiThus,e4.90 m s je1.96 m j =2v xi =(b)20.860 m3.34 m s .The vertical velocity component with which it hits the floor isev yf = v yi + a y t = 0 + 9.80 m s 21.40 mjFGH 3.34 m s IJK = 4.11 m s .Hence, the angle at which the mug strikes the floor is given by = tan 1F v I = tan FG 4.11 IJ =GH v JKH 3.34 Kyfxf150.9 .2.8586P4.12Motion in Two DimensionsThe mug is a projectile from just after leaving the counter until just before it reaches the floor. Takingthe origin at the point where the mug leaves the bar, the coordinates of the mug at any time arex f = v xi t +111a x t 2 = v xi t + 0 and y f = v yi t + a y t 2 = 0 g t 2 .222When the mug reaches the floor, y f = h soh = 1 2gt2which gives the time of impact ast=(a)2h.gSince x f = d when the mug reaches the floor, x f = v xi t becomes d = v xi2hgiving theginitial velocity asv xi = d(b)g.2hJust before impact, the x-component of velocity is stillv xf = v xiwhile the y-component isv yf = v yi + a y t = 0 g2h.gThen the direction of motion just before impact is below the horizontal at an angle of = tan 1FvGG vHyfxfIJJ = tanK1FgGGHd2hgg2hIJJ =Ktan 1FG 2 h IJHdK.Chapter 4P4.13(a)The time of flight of the first snowball is the nonzero root of y f = yi + v yi t1 +bgbg0 = 0 + 25.0 m s sin 70.0 t1 t1 =2( 25.0 m s) sin 70.09.80 m s 212a y t12129.80 m s 2 t12ej= 4.79 s .The distance to your target isbagfx f xi = v xi t1 = 25.0 m s cos 70.0 4.79 s = 41.0 m .Now the second snowball we describe byy f = yi + v yi t 2 +bg1ayt2 22ej20 = 25.0 m s sin 2 t 2 4.90 m s 2 t 2aft 2 = 5.10 s sin 2x f xi = v xi t 2bgafaf41.0 m = 25.0 m s cos 2 5.10 s sin 2 = 128 m sin 2 cos 20.321 = sin 2 cos 2Using sin 2 = 2 sin cos we can solve 0.321 =1sin 2 222 2 = sin 1 0.643 and 2 = 20.0 .(b)afat1 t 2 = 4.79 s 1.75 s = 3.05 s .P4.14fThe second snowball is in the air for time t 2 = 5.10 s sin 2 = 5.10 s sin 20 = 1.75 s , so youthrow it after the first byFrom Equation 4.14 with R = 15.0 m , vi = 3.00 m s , max = 45.0g =vi2 9.00== 0.600 m s 2R 15.08788P4.15Motion in Two Dimensionsh=sobgvi2 sin 2 ivi2 sin 2 i; R=; 3h = R ,g2gb23 vi2 sin 2 i vi sin 2 i=g2gg2 sin 2 i tan i==3 sin 2 i24= 53.1 .thus i = tan 13orFG IJHK*P4.16(a)To identify the maximum height we let i be the launch point and f be the highest point:di+ 2b g gb y22v yf = v yi + 2 a y y f yi0=y max =vi2vi22sin imax0g2sin i.2gTo identify the range we let i be the launch and f be the impact point; where t is not zero:1ay t 2210 = 0 + vi sin i t + g t 222 vi sin it=gy f = yi + v yi t +b g1axt 222 v sin i+ 0.d = 0 + vi cos i igx f = xi + v xi t +For this rock, d = y maxvi2 sin 2 i 2 vi2 sin i cos i=g2gsin i= tan i = 4cos i i = 76.0(b)Since g divides out, the answer is the same on every planet.(c)The maximum range is attained for i = 45 :d max vi cos 45 2 vi sin 45 g== 2.125 .dgvi cos 76 2 vi sin 76So d max =17d.8Chapter 4a f(a)x f = v xi t = 8.00 cos 20.0 3.00 = 22.6 m(b)P4.1789Taking y positive downwards,y f = v yi t +1 2gt2a f 1 a9.80fa3.00f2y f = 8.00 sin 20.0 3.00 +(c)af10.0 = 8.00 sin 20.0 t +2= 52.3 m .a f19.80 t 224.90t 2 + 2.74t 10.0 = 0t=*P4.18a2.74f2.74 2+ 196= 1.18 s9.80We interpret the problem to mean that the displacement from fish to bug isaafaff af2.00 m at 30 = 2.00 m cos30 i + 2.00 m sin30 j = 1.73 m i + 1.00 m j.If the water should drop 0.03 m during its flight, then the fish must aim at a point 0.03 m above thebug. The initial velocity of the water then is directed through the point with displacementa1.73 mfi + a1.03 mf j = 2.015 m at 30.7.For the time of flight of a water drop we have1axt 22x f = xi + v xi t +bg1.73 m = 0 + vi cos 30.7 t + 0 sot=1.73 m.vi cos 30.7The vertical motion is described byy f = yi + v yi t +1ayt 2 .2The drop on its path is13.00 cm = 9.80 m s 22e.73 mjFGH v 1cos 30.7 IJK2.iThus,vi =1.73 m 9.80 m s 2= 2.015 m 12.8 s 1 = 25.8 m s .cos30.7 2 0.03 mej90P4.19Motion in Two Dimensions(a)We use the trajectory equation:y f = x f tan i gx 2f2 vi2 cos 2 i.Withx f = 36.0 m, vi = 20.0 m s, and = 53.0we finde9.80 m s ja36.0 mf = 3.94 m.= a36.0 mf tan 53.02b 20.0 m sg cos a53.0f22yf22The ball clears the bar bya3.94 3.05f m =(b)0.889 m .The time the ball takes to reach the maximum height ist1 =bfga20.0 m s sin53.0vi sin i== 1.63 s .g9.80 m s 2xfThe time to travel 36.0 m horizontally is t 2 =t2 =vix36.0 m= 2.99 s .( 20.0 m s) cos 53.0afSince t 2 > t1 the ball clears the goal on its way down .P4.20bgThe horizontal component of displacement is x f = v xi t = vi cos i t . Therefore, the time required todreach the building a distance d away is t =. At this time, the altitude of the water isvi cos iy f = v yi t +FGHIJ FGK Hgdd1a y t 2 = vi sin ivi cos i22 vi cos iIJKTherefore the water strikes the building at a height h above ground level ofh = y f = d tan i gd 22 vi2 cos 2 i.2.Chapter 4*P4.21(a)For the horizontal motion, we have1axt 2224 m = 0 + vi cos 53 2.2 s + 0x f = xi + v xi t +afafvi = 18.1 m s .(b)As it passes over the wall, the ball is above the street by y f = yi + v yi t +bfagaf 1 e9.8 m s ja2.2 sf22y f = 0 + 18.1 m s sin 53 2.2 s +21ayt 22= 8.13 m .So it clears the parapet by 8.13 m 7 m = 1.13 m .(c)Note that the highest point of the balls trajectory is not directly above the wall. For thewhole flight, we have from the trajectory equationg FGH 2vby f = tan i x f g2icos2Ix JKi2forF.8f GG 2 18.19m sm scosgHba26 m = tan 53 x f 22IJx .53 JKSolving,e0.041 2 m jx12f 1.33 x f + 6 m = 0andxf =afa f .1.33 1.33 2 4 0.0412 6e2 0.0412 m1jThis yields two results:x f = 26.8 m or 5.44 mThe ball passes twice through the level of the roof.It hits the roof at distance from the wall26.8 m 24 m = 2.79 m .2f9192*P4.22Motion in Two DimensionsWhen the bomb has fallen a vertical distance 2.15 km, it has traveled a horizontal distance x f given byxf =a3.25 kmf a2.15 kmf2y f = x f tan 2= 2.437 kmgx 2f2 vi2 cos 2 ie9.8 m s jb2 437 mg2 150 m = b 2 437 mg tan 2b 280 m sg cos 2 150 m = b 2 437 mg tan a371.19 mfe1 + tan j22i22i2ii2 tan 6.565 tan i 4.792 = 0 tan i =FH16.565 2a6.565f 4a1fa4.792f IK = 3.283 3.945 .2Select the negative solution, since i is below the horizontal. tan i = 0.662 , i = 33.5P4.23The horizontal kick gives zero vertical velocity to the rock. Then its time of flight follows fromy f = yi + v yi t +1ayt 2219.80 m s 2 t 22t = 2.86 s .40.0 m = 0 + 0 +ejThe extra time 3.00 s 2.86 s = 0.143 s is the time required for the sound she hears to travel straightback to the player.It covers distanceb343 m sg0.143 s = 49.0 m =awhere x represents the horizontal distance the rock travels.x = 28.3 m = v xi t + 0t 2 v xi =fx 2 + 40.0 m28.3 m= 9.91 m s2.86 s293Chapter 4P4.24From the instant he leaves the floor until just before he lands, the basketball star is a projectile. His22vertical velocity and vertical displacement are related by the equation v yf = v yi + 2 a y y f yi .Applying this to the upward part of his flight gives 0 =2v yid i+ 2e 9.80 m s ja1.85 1.02f m . From this,2v yi = 4.03 m s . [Note that this is the answer to part (c) of this problem.]jaef2For the downward part of the flight, the equation gives v yf = 0 + 2 9.80 m s 2 0.900 1.85 m .Thus the vertical velocity just before he lands isv yf = 4.32 m s.(a)His hang time may then be found from v yf = v yi + a y t :ej4.32 m s = 4.03 m s + 9.80 m s 2 tor t = 0.852 s .(b)Looking at the total horizontal displacement during the leap, x = v xi t becomesa2.80 m = v xi 0.852 sfwhich yields v xi = 3.29 m s .(c)v yi = 4.03 m s . See above for proof.(d)The takeoff angle is: = tan 1F v I = tan F 4.03 m s I =GH v JKGH 3.29 m s JKyi150.8 .xi(e)Similarly for the deer, the upward part of the flight gives22v yf = v yi + 2 a y y f yi :dijaef20 = v yi + 2 9.80 m s 2 2.50 1.20 mso v yi = 5.04 m s .dijaef222For the downward part, v yf = v yi + 2 a y y f yi yields v yf = 0 + 2 9.80 m s 2 0.700 2.50 mand v yf = 5.94 m s.ejThe hang time is then found as v yf = v yi + a y t : 5.94 m s = 5.04 m s + 9.80 m s 2 t andt = 1.12 s .94*P4.25Motion in Two DimensionsThe arrows flight time to the collision point ist=x f xiv xi=b150 m= 5.19 s .45 m s cos 50gThe arrows altitude at the collision isy f = yi + v yi t +b1ayt 22fga= 0 + 45 m s sin 50 5.19 s +(a)ja19.8 m s 2 5.19 s2ef2= 47.0 m .The required launch speed for the apple is given byd22v yf = v yi + 2 a y y f yieija20 = v yi + 2 9.8 m s 2 47 m 0fv yi = 30.3 m s .(b)The time of flight of the apple is given byv yf = v yi + a y t0 = 30.3 m s 9.8 m s 2 tt = 3.10 s .So the apple should be launched after the arrow by 5.19 s 3.10 s = 2.09 s .*P4.26For the smallest impact angle = tan 1Fv I,GH v JKyfxfwe want to minimize v yf and maximize v xf = v xi . The final y-component22of velocity is related to v yi by v yf = v yi + 2 gh , so we want to minimize v yiand maximize v xi . Both are accomplished by making the initial velocityhorizontal. Then v xi = v , v yi = 0 , and v yf = 2 gh . At last, the impactangle is = tan 1Fv I =GH v JKyfxftan 1FGH2 ghvIJK.FIG. P4.26Chapter 4Section 4.4P4.27P4.28Uniform Circular Motionbgv2, T = 24 h 3 600 s h = 86 400 sR2 R 2 ( 6.37 10 6 m)v=== 463 m sT86 400 sba=g266.37 10 m= 0.033 7 m s 2 directed toward the center of Earth .r = 0.500 m;vt =a=P4.30220.0 m sv2ac === 377 m s 2r1.06 mThe mass is unnecessary information.b463 m sga=P4.2995ac =af2 r 2 0.500 m== 10. 47 m s = 10.5 m s60 .0 sT200 reva10. 47v2=R0.5v2ref2= 219 m s 2 inwardjafv = a c r = 3 9.8 m s 2 9.45 m = 16.7 m safEach revolution carries the astronaut over a distance of 2 r = 2 9.45 m = 59.4 m. Then the rotationrate is16.7 m sP4.31(a)v = raafbfbFG 1 rev IJ =H 59.4 m Kgbgb0.281 rev s .ggAt 8.00 rev s , v = 0.600 m 8.00 rev s 2 rad rev = 30.2 m s = 9.60 m s .At 6.00 rev s , v = 0.900 m 6.00 rev s 2 rad rev = 33.9 m s = 10.8 m s .6.00 rev s gives the larger linear speed.b(b)9.60 m sv2=Acceleration =r0.600 m(c)At 6.00 rev s , acceleration =g2= 1.52 10 3 m s 2 .b10.8 m sg0.900 m2= 1.28 10 3 m s 2 .96P4.32Motion in Two DimensionsThe satellite is in free fall. Its acceleration is due to gravity and is by effect a centripetal acceleration.ac = gsov2= g.rSolving for the velocity, v = rg =a6 ,400 + 600fe10 mje8.21 m s j =3v=27.58 10 3 m s2rTandeT = 5.80 10 3 sSection 4.5P4.33j32 r 2 7 ,000 10 m== 5.80 10 3 sv7.58 10 3 m sT=FG 1 min IJ = 96.7 min .H 60 s KTangential and Radial AccelerationWe assume the train is still slowing down at the instant in question.ac =v2= 1.29 m s 2rat =40.0 km h 10 3 m kmv=15.0 stbge1h3 600 sj = 0.741 m se1.29 m s j + e0.741 m s jF a I = tan FG 0.741 IJGH a JKH 1.29 K2 22a = a c + a t2 =at an angle of tan 1jet2 21ca = 1.48 m s 2 inward and 29.9 o backwardP4.34(a)at = 0.600 m s 2(b)4.00 m sv2ar ==r20.0 m(c)2a = a t2 + a r = 1.00 m s 2b = tan 1g2= 0.800 m s 2ar= 53.1 inward from pathat2FIG. P4.33Chapter 4P4.35r = 2.50 m , a = 15.0 m s 2(a)(b)jaefa c = a cos 30.0 o = 15.0 m s 2 cos 30 = 13.0 m s 2v2rso v 2 = ra c = 2.50 m 13.0 m s 2 = 32.5 m 2 s 2ac =ejFIG. P4.35v = 32.5 m s = 5.70 m s(c)2a 2 = a t2 + a r2so a t = a 2 a r =P4.3697e15.0 m s j e13.0 m s j =2 227.50 m s 2(a)See figure to the right.(b)The components of the 20.2 and the 22.5 m s 2 along the rope togetherconstitute the centripetal acceleration:ej af eja c = 22.5 m s 2 cos 90.036.9 + 20.2 m s 2 cos 36.9 = 29.7 m s 2(c)*P4.37v2so v = a c r = 29.7 m s 2 1.50 m = 6.67 m s tangent to circlerv = 6.67 m s at 36.9 above the horizontalaac =fFIG. P4.36atLet i be the starting point and f be one revolution later. The curvilinear motionwith constant tangential acceleration is described by x = v xi t +1axt 221 2at t24 raar2 r = 0 +at =and v xf = v xi + a x t , v f = 0 + a t t =Then tan =4 r t 2at1= 2=a r t 16 2 r 4tFIG. P4.372v 2 16 2 r 24 rf=. The magnitude of the radial acceleration is a r =.rtt 2r = 4.55 .98Motion in Two DimensionsSection 4.6P4.38(a)Relative Velocity and Relative Accelerationeajv H = 0 + a H t = 3.00 i 2.00 j m s 2 5.00 sejfv H = 15.0 i 10.0 j m sejav J = 0 + a j t = 1.00 i + 3.00 j m s 2 5.00 sefjv J = 5.00 i +15.0 j m sejv HJ = v H v J = 15.0 i 10.0 j 5.00 i 15.0 j m sejv HJ = 10.0 i 25.0 j m sv HJ = (10.0) 2 + ( 25.0) 2 m s = 26.9 m s(b)rH = 0 + 0 +ea11a H t 2 = 3.00 i 2.00 j m s 2 5.00 s22ejf2jrH = 37.5 i 25.0 j mf = e12.5 i + 37.5 jj m= r r = e37.5 i 25.0 j 12.5 i 37.5 jj m= e 25.0 i 62.5 jj m= a 25.0 f + a62.5 f m = 67.3 mrJ =rHJrHJrHJ(c)eHj2J22eja HJ = a H a J = 3.00 i 2.00 j 1.00 i 3.00 j m s 2a HJ =*P4.39a11.00 i + 3.00 j m s 2 5.00s2e2.00i 5.00 jj m s2v ce = the velocity of the car relative to the earth.v wc = the velocity of the water relative to the car.v we = the velocity of the water relative to the earth.These velocities are related as shown in the diagram at the right.(a)(b)Since v we is vertical, v wc sin 60.0 = v ce = 50.0 km h orv wc = 57.7 km h at 60.0 west of vertical .Since v ce has zero vertical component,bvcevwe60vwcv we = v ce + v wcFIG. P4.39gv we = v wc cos 60.0 = 57.7 km h cos 60.0 = 28.9 km h downward .Chapter 4P4.40The bumpers are initially 100 m = 0.100 km apart. After time t the bumper of the leading car travels40.0 t, while the bumper of the chasing car travels 60.0t.Since the cars are side by side at time t, we have0.100 + 40.0t = 60.0t ,yieldingt = 5.00 10 3 h = 18.0 s .P4.41Total time in still water t =d 2 000== 1.67 10 3 s .v 1. 20Total time = time upstream plus time downstream:1 000= 1.43 10 3 s(1.20 0.500 )1 000== 588 s .1.20 + 0.500t up =t downTherefore, ttotal = 1.43 10 3 + 588 = 2.02 10 3 s .P4.42v = 150 2 + 30.0 2 = 153 km h = tan 1P4.4399FG 30.0 IJ =H 150 K11.3 north of westFor Alan, his speed downstream is c + v, while his speed upstream is c v .Therefore, the total time for Alan ist1 =2LLLc+=2c+v cv1 v2.cFor Beth, her cross-stream speed (both ways) isc2 v2 .Thus, the total time for Beth is t 2 =Since 1 2Lc2 v2=2Lc1v2c2.v2< 1 , t1 > t 2 , or Beth, who swims cross-stream, returns first.c2100P4.44Motion in Two Dimensions(a)To an observer at rest in the train car, the bolt accelerates downward and toward the rear ofthe train.a=tan =b2.50 m sg + b9.80 m sg22.50 m s 29.80 m s 22= 10.1 m s 2= 0.255 = 14.3 to the south from the vertical(b)P4.45a = 9.80 m s 2 vertically downwardIdentify the student as the S observer and the professor asthe S observer. For the initial motion in S, we havevyvx= tan 60.0 = 3 .Let u represent the speed of S relative to S. Then becausethere is no x-motion in S, we can write v x = v + u = 0 soxthat v x = u = 10.0 m s . Hence the ball is thrownbackwards in S. Then,v y = v = 3 v = 10.0 3 m s .yx2Using v y = 2 gh we finde10.0 3 m sj =h=2e9.80 m s jFIG. P4.452215.3 m .The motion of the ball as seen by the student in S is shown in diagram (b). The view of the professorin S is shown in diagram (c).*P4.46Choose the x-axis along the 20-km distance. The ycomponents of the displacements of the ship andthe speedboat must agree:b26 km hgt sina4015f = b50 km hgt sin 11.0= 12.7 . = sin 150xN254015EyThe speedboat should head15+12.7 = 27.7 east of north .FIG. P4.46Chapter 4101Additional Problemsbg(a)The speed at the top is v x = vi cos i = 143 m s cos 45 = 101 m s .(b)*P4.47In free fall the plane reaches altitude given bydi0 = b143 m s sin 45g + 2e 9.8 m s jd y 31 000 ft iF 3.28 ft IJ = 3.27 10 ft .y = 31 000 ft + 522 mGH 1m K22v yf = v yi + 2 a y y f yi223f(c)fFor the whole free fall motion v yf = v yi + a y tej101 m s = +101 m s 9.8 m s 2 tt = 20.6 s(d)ac =v2rejv = a c r = 0.8 9.8 m s 2 4,130 m = 180 m sP4.48At any time t, the two drops have identical y-coordinates. The distance between the two drops isthen just twice the magnitude of the horizontal displacement either drop has undergone. Therefore,af b g bgd = 2 x t = 2 v xi t = 2 vi cos i t = 2 vi t cos i .P4.49After the string breaks the ball is a projectile, and reaches the ground at time t: y f = v yi t +1.20 m = 0 +19.80 m s 2 t 22ejso t = 0.495 s.Its constant horizontal speed is v x =x 2.00 m== 4.04 m st 0.495 sbg4.04 m sv2so before the string breaks a c = x =r0.300 m2= 54.4 m s 2 .1ayt 22102P4.50Motion in Two Dimensions(a)bgd iy f = tan i x f g2 vi2x2fcos 2 iSetting x f = d cos , and y f = d sin , we havebgbgd sin = tan i d cos Solving for d yields, d =or d =(b)P4.51Settingbg2 vi2bd cos g .22cos iFIG. P4.502 vi2 cos i sin i cos sin cos i2 vi2 cos i sin i g cos 2 g2g cos .bv 2 1 sin dd= 0 leads to i = 45+and d max = id i2g cos 2 g.Refer to the sketch:(b)bg x = v xi t ; substitution yields 130 = vi cos 35.0 t .1 y = v yi t + at 2 ; substitution yields2bagf19.80 t 2 .220.0 = vi sin 35.0 t +Solving the above gives t = 3.81 s .(a)vi = 41.7 m s(c)FIG. P4.51v yf = vi sin i gt , v x = vi cos ia fa fAt t = 3.81 s , v yf = 41.7 sin 35.0 9.80 3.81 = 13.4 m safv x = 41.7 cos 35.0 = 34.1 m s22v f = v x + v yf = 36.7 m s .Chapter 4P4.52(a)The moons gravitational acceleration is the probes centripetal acceleration:(For the moons radius, see end papers of text.)a=v2rv219.80 m s 2 =61.74 10 6 mejv = 2.84 10 6 m 2 s 2 = 1.69 km s(b)P4.53(a)2 rT2 r 2 (1.74 10 6 m)T=== 6.47 10 3 s = 1.80 hv1.69 10 3 m sv=b5.00 m sv2ac ==r1.00 mg2= 25.0 m s 2at = g = 9.80 m s 2(b)See figure to the right.(c)2a = a c + a t2 = = tan 1e25.0 m s j + e9.80 m s jFG a IJ = tanHa Kt2 21cP4.549.80 m s 225.0 m s 22 2= 26.8 m s 2= 21.4FIG. P4.53x f = vix t = vi t cos 40.010.0 m.vi cos 40.0At this time, y f should be 3.05 m 2.00 m = 1.05 m .Thus, when x f = 10.0 m , t =Thus, 1.05 m =bv sin 40.0g10.0 m + 1 e9.80 m s jLM 10.0 m OP .v cos 40.02N v cos 40.0 Q2i2iFrom this, vi = 10.7 m s .i103104P4.55Motion in Two DimensionsThe special conditions allowing use of the horizontal range equation applies.For the ball thrown at 45,D = R 45 =vi2 sin 90.gFor the bouncing ball,v 2 sin 2D = R1 + R 2 = i+gvi 22e jsin 2gwhere is the angle it makes with the ground when thrown and when bouncing.(a)We require:vi2 vi2 sin 2 vi2 sin 2=+gg4g45 = 26.6sin 2 =(b)FIG. P4.55The time for any symmetric parabolic flight is given by1 2gt210 = vi sin i t gt 2 .2y f = v yi t If t = 0 is the time the ball is thrown, then t =2 vi sin iis the time at landing.gSo for the ball thrown at 45.0t 45 =2 vi sin 45.0.gFor the bouncing ball,2 v sin 26.6 2t = t1 + t 2 = i+ge j sin 26.6 = 3v sin 26.6 .vi2igThe ratio of this time to that for no bounce is3 vi sin 26.6 g2 vi sin 45 .0 g=1.34= 0.949 .1.41gChapter 4P4.56105Using the range equation (Equation 4.14)R=vi2 sin( 2 i )gvi2. Given R, this yields vi = gR .gIf the boy uses the same speed to throw the ball vertically upward, thenthe maximum range occurs when i = 45 , and has a value R =v y = gR gt and y = gR t gt 22at any time, t.R, and so the maximum height reached isgAt the maximum height, v y = 0, giving t =y maxP4.57R g= gRg 2F RIGH g JK2= RRR=.22Choose upward as the positive y-direction and leftward as thepositive x-direction. The vertical height of the stone when releasedfrom A or B isavifyi = 1.50 + 1.20 sin 30.0 m = 2.10 m(a)BAThe equations of motion after release at A area30 1.20 mfv y = vi sin 60.0 gt = 1.30 9.80t m s30viv x = vi cos 60.0 = 0.750 m se x = a0.750t f m1.30 a1.30fWhen y = 0 , t =jy = 2.10 +1.30t 4.90t 2 mFIG. P4.57A2+ 41.29.80(b)afaf= 0.800 s. Then, x A = 0.750 0.800 m = 0.600 m .The equations of motion after release at point B areafafv y = vi sin 60.0 gt = 1.30 9.80t m sv x = vi cos 60.0 = 0.750 m sejyi = 2.10 1.30t 4.90t 2 m .When y = 0 , t =b+1.30 1.50 m sv2=r1.20 ma1.30f9.80g2+ 41.2afaf= 0.536 s. Then, x B = 0.750 0.536 m = 0.402 m .2= 1.87 m s 2 toward the center(c)ar =(d)After release, a = g j = 9.80 m s 2 downward106P4.58Motion in Two DimensionsThe football travels a horizontal distanceR=b g = a20.0f sina60.0f = 35.3 m.vi2 sin 2 ig29.80Time of flight of ball isFIG. P4.582 v sin i 2( 20.0) sin 30.0t= i== 2.04 s .g9.80The receiver is x away from where the ball lands and x = 35.3 20.0 = 15.3 m. To cover thisdistance in 2.04 s, he travels with a velocityv=P4.59(a)15.3= 7.50 m s in the direction the ball was thrown .2.041 2g t ; x = vi t2Combine the equations eliminating t:y= FG IJH K1 xy= gvi2b g = FGH 2g y IJK vFrom this, x22.2iFIG. P4.59thus x = vi2 y2( 300 )= 275= 6.80 10 3 = 6.80 km .g9.80(b)The plane has the same velocity as the bomb in the x direction. Therefore, the plane will be3 000 m directly above the bomb when it hits the ground.(c)When is measured from the vertical, tan =therefore, = tan 1F x I = tan F 6 800 I =GH y JKGH 3 000 JK1xy66. 2 .Chapter 4*P4.60(a)We use the approximation mentioned in the problem. The time to travel 200 m horizontally isx200 mt=== 0.200 s . The bullet falls byv x 1,000 m s(c)ja f11a y t 2 = 0 + 9.8 m s 2 0.2 s22e y = v yi t +(b)107The telescope axis must point below the barrel axis0.196 m= 0.056 1 .by = tan 1200 mt=50.0 m= 0.050 0 s . The bullet falls by only1 000 m sy=ja19.8 m s 2 0.05 s2eaf22= 0.196 m .barrel axisbullet pathscope axis50150 200 250FIG. P4.60(b)= 0.0122 m .aff11200 m , the scope axis points to a location 19.6 cm = 4.90 cm above the44barrel axis, so the sharpshooter must aim low by 4.90 cm 1.22 cm = 3.68 cm .At range 50 m =(d)t=150 m= 0.150 s1 000 m sjaaf129.8 m s 2 0.15 s = 0.110 m215019.6 cm 11.0 cm = 3.68 cm .Aim low by200y=(e)t=ef250 m= 0.250 s1 000 m sy=ja19.8 m s 2 0.25 s2eAim high by 30.6 cm f2= 0.306 maf25019.6 cm = 6.12 cm .200(f), (g) Many marksmen have a hard time believing it, butthey should aim low in both cases. As in case (a) above,the time of flight is very nearly 0.200 s and the bulletfalls below the barrel axis by 19.6 cm on its way. The0.0561 angle would cut off a 19.6-cm distance on avertical wall at a horizontal distance of 200 m, but on avertical wall up at 30 it cuts off distance h as shown,where cos 30 = 19.6 cm h , h = 22.6 cm. The marksmanbarrel axisscope3019.6 cmmust aim low by 22.6 cm 19.6 cm = 3.03 cm . Theanswer can be obtained by considering limiting cases.Suppose the target is nearly straight above or belowyou. Then gravity will not cause deviation of the pathof the bullet, and one must aim low as in part (c) tocancel out the sighting-in of the telescope.30h19.6 cmscope axisbullet hits hereFIG. P4.60(fg)108P4.61Motion in Two Dimensions(a)From Part (c), the raptor dives for 6.34 2.00 = 4.34 sundergoing displacement 197 m downward and10.0 4.34 = 43.4 m forward.a fa fdv=t(b)(c)P4.62 = tan 1197 =FG 197 IJ =H 43.4 Ka197f + a43.4f22= 46.5 m s4.3477.61 2gt , t = 6.34 s2FIG. P4.61Measure heights above the level ground. The elevation y b of the ball followsyb = R + 0 with x = vi t so y b = R (a)gx 22 vi21 2gt2.The elevation yr of points on the rock is described by2yr + x 2 = R 2 .We will have y b = y r at x = 0 , but for all other x we require the ball to be above the rock2surface as in y b > y r . Then y b + x 2 > R 2F R gx IGH 2 v JK22+ x2 > R22iR2 gx 2 Rvi2+g 2x44vi4g 2x44vi4+ x2 > R2+ x2 >gx 2 Rvi2.If this inequality is satisfied for x approaching zero, it will be true for all x. If the ballsparabolic trajectory has large enough radius of curvature at the start, the ball will clear thegRwhole rock: 1 > 2vivi > gR .(b)With vi = gR and y b = 0 , we have 0 = R gx 22 gRor x = R 2 .The distance from the rocks base isxR=ej2 1 R .Chapter 4P4.63(a)While on the inclinev 2 vi2 = 2 axfv f vi = ata fa fv 2 0 = 2 4.00 50.0f20.0 0 = 4.00tv f = 20.0 m st = 5.00 s(b)FIG. P4.63Initial free-flight conditions give usv xi = 20.0 cos 37.0 = 16.0 m sandv yi = 20.0 sin 37.0 = 12.0 m sv xf = v xi since a x = 0afaf av yf = 2 a y y + v yi 2 = 2 9.80 30.0 + 12.0a16.0f + a27.1f2v f = v xf 2 + v yf 2 =(c)t1 = 5 s ; t 2 =v yf v yiay=2f2= 27.1 m s= 31.5 m s at 59.4 below the horizontal27.1 + 12.0= 1.53 s9.80t = t1 + t 2 = 6.53 s(d)P4.64a fx = v xi t1 = 16.0 1.53 = 24.5 my 2 = 16 xEquation of bank:Equations of motion:x = vi t1y = g t22a1fa 2fa3 fF IGH JK1 x2g. Equate y2 vi2from the bank equation to y from the equations of motion:Substitute for t from (2) into (3) y = L 1 Fx16 x = M g GMN 2 H vFrom this, x = 0 or x3FIG. P4.64I OP g x 16x = xF g x 16I = 0 .JK PQ 4vGH 4v JKF 10 I = 18.8 m . Also,64v=and x = 4GgH 9.80 JK1 Fx I1 a9.80fa18.8fy = gG J = 2 Hv K2a10.0f = 17.3 m .2222 42i4i2 34i4i1/342222i2109110P4.65Motion in Two Dimensions(a)a f1 21at ; 70.0 = 15.0 t 22270.0 = vi tRoadrunner: x = vi t ;Coyote:x =Solving the above, we getvi = 22.9 m s and t = 3.06 s.(b)At the edge of the cliff,a fa fv xi = at = 15.0 3.06 = 45.8 m s .Substituting into y =1a y t 2 , we find2af19.80 t 22t = 4.52 s100 = x = v xi t +a faf a faf112a x t 2 = 45.8 4.52 s + 15.0 4.52 s .22Solving, x = 360 m .(c)For the Coyotes motion through the aira ft = 0 9.80a 4.52f =v xf = v xi + a x t = 45.8 + 15 4.52 = 114 m sv yf = v yi + a yP4.6644.3 m s .Think of shaking down the mercury in an old fever thermometer. Swing your hand through acircular arc, quickly reversing direction at the bottom end. Suppose your hand moves through onequarter of a circle of radius 60 cm in 0.1 s. Its speed is14a2 fa0.6 mf 9 m s0.1 sand its centripetal acceleration isv 2 ( 9 m s) 2~ 10 2 m s 2 .r0.6 mThe tangential acceleration of stopping and reversing the motion will make the total accelerationsomewhat larger, but will not affect its order of magnitude.Chapter 4P4.67(a) x = v xi t , y = v yi t +1 2gt2a111fd cos 50.0 = 10.0 cos 15.0 tandaf d sin 50.0 = 10.0 sin 15.0 t +af19.80 t 2 .2Solving, d = 43.2 m and t = 2.88 s .(b)Since a x = 0 ,FIG. P4.67v xf = v xi = 10.0 cos 15.0 = 9.66 m sa fv yf = v yi + a y t = 10.0 sin 15.09.80 2.88 = 25.6 m s .Air resistance would decrease the values of the range and maximum height. As an airfoil, hecan get some lift and increase his distance.*P4.68For one electron, we havey = viy t , D = vix t +11a x t 2 a x t 2 , v yf = v yi , and v xf = v xi + a x t a x t .22The angle its direction makes with the x-axis is given by = tan 1v yfv xf= tan 1v yiaxt= tan 1v yi taxt2= tan 1y.2DFIG. P4.68Thus the horizontal distance from the aperture to the virtual source is 2D. The source is atcoordinate x = D .*P4.69(a)The ice chest floats downstream 2 km in time t, so that 2 km = v w t . The upstream motion ofthe boat is described by d = ( v v w )15 min. The downstream motion is described by2 kmd + 2 km = ( v + v w )( t 15 min) . We eliminate t =and d by substitution:vwbv v g15 min + 2 km = bv + v gFGH 2vkm 15 minIJKvva15 minf v a15 minf + 2 km =2 km + 2 km va15 minf v a15 minfvvva30 minf =2 kmvwwwwwww2 kmvw == 4.00 km h .30 min(b)In the reference frame of the water, the chest is motionless. The boat travels upstream for 15 minat speed v, and then downstream at the same speed, to return to the same point. Thus it travelsfor 30 min. During this time, the falls approach the chest at speed v w , traveling 2 km. Thusvw =x2 km== 4.00 km h . t 30 min112*P4.70Motion in Two DimensionsLet the river flow in the x direction.(a)To minimize time, swim perpendicular to the banks in the y direction. You are in thewater for time t in y = v y t , t =(b)80 m= 53.3 s .1.5 m sbgThe water carries you downstream by x = v x t = 2.50 m s 53.3 s = 133 m .(c)vwvwvwvsvsvs + vwvsvs + vwvs + vwTo minimize downstream drift, you should swim so thatyour resultant velocity v s + v w is perpendicular to yourswimming velocity v s relative to the water. This conditionis shown in the middle picture. It maximizes the anglebetween the resultant velocity and the shore. The angle1.5 m sbetween v s and the shore is given by cos =,2.5 m svsvs + vwv w = 2.5 m/s i = 53.1 .(d)Now v y = v s sin = 1.5 m s sin 53.1 = 1.20 m st=y80 m== 66.7 sv y 1.2 m sbg x = v x t = 2.5 m s 1.5 m s cos 53.1 66.7 s = 107 m .Chapter 4*P4.71113Find the highest firing angle H for which the projectile will clear the mountain peak; this willyield the range of the closest point of bombardment. Next find the lowest firing angle; this will yieldthe maximum range under these conditions if both H and L are > 45 ; x = 2500 m, y = 1800 m ,vi = 250 m s .a f1 21gt = vi sin t gt 222x f = v xi t = vi cos ty f = v yi t afThust=xfvi cos .Substitute into the expression for y fxxa f v cos 1 gFGH v cos IJK2y f = vi sin ffii2= x f tan gx 2f2 vi2 cos 2 gx 21f2= tan + 1 so y f = x f tan 2 tan 2 + 1 andbut2cos 2 vie0=gx 2f2 vi2jtan 2 x f tan +gx 2f2 vi2+ yf .Substitute values, use the quadratic formula and findtan = 3.905 or 1.197 , which gives H = 75.6 and L = 50.1 .bgRange at H =vi2 sin 2 H= 3.07 10 3 m from enemy shipg3.07 10 3 2 500 300 = 270 m from shore.bgRange at L =vi2 sin 2 L= 6.28 10 3 m from enemy shipg6.28 10 3 2 500 300 = 3.48 10 3 from shore.Therefore, safe distance is < 270 m or > 3.48 10 3 m from the shore.FIG. P4.71114*P4.72Motion in Two DimensionsWe follow the steps outlined in Example 4.7, eliminating t =d cos to findvi cos vi sin d cos gd 2 cos 2 2= d sin .vi cos 2 vi cos 2 Clearing of fractions,2 vi2 cos sin cos gd cos 2 = 2 vi2 cos 2 sin .To maximize d as a function of , we differentiate through with respect to and setaf2 vi2 cos cos cos + 2 vi2 sin sin cos gdd= 0:dafddcos 2 = 2 vi2 2 cos sin sin .dWe use the trigonometric identities from Appendix B4 cos 2 = cos 2 sin 2 andsin 1sin 2 = 2 sin cos to find cos cos 2 = sin 2 sin . Next,= tan and cot 2 =givecos tan 2afcot 2 = tan = tan 902 so = 902 and = 45.2ANSWERS TO EVEN PROBLEMSP4.2ej(b) v = 18.0 i + a 4.00 9.80t f j ;(c) a = e 9.80 m s j j ;(d) a54.0 mf i a32.1 mf j ;(e) b18.0 m sg i b 25.4 m sg j ;(f) e 9.80 m s j j(a) r = 18.0t i + 4.00t 4.90t 2 j ;P4.8eeP4.10P4.12j(a) v = 5.00 i + 0 j m s ;ee(c) a circle of radius 5.00 m centered at0 , 4.00 maP4.6f2(a) v = 12.0t j m s ; a = 12.0 j m s ;ej3jm, 1.68 10 3 mghorizontally;2h2hbelow the horizontal(b) tan 1d(a) d(b) r = 3.00 i 6.00 j m; v = 12.0 j m s0.600 m s 2P4.16jesin t i + cos t jj ;v = 5.00 m cos t i + sin t j ;a = 5.00 m 2e7.23 10P4.14j+5.00 m sin t i cos t j ;ejFG IJH Kja = 0 i + 5.00 2 j m s 2 ;(b) r = 4.00 m jj(b) r = 10.0 i + 6.00 j m; 7.81 m s2ejv = 5.00 i + 3.00t j m s ;2P4.4e(a) r = 5.00t i + 1.50t 2 j m ;(a) 76.0; (b) the same; (c)P4.1825.8 m sP4.20d tan i e2 vgd 22icos 2 ij17d8Chapter 4P4.2233.5 below the horizontalP4.482vi t cos iP4.24(a) 0.852 s; (b) 3.29 m s ; (c) 4.03 m s;(d) 50.8; (e) 1.12 sP4.50(a) see the solution;1151FGH2ghIJKP4.26tanP4.280.033 7 m s 2 toward the center of theEarthv(b) i = 45+ ; d max =2bvi2 1 sin g cos P4.520.281 rev sP4.327.58 10 3 m s; 5.80 10 3 sP4.34(a) 0.600 m s 2 forward;(b) 0.800 m s 2 inward;(a) 1.69 km s ; (b) 6.47 10 3 sP4.5410.7 m sP4.56P4.30g2R2P4.587.50 m s in the direction the ball wasthrownP4.60(a) 19.6 cm; (b) 0.0561;(c) aim low 3.68 cm; (d) aim low 3.68 cm;(e) aim high 6.12 cm; (f) aim low;(g) aim low(c) 1.00 m s 2 forward and 53.1 inwardP4.36(a) see the solution; (b) 29.7 m s 2 ;(c) 6.67 m s at 36.9 above the horizontalP4.62(a)P4.38(a) 26.9 m s ; (b) 67.3 m;P4.64a18.8 m; 17.3 mfP4.66see the solution; ~ 10 2 m s 2P4.68x = DP4.70(a) at 90 to the bank; (b) 133 m;(c) upstream at 53.1 to the bank; (d) 107 mP4.72see the solutionej(c) 2.00 i 5.00 j m s18.0 sP4.42153 km h at 11.3 north of west2(a) 10.1 m s at 14.3 south from thevertical; (b) 9.80 m s 2 verticallydownwardP4.4627.7 east of northej2 1 R2P4.40P4.44gR ; (b)5The Laws of MotionCHAPTER OUTLINE5.15.25.35.45.55.65.75.8The Concept of ForceNewtons First Law andInertial FramesMassNewtons Second LawThe Gravitational Forceand WeightNewtons Third LawSome Applications ofNewtons LawsForces of FrictionANSWERS TO QUESTIONSQ5.1(a)The force due to gravity of the earth pulling down onthe ballthe reaction force is the force due to gravityof the ball pulling up on the earth. The force of thehand pushing up on the ballreaction force is ballpushing down on the hand.(b)The only force acting on the ball in free-fall is thegravity due to the earth -the reaction force is thegravity due to the ball pulling on the earth.Q5.2The resultant force is zero, as the acceleration is zero.Q5.3Mistake one: The car might be momentarily at rest, in theprocess of (suddenly) reversing forward into backward motion.In this case, the forces on it add to a (large) backward resultant.Mistake two: There are no cars in interstellar space. If the car is remaining at rest, there aresome large forces on it, including its weight and some force or forces of support.Mistake three: The statement reverses cause and effect, like a politician who thinks that hisgetting elected was the reason for people to vote for him.Q5.4When the bus starts moving, the mass of Claudette is accelerated by the force of the back of the seaton her body. Clark is standing, however, and the only force on him is the friction between his shoesand the floor of the bus. Thus, when the bus starts moving, his feet start accelerating forward, butthe rest of his body experiences almost no accelerating force (only that due to his being attached tohis accelerating feet!). As a consequence, his body tends to stay almost at rest, according to Newtonsfirst law, relative to the ground. Relative to Claudette, however, he is moving toward her and fallsinto her lap. (Both performers won Academy Awards.)Q5.5First ask, Was the bus moving forward or backing up? If it was moving forward, the passenger islying. A fast stop would make the suitcase fly toward the front of the bus, not toward the rear. If thebus was backing up at any reasonable speed, a sudden stop could not make a suitcase fly far. Fineher for malicious litigiousness.Q5.6It would be smart for the explorer to gently push the rock back into the storage compartment.Newtons 3rd law states that the rock will apply the same size force on her that she applies on it. Theharder she pushes on the rock, the larger her resulting acceleration.117118The Laws of MotionQ5.7The molecules of the floor resist the ball on impact and push the ball back, upward. The actual forceacting is due to the forces between molecules that allow the floor to keep its integrity and to preventthe ball from passing through. Notice that for a ball passing through a window, the molecular forceswerent strong enough.Q5.8While a football is in flight, the force of gravity and air resistance act on it. When a football is in theprocess of being kicked, the foot pushes forward on the ball and the ball pushes backward on thefoot. At this time and while the ball is in flight, the Earth pulls down on the ball (gravity) and the ballpulls up on the Earth. The moving ball pushes forward on the air and the air backward on the ball.Q5.9It is impossible to string a horizontal cable without its sagging a bit. Since the cable has a mass,gravity pulls it downward. A vertical component of the tension must balance the weight for thecable to be in equilibrium. If the cable were completely horizontal, then there would be no verticalcomponent of the tension to balance the weight.Some physics teachers demonstrate this by asking a beefy student to pull on the ends of acord supporting a can of soup at its center. Some get two burly young men to pull on opposite endsof a strong rope, while the smallest person in class gleefully mashes the center of the rope down tothe table. Point out the beauty of sagging suspension-bridge cables. With a laser and an optical lever,demonstrate that the mayor makes the courtroom table sag when he sits on it, and the judge bendsthe bench. Give them I make the floor sag buttons, available to instructors using this manual.Estimate the cost of an infinitely strong cable, and the truth will always win.Q5.10As the barbell goes through the bottom of a cycle, the lifter exerts an upward force on it, and thescale reads the larger upward force that the floor exerts on them together. Around the top of theweights motion, the scale reads less than average. If the iron is moving upward, the lifter candeclare that she has thrown it, just by letting go of it for a moment, so our answer applies also to thiscase.Q5.11As the sand leaks out, the acceleration increases. With the same driving force, a decrease in the masscauses an increase in the acceleration.Q5.12As the rocket takes off, it burns fuel, pushing the gases from the combustion out the back of therocket. Since the gases have mass, the total remaining mass of the rocket, fuel, and oxidizerdecreases. With a constant thrust, a decrease in the mass results in an increasing acceleration.Q5.13The friction of the road pushing on the tires of a car causes an automobile to move. The push of theair on the propeller moves the airplane. The push of the water on the oars causes the rowboat tomove.Q5.14As a man takes a step, the action is the force his foot exerts on the Earth; the reaction is the force ofthe Earth on his foot. In the second case, the action is the force exerted on the girls back by thesnowball; the reaction is the force exerted on the snowball by the girls back. The third action is theforce of the glove on the ball; the reaction is the force of the ball on the glove. The fourth action is theforce exerted on the window by the air molecules; the reaction is the force on the air moleculesexerted by the window. We could in each case interchange the terms action and reaction.Q5.15The tension in the rope must be 9 200 N. Since the rope is moving at a constant speed, then theresultant force on it must be zero. The 49ers are pulling with a force of 9 200 N. If the 49ers werewinning with the rope steadily moving in their direction or if the contest was even, then the tensionwould still be 9 200 N. In all of these case, the acceleration is zero, and so must be the resultant forceon the rope. To win the tug-of-war, a team must exert a larger force on the ground than theiropponents do.Chapter 5119Q5.16The tension in the rope when pulling the car is twice that in the tug-of-war. One could consider thecar as behaving like another team of twenty more people.Q5.17This statement contradicts Newtons 3rd law. The force that the locomotive exerted on the wall isthe same as that exerted by the wall on the locomotive. The wall temporarily exerted on thelocomotive a force greater than the force that the wall could exert without breaking.Q5.18The sack of sand moves up with the athlete, regardless of how quickly the athlete climbs. Since theathlete and the sack of sand have the same weight, the acceleration of the system must be zero.Q5.19The resultant force doesnt always add to zero. If it did, nothing could ever accelerate. If we choose asingle object as our system, action and reaction forces can never add to zero, as they act on differentobjects.Q5.20An object cannot exert a force on itself. If it could, then objects would be able to acceleratethemselves, without interacting with the environment. You cannot lift yourself by tugging on yourbootstraps.Q5.21To get the box to slide, you must push harder than the maximum static frictional force. Once the boxis moving, you need to push with a force equal to the kinetic frictional force to maintain the boxsmotion.Q5.22The stopping distance will be the same if the mass of the truck is doubled. The stopping distance willdecrease by a factor of four if the initial speed is cut in half.Q5.23If you slam on the brakes, your tires will skid on the road. The force of kinetic friction between thetires and the road is less than the maximum static friction force. Anti-lock brakes work by pumpingthe brakes (much more rapidly that you can) to minimize skidding of the tires on the road.Q5.24With friction, it takes longer to come down than to go up. On the way up, the frictional force and thecomponent of the weight down the plane are in the same direction, giving a large acceleration. Onthe way down, the forces are in opposite directions, giving a relatively smaller acceleration. If theincline is frictionless, it takes the same amount of time to go up as it does to come down.Q5.25(a)The force of static friction between the crate and the bed of the truck causes the crate toaccelerate. Note that the friction force on the crate is in the direction of its motion relative tothe ground (but opposite to the direction of possible sliding motion of the crate relative tothe truck bed).(b)It is most likely that the crate would slide forward relative to the bed of the truck.Q5.26In Question 25, part (a) is an example of such a situation. Any situation in which friction is the forcethat accelerates an object from rest is an example. As you pull away from a stop light, friction is theforce that accelerates forward a box of tissues on the level floor of the car. At the same time, frictionof the ground on the tires of the car accelerates the car forward.120The Laws of MotionSOLUTIONS TO PROBLEMSThe following problems cover Sections 5.15.6.Section 5.1The Concept of ForceSection 5.2Newtons First Law and Inertial FramesSection 5.3MassSection 5.4Newtons Second LawSection 5.5The Gravitational Force and WeightSection 5.6Newtons Third LawP5.1For the same force F, acting on different massesF = m 1 a1andF = m2 a2(a)m1a1= 2 =3m2a1(b)F = m1 + m 2 a = 4m1 a = m1 3.00 m s 2afca = 0.750 m s*P5.2v f = 880 m s, m = 25.8 kg , x f = 6 mv 2 = 2 ax f = 2 x ffF=P5.32mv 2f2x fFG F IJH mK= 1.66 10 6 N forwardm = 3.00 kgeja = 2.00 i + 5.00 j m s 2e6.00i + 15.0 jj Na6.00f + a15.0f N = F = ma =F =2216.2 NhChapter 5P5.4Fg = weight of ball = mgv release = v and time to accelerate = t :a=(a)v v v= = it t tDistance x = vt :x=(b)Fp Fg j =Fg vgtFG v IJ t =H 2KiFg vFp =P5.5vt2egti + Fg jjm = 4.00 kg , v i = 3.00 i m s , v 8 = 8.00 i + 10.0 j m s , t = 8.00 sa=v 5.00 i + 10.0 j=m s2t8.00F = ma =e2.50i + 5.00 jj N22F = ( 2.50) +(5.00) = 5.59 NP5.6(a)Let the x-axis be in the original direction of the molecules motion.ev f = vi + at: 670 m s = 670 m s + a 3.00 10 13 sja = 4. 47 10 15 m s 2(b)For the molecule, F = ma . Its weight is negligible.ejFwall on molecule = 4.68 10 26 kg 4.47 10 15 m s 2 = 2.09 10 10 NFmolecule on wall = +2.09 10 10 N121122P5.7The Laws of Motion(a) F = ma and v 2 = vi2 + 2 ax f or a =fv 2 vi2f2x f.Therefore,F = mev2f vi2j2x f F = 9.11 10 31(b)LMe7.00 10kg N5j e3.00 102b0.050 0 mgm s225m s2j OPQ2= 3.64 10 18 N .The weight of the electron ischchFg = mg = 9.111031 kg 9.80 m s 2 = 8.93 1030 NThe accelerating force is 4.08 10 11 times the weight of the electron.P5.8afFg = mg = 120 lb = 4.448 N lb (120 lb)= 534 N(b)P5.9(a)m=Fgg=534 N= 54.5 kg9.80 m s 2Fg = mg = 900 N , m =900 N= 91.8 kg9.80 m s 2cF hgP5.10on Jupiterch= 91.8 kg 25.9 m s 2 = 2.38 kNImagine a quick trip by jet, on which you do not visit the rest room and your perspiration is justcanceled out by a glass of tomato juice. By subtraction, Fg = mg p and Fg = mg C givec hcpc hChFg = m g p g C .For a person whose mass is 88.7 kg, the change in weight isbgFg = 88.7 kg 9.809 5 9.780 8 = 2.55 N .A precise balance scale, as in a doctors office, reads the same in different locations because itcompares you with the standard masses on its beams. A typical bathroom scale is not precise enoughto reveal this difference.Chapter 5P5.11(a) F = F1 + F2 = e 20.0 i + 15.0 jj N F = ma:20.0 i + 15.0 j = 5.00aeja = 4.00 i + 3.00 j m s 2ora = 5.00 m s 2 at = 36.9(b)F2 x = 15.0 cos 60.0 = 7.50 NFIG. P5.11F2 y = 15.0 sin 60.0 = 13.0 NejF2 = 7.50 i + 13.0 j N F = F1 + F2 = e27.5 i + 13.0 jj N = ma = 5.00aa=P5.12e5.50i + 2.60 jj m s2= 6.08 m s 2 at 25.3We find acceleration:r f ri = v i t +1 2at2af = 0.720 s aa = e5.83 i 4.58 jj m s .14.20 m i 3.30 mj = 0+ a 1.20 s2222Now F = ma becomesFg + F2 = maebjgejF2 = 2.80 kg 5.83 i 4.58 j m s 2 + 2.80 kg 9.80 m s 2 jF2 =P5.13(a)e16.3 i + 14.6 jj N.You and the earth exert equal forces on each other: m y g = M e a e . If your mass is 70.0 kg,a70.0 kg fc9.80 m s h ==2ae(b)5.98 10 24 kg~ 1022 m s 2 .You and the planet move for equal times intervals according to x =50.0 cm high,2xyayxe ==1 2at . If the seat is22xeaeafmy70.0 kg 0.500 mae~ 10 23 m .xy =xy =ayme5.98 10 24 kg123124P5.14The Laws of Motion F = ma readse2.00 i + 2.00 j + 5.00i 3.00 j 45.0ij N = me3.75 m s ja2where a represents the direction of ae42.0 i 1.00 jj N = me3.75 m s ja2FG 1.00 IJ below the x-axisH 42.0 K F = 42.0 N at 181 = mc3.75 m s ha .F =22(42.0) +(1.00) N at tan12For the vectors to be equal, their magnitudes and their directions must be equal.(a) a is at 181 counterclockwise from the x-axis(b)m=(d)v f = v i + at = 0 + 3.75 m s 2 at 181 10.0 s so v f = 37.5 m s at 18142.0 N= 11.2 kg3.75 m s 2ejv f = 37.5 m s cos 181 i + 37.5 m s sin 181 j so v f =(c)v f = 37.5 2 + 0.893 2 m s = 37.5 m s(a)15.0 lb up(b)5.00 lb up(c)P5.150Section 5.7P5.16e37.5 i 0.893 jj m sSome Applications of Newtons Lawsdydx= 10t , v y == 9t 2dtdtdv ydv x= 10 , a y == 18tax =dtdtvx =At t = 2.00 s , a x = 10.0 m s 2 , a y = 36.0 m s 2 Fx = ma x : 3.00 kg e10.0 Fy = ma y : 3.00 kg e36.0jm s j = 108 Nm s 2 = 30.0 N2F=Fx2 + Fy2 = 112 NChapter 5P5.17m = 1.00 kg50.0 mmg = 9.80 N0.200 mtan =25.0 m = 0.4580.200 mTTBalance forces,mg2T sin = mgT=9.80 N= 613 N2 sin FIG. P5.17T3 = Fg(1)T1 sin 1 + T2 sin 2 = Fg(2)T1 cos 1 = T2 cos 2P5.18(3)21Eliminate T2 and solve for T1T1 =bsinFg cos 21cos 2 + cos 1 sin 2Fgg=Fg cos 2bsin 1 + 2T3 = Fg = 325 NFG cos 25.0 IJ = 296 NH sin 85.0 KF cos IJ = 296 NFG cos 60.0 IJ ==T GH cos 25.0 KH cos KP5.19112See the solution for T1 in Problem 5.18.T11T1 = FgT2g2T3163 NFIG. P5.18T2125126P5.20The Laws of Motion(a)An explanation proceeding from fundamental physical principles willbe best for the parents and for you. Consider forces on the bit of stringtouching the weight hanger as shown in the free-body diagram:Horizontal Forces:Vertical Forces: Fx = ma x : Tx + T cos = 0 Fy = ma y : Fg + T sin = 0FIG. P5.20You need only the equation for the vertical forces to find that the tension in the string isFg. The force the child feels gets smaller, changing from T to T cos , whilegiven by T =sin the counterweight hangs on the string. On the other hand, the kite does not notice what youare doing and the tension in the main part of the string stays constant. You do not need alevel, since you learned in physics lab to sight to a horizontal line in a building. Share withthe parents your estimate of the experimental uncertainty, which you make by thinkingcritically about the measurement, by repeating trials, practicing in advance and looking forvariations and improvements in technique, including using other observers. You will thenbe glad to have the parents themselves repeat your measurements.Fge(b)P5.21T=(a)j=Isolate either masssin =0.132 kg 9.80 m s 2sin 46.31.79 NT + mg = ma = 0T = mg .The scale reads the tension T,soFIG. P5.21(a)ejT = mg = 5.00 kg 9.80 m s 2 = 49.0 N .(b)Isolate the pulleyT2 + 2T1 = 0T2 = 2 T1 = 2mg = 98.0 N .(c) F = n + T + mg = 0FIG. P5.21(b)Take the component along the inclinen x + Tx + mg x = 0or0 + T mg sin 30.0 = 0T = mg sin 30.0 == 24.5 N .a fmg 5.00 9.80=22FIG. P5.21(c)Chapter 5P5.22127The two forces acting on the block are the normal force, n, and theweight, mg. If the block is considered to be a point mass and the xaxis is chosen to be parallel to the plane, then the free bodydiagram will be as shown in the figure to the right. The angle isthe angle of inclination of the plane. Applying Newtons secondlaw for the accelerating system (and taking the direction up theplane as the positive x direction) we haveFIG. P5.22 Fy = n mg cos = 0: n = mg cos Fx = mg sin = ma : a = g sin (a)When = 15.0a = 2.54 m s 2(b)Starting from restdiv 2 = vi2 + 2 a x f xi = 2 ax ffejafv f = 2 ax f = 2 2.54 m s 2 2.00 m = 3.18 m sP5.23Choose a coordinate system with i East and j North. F = ma = 1.00 kg e10.0jm s 2 at 30.0a5.00 Nf j + F = a10.0 Nf30.0 = a5.00 Nf j + a8.66 Nfi1 F1 = 8.66 N (East )*P5.24FIG. P5.23First, consider the block moving along the horizontal. The onlyforce in the direction of movement is T. Thus, Fx = maa fT = 5 kg an(1)Next consider the block that moves vertically. The forces on it arethe tension T and its weight, 88.2 N.We have5 kg+xT49 NT+y9 kgFg = 88.2 NFIG. P5.24 Fy = maa f88.2 N T = 9 kg a(2)Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can beadded to give 88.2 N = 14 kg a. Thenbga = 6.30 m s 2 and T = 31.5 N .128P5.25The Laws of MotionAfter it leaves your hand, the blocks speed changes onlybecause of one component of its weight: Fx = ma xTaking v f mg sin 20.0 = madi= 0 , v = 5.00 m s, and a = g sina 20.0f givesv 2 = vi2 + 2 a x f xi .fia2fc0 = (5.00) 2(9.80) sin 20.0 x f 0horxf =P5.26FIG. P5.2525.0= 3.73 m .2(9.80) sin 20.0afm1 = 2.00 kg , m 2 = 6.00 kg , = 55.0(a) Fx = m 2 g sin T = m 2 aandT m1 g = m1 aa=am 2 g sin m1 g= 3.57 m s 2m1 + m 2FIG. P5.26f(b)(c)*P5.27T = m1 a + g = 26.7 NSince vi = 0 , v f = at = 3.57 m s 2 ( 2.00 s)= 7.14 m s .chWe assume the vertical bar is in compression, pushing upon the pin with force A, and the tilted bar is in tension,exerting force B on the pin at 50 . Fx = 0:2 500 N cos 30+ B cos 50 = 0 Fy = 0:302 500 N50A2 500 N sin 30+ A 3.37 10 3 N sin 50 = 0B = 3.37 10 3 NA = 3.83 10 3 NPositive answers confirm thatB is in tension and A is in compression.B2 500 N cos30B cos50A2 500 N sin30FIG. P5.27B sin50Chapter 5P5.28First, consider the 3.00 kg rising mass. The forces on it arethe tension, T, and its weight, 29.4 N. With the upwarddirection as positive, the second law becomes Fy = ma y : T 29.4 N = a3.00 kg fa(1)The forces on the falling 5.00 kg mass are its weight and T,and its acceleration is the same as that of the rising mass.Calling the positive direction down for this mass, we have Fy = ma y : 49 N T = a5.00 kg faFIG. P5.28(2)Equations (1) and (2) can be solved simultaneously by adding them:af afT 29.4 N + 49.0 N T = 3.00 kg a + 5.00 kg a(b)This gives the acceleration asa=(a)19.6 N= 2.45 m s 2 .8.00 kgThenafchT 29.4 N = 3.00 kg 2.45 m s 2 = 7.35 N .The tension isT = 36.8 N .(c)Consider either mass. We havey = vi t +*P5.2911 22at = 0 + 2.45 m s 2 (1.00 s) = 1.23 m .22chAs the man rises steadily the pulley turns steadily and the tension inthe rope is the same on both sides of the pulley. Choose man-pulleyand-platform as the system:T Fy = ma y+T 950 N = 0T = 950 N .The worker must pull on the rope with force 950 N .950 NFIG. P5.29129130*P5.30The Laws of MotionBoth blocks move with acceleration a =a=(a)FG mHmIJK m1g:2 + m12F 7 kg 2 kg I 9.8 m sGH 7 kg + 2 kg JK2= 5.44 m s 2 .Take the upward direction as positive for m1 .di22v xf = v xi + 2 a x x f x i :b0 = 2.4 m sxf = g + 2e5.44 m s jdx 0i225.76 m 2 s 2e2 5.44 m s 2jf= 0.529 mx f = 0.529 m below its initial level(b)jaev xf = v xi + a x t: v xf = 2.40 m s + 5.44 m s 2 1.80 sfv xf = 7. 40 m s upwardP5.31Forces acting on 2.00 kg block:T m1 g = m 1 a(1)Forces acting on 8.00 kg block:Fx T = m 2 a(a)(2)Eliminate T and solve for a:a=Fx m1 gm1 + m 2a > 0 for Fx > m1 g = 19.6 N .(b)Eliminate a and solve for T:T=am1Fx + m 2 gm1 + m 2fFIG. P5.31T = 0 for Fx m 2 g = 78.4 N .(c)Fx , Nax , m s 210078.450.0050.010012.59.806.961.963.048.04Chapter 5*P5.32(a)For force components along the incline, with the upward direction taken as positive, Fx = ma x : mg sin = ma xeja x = g sin = 9.8 m s 2 sin 35 = 5.62 m s 2 .For the upward motion,d22v xf = v xi + 2 a x x f xib0= 5 m sxf =(b)ig + 2e5.62 m s jdx 0i2225 m 2 s 2ej2 5.62 m s 2f= 2.22 m .The time to slide down is given byx f = xi + v xi t +1axt 220 = 2.22 m + 0 +t=a15.62 m s 2 t 22ejf = 0.890 s .2 2.22 m5.62 m s2For the second particle,x f = xi + v xi t +1axt 22af eja0 = 10 m + v xi 0.890 s + 5.62 m s 2 0.890 sv xi =10 m + 2.22 m= 8.74 m s0.890 sspeed = 8.74 m s .f2131132P5.33The Laws of MotionFirst, we will compute the needed accelerations:a1fa 2fa3 fa4fay = 0v yf v yi 1. 20 m s 0=ay =t0.800 s= 1.50 m s 2Before it starts to move:During the first 0.800 s:While moving at constant velocity: a y = 0v yf v yi 0 1.20 m s=During the last 1.50 s:ay =1.50 st= 0.800 m s 2Newtons second law is:FIG. P5.33 Fy = ma ybgej bgS = 706 N + b72.0 kg ga+S 72.0 kg 9.80 m s 2 = 72.0 kg a y(a)When a y = 0 , S = 706 N .(d)P5.34When a y = 1.50 m s 2 , S = 814 N .(c)When a y =0.800 m s 2 , S = 648 N .(a)Pulley P1 has acceleration a 2 .Since m1 moves twice the distance P1 moves in the sametime, m1 has twice the acceleration of P1 , i.e., a1 = 2 a 2 .(b).When a y = 0 , S = 706 N .(b)yFrom the figure, and using F = ma:m 2 g T2 = m 2 a 2T1 = m1 a1 = 2m1 a 2T2 2T1 = 0a1fa 2fa3 fFIG. P5.34Equation (1) becomes m 2 g 2T1 = m 2 a 2 . This equation combined with Equation (2) yieldsFGHIJKT1m2m1 + 2 = m 2 gm12T1 =(c)m1 m 2m1 m 2g and T2 =g .2m1 + 1 m 2m1 + 1 m 224From the values of T1 and T2 we find thata1 =m2 gT1=2m1 + 1 m 2m12and a 2 =m2 g1a1 =.24m 1 + m 2Chapter 5Section 5.8*P5.35Forces of Friction+y+yn ground = Fg /2 = 85.0 lbn tip22.0 22.0F2F1f+x+xF = 45.8 lbFg = 170 lbFree-Body Diagram of Person22.0Free-Body Diagram of Crutch TipFIG. P5.35From the free-body diagram of the person, Fx = F1 sina22.0f F2 sina22.0f = 0 ,which givesF1 = F2 = F .Then, Fy = 2 F cos 22.0+85.0 lbs 170 lbs = 0 yields F = 45.8 lb.(a)Now consider the free-body diagram of a crutch tip. Fx = f ( 45.8 lb) sin 22.0= 0 ,orf = 17. 2 lb . Fy = n tip ( 45.8 lb) cos 22.0= 0 ,which givesn tip = 42.5 lb .For minimum coefficient of friction, the crutch tip will be on the verge of slipping, sof17.2 lbf = f s max = s n tip and s === 0.404 .n tip 42.5 lba f(b)As found above, the compression force in each crutch isF1 = F2 = F = 45.8 lb .133134P5.36The Laws of MotionFor equilibrium: f = F and n = Fg . Also, f = n i.e.,=s =fF=n Fg75.0 N= 0.30625.0 9.80 Na fFIG. P5.36andk =P5.37 Fy = ma y :60.0 N= 0.245 .25.0(9.80) N+n mg = 0fs sn = s mgThis maximum magnitude of static friction acts so long as the tires roll without skidding. Fx = ma x : f s = maThe maximum acceleration isa = s g .The initial and final conditions are: x i = 0 , vi = 50.0 mi h = 22.4 m s , v f = 0div 2 = vi2 + 2 a x f xi : vi2 = 2 s gx ff(a)xf =vi22 gxf =a22.4 m sf =2(0.100 )c9.80 m s hxf =vi22 g2(b)2a22.4 m sf =2(0.600)c9.80 m s h256 m2xf =242.7 mChapter 5P5.38If all the weight is on the rear wheels,(a)F = ma: s mg = maButx =so s =2 x:gt 2s =(b)*P5.39(a)at 2 s gt 2=22afb2 0.250 mi 1 609 m mie9.80 m s ja4.96 sf22g=3.34 .Time would increase, as the wheels would skid and only kinetic friction would act; orperhaps the car would flip over.The person pushes backward on the floor. The floor pushes forwardon the person with a force of friction. This is the only horizontalforce on the person. If the persons shoe is on the point of slippingthe static friction force has its maximum value. Fx = ma x : Fy = ma y :f = sn = ma xn mg = 0ma x = s mgx f = xi + v xi t +1ax t 22eejja x = s g = 0.5 9.8 m s 2 = 4.9 m s 213 m = 0 + 0 + 4.9 m s 2 t 22FIG. P5.39t = 1.11 s(b)P5.40xf =2x f2(3 m)1 s gt 2 , t === 0.875 s2s g(0.8) 9.8 m s 2chm suitcase = 20.0 kg , F = 35.0 N Fx = ma x : Fy = ma y :(a)20.0 N + F cos = 0+n + F sin Fg = 0F cos = 20.0 Ncos =20.0 N= 0.57135.0 N = 55.2(b)n = Fg F sin = 196 35.0(0.821) Nn = 167 NFIG. P5.40135136P5.41The Laws of Motionm = 3.00 kg , = 30.0 , x = 2.00 m, t = 1.50 s(a)x=1 2at :2af12a 1.50 s24.00a== 1.78 m s 221.502.00 m =FIG. P5.41a f F = n + f + mg = m a :Along x: 0 f + mg sin 30.0 = mabf = m g sin 30.0 agAlong y: n + 0 mg cos 30.0 = 0n = mg cos 30.0aff m g sin 30.0aa, k = tan 30.0== 0.368nmg cos 30.0g cos 30.0(b)k =(c)f = m g sin 30.0a , f = 3.00 9.80 sin 30.01.78 = 9.37 N(d)v 2 = vi2 + 2 a x f xifaafcfhwherex f xi = 2.00 ma fa fv 2 = 0 + 2 1.78 2.00 = 7.11 m 2 s 2fv f = 7.11 m 2 s 2 = 2.67 m sChapter 5*P5.42First we find the coefficient of friction: Fy = 0:n+n mg = 0f = sn = s mg Fx = ma x : s mg = v2fmvi22 x=vi2+ 2 a x x = 0nfbgjaefmgmg sin10288 ft sv2s = i == 0.9812 gx 2 32.1 ft s 2 123 ftfmg cos10FIG. P5.42Now on the slope Fy = 0: Fx = ma x :+n mg cos 10 = 0f s = sn = s mg cos 10 s mg cos 10+ mg sin 10 = x ==P5.43mvi22 xvi22 g s cos 10 sin 10bgb88 ft sg=2e32.1 ft s ja0.981 cos 10 sin 10f22152 ft .T f k = 5.00 a (for 5.00 kg mass)9.00 g T = 9.00 a (for 9.00 kg mass)Adding these two equations gives:a f137a fa f9.00 9.80 0.200 5.00 9.80 = 14.0 aa = 5.60 m s 2 T = 5.00 5.60 + 0.200 5.00 9.80a fa fa f= 37.8 NFIG. P5.43138P5.44The Laws of MotionLet a represent the positive magnitude of the acceleration a j ofnm1 , of the acceleration a i of m 2 , and of the acceleration +aj of m 3 .Call T12 the tension in the left rope and T23 the tension in the cordon the right.For m1 , Fy = ma y Fx = ma xT12 + k n + T23 = m 2 aand Fy = ma yn m2 g = 0for m 3 , Fy = ma yT23+T12 m1 g = m1 aFor m 2 ,T12T23 m 3 g = +m 3 af = knm2 gT12T23m1 gm3 gwe have three simultaneous equationsbg 0.350a9.80 N f T = b1.00 kg ga+T 19.6 N = b 2.00 kg ga .T12 + 39.2 N = 4.00 kg a+T122323(a)FIG. P5.44Add them up:af+39.2 N 3. 43 N 19.6 N = 7.00 kg aa = 2.31 m s 2 , down for m1 , left for m 2 , and up for m 3 .(b)afcNow T12 + 39. 2 N = 4.00 kg 2.31 m s 2hT12 = 30.0 Nafcand T23 19.6 N = 2.00 kg 2.31 m s 2hT23 = 24.2 N .P5.45(a)See Figure to the right(b)68.0 T m 2 g = m 2 a (Block #2)T m1 g = m1 a (Block #1)Tm1n1m1Adding,g bg68.0a=bm + m g g =11.29 m s 22T = m1 a + m1 g = 27. 2 NTm2f2 = k n 2m1 g = 118 N68.0 m1 + m 2 g = m1 + m 2 aFn2Tf1 = k n 1bm2m2 g = 176 NFIG. P5.45FChapter 5P5.46(Case 1, impending upward motion)Setting Fx = 0:P cos 50.0n = 0fs , max = sn:fs , max = s P cos 50.0af= 0.250 0.643 P = 0.161 PSetting Fy = 0:a fP sin 50.00.161P 3.00 9.80 = 0Pmax = 48.6 N(Case 2, impending downward motion)As in Case 1,FIG. P5.46fs, max = 0.161PSetting Fy = 0:a fP sin 50.0+0.161P 3.00 9.80 = 0Pmin = 31.7 N*P5.47yWhen the sled is sliding uphill Fy = ma y :+n mg cos = 0 Fx = ma x :+ mg sin + k mg cos = ma upnf = k n = k mg cos mg sin v f = 0 = vi + a up t upmg cos vi = a up t up1vi + v f t up2112x = a up t up + 0 t up = a up t up22x =dexiFIG. P5.47jWhen the sled is sliding down, the direction of the friction force is reversed:mg sin k mg cos = ma downx=12a down t down .2Nowt down = 2t up112a up tup = a down 2t up22a up = 4a downe jb2g sin + k g cos = 4 g sin k g cos 5 k cos = 3 sin k =FG 3 IJ tanH 5Kgf139140*P5.48The Laws of MotionSince the board is in equilibrium, Fx = 0 and we see that the normalforces must be the same on both sides of the board. Also, if theminimum normal forces (compression forces) are being applied, theboard is on the verge of slipping and the friction force on each side isa ff = fsmaxnffn= sn .The board is also in equilibrium in the vertical direction, so Fy = 2 f Fg = 0 , orf=Fg2.Fg = 95.5 NFIG. P5.48The minimum compression force needed is thenn=*P5.49(a)fsa=Fg2 s=95.5 N= 72.0 N .2(0.663)fn + F sin 15 75 N cos 25 = 0nF15f s, max n = 67.97 0.259 Ffs , max = s n = 24.67 0.094F25For equilibrium: F cos 15+24.67 0.094F 75 sin 25= 0 .This gives F = 8.05 N .75 NFIG. P5.49(a)(b)nF cos 15( 24.67 0.094F ) 75 sin 25 = 0 .FThis gives F = 53.2 N .15f s, max2575 NFIG. P5.49(b)(c)f k = k n = 10.6 0.040 F . Since the velocity is constant, the netforce is zero:F cos 15(10.6 0.040 F ) 75 sin 25 = 0 .This gives F = 42.0 N .nF15fk2575 NFIG. P5.49(c)Chapter 5*P5.50We must consider separately the disk when it is in contact with the roofand when it has gone over the top into free fall. In the first case, we takex and y as parallel and perpendicular to the surface of the roof: Fy = ma y :+n mg cos = 0n = mg cos then friction is f k = k n = k mg cos Fx = ma x :FIG. P5.50 f k mg sin = ma xafa x = k g cos g sin = 0.4 cos 37 sin 37 9.8 m s 2 = 9.03 m s 2The Frisbee goes ballistic with speed given byi bd22v xf = v xi + 2 a x x f xi = 15 m sg + 2e9.03 m s ja10 m 0f = 44.4 m222v xf = 6.67 m sFor the free fall, we take x and y horizontal and vertical:d22v yf = v yi + 2 a y y f yibig + 2e9.8 m s jdyb4.01 m sg = 6.84 m= 6.02 m +0 = 6.67 m s sin 3722f 10 m sin 37i2yf19.6 m s 2Additional ProblemsP5.51(a)see figure to the right(b)First consider Pat and the chair as the system.Note that two ropes support the system, andT = 250 N in each rope. Applying F = ma2T 480 = ma , where m =480= 49.0 kg .9.80FIG. P5.51Solving for a givesa=(c)500 480= 0.408 m s 2 .49.0 F = ma on Pat:320 F = n + T 320 = ma , where m = 9.80 = 32.7 kgn = ma + 320 T = 32.7(0.408)+ 320 250 = 83.3 N .s2141142P5.52The Laws of Motion F = ma gives the objects accelerationa= F = e8.00 i 4.00tjj N2.00 kgmej eja = 4.00 m s 2 i 2.00 m s 3 t j =dv.dtIts velocity iszzvtdv = v v i = v 0 = adtvi0zetj ev = e 4.00t m s ji e1.00tv=02(a)jm s jj.4.00 m s 2 i 2.00 m s 3 t j dt232We require v = 15.0 m s , v = 225 m 2 s 216.0t 2 m 2 s 4 + 1.00t 4 m 2 s 6 = 225 m 2 s 21.00t 4 + 16.0 s 2 t 2 225 s 4 = 0t2 =16.0 a16.0f 4a225f = 9.00 s22.00t = 3.00 s .Take ri = 0 at t = 0. The position isz zett00r = vdt =er = 4.00 m s 2at t = 3 s we evaluate.e18.0i 9.00 jj m(c)r=(b)So r = (18.0) +(9.00) m = 20.1 m22j ej4.00t m s 2 i 1.00t 2 m s 3 j dt23j t2 i e1.00 m s j t3 j32Chapter 5*P5.53(a)ySituation A Fx = ma x : Fy = ma y :FA + sn mg sin = 0+n mg cos = 0aFA = mg sin s cos fsFAmg cos FIG. P5.53(a)f.ySituation B Fx = ma x : Fy = ma y :FB cos + s n mg sin = 0FBfsaamg sin s cos mg cos mg sin FIG. P5.53(b)FB cos + s mg cos + s FB sin mg sin = 0FB =fcos + s sin fFA = 2 kg 9.8 m s 2 sin 250.16 cos 25 = 5.44 NFB =af19.6 N 0.278= 5.59 Ncos 25+0.16 sin 25Student A need exert less force.(d)FB =FAF= Acos 25+0.38 sin 25 1.07Student B need exert less force.b gP Q = b3 kg gaQ = b 4 kg gaP5.5418 N P = 2 kg aa fAdding gives 18 N = 9 kg a soFIG. P5.54a = 2.00 m s 2 .(b)ejQ = 4 kg 2 m s 2 = 8.00 N net force on the 4 kgejP 8 N = 3 kg 2 m s 2 = 6.00 N net force on the 3 kg and P = 14 Nej18 N 14 N = 2 kg 2 m s 2 = 4.00 N net force on the 2 kgcontinued on next pagexn FB sin + n mg cos = 0Substitute n = mg cos + FB sin to find(c)xnmg sin Eliminate n = mg cos to solve for(b)143144The Laws of Motion(c)(d)P5.55From above, Q = 8.00 N and P = 14.0 N .The 3-kg block models the heavy block of wood. The contact force on your back isrepresented by Q, which is much less than the force F. The difference between F and Q isthe net force causing acceleration of the 5-kg pair of objects. The acceleration is real andnonzero, but lasts for so short a time that it never is associated with a large velocity. Theframe of the building and your legs exert forces, small relative to the hammer blow, to bringthe partition, block, and you to rest again over a time large relative to the hammer blow.This problem lends itself to interesting lecture demonstrations. One person can hold a leadbrick in one hand while another hits the brick with a hammer.(a)First, we note that F = T1 . Next, we focus on themass M and write T5 = Mg . Next, we focus on thebottom pulley and write T5 = T2 + T3 . Finally, wefocus on the top pulley and write T4 = T1 + T2 + T3 .Since the pulleys are not starting to rotate and arefrictionless, T1 = T3 , and T2 = T3 . From thisMginformation, we have T5 = 2T2 , soT2 =.2Then T1 = T2 = T3 =Mg3 Mg, and T4 =, and22T5 = Mg .(b)Since F = T1 , we have F =Mg.2FIG. P5.55P5.56We find the divers impact speed by analyzing his free-fall motion:chv 2 = vi2 + 2 ax = 0 + 2 9.80 m s 2 (10.0 m) so v f = 14.0 m s.fNow for the 2.00 s of stopping, we have v f = vi + at :a0 = 14.0 m s + a 2.00 sf2a = +7.00 m s .Call the force exerted by the water on the diver R. Usingej Fy = ma ,e+ R 70.0 kg 9.80 m s 2 = 70.0 kg 7.00 m s 2R = 1.18 kN .jChapter 5P5.57(a)145The crate is in equilibrium, just before it starts tomove. Let the normal force acting on it be n andthe friction force, fs .Resolving vertically:n = Fg + P sin FIG. P5.57Horizontally:P cos = fsBut,fs sni.e.,cP cos s Fg + P sin horafP cos s sin s Fg .Divide by cos :afP 1 s tan s Fg sec .ThenPminimum =(b)P= s Fg sec 1 s tan .0.400(100 N ) sec 1 0.400 tan b gPa N f deg0.0015.030.045.060.040.046.460.194.3260If the angle were 68.2 or more, the expression for P would go to infinity and motion wouldbecome impossible.146P5.58The Laws of Motion(a)Following the in-chapter Example about a block on a frictionless incline, we havecha = g sin = 9.80 m s 2 sin 30.0a = 4.90 m s 2(b)The block slides distance x on the incline, with sin 30.0 =chc0.500 mxhx = 1.00 m: v 2 = vi2 + 2 a x f xi = 0 + 2 4.90 m s 2 (1.00 m)fv f = 3.13 m s after time t s =(c)Now in free fall y f yi = v yi t +2x fvf=2(1.00 m)3.13 m s= 0.639 s .1ayt 2 :2bg2.00 = 3.13 m s sin 30.0 t e4.90 m s jt + b1.56 m sgt 2.00 m = 0219.80 m s 2 t 22ej2t=1.56 m s b1.56 m sg 4e4.90 m s ja2.00 mf29.80 m s 2Only one root is physicalt = 0.499 sbgafx f = v x t = 3.13 m s cos 30.0 0.499 s = 1.35 m(d)total time = t s + t = 0.639 s + 0.499 s = 1.14 s(e)The mass of the block makes no difference.2Chapter 5P5.59With motion impending,n + T sin mg = 0bf = s mg T sin gandT cos s mg + sT sin = 0FIG. P5.59soT= s mg.cos + s sin To minimize T, we maximize cos + s sin bgdcos + s sin = 0 = sin + s cos .d(a)(b)*P5.60 = tan1 s = tan1 0.350 = 19.3T=afc0.350 1.30 kg 9.80 m s 2cos 19.3+0.350 sin 19.3(a)afchSee Figure (b) to the right.(c)4.21 NSee Figure (a) to the right.(b)h=For the pin, Fy = ma y :mg = 36.4 kg 9.8 m s 2 = 357 NC cos 357 N = 0357 N.C=cos For the foot,FIG. P5.60(a) Fy = ma y :FIG. P5.60(b)+n B C cos = 0n B = 357 N .(d)For the foot with motion impending, Fx = ma x :+ f s C sin s = 0 sn B = C sin s357 N cos s sin sC sin ss === tan s .nB357 Nb(e)The maximum coefficient is s = tan s = tan 50.2 = 1.20 .g147148P5.61The Laws of Motion F = maFor m1 :For m 2 :T = m1 aT m2 g = 0Eliminating T,a=m2 gm1For all 3 blocks:FIG. P5.61afF = M + m1 + m 2 a =afP5.62e j1221a ft2 s2tsaM + m + m fFGH m g IJKmxm0001.021.04 00.1001.532.34 10.2002.014.04 00.3502.646.97 00.5003.3010.890.7503.7514.061.00FIG. P5.62From x =1 21at the slope of a graph of x versus t 2 is a , and22eja = 2 slope = 2 0.071 4 m s 2 = 0.143 m s 2 .From a = g sin ,a = 9.80 m s 2FG 1.77 4 IJ = 0.137 m sH 127.1 K2, different by 4%.The difference is accounted for by the uncertainty in the data, which we may estimate from the thirdpoint asbga f = 18%.0.350 0.071 4 4.040.350Chapter 5P5.63af(1)m1 a A = T a =T+Am1(2)MA = R x = T A =TM(3)m2 a = m2 g T T = m2 g a(a)bgFIG. P5.63Substitute the value for a from (1) into (3) and solve for T:LM F T + AI OP .N GH m JK QT = m2 g 1Substitute for A from (2):LM F T + T I OP =N GH m M JK QT = m2 g (b)1From (2), A =112a A =am 2 g m1 + Mafm1 M + m 2 M + m1f.T, Substitute the value of T:MA=(d)LMm MN m M + m amSolve (3) for a and substitute value of T:a=(c)m2 gMm 2 gm1 M + m 2 m1 + Mafm1 m 2 gm1 M + m 2 m1 + Maf.1OP+ Mf Q.149150P5.64The Laws of Motion(a), (b) Motion impendingn = 49.0 Nn = 49.0 Nf s1P5.00 kg15.0 kgf s1f s2Fg = 49.0 N147 N196 Nafs1 = n = 14.7 Nffs2 = 0.500 196 N = 98.0 NFIG. P5.64P = f s1 + f s 2 = 14.7 N + 98.0 N = 113 N(c)Once motion starts, kinetic friction acts.afaf bg112.7 N 0.100 49.0 N 0.400 196 N = 15.0 kg a 2a 2 = 1.96 m s 2af bg0.100 49.0 N = 5.00 kg a1a1 = 0.980 m s 2*P5.65(a)2Let x represent the position of the glider along the air track. Then z 2 = x 2 + h0 ,dx 1 2dzdz2 122 1 2is the rate at which string passes= z h02zx = z 2 h0, vx =. Nowdt 2dtdtover the pulley, so it is equal to v y of the counterweight.ejej a fc2v x = z z 2 h0(b)(c)ax =h1 2v y = uv ydv ydv xdudat release from rest, v y = 0 and a x = ua y .= uv y = u+ vydtdtdtdt80.0 cm2, z = 1.60 m , u = z 2 h0zFor the counterweightsin 30.0 =e Fy = ma y :j1 2ez = 1.6 2 0.8 2j a1.6f = 1.15 .1 2T 0.5 kg 9.8 m s 2 = 0.5 kga ya y = 2T + 9.8For the glider Fx = ma x :afT cos 30 = 1.00 kg a x = 1.15 a y = 1.15 2T + 9.8 = 2.31T + 11.3 N3.18T = 11.3 NT = 3.56 NChapter 5*P5.66The upward acceleration of the rod is described byy f = yi + v yi t +1ayt 221a y 8 10 3 s2a y = 31.2 m s 2e1 10 3 m = 0 + 0 +j2The distance y moved by the rod and the distance xmoved by the wedge in the same time are relatedyy. Then their speeds andby tan 15 = x =xtan 15accelerations are related byFIG. P5.66dydx1=dt tan 15 dtandd2xdt2=FGHIJKd2y11=31.2 m s 2 = 117 m s 2 .2tan 15 dttan 15The free body diagram for the rod is shown. Here H and H are forces exerted by the guide. Fy = ma y :n cos 15 mg = ma yejen cos 150.250 kg 9.8 m s 2 = 0.250 kg 31.2 m s 210.3 N= 10.6 Nn=cos 15jFor the wedge, Fx = Ma x :en sin 15+ F = 0.5 kg 117 m s 2afjF = 10.6 N sin 15+58.3 N = 61.1 N*P5.67(a)Consider forces on the midpoint of the rope. It is nearly inequilibrium just before the car begins to move. Take the y-axisin the direction of the force you exert: Fy = ma y :T sin + f T sin = 0T=(b)T=100 N= 410 N2 sin 7f.2 sin FIG. P5.67151152P5.68The Laws of MotionSince it has a larger mass, we expect the 8.00-kg block to movedown the plane. The acceleration for both blocks should have thesame magnitude since they are joined together by a non-stretchingstring. Define up the left hand plane as positive for the 3.50-kgobject and down the right hand plane as positive for the 8.00-kgobject. F1 = m1 a1 : F2 = m 2 a 2 : m1 g sin 35.0+T = m1 aFIG. P5.68m 2 g sin 35.0T = m 2 aanda fa fa8.00fa9.80f sin 35.0T = 8.00a . 3.50 9.80 sin 35.0+T = 3.50 aAdding, we obtainaf+45.0 N 19.7 N = 11.5 kg a .(b)Thus the acceleration isa = 2.20 m s 2 .By substitution,afch19.7 N + T = 3.50 kg 2.20 m s 2 = 7.70 N .(a)The tension isT = 27.4 N .P5.69a = 5.00 m s 2Choose the x-axis pointing down the slope.av f = vi + at: 30.0 m s = 0 + a 6.00 s2fa = 5.00 m s .Consider forces on the toy. Fx = ma x :emg sin = m 5.00 m s 2j = 30.7 Fy = ma y : mg cos + T = 0T = mg cos = 0.100 9.80 cos 30.7T = 0.843 Nafa fFIG. P5.69Chapter 5*P5.70153Throughout its up and down motion after release the block has Fy = ma y :+n mg cos = 0n = mg cos .Let R = R x i + R y j represent the force of table on incline. We have Fx = ma x :+ R x n sin = 0R x = mg cos sin Fy = ma y : Mg n cos + Ry = 0R y = Mg + mg cos 2 .ejR = mg cos sin to the right + M + m cos 2 g upward*P5.71FIG. P5.70Take +x in the direction of motion of the tablecloth. For the mug: Fx = ma x0.1 N = 0.2 kg a xa x = 0.5 m s 2 .Relative to the tablecloth, the acceleration of the mug is 0.5 m s 2 3 m s 2 = 2.5 m s 2 . The mugreaches the edge of the tablecloth after time given by x = v xi t +1axt 2212.5 m s 2 t 22t = 0.490 s .e0.3 m = 0 +jThe motion of the mug relative to tabletop is over distanceja11a x t 2 = 0.5 m s 2 0.490 s22eThe tablecloth slides 36 cm over the table in this process.f2= 0.060 0 m .154P5.72The Laws of Motion Fy = ma y : n mg cos = 0ora fn = a82.3 N f cos n = 8.40 9.80 cos Fx = ma x : mg sin = maora = g sin eja = 9.80 m s 2 sin , deg n , N0.0082.3a, m s 20.005.0082.00.85410.081.11.7015.079.52.5420.077.43.3525.074.64.1430.071.34.9035.067.45.6240.063.1FIG. P5.726.3045.058.26.9350.052.97.5155.047.28.0360.041.28.4965.034.88.8870.028.29.2175.021.39.4780.014.39.6585.07.179.7690.00.009.80At 0, the normal force is the full weight and the acceleration is zero. At 90, the mass is in free fallnext to the vertical incline.Chapter 5P5.73(a)Apply Newtons second law to two pointswhere butterflies are attached on either halfof mobile (other half the same, by symmetry)(1)(2)(3)(4)T2 cos 2 T1 cos 1 = 0T1 sin 1 T2 sin 2 mg = 0T2 cos 2 T3 = 0T2 sin 2 mg = 0Substituting (4) into (2) for T2 sin 2 ,T1 sin 1 mg mg = 0 .FIG. P5.69ThenT1 =2mg.sin 1Substitute (3) into (1) for T2 cos 2 :T3 T1 cos 1 = 0 , T3 = T1 cos 1Substitute value of T1 :T3 = 2mg2mgcos 1== T3 .sin 1tan 1From Equation (4),T2 =(b)mg.sin 2Divide (4) by (3):mgT2 sin 2.=T2 cos 2T3Substitute value of T3 :tan 2 =FGHmg tan 1tan 1, 2 = tan122mgIJK.Then we can finish answering part (a):T2 =(c)mgsinbtan1 12tan 1g.D is the horizontal distance between the points at which the two ends of the string areattached to the ceiling.D = 2 cos 1 + 2 cos 2 + and L = 5D=RSTLMNFGH1L2 cos 1 + 2 cos tan1 tan 125IJ OP + 1UKQ VW155156The Laws of MotionANSWERS TO EVEN PROBLEMSP5.21.66 10 6 N forwardP5.4P5.42152 ftFg vvt(a) ; (b)i + Fg jgt2P5.44(a) 2.31 m s 2 down for m1 , left for m 2 andup for m 3 ; (b) 30.0 N and 24.2 N(a) 4.47 10 15 m s 2 away from the wall;(b) 2.09 10 10 N toward the wallP5.46Any value between 31.7 N and 48.6 NP5.4872.0 NP5.8(a) 534 N down; (b) 54.5 kgP5.506.84 mP5.102.55 N for an 88.7 kg personP5.52(a) 3.00 s; (b) 20.1 m; (c) 18.0 i 9.00 j mP5.12e16.3i + 14.6 jj NP5.54P5.14(a) 181; (b) 11.2 kg; (c) 37.5 m s ;(a) 2.00 m s 2 to the right;(b) 8.00 N right on 4 kg;6.00 N right on 3 kg; 4 N right on 2 kg;(c) 8.00 N between 4 kg and 3 kg;14.0 N between 2 kg and 3 kg;(d) see the solutionP5.6F IGH JKej(d) 37.5 i 0.893 j m sejP5.16112 NP5.18T1 = 296 N ; T2 = 163 N ; T3 = 325 NP5.561.18 kNP5.20(a) see the solution; (b) 1.79 NP5.58P5.22(a) 2.54 m s 2 down the incline;(b) 3.18 m s(a) 4.90 m s 2 ; (b) 3.13 m s at 30.0 belowthe horizontal; (c) 1.35 m; (d) 1.14 s; (e) NoP5.60(a) and (b) see the solution; (c) 357 N;(d) see the solution; (e) 1.20P5.24see the solution; 6.30 m s 2 ; 31.5 NP5.62see the solution; 0.143 m s 2 agrees withP5.26(a) 3.57 m s 2 ; (b) 26.7 N; (c) 7.14 m sP5.28(a) 36.8 N; (b) 2.45 m s 2 ; (c) 1.23 mP5.30(a) 0.529 m; (b) 7.40 m s upwardP5.32(a) 2.22 m; (b) 8.74 m sP5.34(a) a1 = 2 a 2 ;m1 m 2 gm m g; T2 = 1 2 ;(b) T1 =m2m2m1 + 2m1 + 42m2 gm2 g; a2 =(c) a1 =m24m1 + m 22m1 + 2P5.36 s = 0.306 ; k = 0.245P5.38(a) 55.2; (b) 167 NP5.64(a) see the solution;(b) on block one:49.0 N j 49.0 N j + 14.7 N i ;on block two: 49.0 N j 14.7 N i 147 N j+196 N j 98.0 N i + 113 N i ;(c) for block one: 0.980 i m s 2 ;for block two: 1.96 m s 2 iP5.6661.1 NP5.68(a) 2.20 m s 2 ; (b) 27.4 NP5.70mg cos sin to the right(a) 3.34; (b) Time would increaseP5.400.137 m s 2ej+ M + m cos 2 g upwardP5.72see the solution6Circular Motion and OtherApplications of Newtons LawsCHAPTER OUTLINE6.1Newtons Second LawApplied to Uniform CircularMotionNonuniform Circular MotionMotion in AcceleratedFramesMotion in the Presence ofResistive ForcesNumerical Modeling inParticle DynamicsANSWERS TO QUESTIONS6.46.5Q6.4Mud flies off a rapidly spinning tire because the resultant forceis not sufficient to keep it moving in a circular path. In this case,the force that plays a major role is the adhesion between themud and the tire.Q6.2The spring will stretch. In order for the object to move in acircle, the force exerted on the object by the spring must have amv 2. Newtons third law says that the force exerted onsize ofrthe object by the spring has the same size as the force exertedby the object on the spring. It is the force exerted on the springthat causes the spring to stretch.Q6.36.26.3Q6.1Driving in a circle at a constant speed requires a centripetalacceleration but no tangential acceleration.(a)The object will move in a circle at a constant speed.(b)The object will move in a straight line at a changing speed.Q6.5The speed changes. The tangential force component causes tangential acceleration.Q6.6Consider the force required to keep a rock in the Earths crust moving in a circle. The size of theforce is proportional to the radius of the circle. If that rock is at the Equator, the radius of the circlethrough which it moves is about 6400 km. If the rock is at the north pole, the radius of the circlethrough which it moves is zero!Q6.7Consider standing on a bathroom scale. The resultant force on you is your actual weight minus thenormal force. The scale reading shows the size of the normal force, and is your apparent weight. Ifyou are at the North or South Pole, it can be precisely equal to your actual weight. If you are at theequator, your apparent weight must be less, so that the resultant force on you can be a downwardforce large enough to cause your centripetal acceleration as the Earth rotates.Q6.8A torque is exerted by the thrust force of the water times the distance between the nozzles.157158Circular Motion and Other Applications of Newtons LawsQ6.9I would not accept that statement for two reasons. First, to be beyond the pull of gravity, onewould have to be infinitely far away from all other matter. Second, astronauts in orbit are moving ina circular path. It is the gravitational pull of Earth on the astronauts that keeps them in orbit. In thespace shuttle, just above the atmosphere, gravity is only slightly weaker than at the Earths surface.Gravity does its job most clearly on an orbiting spacecraft, because the craft feels no other forces andis in free fall.Q6.10This is the same principle as the centrifuge. All the material inside the cylinder tends to move alonga straight-line path, but the walls of the cylinder exert an inward force to keep everything movingaround in a circular path.Q6.11The ball would not behave as it would when dropped on the Earth. As the astronaut holds the ball,she and the ball are moving with the same angular velocity. The ball, however, being closer to thecenter of rotation, is moving with a slower tangential velocity. Once the ball is released, it actsaccording to Newtons first law, and simply drifts with constant velocity in the original direction ofits velocity when releasedit is no longer attached to the rotating space station. Since the ballfollows a straight line and the astronaut follows a circular path, it will appear to the astronaut thatthe ball will fall to the floor. But other dramatic effects will occur. Imagine that the ball is held sohigh that it is just slightly away from the center of rotation. Then, as the ball is released, it will movevery slowly along a straight line. Thus, the astronaut may make several full rotations around thecircular path before the ball strikes the floor. This will result in three obvious variations with theEarth drop. First, the time to fall will be much larger than that on the Earth, even though the feet ofthe astronaut are pressed into the floor with a force that suggests the same force of gravity as onEarth. Second, the ball may actually appear to bob up and down if several rotations are made whileit falls. As the ball moves in a straight line while the astronaut rotates, sometimes she is on the sideof the circle on which the ball is moving toward her and other times she is on the other side, wherethe ball is moving away from her. The third effect is that the ball will not drop straight down to herfeet. In the extreme case we have been imagining, it may actually strike the surface while she is onthe opposite side, so it looks like it ended up falling up. In the least extreme case, in which only aportion of a rotation is made before the ball strikes the surface, the ball will appear to movebackward relative to the astronaut as it falls.Q6.12The water has inertia. The water tends to move along a straight line, but the bucket pulls it in andaround in a circle.Q6.13There is no such force. If the passenger slides outward across the slippery car seat, it is because thepassenger is moving forward in a straight line while the car is turning under him. If the passengerpushes hard against the outside door, the door is exerting an inward force on him. No object isexerting an outward force on him, but he should still buckle his seatbelt.Q6.14Blood pressure cannot supply the force necessary both to balance the gravitational force and toprovide the centripetal acceleration, to keep blood flowing up to the pilots brain.Q6.15The person in the elevator is in an accelerating reference frame. The apparent acceleration due togravity, g, is changed inside the elevator. g= g aQ6.16When you are not accelerating, the normal force and your weight are equal in size. Your bodyinterprets the force of the floor pushing up on you as your weight. When you accelerate in anelevator, this normal force changes so that you accelerate with the elevator. In free fall, you arenever weightless since the Earths gravity and your mass do not change. It is the normal forceyourapparent weightthat is zero.Chapter 6Q6.17159From the proportionality of the drag force to the speed squared and from Newtons second law, wederive the equation that describes the motion of the skydiver:mdv ydt= mg D A 2vy2where D is the coefficient of drag of the parachutist, and A is the projected area of the parachutistsbody. At terminal speed,ay =dv ydtF 2mg I= 0 and V GH D A JKT12.When the parachute opens, the coefficient of drag D and the effective area A both increase, thusreducing the speed of the skydiver.Modern parachutes also add a third term, lift, to change the equation tomdv ydt= mg D A 2 L A 2vy vx22where v y is the vertical velocity, and v x is the horizontal velocity. The effect of lift is clearly seen inthe paraplane, an ultralight airplane made from a fan, a chair, and a parachute.Q6.18The larger drop has higher terminal speed. In the case of spheres, the text demonstrates thatterminal speed is proportional to the square root of radius. When moving with terminal speed, anobject is in equilibrium and has zero acceleration.Q6.19Lower air density reduces air resistance, so a tank-truck-load of fuel takes you farther.Q6.20Suppose the rock is moving rapidly when it enters the water. The speed of the rock decreases until itreaches terminal velocity. The acceleration, which is upward, decreases to zero as the rockapproaches terminal velocity.Q6.21The thesis is false. The moment of decay of a radioactive atomic nucleus (for example) cannot bepredicted. Quantum mechanics implies that the future is indeterminate. On the other hand, oursense of free will, of being able to make choices for ourselves that can appear to be random, may bean illusion. It may have nothing to do with the subatomic randomness described by quantummechanics.160Circular Motion and Other Applications of Newtons LawsSOLUTIONS TO PROBLEMSSection 6.1P6.1Newtons Second Law Applied to Uniform Circular Motionm = 3.00 kg , r = 0.800 m. The string will break if the tension exceedsthe weight corresponding to 25.0 kg, soa fTmax = Mg = 25.0 9.80 = 245 N .When the 3.00 kg mass rotates in a horizontal circle, the tensioncauses the centripetal acceleration,a f3.00 v 2mv 2=.r0.800soT=Thenv2 =and0 v 65.3orP6.2af aa ff0.800 T0.800 Tmax 0.800 245rT=== 65.3 m 2 s 2m3.003.003.00FIG. P6.10 v 8.08 m s .v2, both m and r are unknown but remain constant. Therefore, F is proportional to v 2r218.0and increases by a factor ofas v increases from 14.0 m/s to 18.0 m/s. The total force at the14.0higher speed is thenInF = mFG IJH KF 18.0 I a130 Nf = 215 N . Ffast = G 14.0 JH KSymbolically, writeDividing givesF mI Fslow = G r J b14.0H K Ffast Fslow=FG 18.0 IJH 14.0 Kmsg22andF mI Ffast = G r J b18.0H Kg2ms .2, orF 18.0 I2F 18.0 I a130 Nf = Ffast = G 14.0 J Fslow = G 14.0 JH KH K2215 N .This force must be horizontally inward to produce the drivers centripetal acceleration.Chapter 6ejej2P6.4(a)(b)P6.39.11 0 31 kg 2.20 10 6 m smv 2=F=r0.530 10 10 m2.20 10 6 m sv2== 9.13 10 22 m s 2 inwarda=r0.530 10 10 mej2Neglecting relativistic effects. F = ma c =e(a)(b)mv 2re2.998 10 m sjkg ja0.480 mf7F = 2 1.661 10P6.5= 8.32 10 8 N inward272= 6.22 10 12 Nstatic frictione jma i = f i + nj + mg j Fy = 0 = n mgv2= f = n = mg .r250.0 cm sv2== 0.085 0 .Then =rg30.0 cm 980 cm s 2thus n = mg and Fr = mbaP6.6(a)P6.7fe Fy = ma y , mg moon down =v = g moon r =(b)gmv 2downre1.52 m s je1.7 102ej6jm + 100 10 3 m = 1.65 10 3 m sj2 1.8 10 6 m2rv=,T== 6.84 10 3 s = 1.90 hT1.65 10 3 m sn = mg since a y = 0The force causing the centripetal acceleration is the frictional force f.From Newtons second law f = ma c =mv 2.rBut the friction condition is f sni.e.,FIG. P6.7mv 2 s mgrafev s rg = 0.600 35.0 m 9.80 m s 2jv 14.3 m s161162Circular Motion and Other Applications of Newtons Laws2v=rb86.5 km hge1h3 600 sje1 000 m1 km2j F 1g I=GH 9.80 m s JKP6.8a=P6.9T cos 5.00 = mg = 80.0 kg 9.80 m s 2261.0 mbge0.966 gja68.6 Nfi + a784 Nf j(a)T = 787 N : T =(b)T sin 5.00 = ma c : a c = 0.857 m s 2 toward the center ofthe circle.The length of the wire is unnecessary information. Wecould, on the other hand, use it to find the radius of thecircle, the speed of the bob, and the period of the motion.P6.10(b)v=235 m= 6.53 m s36.0 sThe radius is given by12r = 235 m4r = 150 m(a)F v I toward centerGH r JKb6.53 m sg at 35.0 north of west=2ar =2150 mejee jj= 0.285 m s 2 cos 35.0 i + sin 35.0 j= 0.233 m s 2 i + 0.163 m s 2 j(c)a==dvf viite6.53 m s j 6.53 m s ij36.0 s= 0.181 m s 2 i + 0.181 m s 2 jFIG. P6.9Chapter 6b gejFg = mg = 4 kg 9.8 m s 2 = 39.2 NTa1.5 m2m = 48.6r = 2 m cos 48.6 = 1.32 m*P6.11163sin =39.2 Na fmv 2 Fx = ma x = rTa cos 48.6+Tb cos 48.6 =Ta + Tb =Tbb gb4 kg 6 m s109 N= 165 Ncos 48.6gforcesacv21.32 mmotionFIG. P6.11 Fy = ma y+Ta sin 48.6Tb sin 48.639.2 N = 0Ta Tb =(a)39. 2 N= 52.3 Nsin 48.6To solve simultaneously, we add the equations in Ta and Tb :Ta + Tb + Ta Tb = 165 N + 52.3 NTa =(b)*P6.12217 N= 108 N2Tb = 165 N Ta = 165 N 108 N = 56.2 Nv2. Let f represent the rotation rate. Each revolution carries each bit of metal through distancer2r , so v = 2rf andac =ac =v2= 4 2 rf 2 = 100 g .rA smaller radius implies smaller acceleration. To meet the criterion for each bit of metal we considerthe minimum radius:F 100 g IJf =GH 4 r K212F 100 9.8 m s I=GH 4 a0.021 mf JK2212= 34.4FGHIJK1 60 s= 2.06 10 3 rev min .s 1 min164Circular Motion and Other Applications of Newtons LawsSection 6.2P6.13Nonuniform Circular MotionM = 40.0 kg , R = 3.00 m, T = 350 N(a)TMv 2RRM F = 2T Mg =gFGH IJKF 3.00 IJ = 23.1 em s j= 700 a 40.0fa9.80f GH 40.0 Kbv 2 = 2T Mgv222n Mg = F =n = Mg +P6.14(a)child + seatMv 2Rchild aloneFIG. P6.13(a)FIG. P6.13(b)IJKFGHMv 223.1= 40.0 9.80 += 700 NR3.00Consider the forces acting on the system consisting of the child and the seat: Fy = ma y 2T mg = mv2 = Rv=(b)FG 2T gIJHm KF 2T gIJRGHm Kv2RConsider the forces acting on the child alone:F Fy = ma y n = mG g +Hand from above, v 2 = Rv2RIJKFG 2T gIJ , soHm KFGHn=m g+P6.15nMgv = 4.81 m s(b)MgTIJK2T g = 2T .mLet the tension at the lowest point be T. F = ma:T mg = ma c =FGHmv 2rIJKLb8.00 m sgT = b85.0 kg gM9.80 m s +10.0 mMNT=m g+v2r22OPPQ = 1.38 kN > 1 000 NHe doesnt make it across the river because the vine breaks.FIG. P6.15Chapter 6b(a)(b)P6.164.00 m sv2=ac =r12.0 mg22a = a c + a t2a=a1.33f + a1.20f2= 1.33 m s 22at an angle = tan 1= 1.79 m s 2FG a IJ =Ha KcFIG. P6.1648.0 inwardtP6.17 Fy =mv 2= mg + nrBut n = 0 at this minimum speed condition, somv 2= mg v = gr =rP6.18e9.80 m s ja1.00 mf =23.13 m s .At the top of the vertical circle,T=ma(a)f a4..00f a0.400fa9.80f =0 500v2 mgR2or T = 0.400P6.19FIG. P6.178.88 Nv = 20.0 m s,BCn = force of track on roller coaster, and15 m10 mR = 10.0 m .AMv 2 F = R = n MgFIG. P6.19From this we findn = Mg +bge500 kg 20.0 m s 2Mv 2= 500 kg 9.80 m s 2 +R10.0 mbgejn = 4 900 N + 20 000 N = 2.49 10 4 N(b)At B, n Mg = Mv 2RThe max speed at B corresponds ton=0 Mg = 2Mv max v max = Rg = 15.0 9.80 = 12.1 m sRa fj165166Circular Motion and Other Applications of Newtons Lawsbeg2(a)(b)P6.20v2ac =rLet n be the force exerted by the rail.13.0 m sv2== 8.62 mr=a c 2 9.80 m s 2jNewtons law givesMv 2Mg + n =rv2n=M g = M 2 g g = Mg , downwardrIJKFGH(c)ac =v2rac =b13.0 m sgbFIG. P6.20g220.0 m= 8.45 m s 2If the force exerted by the rail is n1thenMv 2= Ma crn1 = M a c g which is < 0 since a c = 8.45 m s 2n1 + Mg =bgThus, the normal force would have to point away from the center of the curve. Unless theyhave belts, the riders will fall from the cars. To be safe we must require n1 to be positive.Then a c > g . We needv2> g or v > rg =rSection 6.3P6.21(a)a20.0 mfe9.80 m s j , v > 14.0 m s .2Motion in Accelerated FramesT18.0 N Fx = Ma , a = M = 5.00 kg =3.60 m s 2to the right.(b)If v = const, a = 0, so T = 0 (This is alsoan equilibrium situation.)(c)Someone in the car (noninertial observer)claims that the forces on the mass along xare T and a fictitious force (Ma). Someoneat rest outside the car (inertial observer)claims that T is the only force on M in thex-direction.5.00 kgFIG. P6.21167Chapter 6*P6.22We adopt the view of an inertial observer. If it is on the verge of sliding, thecup is moving on a circle with its centripetal acceleration caused by friction. Fy = ma y :f=f+n mg = 0 Fx = ma x :nmv 2= sn = s mgrjaemgFIG. P6.22fv = s gr = 0.8 9.8 m s 2 30 m = 15.3 m sIf you go too fast the cup will begin sliding straight across the dashboard to the left.P6.23The only forces acting on the suspended object are the force of gravity mgand the force of tension T, as shown in the free-body diagram. ApplyingNewtons second law in the x and y directions, Fx = T sin = ma Fy = T cos mg = 0or(a)(1)T cos = mg(2)T cos T sin mgFIG. P6.23Dividing equation (1) by (2) givestan =a 3.00 m s 2== 0.306 .g 9.80 m s 2Solving for , = 17.0(b)From Equation (1),T=*P6.24afca fh0.500 kg 3.00 m s 2ma== 5.12 N .sin sin 17.0The water moves at speedv=af2r 2 0.12 m== 0.104 m s .T7.25 sThe top layer of water feels a downward force of gravity mg and an outward fictitious force in theturntable frame of reference,bmv 2 m 0.104 m s=r0.12 mg2= m9.01 10 2 m s 2 .It behaves as if it were stationary in a gravity field pointing downward and outward attan 10.090 1 m s 29.8 m s 2= 0.527 .Its surface slopes upward toward the outside, making this angle with the horizontal.168P6.25Circular Motion and Other Applications of Newtons LawsFmax = Fg + ma = 591 NFmin = Fg ma = 391 N(a)Adding, 2 Fg = 982 N, Fg = 491 N(b)Since Fg = mg , m =(c)Subtracting the above equations,491 N= 50.1 kg9.80 m s 2 a = 2.00 m s 22ma = 200 NP6.26(a) Fr = marmg =g=T=(b)*P6.27FGHmv 2 m 2R=RR TIJK24 2 RT24 2 R6.37 10 6 m= 2= 5.07 10 3 s = 1.41 hg9.80 m s 2speed increase factor =FGHIJKv newT2R Tcurrent24.0 h== current == 17.12R1.41 hv current TnewTnewThe car moves to the right with acceleration a. We find the acceleration of a b of the block relative tothe Earth. The block moves to the right also. Fy = ma y : Fx = ma x :+n mg = 0 , n = mg , f = k mg+ k mg = ma b , a b = k gThe acceleration of the block relative to the car is a b a = k g a . In this frame the block starts fromrest and undergoes displacement and gains speed according tod22v xf = v xi + 2 a x x f xi2v xf(a)d bv = 2 a kgcontinued on next pagegi12bgaif bg= 0 + 2 kg a 0 = 2 a kg .to the leftChapter 6(b)169The time for which the box slides is given by1v xi + v xf t21 =0 2 a kg2diLM d bgi OQPtNF 2 I .t=GH a g JKx =1212kThe car in the Earth frame acquires finals speed v xf = v xiF 2 I+ at = 0 + aGH a g JK12. The speedkof the box in the Earth frame is thenbga 2 f b a g g + a 2 f=ba g gv be = v bc + v ce = 2 a k g121212k+aa12*P6.28b k g22 a kgg12=12k12k2 k g.vConsider forces on the backpack as it slides in the Earth frame of reference. Fy = ma y : Fx = ma x :We solve for k : vt L =bgb+n mg = ma , n = m g + a , f k = k m g + abg k m g + a = ma xThe motion across the floor is described by L = vt +P6.29=12kk=F 2 IGH a g JK ga2 fba g gafbgg11a x t 2 = vt k g + a t 2 .222 vt L1= k .k g + a t2 ,2g + a t2bgbgIn an inertial reference frame, the girl is accelerating horizontally inward atb5.70 m sv2=r2.40 mg2= 13.5 m s 2In her own non-inertial frame, her head feels a horizontally outward fictitious force equal to its masstimes this acceleration. Together this force and the weight of her head add to have a magnitudeequal to the mass of her head times an acceleration ofg2 +Fv IGH r JK22=a9.80f + a13.5f22m s 2 = 16.7 m s 216.7= 1.71 .9.80Thus, the force required to lift her head is larger by this factor, or the required force isThis is larger than g by a factor ofafF = 1.71 55.0 N = 93.8 N .170*P6.30Circular Motion and Other Applications of Newtons Laws(a)The chunk is at radius r =0.137 m + 0.080 m= 0.054 2 m . Its speed is4v=20 0002r= 2 0.054 2 m= 114 m s60 sTbgand its accelerationac =bg2114 m sv2== 2.38 10 5 m s 2 horizontally inwardr0.054 2 m= 2.38 10 5 m s 2(b)F g I=GH 9.8 m s JK22.43 10 4 g .In the frame of the turning cone, the chunk feels amv 2. In this frame itshorizontally outward force ofr3.3 cmacceleration is up along the cone, at tan 1 a13 .7 8 f cm = 49.2 .e+n jemv 2rf49.22Take the y axis perpendicular to the cone: Fy = ma y :anFIG. P6.30(b)mv 2sin 49.2 = 0rjn = 2 10 3 kg 2.38 10 5 m s 2 sin 49.2 = 360 N(c)aff = k n = 0.6 360 N = 216 N Fx = ma x :e2 103je2mvcos 49. 2 f = ma xrjejkg 2.38 10 5 m s 2 cos 49.2216 N = 2 10 3 kg a xa x = 47.5 10 4 m s 2 radially up the wall of the coneP6.31F 4 R I cos 35.0 = 0.027 6 m sGH T JKWe take the y axis along the local vertical.ba g = 9.80 ba g = 9.78 m sba g = 0.015 8 m sar =22net ynet x = arctan2eN2r y2ax= 0.092 8ay35.0 a r(exaggerated size)g0a net35.0EquatorFIG. P6.31Chapter 6Section 6.4P6.32Motion in the Presence of Resistive Forcesm = 80.0 kg , vT = 50.0 m s , mg =(a)2DAvT DA mg= 2 = 0.314 kg m22vTv = 30.0 m sAta=g(b)171DAv 22m= 9.80 a0.314fa30.0f80.02= 6. 27 m s 2 downwardAt v = 50.0 m s , terminal velocity has been reached. Fy = 0 = mg Rbgej R = mg = 80.0 kg 9.80 m s 2 = 784 N directed up(c)v = 30.0 m sDAv 2= 0.314 30.02AtaP6.33(a)fa f2= 283 N upwarda = g bvWhen v = vT , a = 0 and g = bvTb=gvTThe Styrofoam falls 1.50 m at constant speed vT in 5.00 s.y 1.50 m== 0.300 m s5.00 stThus,Thenb=(b)At t = 0 , v = 0 anda = g = 9.80 m s 2 down(c)P6.34vT =When v = 0.150 m s, a = g bv = 9.80 m s 2 32.7 s 1 0.150 m s = 4.90 m s 2(a)=9.80 m s 2= 32.7 s 10.300 m sjbegdownm12, A = 0.020 1 m 2 , R = air ADvT = mgV2m = beadV = 0.830 g cm 3LM 4 a8.00 cmf OP = 1.78 kgN3Q3Assuming a drag coefficient of D = 0.500 for this spherical object, and taking the density ofair at 20C from the endpapers, we havevT =(b)v2f=vi2bgej =0.500e1.20 kg m je0.020 1 m j2 1.78 kg 9.80 m s 23b53.8 m sg ==+ 2 gh = 0 + 2 gh : h =2 g 2e9.80 m s jv2f222148 m53.8 m s172P6.35Circular Motion and Other Applications of Newtons LawsSince the upward velocity is constant, the resultant force on the ball is zero. Thus, the upwardapplied force equals the sum of the gravitational and drag forces (both downward):F = mg + bv .The mass of the copper ball isFG IJ eHK4r 34= 8.92 10 3 kg m3 2.00 10 2 m33m=jej3= 0.299 kg .The applied force is thenafa f afejF = mg + bv = 0.299 9.80 + 0.950 9.00 10 2 = 3.01 N .P6.36 Fy = ma y+T cos 40.0 mg = 0b620 kgge9.80 m s j = 7.93 102T=3cos 40.0Fx = ma xN R + T sin 40.0 = 0ejR = 7.93 10 3 N sin 40.0 = 5.10 10 3 N =D=P6.37(a)e2 5.10 10 3 NjFHkg m s 2NIK1DAv 222R=Av 21. 20 kg m 2 3.80 m 2 40.0 m sejejb2= 1. 40R = vT b = mgAt terminal velocity,b =(b)gFIG. P6.36ejej3.00 10 3 kg 9.80 m s 2mg== 1.47 N s mvT2.00 10 2 m sIn the equation describing the time variation of the velocity, we haveev = vT 1 e bt mjv = 0.632 vT when e bt m = 0.368FG m IJ lna0.368f =HbK2.04 10 3 sor at time(c)P6.38t=At terminal velocity,R = vT b = mg = 2.94 10 2 NThe resistive force isafejb11DAv 2 = 0.250 1.20 kg m3 2.20 m 2 27.8 m s22R = 255 NR=a=255 NR== 0.212 m s 2m1200 kgjeg2Chapter 6P6.39afav t = vi e ct(a)fv 20.0 s = 5.00 = vi e 20 .0 c , vi = 10.0 m s .(b)At t = 40.0 sFG 1 IJH 2Kv = b10.0 m sge(c)v = vi e cts=So 5.00 = 10.0 e 20 .0 c and 20.0 c = lnP6.40 F = ma kmv 2 = m kdt =dvv2c=40 .0 cc h=ln1220.0b3.47 10 2 s 1fga= 10.0 m s 0.250 = 2.50 m sdv= cvi e ct = cvdtdvdtz ztv0v0 k dt = v 2 dva fk t 0 =v 11v=v01 1+v v01 + v 0 kt1 1=+ kt =v v0v0v0v=1 + v 0 kt*P6.41(a)From Problem 40,v=v0dx=dt 1 + v 0 ktz zxtdx = v 000ztdt1 v 0 kdt=1 + v 0 kt k 0 1 + v 0 ktbbggt1ln 1 + v 0 kt0k1x 0 = ln 1 + v 0 kt ln 1k1x = ln 1 + v 0 ktkxx0 =b(b)*P6.42bggWe have ln 1 + v 0 kt = kxv0v0= kx = v 0 e kx = v1 + v 0 kt = e kx so v =1 + v 0 kt eWe write kmv 2 = 1DAv 2 so2k=ejej332DA 0.305 1.20 kg m 4.2 10 m== 5.3 10 3 m2m2 0.145 kgbbgv = v 0 e kx = 40.2 m s ee 5.3 103gjam 18 .3 mf=36.5 m s173174P6.43Circular Motion and Other Applications of Newtons Lawsafaf1DAv 2 , we estimate that D = 1.00 , = 1.20 kg m3 , A = 0.100 m 0.160 m = 1.60 10 2 m 22and v = 27.0 m s. The resistance force is thenIn R =R=a fejb11.00 1.20 kg m3 1.60 10 2 m 2 27.0 m s2jeg2= 7.00 NorR ~ 10 1 NSection 6.5Numerical Modeling in Particle Dynamicsaf af afNote: In some problems we compute each new position as x t + t = x t + v t + t t , rather thanx t + t = x t + v t t as quoted in the text. This method has the same theoretical validity as that presented inthe text, and in practice can give quicker convergence.aP6.44f af af(a)(b)At v = vT , a = 0, mg + bvT = 0afts00.0050.010.015vT =xma fvms221.999 7551.999 300.0490.095 550.139 77ejej3.00 10 3 kg 9.80 m s 2mg== 0.980 m sb3.00 10 2 kg sa fb gF mN29.427.9326.53425.2ea m s2j9.89.318.844 58.40. . . we list the result after each tenth iteration0.050.10.150.20.250.30.350.40.450.50.550.60.651.9901.9651.9301.8891.8451.7991.7521.7041.651.611.561.511.460.3930.6290.7700.8540.9040.9350.9530.9640.9700.9740.9770.9780.97917.610.56.313.782.261.350.8110.4860.2910.1740.1100.062 40.037 45.873.512.101.260.7540.4510.2700.1620.096 90.058 00.034 70.020 80.012 5Terminal velocity is never reached. The leaf is at 99.9% of vT after 0.67 s. The fall to theground takes about 2.14 s. Repeating with t = 0.001 s , we find the fall takes 2.14 s.Chapter 6P6.45(a)When v = vT , a = 0,2 F = mg + CvT = 0mg=vT = Caf(b)tsa fe4.80 104jekg 9.80 m s 22.50 105kg ma fb gj=13.7 m sevmsF mNa m s201.963.885.683 27.306 88.710 79.880 310.82311.56312.1312.56xm000.3921.1682.303.775.517.489.6511.9614.400.20.40.60.81.01.21.41.61.82175 4.704 4.608 4.327 63.896 53.369 32.807 12.263 51.775 31.361 61.030.7629.89.599 99.015 98.117 87.019 35.848 14.715 63.698 62.836 62.141.590.1540.029 10.005 42j0.3210.060 60.011 3. . . listing results after each fifth step13.4913.6713.7127.441.054.7345The hailstone reaches 99% of vT after 3.3 s, 99.95% of vT after 5.0 s, 99.99% of vT after 6.0 s,99.999% of vT after 7.4 s.P6.46(a)At terminal velocity,2 F = 0 = mg + CvTC=(b)(c)emg2vTb0.142 kgge9.80 m s j =b42.5 m sg2=jb2Cv 2 = 7.70 10 4 kg m 36.0 m sg27.70 10 4 kg m= 0.998 NElapsedTime (s)Altitude(m)Speed(m/s)ResistanceForce (N)NetForce (N)0.000 000.050 002.950 003.000 003.050 006.250 006.300 000.000 001.757 92Accelerationm s236.000 0035.158 420.998 490.952 352.390 092.343 9516.831 5816.506 6748.623 2748.640 0048.632 240.824 940.334 760.155 270.000 520.000 090.000 021.392 121.391 691.391 589.803 699.800 619.799 871.250 850.106 5226.852 9727.147 360.555 550.567 800.836 050.823 805.887 695.801 44ejMaximum height is about 49 m . It returns to the ground after about 6.3 s with a speedof approximately 27 m s .176P6.47Circular Motion and Other Applications of Newtons Laws(a)At constant velocitymg=vT = CvT = (b)(a)b50.0 kg ge9.80 m s j =20.200 kg mb50.0 kg gb9.80 m sg =20.0 kg m49.5 m s with chute closed and4.95 m s with chute open.We use time increments of 0.1 s for 0 < t < 10 s , then 0.01 s for 10 s < t < 12 s , and then 0.1 sagain.time(s)012471010.110.31112501001456.482 F = 0 = mg + CvTheight(m)10009959809298126746716696656594712240velocity(m/s)09.718.632.743.747.716.78.025.094.954.954.954.95We use a time increment of 0.01 s.time(s)00.1000.2000.4001.001.922.004.005.006.85x(m)07.8114.927.151.970.070.980.481.481.8y(m)05.4310.218.332.738.538.526.717.70(b)range = 81.8 m(c)So we have maximum range at = 15.9with 30.035.025.020.015.010.017.016.015.515.816.115.9we find range86.410 m81.8 m90.181 m92.874 m93.812 m90.965 m93.732 m93.839 8 m93.829 m93.839 m93.838 m93.840 2 mChapter 6P6.49(a) F = mg + Cv 2 = 0 . Thus,At terminal speed,C=(b)mgv2b0.046 0 kgge9.80 m s j =b44.0 m sg2=22.33 10 4 kg mWe set up a spreadsheet to calculate the motion, try different initial speeds, and home in on53 m s as that required for horizontal range of 155 m, thus:Timet (s)x(m)0.000 00.002 72.501 62.504 32.506 93.423 83.426 53.429 15.151 65.154 3(c)1770.000 00.121 1vx(m/s)eaxm s2jy(m)vyayem s jv=22vx+2vytan 1Fv IGH v JKyx(m/s)(deg)45.687 0 10.565 9 0.000 0 27.451 5 13.614 645.659 0 10.552 9 0.072 7 27.415 5 13.604 653.300 053.257 431.000 030.982 290.194 6 28.937 5 4.238 8 32.502 4 0.023 5 9.800 090.271 3 28.926 3 4.235 5 32.502 4 0.002 4 9.800 090.348 0 28.915 0 4.232 2 32.502 4 0.028 4 9.800 028.937 528.926 328.915 10.046 60.004 80.056 3115.229 8 25.492 6 3.289 6 28.397 2 8.890 5 9.399 9115.297 4 25.483 9 3.287 4 28.373 6 8.915 4 9.397 7115.364 9 25.475 1 3.285 1 28.350 0 8.940 3 9.395 426.998 426.998 426.998 419.226 219.282 219.338 2154.996 8 20.843 8 2.199 2 0.005 9 23.308 7 7.049 8155.052 0 20.838 0 2.198 0 0.055 9 23.327 4 7.045 431.269 231.279 248.195 448.226 2(m/s)Similarly, the initial speed is 42 m s . The motion proceeds thus:Timet (s)x(m)0.000 0 0.000 00.003 5 0.100 62.740 5 66.307 82.744 0 66.379 72.747 5 66.451 63.146 5 74.480 53.150 0 74.549 53.153 5 74.618 55.677 0 118.969 75.680 5 119.024 8vx(m/s)eaxm s2jy(m)vyayem s j2v=2vx+2vytan 1Fv IGH v JKyx(m/s)(deg)0.000 0 30.826 6 14.610 30.107 9 30.775 4 14.594 342.150 042.102 647.000 046.967 120.548 4 2.137 4 39.485 4 0.026 0 9.800 020.541 0 2.135 8 39.485 5 0.008 3 9.800 020.533 5 2.134 3 39.485 5 0.042 6 9.800 020.548 520.541 020.533 50.072 50.023 10.118 819.715 6 1.967 6 38.696 3 3.942 3 9.721 319.708 7 1.966 2 38.682 5 3.976 4 9.720 019.701 8 1.964 9 38.668 6 4.010 4 9.718 620.105 820.105 820.105 811.307 711.406 711.505 615.739 4 1.254 0 0.046 5 25.260 0 6.570 115.735 0 1.253 3 0.041 9 25.283 0 6.564 229.762 329.779 558.073 158.103 728.746 2 4.182 928.731 6 4.178 7(m/s)The trajectory in (c) reaches maximum height 39 m, as opposed to 33 m in (b). In both, theball reaches maximum height when it has covered about 57% of its range. Its speed is aminimum somewhat later. The impact speeds are both about 30 m/s.178Circular Motion and Other Applications of Newtons LawsAdditional Problems*P6.50When the cloth is at a lower angle , the radialcomponent of F = ma reads68Rmv 2n + mg sin =.rmgAt = 68.0 , the normal force drops to zero andv2.g sin 68 =rv = rg sin 68 =ppmg sin68mg cos68FIG. P6.50a0.33 mfe9.8 m s j sin 68 = 1.73 m s2The rate of revolution isbangular speed = 1.73 m s*P6.51(a)bv = 30 km hgFGH 12rev IJK FGH 2 a02.r mf IJK =r330.835 rev s = 50.1 rev min .1hgFGH 3 600 s IJK FGH 1 1000 m IJK = 8.33 m skm Fy = ma y : +n mg = mv 2rLF vIb8.33 m sgn = mG g J = 1 800 kg M9.8 m s 20. 4 mMNH rK22n2OPPQmgFIG. P6.51= 1.15 10 4 N up(b)Take n = 0 . Then mg =mv 2.rv = gr =P6.52(a) Fy = ma y =mg n =(b)mv 2Re9.8 m s ja20.4 mf =2mv 2Rn = mg When n = 0 ,mg =Then,v=mv 2Rmv 2RgR .14.1 m s = 50.9 km hChapter 6(a)slope =(b)*P6.53slope =(c)(d)0.160 N 0= 0.016 2 kg m9.9 m 2 s 2Rv2=12DAv 2v1D A2=21DA = 0.016 2 kg m22 0.016 2 kg mD=1.20 kg m3 0.105 me179bgjaf2= 0.778ejejFrom the table, the eighth point is at force mg = 8 1.64 10 3 kg 9.8 m s 2 = 0.129 N andbg2horizontal coordinate 2.80 m s . The vertical coordinate of the line is here0.129 N 0.127 N2= 1.5%.0.016 2 kg m 2.8 m s = 0.127 N . The scatter percentage is0.127 Nbgbg(e)P6.54The interpretation of the graph can be stated thus: For stacked coffee filters falling atterminal speed, a graph of air resistance force as a function of squared speed demonstratesthat the force is proportional to the speed squared within the experimental uncertaintyestimated as 2%. This proportionality agrees with that described by the theoretical equation1R = DAv 2 . The value of the constant slope of the graph implies that the drag coefficient2for coffee filters is D = 0.78 2% .(a)While the car negotiates the curve, the accelerometer is at the angle .mv 2rHorizontally:T sin =Vertically:T cos = mgwhere r is the radius of the curve, and v is the speed of the car.tan =By division,v2= g tan :Then a c =rv2rgea c = 9.80 m s2j tan 15.0a c = 2.63 m s 2b23.0 m sgr=(b)v2r=ac(c)v 2 = rg tan = 201 m 9.80 m s 2 tan 9.0022.63 m s 2afejv = 17.7 m s= 201 mFIG. P6.54180P6.55Circular Motion and Other Applications of Newtons LawsTake x-axis up the hill Fx = ma x :+T sin mg sin = maTa = sin g sin m Fy = ma y : +T cos mg cos = 0mg cos T=cos g cos sin g sin a=cos a = g cos tan sin b*P6.56(a)af2 7.46 m= 1.23 m s . The38 stotal force on it must add toThe speed of the bag isma c =b30 kg gb1.23 m sgfsynac2mg= 6.12 N7.46 mgFIG. P6.56 Fx = ma x : Fy = ma y :fs cos 20 n sin 20 = 6.12 Nbgejfs sin 20 + n cos 20 30 kg 9.8 m s 2 = 0f cos 20 6.12 Nn= ssin 20Substitute:fs sin 20 + fscos 2 20cos 20 6.12 N= 294 Nsin 20sin 20f s 2.92 = 294 N + 16.8 Nafa ffs = 106 N(b)abfgb2 7.94 m= 1.47 m s34 s230 kg 1.47 m s= 8.13 Nma c =7.94 mfs cos 20 n sin 20 = 8.13 Nv=gfs sin 20 + n cos 20 = 294 Nfs cos 20 8.13 Nsin 20cos 2 20cos 20 8.13 N= 294 Nfs sin 20 + fssin 20sin 20fs 2.92 = 294 N + 22.4 Nn=aa ffs = 108 Nn=fa108 Nf cos 20 8.13 N = 273 Nsin 20fs 108 N= 0.396s = =n 273 NxChapter 6P6.57(a)Since the centripetal acceleration of a person is downward (towardthe axis of the earth), it is equivalent to the effect of a fallingelevator. Therefore,Fg = Fg (b)181mv 2or Fg > Fgra fAt the poles v = 0 and Fg = Fg = mg = 75.0 9.80 = 735 N down.bFIG. P6.57gAt the equator, Fg = Fg ma c = 735 N 75.0 0.033 7 N = 732 N down.(a)(b)Since the object of mass m 2 is in equilibrium, Fy = T m 2 g = 0orP6.58T = m2 g .The tension in the string provides the required centripetal acceleration of the puck.Thus,P6.59(a)FromFc =we have(c)Fc = T = m 2 g .v=bv = 300 mi hm1 v 2RRFc=m1FG m IJ gRHm K2.188 ftgFGH 60.0.0 mi sh IJK = 440 ft sAt the lowest point, his seat exerts an upward force; therefore, his weight seems to increase.His apparent weight isFg = mg + m(b)FGHv2160= 160 +r32.0IJ a440fK 1 2002= 967 lb .At the highest point, the force of the seat on the pilot is directed down andFg = mg mv2= 647 lb .rSince the plane is upside down, the seat exerts this downward force.(c)mv 2. If we vary the aircrafts R and v such that the above is true,Rthen the pilot feels weightless.When Fg = 0 , then mg =182P6.60Circular Motion and Other Applications of Newtons LawsFor the block to remain stationary,e Fy = 0 and Fx = mar .jejn1 = m p + m b g so f s1n1 = s1 m p + m b g .mb gAt the point of slipping, the required centripetal force equals themaximum friction force:e2v maxn1j r = em + m j ga0.750fa0.120fa9.80f = 0.939 m s . mp + mbor v max = s1 rg =s1pmp gfpbfFor the penny to remain stationary on the block:mb g Fy = 0 n 2 m p g = 0 or n 2 = m p g Fx = ma r f p = m pandn22v.rfpWhen the penny is about to slip on the block, f p = f p , max = s 2 n 2or s 2 m p g = m pmp g2v maxrmp gv max = s 2 rg =a0.520fa0.120fa9.80f = 0.782 m sFIG. P6.60This is less than the maximum speed for the block, so the penny slips before the block starts to slip.The maximum rotation frequency isMax rpm =P6.61v=agLM aNOPFG 60 s IJ =f QH 1 min Kv max1 rev= 0.782 m s2r2 0.120 mbf2r 2 9.00 m== 3.77 m sT15.0 safv2= 1.58 m s 2r(a)ar =(b)Flow = m g + a r = 455 N(c)Fhigh(d)P6.6262.2 rev min .2Fmid = m g 2 + a r = 397 N upward and at = tan 1b g= mb g a g =r328 Nar1.58= tan 1= 9.15 inward .g9.8Standing on the inner surface of the rim, and moving with it, each person will feel a normal forceexerted by the rim. This inward force causes the 3.00 m s 2 centripetal acceleration:ac =v2:rThe period of rotation comes from v =so the frequency of rotation ise3.00 m s ja60.0 mf = 13.4 m s2r 2 a60.0 mf== 28.1 sT=v13. 4 m s111 F 60 s I=f= =G J = 2.14 rev minT 28.1 s 28.1 s H 1 min Kv = ac r =2r:T2.Chapter 6P6.63(a)The mass at the end of the chain is in vertical equilibrium.TThus T cos = mg .l = 2.50 mmv 2Horizontally T sin = ma r =rR = 4.00 mafr = a 2.50 sin 28.0+4.00f m = 5.17 mrr = 2.50 sin + 4.00 mv2.Then a r =5.17 mBy division tan =183mgFIG. P6.63arv2=g 5.17 ga fa fafv 2 = 5.17 g tan = 5.17 9.80 tan 28.0 m 2 s 2v = 5.19 m s(b)T cos = mgbgej50.0 kg 9.80 m s 2mgT=== 555 Ncos cos 28.0P6.64(a)The putty, when dislodged, rises and returns to the original level in time t. To find t, we use2vwhere v is the speed of a point on the rim of the wheel.v f = vi + at : i.e., v = + v gt or t =g2R2 v 2RIf R is the radius of the wheel, v ==, so t =.tgvThus, v 2 = Rg and v = Rg .(b)The putty is dislodged when F, the force holding it to the wheel isF=P6.65(a)mv 2Rn=f mg = 0f = snT=(b)v=f2RT4 2 R sgT = 2.54 s#mv 2= m g .RnFGHIJKrev 1 rev 60 srev== 23.6min 2.54 s minminmgFIG. P6.65184P6.66Circular Motion and Other Applications of Newtons LawsLet the xaxis point eastward, the y-axis upward, and the z-axis point southward.(a)vi2 sin 2 igThe initial speed of the ball is thereforeThe range is Z =vi =gZ=sin 2 ia9.80fa285f = 53.0 m ssin 96.0The time the ball is in the air is found from y = viy t +b1a y t 2 as2f egaj0 = 53.0 m s sin 48.0 t 4.90 m s 2 t 2giving t = 8.04 s .ej62R e cos i 2 6.37 10 m cos 35.0== 379 m s86 400 s86 400 s(b)vix =(c)360 of latitude corresponds to a distance of 2R e , so 285 m is a change in latitude of =FG S IJ a360f = FG 285 m IJ a360f = 2.56 10GH 2 e6.37 10 mj JKH 2R K6e3degreesThe final latitude is then f = i = 35.00.002 56 = 34.997 4 .The cup is moving eastward at a speed v fx =2R e cos f86 400 s, which is larger than the eastwardvelocity of the tee byv x = v fx v fi ==2R e2R ecos f cos i =cos i cos i86 400 s86 400 sb2R ecos i cos + sin i sin cos i86 400 sSince is such a small angle, cos 1 and v x v x (d)b g ege2R esin i sin .86 400 sj sin 35.0 sin 0.002 56 =2 6.37 10 6 m86 400 sjafx = v x t = 1.19 10 2 m s 8.04 s = 0.095 5 m = 9.55 cm1.19 10 2 m sChapter 6P6.67(a)If the car is about to slip down the incline, f is directed upthe incline. Fy = n cos + f sin mg = 0 wheren=bgbgt2v minyieldsRbRg tan sv min =nfmg s mgand f =.cos 1 + s tan cos 1 + s tan Fx = n sin f cos = mThen,f = sn givestmgn cos gf sin .1 + s tan f cos n sin When the car is about to slip up the incline, f is directeddown the incline. Then, Fy = n cos f sin mg = 0with f = sn yieldsmg s mgmgn=and f =.cos 1 s tan cos 1 s tan bIn this case,g Fx = n sin + f cos = mIf v min =(c)v min =bRg tan sb1 + s tan 1 s tan g = 0 , thengtn2v max, which givesRRg tan + sv max =(b)bgf.tmg s = tan .a100 mfe9.80 m s jatan 10.00.100f =1 + a0.100 f tan 10.0a100 mfe9.80 m s jatan 10.0+0.100f =1 a0.100f tan 10.0n cos 2n sin 8.57 m sf cos 2v max =16.6 m sf sin mgFIG. P6.67185186P6.68Circular Motion and Other Applications of Newtons Laws(a)The bead moves in a circle with radius v = R sin at a speedofv=2r 2R sin =TTThe normal force hasan inward radial component of n sin and an upward component of n cos Fy = ma y :n cos mg = 0orn=Then Fx = n sin = mmgcos FIG. P6.68(a)FG mg IJ sin = m FG 2R sin IJKH cos KR sin HTv2becomesrwhich reduces tog sin 4 2 R sin =cos T2This has two solutions:sin = 0 = 0andcos =2gT 24 2 R(1)(2)If R = 15.0 cm and T = 0.450 s, the second solution yieldse9.80 m s ja0.450 sfcos =4 a0.150 mf222= 0.335 and = 70.4Thus, in this case, the bead can ride at two positions = 70. 4 and = 0 .(b)At this slower rotation, solution (2) above becomese9.80 m s ja0.850 sfcos =4 a0.150 mf222= 1.20 , which is impossible.In this case, the bead can ride only at the bottom of the loop, = 0 . The loops rotationmust be faster than a certain threshold value in order for the bead to move away from thelowest position.Chapter 6P6.69At terminal velocity, the accelerating force of gravity is balanced by frictional drag: mg = arv + br 2 v 2(a)ej ejmg = 3.10 10 9 v + 0.870 10 10 v 2LM 4 e10 mj OPN3Q= e3.10 10 jv + e0.870 10 jv4.11 10 1135m = V = 1 000 kg m3For water,9102Assuming v is small, ignore the second term on the right hand side: v = 0.013 2 m s .(b)ej ejmg = 3.10 10 8 v + 0.870 10 8 v 2Here we cannot ignore the second term because the coefficients are of nearly equalmagnitude.ej eja3.10f + 4a0.870fa4.11f = 1.03 m s2a0.870f4.11 10 8 = 3.10 10 8 v + 0.870 10 8 v 2v=(c)e23.10 j ejmg = 3.10 10 7 v + 0.870 10 6 v 2Assuming v > 1 m s , and ignoring the first term:ej4.11 10 5 = 0.870 10 6 v 2P6.70187v=FG mg IJ LM1 expFG bt IJ OP where expaxf = eH b KN H m KQv = 6.87 m sxis the exponential function.mgbAt t ,v vT =At t = 5.54 s0.500 vT = vT 1 expLMNMF ba5.54 sf I OPGH 9.00 kg JK QPF ba5.54 sf I = 0.500 ;GH 9.00 kg JK b a5.54 sf= ln 0.500 = 0.693 ;exp9.00 kgb=b9.00 kg ga0.693f = 1.13 m s5.54 sb9.00 kg ge9.80 m s j =2mgb(a)vT =vT =(b)0.750 vT = vT 1 expLMNFG 1.13t IJ OPH 9.00 s K Q1.13 kg sFG 1.13t IJ = 0.250H 9.00 s K9.00aln 0.250 ft=s=exp1.13continued on next page11.1 s78.3 m s188Circular Motion and Other Applications of Newtons Laws(c)FG IJ LM1 expFG bt IJ OP ;H K N H mKQmgdx=dtbz z FGHxdx =x0t0mgbIJ LM1 expFG bt IJ OPdtKN H m KQFGHI FG IJ = mgt + F m g I LexpFG bt IJ 1OJK H K b GH b JK MN H m K PQF b9.00 kg g e9.80 m s j I5.54 sJ expa0.693f 1x = 9.00 kg e9.80 m s j+G1.13 kg s GH b1.13 m sg JKx = 434 m + 626 ma 0.500f = 121 mmgtm2 g bt+x x0 =exp2bmbt22022At t = 5.54 s ,P6.7122 Fy = L y Ty mg = L cos 20.0T sin 20.07.35 N = ma y = 0 Fx = L x + Tx = L sin 20.0+T cos 20.0 = mbgv2r235.0 m sv2= 0.750 kg= 16.3 Nmr60.0 m cos 20.0af L sin 20.0+T cos 20.0 = 16.3 NL cos 20.0T sin 20.0 = 7.35 Ncos 20.016.3 N=sin 20.0 sin 20.0sin 20.07.35 N=LTcos 20.0 cos20.016.3 N7.35 NT cot 20.0+ tan 20.0 =sin 20.0 cos 20.0T 3.11 = 39.8 NL+Taa fT = 12.8 NfFIG. P6.71Chapter 6a f18.93.0042.14.00(a)af2.00P6.7273.8(b)tsdm4.881.007006006.00 1547.00 1995008.00 2469.00 29645213.050514.055815.061116.066417.071718.077019.082320.0(c)39912.040034711.0d (m)9008005.00 11210.087630020010000246810 12 14 16 18 20t (s)A straight line fits the points from t = 11.0 s to 20.0 s quite precisely. Its slope is the terminalspeed.vT = slope =876 m 399 m= 53.0 m s20.0 s 11.0 sdvdx=0k= kvdtdtv = vi kx implies the acceleration isa=Then the total force is*P6.73189 F = ma = ma kvf F = kmvThe resistive force is opposite to the velocity:.ANSWERS TO EVEN PROBLEMSP6.46.22 10 12 NP6.60.966 g(a)(a) 1.33 m s 2 ; (b) 1.79 m s 2 forward and48.0 inwardP6.188.88 N(a) 1.65 km s ; (b) 6.84 10 3 sP6.82.06 10 3 rev minP6.16215 N horizontally inwardP6.12P6.14P6.2P6.10ej(c) e 0.181 i + 0.181 jj m s(a) 0.233 i + 0.163 j m s 2 ; (b) 6.53 m s ;2RFG 2T gIJ ; (b) 2T upwardHm K190Circular Motion and Other Applications of Newtons LawsP6.46(a) 7.70 10 4 kg m; (b) 0.998 N;(c) The ball reaches maximum height 49 m.Its flight lasts 6.3 s and its impact speed is27 m s .15.3 m s Straight across the dashboard tothe leftP6.48(a) see the solution; (b) 81.8 m; (c) 15.9P6.240.527P6.500.835 rev sP6.26(a) 1.41 h; (b) 17.1P6.52(a) mg P6.28k =P6.54(a) 2.63 m s 2 ; (b) 201 m; (c) 17.7 m sP6.30(a) 2.38 10 5 m s 2 horizontally inward= 2.43 10 4 g ; (b) 360 N inwardperpendicular to the cone;(c) 47.5 10 4 m s 2P6.56(a) 106 N; (b) 0.396P6.58(a) m 2 g ; (b) m 2 g ; (c)P6.32(a) 6.27 m s 2 downward ; (b) 784 N up;(c) 283 N upP6.6062.2 rev minP6.622.14 rev minP6.34(a) 53.8 m s ; (b) 148 mP6.64P6.361.40(a) v = Rg ; (b) m gP6.380. 212 m s 2P6.66(a) 8.04 s; (b) 379 m s; (c) 1.19 cm s ;(d) 9.55 cmP6.40see the solutionP6.68(a) either 70.4 or 0; (b) 0P6.4236.5 m sP6.70(a) 78.3 m s ; (b) 11.1 s; (c) 121 mP6.44(a) 0.980 m s ; (b) see the solutionP6.72(a) and (b) see the solution; (c) 53.0 m sP6.20(a) 8.62 m; (b) Mg downward;(c) 8.45 m s 2 , Unless they are belted in,the riders will fall from the cars.P6.22a2 vt Lb g + agtf2mv 2; (b) v = gRRFG m IJ gRHm K217Energy and Energy TransferCHAPTER OUTLINE7.17.27.37.47.57.67.77.87.9Systems and EnvironmentsWork Done by a ConstantForceThe Scalar Product of TwoVectorsWork Done by a VaryingForceKinetic Energy and theWork-Kinetic EnergyTheoremThe Non-IsolatedSystemConservation ofEnergySituations Involving KineticFrictionPowerEnergy and the AutomobileANSWERS TO QUESTIONSQ7.1The force is perpendicular to every increment of displacement.Therefore, F r = 0 .Q7.2(a)Positive work is done by the chicken on the dirt.(b)No work is done, although it may seem like there is.(c)Positive work is done on the bucket.(d)Negative work is done on the bucket.(e)Negative work is done on the persons torso.Q7.3Yes. Force times distance over which the toe is in contact withthe ball. No, he is no longer applying a force. Yes, both airfriction and gravity do work.Q7.4Force of tension on a ball rotating on the end of a string. Normal force and gravitational force on anobject at rest or moving across a level floor.Q7.5(a)Tension(c)Positive in increasing velocity on the downswing.Negative in decreasing velocity on the upswing.(b)Air resistanceQ7.6No. The vectors might be in the third and fourth quadrants, but if the angle between them is lessthan 90 their dot product is positive.Q7.7The scalar product of two vectors is positive if the angle between them is between 0 and 90. Thescalar product is negative when 90 < < 180 .Q7.8If the coils of the spring are initially in contact with one another, as the load increases from zero, thegraph would be an upwardly curved arc. After the load increases sufficiently, the graph will belinear, described by Hookes Law. This linear region will be quite large compared to the first region.The graph will then be a downward curved arc as the coiled spring becomes a completely straightwire. As the load increases with a straight wire, the graph will become a straight line again, with asignificantly smaller slope. Eventually, the wire would break.Q7.9k = 2 k . To stretch the smaller piece one meter, each coil would have to stretch twice as much as onecoil in the original long spring, since there would be half as many coils. Assuming that the spring isideal, twice the stretch requires twice the force.191192Energy and Energy TransferQ7.10Kinetic energy is always positive. Mass and squared speed are both positive. A moving object canalways do positive work in striking another object and causing it to move along the same directionof motion.Q7.11Work is only done in accelerating the ball from rest. The work is done over the effective length of thepitchers armthe distance his hand moves through windup and until release.Q7.12Kinetic energy is proportional to mass. The first bullet has twice as much kinetic energy.Q7.13The longer barrel will have the higher muzzle speed. Since the accelerating force acts over a longerdistance, the change in kinetic energy will be larger.Q7.14(a)Kinetic energy is proportional to squared speed. Doubling the speed makes an object'skinetic energy four times larger.(b)If the total work on an object is zero in some process, its speed must be the same at the finalpoint as it was at the initial point.Q7.15The larger engine is unnecessary. Consider a 30 minute commute. If you travel the same speed ineach car, it will take the same amount of time, expending the same amount of energy. The extrapower available from the larger engine isnt used.Q7.16If the instantaneous power output by some agent changes continuously, its average power in aprocess must be equal to its instantaneous power at least one instant. If its power output is constant,its instantaneous power is always equal to its average power.Q7.17It decreases, as the force required to lift the car decreases.Q7.18As you ride an express subway train, a backpack at your feet has no kinetic energy as measured byyou since, according to you, the backpack is not moving. In the frame of reference of someone on theside of the tracks as the train rolls by, the backpack is moving and has mass, and thus has kineticenergy.Q7.19The rock increases in speed. The farther it has fallen, the more force it might exert on the sand at thebottom; but it might instead make a deeper crater with an equal-size average force. The farther itfalls, the more work it will do in stopping. Its kinetic energy is increasing due to the work that thegravitational force does on it.Q7.20The normal force does no work because the angle between the normal force and the direction ofmotion is usually 90. Static friction usually does no work because there is no distance throughwhich the force is applied.Q7.21An argument for: As a glider moves along an airtrack, the only force that the track applies on theglider is the normal force. Since the angle between the direction of motion and the normal force is90, the work done must be zero, even if the track is not level.Against: An airtrack has bumpers. When a glider bounces from the bumper at the end of theairtrack, it loses a bit of energy, as evidenced by a decreased speed. The airtrack does negative work.Q7.22Gaspard de Coriolis first stated the work-kinetic energy theorem. Jean Victor Poncelet, an engineerwho invaded Russia with Napoleon, is most responsible for demonstrating its wide practicalapplicability, in his 1829 book Industrial Mechanics. Their work came remarkably late compared to theelucidation of momentum conservation in collisions by Descartes and to Newtons MathematicalPrinciples of the Philosophy of Nature, both in the 1600s.Chapter 7SOLUTIONS TO PROBLEMSSection 7.1Systems and EnvironmentsSection 7.2Work Done by a Constant ForceP7.1W = Fr cos = 16.0 N 2.20 m cos 25.0 = 31.9 Ja(a)faf(b), (c) The normal force and the weight are both at 90 to the displacement in any time interval.Both do 0 work. W = 31.9 J + 0 + 0 =(d)P7.231.9 JThe component of force along the direction of motion isafF cos = 35.0 N cos 25.0 = 31.7 N .The work done by this force isafafafW = F cos r = 31.7 N 50.0 m = 1.59 10 3 J .P7.3Method One.Let represent the instantaneous angle the rope makes with the vertical asit is swinging up from i = 0 to f = 60 . In an incremental bit of motionfrom angle to + d , the definition of radian measure implies thatafr = 12 m d . The angle between the incremental displacement and thebgforce of gravity is = 90+ . Then cos = cos 90+ = sin .The work done by the gravitational force on Batman iszzf = 60 i =0W = F cos dr =bgaFIG. P7.3fmg sin 12 m df z sin d = b80 kg ge9.8 m s ja12 mfb cos g= a 784 N fa12 mfa cos 60+1f = 4.70 10 Ja= mg 12 m60 2060 03Method Two.bgejThe force of gravity on Batman is mg = 80 kg 9.8 m s 2 = 784 N down. Only his verticaldisplacement contributes to the work gravity does. His original y-coordinate below the tree limb is12 m. His final y-coordinate is 12 m cos 60 = 6 m . His change in elevation isafaf6 m 12 m = 6 m . The work done by gravity isafa fW = Fr cos = 784 N 6 m cos 180 = 4.70 kJ .193194Energy and Energy Transferja fa fe(a)W = mgh = 3.35 10 5 9.80 100 J = 3.28 10 2 J(b)P7.4Since R = mg , Wair resistance = 3.28 10 2 JSection 7.3The Scalar Product of Two VectorsP7.5A = 5.00 ; B = 9.00 ; = 50.0A B = AB cos = 5.00 9.00 cos 50.0 = 28.9P7.6A B = A x i + A y j + A z k B x i + B y j + Bz ka fa fejejA B = A B e i i j + A B e i jj + A B e i k j+ A B e j i j + A B e j jj + A B e j k j+ A B ek i j + A B ek jj + A B ek k jxxxyx zy xyyyz xz yzz zA B = A x Bx + A y B y + A z Bza fa fafa fP7.8(a)W = F r = Fx x + Fy y = 6.00 3.00 N m + 2.00 1.00 N m = 16.0 J(b)P7.7 = cos 116FG F r IJ = cos=H Fr Ka6.00f + a2.00f jea3.00f + a1.00f je1222236.9We must first find the angle between the two vectors. It is: = 36011890.0132 = 20.0ThenafbgF v = Fv cos = 32.8 N 0.173 m s cos 20.0or F v = 5.33P7.9(a)N mJ= 5.33 = 5.33 WssA = 3.00 i 2.00 jB = 4.00 i 4.00 j(b) = cos 1B = 3.00 i 4.00 j + 2.00kA = 2.00 i + 4.00 j(c)FIG. P7.8cos =a fa fAB6.00 16.0=AB20.0 29.0A = i 2.00 j + 2.00kB = 3.00 j + 4.00kA B12.0 + 8.00= cos 1= 11.3AB13.0 32.0 = cos 1a fa fFG A B IJ = cos FG 6.00 + 8.00 IJ =H AB KH 9.00 25.0 K1 = 15682.3Chapter 7P7.10ej eA B = 3.00 i + j k i + 2.00 j + 5.00kjA B = 4.00 i j 6.00kaf ejeajf afC A B = 2.00 j 3.00k 4.00 i j 6.00k = 0 + 2.00 + +18.0 = 16.0Section 7.4Work Done by a Varying ForcezfP7.11W = Fdx = area under curve from xi to x fix f = 8.00 mxi = 0(a)W = area of triangle ABC =W0 8 =FG 1 IJ 8.00 m 6.00 N =H 2K24.0 JFIG. P7.11x f = 10.0 mxi = 8.00 m(b)FG 1 IJ AC altitude,H 2KFG 1 IJ CE altitude,H 2KF 1I= G J a 2.00 mf a 3.00 N f = 3.00 JH 2KW = area of CDE =W8 10aP7.12fW0 10 = W0 8 + W810 = 24.0 + 3.00 = 21.0 J(c)afFx = 8 x 16 N(a)See figure to the right(b)Wnet =afaf afaf 2.00 m 16.0 N1.00 m 8.00 N+= 12.0 J22FIG. P7.12195196P7.13Energy and Energy TransferzW = Fx dxand W equals the area under the Force-Displacement curveFor the region 0 x 5.00 m ,(a)W=a3.00 Nfa5.00 mf =27.50 JFor the region 5.00 x 10.0 ,(b)aFIG. P7.13fafW = 3.00 N 5.00 m = 15.0 JFor the region 10.0 x 15.0 ,(c)W=a3.00 Nfa5.00 mf =27.50 JFor the region 0 x 15.0(d)afW = 7.50 + 7.50 + 15.0 J = 30.0 JzfP7.14W = F dr =izb5m0P7.15k=ze5mj4x i + 3 y j N dx i0x24 N m xdx + 0 = 4 N m2gbg5m= 50.0 J0a fa f4.00 9.80 NF Mg=== 1.57 10 3 N myy2.50 10 2 m(a)For 1.50 kg mass y =(b)a fa f1.50 9.80mg== 0.938 cmk1.57 10 3Work =1 2ky21Work = 1.57 10 3 N m 4.00 10 2 m2eP7.16(a)jej= 1.25 JSpring constant is given by F = kxk=(b)2Work = Favg x =afafaf230 NF== 575 N mx0.400 ma1230 N 0.400 m = 46.0 J2fChapter 7*P7.17(a)bFapplied = k leaf x + k helper x h = k x + k h x y 0197gbgNNx + 3.60 10 5x 0.5 mmm6.8 10 5 Nx == 0.768 m8.85 10 5 N m5 10 5 N = 5.25 10 5(b)W=FGHIJ aKfN1112k x 2 + khxh =5.25 10 50.768 mm2222+afN13.60 10 50.268 mm22= 1.68 10 5 JzfP7.18(a)W = F drW=i0.600 mzej15 000 N + 10 000 x N m 25 000 x 2 N m 2 dx cos 0010 000 x 2 25 000 x 3W = 15 000 x +230.600 m0W = 9.00 kJ + 1.80 kJ 1.80 kJ = 9.00 kJ(b)Similarly,b10.0 kN mga1.00 mf e25.0 kN m ja1.00 mfW = a15.0 kN fa1.00 mf +23223W = 11.7 kJ , larger by 29.6%P7.19af12k 0.100 m2 k = 800 N m and to stretch the spring to 0.200 m requires4.00 J =W =P7.20(a)a fa1800 0.2002f2 4.00 J = 12.0 JThe radius to the object makes angle with the horizontal, soits weight makes angle with the negative side of the x-axis,when we take the xaxis in the direction of motion tangent tothe cylinder. Fx = ma xF mg cos = 0F = mg cos FIG. P7.20zf(b)W = F driWe use radian measure to express the next bit of displacement as dr = Rd in terms of thenext bit of angle moved through:z 2W=0mg cos Rd = mgR sin a fW = mgR 1 0 = mgR 20198*P7.21Energy and Energy TransferThe same force makes both light springs stretch.(a)The hanging mass moves down byx = x1 + x 2 =FGHmg mg11+= mg+k1k2k1 k 2= 1.5 kg 9.8 m s 2(b)IJKF 1m + 1m I=GH 1 200 N 1 800 N JK2.04 10 2 mWe define the effective spring constant asFGbg HF 1m + 1m I ==GH 1 200 N 1 800 N JKk=mgF11==+k1 k 2x mg 1 k1 + 1 k 2IJK11*P7.22720 N mSee the solution to problem 7.21.FG 1 + 1 IJHk k KF1 1I(b)k=G + JHk k KL F O N kg m s =k =M P= =NxQ m m(a)x = mg1211P7.2322kgs2Section 7.5Kinetic Energy and the Work-Kinetic Energy TheoremSection 7.6The Non-Isolated SystemConservation of EnergyP7.24(a)KA =(b)12mv B = K B : v B =2(c)2 W = K = K B K A = 2 mevB v 2 j = 7.50 J 1.20 J =A(a)K=11mv 2 = 0.300 kg 15.0 m s22(b)K=10.300 30.02P7.25bgb10.600 kg 2.00 m s22K B=mg2= 1.20 Ja2fa7.50f =0.6005.00 m s1bafa fgb2=ag2= 33.8 Jfa f a4f = 4a33.8f =10.300 15.026.30 J2135 JChapter 7P7.26e199jv i = 6.00 i 2.00 j = m s(a)22vi = vix + viy = 40.0 m sKi =(b)bge11mvi2 = 3.00 kg 40.0 m 2 s 2 = 60.0 J22jv f = 8.00 i + 4.00 jv 2 = v f v f = 64.0 + 16.0 = 80.0 m 2 s 2f13.00K = K f K i = m v 2 vi2 =80.0 60.0 = 60.0 Jf22eP7.27a fjConsider the work done on the pile driver from the time it starts from rest until it comes to rest atthe end of the fall. Let d = 5.00 m represent the distance over which the driver falls freely, andh = 0.12 m the distance it moves the piling.11mv 2 mvi2f22mg h + d cos 0+ F d cos 180 = 0 0 . W = K :Wgravity + Wbeam =b ga fd ia fbmg gah + df = b2 100 kgge9.80 m s ja5.12 mf =F=so2Thus,driver is upward .P7.28(a)a fa f(a)a fKi + W = K f =0 + W =(b)(c)F=faf1mv 2 0 = W = (area under curve from x = 0 to x = 15.0 m)f22 area2 30.0 J== 3.87 m sm4.00 kgK = K f K i =vf =P7.29a1mv 2 0 = W = (area under curve from x = 0 to x = 10.0 m)f22 area2 22.5 J== 3.35 m sm4.00 kgK = K f K i =vf =(c)af1mv 2f2jb115.0 10 3 kg 780 m s2eg2= 4.56 kJ4.56 10 3 JW== 6.34 kNr cos 0.720 m cos 0afv vb780 m sg 0 = 422 km s=a=2x2a0.720 mf F = ma = e15 10 kg je422 10 m s j =2f22i2f(d)8.78 10 5 N . The force on the pile1mv 2 0 = W = (area under curve from x = 0 to x = 5.00 m)f22 area2 7.50 J== 1.94 m sm4.00 kgK = K f K i =vf =(b)0.120 md3326.34 kN200P7.30Energy and Energy Transfer(a)eKf =(b)jv f = 0.096 3 10 8 m s = 2.88 10 7 m s11mv 2 = 9.11 10 31 kg 2.88 10 7 m sf22eKi + W = K f :jej2= 3.78 10 16 J0 + Fr cos = K fafF 0.028 m cos 0 = 3.78 10 16 JF = 1.35 10 14 NF =1.35 10 14 N= 1. 48 10 +16 m s 29.11 10 31 kg(c) F = ma ;a=(d)v xf = v xi + a x t2.88 10 7 m s = 0 + 1.48 10 16 m s 2 tmejt = 1.94 10 9 s1v xi + v xf t210.028 m = 0 + 0 + 2.88 10 7 m s t2x f = xi +Check:diejt = 1.94 10 9 sSection 7.7P7.31Situations Involving Kinetic Friction Fy = ma y :n 392 N = 0n = 392 Na fa f= Fr cos = a130 fa5.00f cos 0 =f k = k n = 0.300 392 N = 118 N(a)WF(b)650 JEint = f k x = 118 5.00 = 588 Ja fa fa fa f(c)Wn = nr cos = 392 5.00 cos 90 = 0(d)W g = mgr cos = 392 5.00 cos 90 = 0(e)K = K f K i = Wother Eint1mv 2 0 = 650 J 588 J + 0 + 0 = 62.0 Jf2(f)vf =a fa f a f2K fm=af2 62.0 J= 1.76 m s40.0 kgFIG. P7.31Chapter 7P7.32(a)vf =so(b)a fe1 2 1 2 1kx i kx f = 500 5.00 10 2222111Ws = mv 2 mvi2 = mv 2 0ff222Ws =2j2c W h = 2a0.625f m s =2.00m11mvi2 f k x + Ws = mv 2f22afa fa fb g10.282 J = b 2.00 kg gv22a0.282 fv =m s = 0.531 m s0 0.350 2.00 9.80 0.050 0 J + 0.625 J = 0 = 0.625 J0.791 m s1mv 2f22ffP7.33(a)2.00af= b10.0 kg gd9.80 m s ia5.00 mf cos 110 =W g = mg cos 90.0+Wg(b)FIG. P7.322168 Jf k = k n = k mg cos Eint = f k = k mg cos afafa fa fEint = 5.00 m 0.400 10.0 9.80 cos 20.0 = 184 J(c)a fa fWF = F = 100 5.00 = 500 J(d)K = Wother Eint = WF + W g Eint = 148 J(e)K =11mv 2 mvi2f222 K2 148+ vi2 =+ 1.50vf =10.0ma fP7.34FIG. P7.33 Fy = ma y :a f a fa2= 5.65 m sfn + 70.0 N sin 20.0147 N = 0n = 123 Nf k = k n = 0.300 123 N = 36.9 Nafafafafafafaf(a)W = Fr cos = 70.0 N 5.00 m cos 20.0 = 329 J(b)W = Fr cos = 123 N 5.00 m cos 90.0 = 0 J(c)W = Fr cos = 147 N 5.00 m cos 90.0 = 0(d)Eint = Fx = 36.9 N 5.00 m = 185 J(e)K = K f K i = W Eint = 329 J 185 J = +144 JafafFIG. P7.34201202P7.35Energy and Energy Transfer k = 0.1001K i f k x + Wother = K f :mvi2 f k x = 0222.00 m sv21x = i == 2.04 mmvi2 = k mgx22 k g 2 0.100 9.80vi = 2.00 m sbaSection 7.8*P7.36Pav =gfa fPowerbW K f mv 2 0.875 kg 0.620 m s===2 tt t2 21 10 3 seWtP=jag2= 8.01 Wfafmgh 700 N 10.0 m== 875 Wt8.00 sP7.37Power =P7.38A 1 300-kg car speeds up from rest to 55.0 mi/h = 24.6 m/s in 15.0 s. The output work of the engine isequal to its final kinetic energy,bgb11 300 kg 24.6 m s2with power P =P7.39(a)g2= 390 kJ390 000 J~ 10 4 W around 30 horsepower.15.0 s W = K , but K = 0 because he moves at constant speed. The skier rises a verticalafdistance of 60.0 m sin 30.0 = 30.0 m . Thus,bjagefWin = Wg = 70.0 kg 9.8 m s 2 30.0 m = 2.06 10 4 J = 20.6 kJ .(b)The time to travel 60.0 m at a constant speed of 2.00 m/s is 30.0 s. Thus,Pinput =P7.40(a)W 2.06 10 4 J== 686 W = 0.919 hp .t30.0 sThe distance moved upward in the first 3.00 s isy = vt =LM 0 + 1.75 m s OPa3.00 sf = 2.63 m .N 2 QThe motor and the earths gravity do work on the elevator car:11mvi2 + Wmotor + mgy cos 180 = mv 2f2212Wmotor = 650 kg 1.75 m s 0 + 650 kg g 2.63 m = 1.77 10 4 J2bAlso, W = P t so P =(b)gbggabfW 1.77 10 4 J== 5.91 10 3 W = 7.92 hp.t3.00 sbgWhen moving upward at constant speed v = 1.75 m s the applied force equals thebgej = 6.37 10 N . Therefore,P = Fv = e6.37 10 N jb1.75 m sg = 1.11 10weight = 650 kg 9.80 m s2334W = 14.9 hp .Chapter 7P7.41203energy = power timeFor the 28.0 W bulb:afejEnergy used = 28.0 W 1.00 10 4 h = 280 kilowatt hrsafbgtotal cost = $17.00 + 280 kWh $0.080 kWh = $39.40For the 100 W bulb:afejEnergy used = 100 W 1.00 10 4 h = 1.00 10 3 kilowatt hrs1.00 10 4 h= 13.3# bulb used =750 h bulbbg ejbgtotal cost = 13.3 $0.420 + 1.00 10 3 kWh $0.080 kWh = $85.60Savings with energy-efficient bulb = $85.60 $39.40 = $46.20*P7.42(a)FG 454 g IJ FG 9 kcal IJ FG 4 186 J IJ = 1.71 10H 1 lb K H 1 g K H 1 kcal KBurning 1 lb of fat releases energy1 lbThe mechanical energy output isJ.e1.71 10 Jja0.20f = nFr cos .Then73.42 10 6 J = nmgy cos 07b gejbJ = ne5.88 10 Jjgaf3.42 10 6 J = n 50 kg 9.8 m s 2 80 steps 0.150 m3.42 10 633.42 10 6 J= 582 .5.88 10 3 JThis method is impractical compared to limiting food intake.where the number of times she must climb the steps is n =(b)Her mechanical power output isP=*P7.43(a)(b)FGHIJK1 hpW 5.88 10 3 J== 90.5 W = 90.5 W= 0.121 hp .t65 s746 WIJ FG 1 kcal IJ FG 1.30 10 J IJ =K H 4 186 J K H 1 gal K1 h F 10 mi I F 1 kcal I F 1.30 10 J IFor bicyclingG JH KH400 kcal H h K G 4 186 J J G 1 gal JK = 776 mi gal .The fuel economy for walking isFGH81h3 mi220 kcalh8423 mi gal .204Energy and Energy TransferSection 7.9P7.44Energy and the AutomobileAt a speed of 26.8 m/s (60.0 mph), the car described in Table 7.2 delivers a power of P1 = 18.3 kW tothe wheels. If an additional load of 350 kg is added to the car, a larger output power ofP2 = P1 + (power input to move 350 kg at speed v)will be required. The additional power output needed to move 350 kg at speed v is:b g bgPout = f v = r mg v .Assuming a coefficient of rolling friction of r = 0.016 0 , the power output now needed from theengine isbgbgejbgP2 = P1 + 0.016 0 350 kg 9.80 m s 2 26.8 m s = 18.3 kW + 1.47 kW .With the assumption of constant efficiency of the engine, the input power must increase by thesame factor as the output power. Thus, the fuel economy must decrease by this factor:P18bfuel economyg = FGH P IJK bfuel economyg = FGH 18.3 +.31.47 IJK b6.40 km Lgor bfuel economy g = 5.92 km L .12122P7.45(a)fuel needed ==(b)(c)12mv 2 1 mvi2f2useful energy per gallonb900 kggb24.6 m sg =a0.150fe1.34 10 J galj=bmv 2 0feff. energy content of fuel21281.35 10 2 gal73.8power =FG 1 gal IJ FG 55.0 mi IJ FG 1.00 h IJ FG 1.34 10H 38.0 mi K H 1.00 h K H 3 600 s K H 1 galAdditional ProblemsP7.4612bgbgAt apex, v = b 40.0 m sg cos 30.0 i + 0 j = b34.6 m sgi11And K = mv = b0.150 kg gb34.6 m sg = 90.0 J22At start, v = 40.0 m s cos 30.0 i + 40.0 m s sin 30.0 j228JI a0.150f =JK8.08 kWgChapter 7P7.47bgbConcentration of Energy output = 0.600 J kg step 60.0 kgbgb1 stepgFGH 1.50 m IJK = 24.0 J mgF = 24.0 J m 1 N m J = 24.0 NP = Fvaf70.0 W = 24.0 N vv = 2.92 m sP7.48(a)a fa fA i = A 1 cos . But also, A i = A x .a Afa1f cos = AThus,Similarly,cos =andcos =Ax.AAyAAzA222A = Ax + Ay + Az .where(b)P7.49or cos =xcos 2 + cos 2 + cos 2 =(a)FG A IJ + FG A IJ + FG A IJH AK H AK H AK2y22A2=1A2x = t + 2.00t 3xz=Therefore,dx= 1 + 6.00t 2dt11K = mv 2 = 4.00 1 + 6.00t 222v=a fea12.0tf m sF = ma = 4.00a12.0t f = a 48.0t f Ndv=dt2(b)a=(c)P = Fv = 48.0t 1 + 6.00t 2 =(d)W=azfeze2.002 .0000Pdt =j e48.0t + 288t j W3j48.0t + 288 t 3 dt = 1 250 Jj = e2.00 + 24.0t22j+ 72.0t 4 J205206*P7.50Energy and Energy Transfer(a)We writeF = ax baf5 000 N = aa0.315 mfF 0.315 IJ = 2.445=GH 0.129 Kb1 000 N = a 0.129 mbbbln 5 = b ln 2.44ln 5= 1.80 = bln 2.441 000 Na== 4.01 10 4 N m1.8 = a1.800.129 mb=aW=zz0. 25 m0. 25 m0(b)0Fdx == 4.01 1044.01 10 4N x 2 .8m1.8 2.8fN 1.8x dxm1.80. 25 m= 4.01 1040afN 0.25 m2.8m1.82.8= 294 J*P7.51The work done by the applied force iszfW = Fapplied dx =zx maxe0izzx maxj k1 x + k 2 x 2 dxx maxx2= k 1 x dx + k 2 x dx = k1200= k1P7.52(a)2x max2+ k22x max0x3+ k23xmax03x max3The work done by the traveler is mghs N where N is the number of steps he climbs duringthe ride.N = (time on escalator)(n)wherehatime on escalatorf = vertical velocity of personandvertical velocity of person = v + nhsThen,N=nhv + nhsand the work done by the person becomes Wperson =continued on next pagemgnhhsv + nhsChapter 7(b)207The work done by the escalator isbga f afbga fWe = power time = force exerted speed time = mgvthas above.v + nhswheret=Thus,We =mgvh.v + nhsAs a check, the total work done on the persons body must add up to mgh, the work anelevator would do in lifting him. W = Wperson + We =It does add up as follows:(a)(b)*P7.54K =2Wand v =m2WmW = F d = Fx d Fx =WdDuring its whole motion from y = 10.0 m to y = 3.20 mm, the force of gravity and the force of theplate do work on the ball. It starts and ends at restKi + W = K f0 + Fg y cos 0+ Fp x cos 180 = 0bg bg5 kg e9.8 m s ja10 mf== 1.53 10mg 10.003 2 m Fp 0.003 20 m = 02FpP7.55g1mv 2 0 = W , so2v2 =P7.53bmgnhhsmgvh mgh nhs + v+== mghv + nhs v + nhsv + nhsFt =mOPa3.00 sf =PQ240 WP = Fv = F vi + at = F 0 +(b)P=2mIJ FG F IJ tK HmKg FGHb(a)LM a20.0 NfMN 5.00 kg3.2 10325N upward208Energy and Energy Transferzf*P7.56(a)W1 = F1 dx =zxi 1 + x ai(b)W2 =k1 x dx =xi 1z xi 2 + x ak 2 x dx = xi 2(c)1k 1 x i1 + x a2b1k 2 xi 2 + x a2bg2g2 xi2 =1 xi22 =ejk1 xi1k21122k1 x a + k1 x a xi1 + k 2 x a k 2 x a x i 222k x1122= k 1 x a + k 2 x a + k1 x a xi1 k 2 x a 1 i122k212= k1 + k 2 x a2W1 + W2 =gz zet(a)jBefore the horizontal force is applied, the springs exert equal forces: k 1 xi1 = k 2 xi 2b*P7.57e12k 2 x a 2 x a xi 22xi 2 =(d)12k 1 x a + 2 x a x i12t00v = a dt =j1.16t 0.21t 2 + 0.24t 3 dtt2t3t4= 1.16 0.21 + 0. 24234t= 0.58t 2 0.07t 3 + 0.06t 40At t = 0 , vi = 0. At t = 2.5 s ,ja f e0.07 m s ja2.5 sf + e0.06 m s ja2.5 sfev f = 0.58 m s 3 2.5 s23454= 4.88 m sKi + W = K f0+W =(b)b11mv 2 = 1 160 kg 4.88 m sf22g2= 1.38 10 4 JAt t = 2.5 s ,ejeja f + e0.240 m s ja2.5 sfa = 1.16 m s 3 2.5 s 0.210 m s 4 2.5 s253Through the axles the wheels exert on the chassis force F = ma = 1 160 kg 5.34m s 2 = 6.19 10 3 Nand inject powerbgP = Fv = 6.19 10 3 N 4.88 m s = 3.02 10 4 W .= 5.34 m s 2 .Chapter 7P7.58(a)The new length of each spring isx 2 + L2 , so its extension isx 2 + L2 L and the force it exerts is kFHIKx 2 + L2 L toward itsfixed end. The y components of the two spring forces add tozero. Their x components add toF = 2 ikFHx 2 + L2 LIKxx 2 + L2FGH= 2 kx i 1 zLx 2 + L2zf(b)0W = Fx dxFIG. P7.58.FGHI dxJx +L Kex + L j+ kLb1 2gW = 2 kx 1 iAz0W = 2 k x dx + kLAze0Aj2 1 2x2 + Lx2W = 2 k22 x dx0AL22202 12AW = 2 kL2 + kA 2 2 kL A 2 + L2W = 0 + kA 2 + 2 kL2 2 kL A 2 + L2*P7.59IJKFor the rocket falling at terminal speed we have F = ma+ R Mg = 0Mg =(a)12DAvT2For the rocket with engine exerting thrust T and flying up at the same speed, F = ma+T Mg R = 0T = 2 MgThe engine power is P = Fv = TvT = 2 MgvT .(b)For the rocket with engine exerting thrust Tb and flying down steadily at 3vT ,12Rb = DA 3 vT = 9 Mg2b g F = maTb Mg + 9 Mg = 0Tb = 8 MgThe engine power is P = Tv = 8 Mg 3 vT = 24MgvT .209210P7.60Energy and Energy Transfer(a)a fej e20.5i + 14.3 jj N= a 42.0 N fecos 150 i + sin 150 jj = e36.4i + 21.0 jj NF1 = 25.0 N cos 35.0 i + sin 35.0 j =F2e15.9 i + 35.3 jj N(b) F = F1 + F2 =(c)a=(d)v f = v i + at = 4.00 i + 2.50 j m s + 3.18 i + 7.07 j m s 2 3.00 sF =e3.18 i + 7.07 jj m sevf =(e)m2jejajee5.54i + 23.7 jj m sr f = ri + v i t +e1 2at2f 1 e3.18i + 7.07 jjem s ja3.00 sf2jb gae2.30i + 39.3 jj m2r f = 0 + 4.00 i + 2.50 j m s 3.00 s +r = r f =(f)Kf =(g)b11mv 2 = 5.00 kgf22g a5.54f + a23.7f em s j =2222Kf =1mvi2 + F r212K f = 5.00 kg 4.00 + 2.502K f = 55.6 J + 1 426 J = 1.48 kJbP7.61f(a)1.48 kJg a f a f bm sg + a15.9 Nfa2.30 mf + a35.3 Nfa39.3 mf W = K :22Ws + W g = 0afj af a1 2kxi 0 + mgx cos 90+60 = 02121.40 10 3 N m 0.100 0.200 9.80 sin 60.0 x = 02x = 4.12 me(b) W = K + Eint :fa fafWs + W g Eint = 01 2kxi + mgx cos 150 k mg cos 60 x = 02121.40 10 3 N m 0.100 0.200 9.80 sin 60.0 x 0.200 9.80 0.400 cos 60.0 x = 02x = 3.35 mej af afa fafafa fafafChapter 7P7.62(a)211a f Lammf FaNf LammfFN2.0015.014.01124.0032.016.01266.0049.018.01498.0064.020.017510.079.022.019012.098.0FIG. P7.62(b)A straight line fits the first eight points, together with the origin. By least-square fitting, itsslope is0.125 N mm 2% = 125 N m 2%In F = kx , the spring constant is k =(c)P7.63bF, the same as the slope of the F-versus-x graph.xfgaF = kx = 125 N m 0.105 m = 13.1 NK i + Ws + W g = K f1111mvi2 + kx i2 kx 2 + mgx cos = mv 2ff222211FIG. P7.630 + kxi2 0 + mgxi cos 100 = mv 2f22111.20 N cm 5.00 cm 0.050 0 m 0.100 kg 9.80 m s 2 0.050 0 m sin 10.0 = 0.100 kg v 222320.150 J 8.51 10 J = 0.050 0 kg vbv=P7.64gafbbg bggejbgbg0.141= 1.68 m s0.050 0bgea6.00f a8.00f jbm sgbge11m v 2 vi2 : Eint = 0.400 kgf22ej22(a)Eint = K = (b)Eint = fr = k mg 2r :5.60 J = k 0.400 kg 9.80 m s 2 2 1.50 mThus,2 k = 0.152 .(c)a fj a= 5.60 JfAfter N revolutions, the object comes to rest and K f = 0 .1mvi22Thus,Eint = K = 0 + K i =or k mg N 2r =This givesb8.00 m sg==N= mg a 2r f a0.152fe9.80 m s j2 a1.50 mfa f12kmvi21mvi2 .221222.28 rev .212P7.65Energy and Energy TransferIf positive F represents an outward force, (same as direction as r), thenzzerffW = F dr =ij2 F0 13 r 13 F0 7 r 7 drri2 F 13 r 12 F0 7 r 6W= 0126W= F013er f12 ri126W = 1.03 1077r f6rfrij + F er ri6076f ri66j= F06 1.89 10134r f127r f6 ri6 F0 13 12r f ri126 ri12W = 1.03 10 77 1.88 10 6 2.44 10 6 10 60 1.89 10 134 3.54 10 12 5.96 10 8 10 120W = 2.49 10 21 J + 1.12 10 21 J = 1.37 10 21 JP7.66Pt = W = K =a m f v22=The density ism m=.vol AxSubstituting this into the first equation and solving for P , since3for a constant speed, we getP=Av2Also, since P = Fv,F=x= v,tFIG. P7.66Av 2.2.Our model predicts the same proportionalities as the empirical equation, and gives D = 1 for thedrag coefficient. Air actually slips around the moving object, instead of accumulating in front of it.For this reason, the drag coefficient is not necessarily unity. It is typically less than one for astreamlined object and can be greater than one if the airflow around the object is complicated.z23 .7P7.67We evaluate375dxby calculating12.8 x + 3.75 x3af + 375a0.100f + 375a0.100f = 0.806a12.8f + 3.75a12.8f a12.9f + 3.75a12.9f a23.6f + 3.75a23.6f375 0.100333andaf + 375a0.100f + 375a0.100f = 0.791 .a12.9f + 3.75a12.9f a13.0f + 3.75a13.0f a23.7f + 3.75a23.7f375 0.100333The answer must be between these two values. We may find it more precisely by using a value forx smaller than 0.100. Thus, we find the integral to be 0.799 N m .Chapter 7*P7.68P=1Dr 2 v 32jaf b8 m sg11 1.20 kg m3 1.5 m2e(a)Pa =(b)324 m sPb v b= 3 =Pa v a8 mseFGHIJK23= 2.17 10 3 W3= 3 3 = 27jPb = 27 2.17 10 3 W = 5.86 10 4 WP7.69(a)The suggested equation Pt = bwd implies all of thefollowing cases:(1)(3)FG w IJ a2dfH 2KF t I F d IP G J = bwG JH 2 K H 2KPt = b(2)and(4)FG t IJ = bFG w IJ dH 2 K H 2KFG P IJ t = bFG w IJ dH 2K H 2KPv = constantndfk = k nFwThese are all of the proportionalities Aristotle lists.FIG. P7.69(b)For one example, consider a horizontal force F pushing an object of weight w at constantvelocity across a horizontal floor with which the object has coefficient of friction k . F = ma implies that:+n w = 0 and F k n = 0so that F = k wAs the object moves a distance d, the agent exerting the force does workW = Fd cos = Fd cos 0 = k wd and puts out power P =WtThis yields the equation Pt = k wd which represents Aristotles theory with b = k .Our theory is more general than Aristotles. Ours can also describe accelerated motion.*P7.70(a)So long as the spring force is greater than the friction force,the block will be gaining speed. The block slows down whenthe friction force becomes the greater. It has maximumspeed when kx a f k = ma = 0.e1.0 10(b)3jN m x a 4.0 N = 00x = 4.0 10 3 m0By the same logic,e1.0 103jN m x b 10.0 N = 0x = 1.0 10 2 mFIG. P7.70213214Energy and Energy TransferANSWERS TO EVEN PROBLEMSP7.21.59 10 3 JP7.445.92 km LP7.4(a) 3.28 10 2 J ; (b) 3.28 10 2 JP7.4690.0 JP7.6see the solutionP7.85.33 WP7.1016.0P7.12(a) see the solution; (b) 12.0 JP7.48AyAxA; cos =; cos = z ;AAA(b) see the solution(a) cos =P7.50(a) a =P7.52(a)40.1 kN; b = 1.80 ; (b) 294 Jm 1.8mgnhhsmgvh; (b)v + nhsv + nhsP7.1450.0 JP7.16(a) 575 N m ; (b) 46.0 JP7.541.53 10 5 N upwardP7.18(a) 9.00 kJ; (b) 11.7 kJ, larger by 29.6%P7.56see the solutionP7.20(a) see the solution; (b) mgRP7.58(a) see the solution;P7.22mg mg11++; (b)(a)k1 k 2k1k2FGHIJK(b) 2 kL2 + kA 2 2 kL A 2 + L21P7.24ej(a) F1 = 20.5 i + 14.3 j N ;ejF2 = 36.4i + 21.0 j N ;(a) 1.20 J; (b) 5.00 m s ; (c) 6.30 JP7.26P7.60(a) 60.0 J; (b) 60.0 Jej(c) e 3.18 i + 7.07 jj m s ;(d) e 5.54i + 23.7 jj m s ;(e) e 2.30 i + 39.3 jj m ; (f) 1.48 kJ; (g) 1.48 kJ(b) 15.9 i + 35.3 j N ;2P7.28(a) 1.94 m s ; (b) 3.35 m s ; (c) 3.87 m sP7.30(a) 3.78 10 16 J ; (b) 1.35 10 14 N ;(c) 1.48 10 +16 m s 2 ; (d) 1.94 nsP7.32(a) 0.791 m s; (b) 0.531 m sP7.34(a) 329 J; (b) 0; (c) 0; (d) 185 J; (e) 144 JP7.368.01 WP7.38~ 10 4 WP7.40(a) 5.91 kW; (b) 11.1 kWP7.42No. (a) 582; (b) 90.5 W = 0.121 hpP7.62(a) see the solution; (b) 125 N m 2% ;(c) 13.1 NP7.64(a) 5.60 J; (b) 0.152; (c) 2.28 revP7.66see the solutionP7.68(a) 2.17 kW; (b) 58.6 kWP7.70(a) x = 4.0 mm ; (b) 1.0 cm8Potential EnergyCHAPTER OUTLINE8.18.2Potential Energy of a SystemThe IsolatedSystemConservation ofMechanical EnergyConservative andNonconservative ForcesChanges in MechanicalEnergy for NonconservativeForcesRelationship BetweenConservative Forces andPotential EnergyEnergy Diagrams and theEquilibrium of a SystemANSWERS TO QUESTIONS8.48.58.6Q8.4The final speed of the children will not depend on the slidelength or the presence of bumps if there is no friction. If there isfriction, a longer slide will result in a lower final speed. Bumpswill have the same effect as they effectively lengthen thedistance over which friction can do work, to decrease the totalmechanical energy of the children.Q8.2Total energy is the sum of kinetic and potential energies.Potential energy can be negative, so the sum of kinetic pluspotential can also be negative.Q8.38.3Q8.1Both agree on the change in potential energy, and the kineticenergy. They may disagree on the value of gravitationalpotential energy, depending on their choice of a zero point.(a)mgh is provided by the muscles.(b)No further energy is supplied to the object-Earth system, but some chemical energy must besupplied to the muscles as they keep the weight aloft.(c)The object loses energy mgh, giving it back to the muscles, where most of it becomes internalenergy.Q8.5Lift a book from a low shelf to place it on a high shelf. The net change in its kinetic energy is zero,but the book-Earth system increases in gravitational potential energy. Stretch a rubber band toencompass the ends of a ruler. It increases in elastic energy. Rub your hands together or let a pearldrift down at constant speed in a bottle of shampoo. Each system (two hands; pearl and shampoo)increases in internal energy.Q8.6Three potential energy terms will appear in the expression of total mechanical energy, one for eachconservative force. If you write an equation with initial energy on one side and final energy on theother, the equation contains six potential-energy terms.215216Q8.7Potential Energy(a)It does if it makes the objects speed change, but not if it only makes the direction of thevelocity change.(b)Yes, according to Newtons second law.Q8.8The original kinetic energy of the skidding can be degraded into kinetic energy of random molecularmotion in the tires and the road: it is internal energy. If the brakes are used properly, the sameenergy appears as internal energy in the brake shoes and drums.Q8.9All the energy is supplied by foodstuffs that gained their energy from the sun.Q8.10Elastic potential energy of plates under stress plus gravitational energy is released when the platesslip. It is carried away by mechanical waves.Q8.11The total energy of the ball-Earth system is conserved. Since the system initially has gravitationalenergy mgh and no kinetic energy, the ball will again have zero kinetic energy when it returns to itsoriginal position. Air resistance will cause the ball to come back to a point slightly below its initialposition. On the other hand, if anyone gives a forward push to the ball anywhere along its path, thedemonstrator will have to duck.Q8.12Using switchbacks requires no less work, as it does not change the change in potential energy fromtop to bottom. It does, however, require less force (of static friction on the rolling drive wheels of acar) to propel the car up the gentler slope. Less power is required if the work can be done over alonger period of time.Q8.13There is no work done since there is no change in kinetic energy. In this case, air resistance must benegligible since the acceleration is zero.Q8.14There is no violation. Choose the book as the system. You did work and the earth did work on thebook. The average force you exerted just counterbalanced the weight of the book. The total work onthe book is zero, and is equal to its overall change in kinetic energy.Q8.15Kinetic energy is greatest at the starting point. Gravitational energy is a maximum at the top of theflight of the ball.Q8.16Gravitational energy is proportional to mass, so it doubles.Q8.17In stirring cake batter and in weightlifting, your body returns to the same conformation after eachstroke. During each stroke chemical energy is irreversibly converted into output work (and internalenergy). This observation proves that muscular forces are nonconservative.Chapter 8Q8.18Let the gravitational energy be zero at the lowest point in themotion. If you start the vibration by pushing down on the block (2),its kinetic energy becomes extra elastic potential energy in thespring ( Us ). After the block starts moving up at its lower turningpoint (3), this energy becomes both kinetic energy (K) andgravitational potential energy ( U g ), and then just gravitationalenergy when the block is at its greatest height (1). The energy thenturns back into kinetic and elastic potential energy, and the cyclerepeats.Q8.19217FIG. Q8.18(a)Kinetic energy of the running athlete is transformed into elastic potential energy of the bentpole. This potential energy is transformed to a combination of kinetic energy andgravitational potential energy of the athlete and pole as the athlete approaches the bar. Theenergy is then all gravitational potential of the pole and the athlete as the athlete hopefullyclears the bar. This potential energy then turns to kinetic energy as the athlete and pole fallto the ground. It immediately becomes internal energy as their macroscopic motion stops.(b)Rotational kinetic energy of the athlete and shot is transformed into translational kineticenergy of the shot. As the shot goes through its trajectory as a projectile, the kinetic energyturns to a mix of kinetic and gravitational potential. The energy becomes internal energy asthe shot comes to rest.(c)Kinetic energy of the running athlete is transformed to a mix of kinetic and gravitationalpotential as the athlete becomes projectile going over a bar. This energy turns back intokinetic as the athlete falls down, and becomes internal energy as he stops on the ground.The ultimate source of energy for all of these sports is the sun. See question 9.Q8.20Chemical energy in the fuel turns into internal energy as the fuel burns. Most of this leaves the carby heat through the walls of the engine and by matter transfer in the exhaust gases. Some leaves thesystem of fuel by work done to push down the piston. Of this work, a little results in internal energyin the bearings and gears, but most becomes work done on the air to push it aside. The work on theair immediately turns into internal energy in the air. If you use the windshield wipers, you takeenergy from the crankshaft and turn it into extra internal energy in the glass and wiper blades andwiper-motor coils. If you turn on the air conditioner, your end effect is to put extra energy out intothe surroundings. You must apply the brakes at the end of your trip. As soon as the sound of theengine has died away, all you have to show for it is thermal pollution.Q8.21A graph of potential energy versus position is a straight horizontal line for a particle in neutralequilibrium. The graph represents a constant function.Q8.22The ball is in neutral equilibrium.Q8.23The ball is in stable equilibrium when it is directly below the pivot point. The ball is in unstableequilibrium when it is vertically above the pivot.218Potential EnergySOLUTIONS TO PROBLEMSSection 8.1Potential Energy of a SystemP8.1With our choice for the zero level for potential energy when the caris at point B,(a)UB = 0 .When the car is at point A, the potential energy of the car-Earthsystem is given byFIG. P8.1U A = mgywhere y is the vertical height above zero level. With 135 ft = 41.1 m , this height is found as:afy = 41.1 m sin 40.0 = 26.4 m .Thus,bgejafU A = 1 000 kg 9.80 m s 2 26.4 m = 2.59 10 5 J .The change in potential energy as the car moves from A to B isU B U A = 0 2.59 10 5 J = 2.59 10 5 J .(b)With our choice of the zero level when the car is at point A, we have U A = 0 . The potentialenergy when the car is at point B is given by U B = mgy where y is the vertical distance ofpoint B below point A. In part (a), we found the magnitude of this distance to be 26.5 m.Because this distance is now below the zero reference level, it is a negative number.Thus,bgejafU B = 1 000 kg 9.80 m s 2 26.5 m = 2.59 10 5 J .The change in potential energy when the car moves from A to B isU B U A = 2.59 10 5 J 0 = 2.59 10 5 J .Chapter 8P8.2(a)219We take the zero configuration of systempotential energy with the child at thelowest point of the arc. When the stringis held horizontal initially, the initialposition is 2.00 m above the zero level.Thus,afafU g = mgy = 400 N 2.00 m = 800 J .(b)From the sketch, we see that at an angleof 30.0 the child is at a vertical height of2.00 m 1 cos 30.0 above the lowestpoint of the arc. Thus,afafaFIG. P8.2fafafU g = mgy = 400 N 2.00 m 1 cos 30.0 = 107 J .(c)The zero level has been selected at the lowest point of the arc. Therefore, U g = 0 at thislocation.*P8.3The volume flow rate is the volume of water going over the falls each second:afbg3 m 0.5 m 1.2 m s = 1.8 m3 sThe mass flow rate ismV= = 1 000 kg m3 1.8 m3 s = 1 800 kg sttejejIf the stream has uniform width and depth, the speed of the water below the falls is the same as thespeed above the falls. Then no kinetic energy, but only gravitational energy is available forconversion into internal and electric energy.ja fenergy mgy m== gy = 1 800 kg s 9.8 m s 2 5 m = 8.82 10 4 J stttThe output power is Puseful = efficiency Pin = 0.25 8.82 10 4 W = 2.20 10 4 WThe input power is Pin =bgbegejThe efficiency of electric generation at Hoover Dam is about 85%, with a head of water (verticaldrop) of 174 m. Intensive research is underway to improve the efficiency of low head generators.Section 8.2*P8.4(a)The Isolated SystemConservation of Mechanical EnergyOne child in one jump converts chemical energy into mechanical energy in the amount thather body has as gravitational energy at the top of her jump:mgy = 36 kg 9.81 m s 2 0.25 m = 88.3 J . For all of the jumps of the children the energy iseja f12e1.05 10 j88.3 J = 1.11 10 J .6(b)90.011.11 10 9 J = 1.11 10 5 J , making the Richter100log E 4.8 log 1.11 10 5 4.8 5.05 4.8magnitude=== 0.2 .1.51.51.5The seismic energy is modeled as E =220P8.5Potential Energya f 1 mv21g a3.50 Rf = 2 g a Rf + v2Ui + K i = U f + K f :mgh + 0 = mg 2 R +22v = 3.00 gRF = m2v:Rn + mg = mv2RLM v g OP = m L 3.00 gR g O = 2.00mgN R Q MN R PQn = 2.00e5.00 10 kg je9.80 m s j2n=m32= 0.098 0 N downwardP8.6FIG. P8.5K i + Ui = K f + U f12m 6.00 m s + 0 = 0 + m 9.80 m s 2 y2From leaving ground to the highest point,bgeb6.00 m sg =y =a2fe9.80 m s jj2The mass makes no difference:*P8.7(a)21.84 m1111mvi2 + kx i2 = mv 2 + kx 2ff22221120 + 10 N m 0.18 m = 0.15 kg v 2 + 0f22f bgagF 10 N I FG 1 kg m IJ == a0.18 mf GH 0.15 kg m JK H 1 N s Kbvf(b)2K i + U si = K f + U sf10 + 10 N m 0.18 m2b0.162 J =vf =fgaabg2b10.15 kg v 2f21+ 10 N m 0. 25 m 0.18 m2=b10.15 kg v 2 + 0.024 5 Jf2fgga1.47 m s2 0.138 J= 1.35 m s0.15 kgf2FIG. P8.7Chapter 8*P8.8The energy of the car is E =2211mv 2 + mgy21mv 2 + mgd sin where d is the distance it has moved along the track.2dEdvP== mv+ mgv sin dtdtE=(a)When speed is constant,jbegP = mgv sin = 950 kg 9.80 m s 2 2.20 m s sin 30 = 1.02 10 4 W(b)2. 2 m s 0dv=a== 0.183 m s 2dt12 sMaximum power is injected just before maximum speed is attained:bgejP = mva + mgv sin = 950 kg 2.2 m s 0.183 m s 2 + 1.02 10 4 W = 1.06 10 4 W(c)*P8.9(a)At the top end,112.20 m smv 2 + mgd sin = 950 kg22FG bHg + e9.80 m s j1 250 m sin 30IJK =225.82 10 6 JEnergy of the object-Earth system is conserved as the object moves between the releasepoint and the lowest point. We choose to measure heights from y = 0 at the top end of thestring.eK + U j = eK + U j :ggi0 + mgyi =f1mv 2 + mgy ff2e9.8 m s ja2 m cos 30f = 1 v + e9.8 m s ja2 mf2v = 2e9.8 m s ja 2 mfa1 cos 30f = 2.29 m s22f2f(b)Choose the initial point at = 30 and the final point at = 15 :aff2 gLacos 15 cos 30f = 2e9.8 m s ja 2 mfacos 15 cos 30f =0 + mg L cos 30 =vf =P8.102a1mv 2 + mg L cos 15f221.98 m sChoose the zero point of gravitational potential energy of the object-spring-Earth system as theconfiguration in which the object comes to rest. Then because the incline is frictionless, we haveEB = E A :K B + U gB + U sB = K A + U gA + U sAafor0 + mg d + x sin + 0 = 0 + 0 +Solving for d givesd=kx 2x .2mg sin 1 2kx .2222P8.11Potential EnergyFrom conservation of energy for the block-spring-Earth system,U gt = U si ,orb0.250 kg ge9.80 m s jh = FGH 1 IJK b5 000 N mga0.100 mf222This gives a maximum height h = 10.2 m .P8.12(a)FIG. P8.11The force needed to hang on is equal to the force F thetrapeze bar exerts on the performer.From the free-body diagram for the performers body, asshown,F mg cos = mv2orF = mg cos + mv2FIG. P8.12Apply conservation of mechanical energy of the performer-Earth system as the performermoves between the starting point and any later point:mgSolve for(b)mv 2bg cos i = mgaf cos +1mv 22band substitute into the force equation to obtain F = mg 3 cos 2 cos iAt the bottom of the swing, = 0 sobF = mg 3 2 cos ibgF = 2mg = mg 3 2 cos iwhich gives i = 60.0 .gg.223Chapter 8P8.13Using conservation of energy for the system of the Earth and the two objects(a)b5.00 kg gga4.00 mf = b3.00 kg gga4.00 mf + 1 a5.00 + 3.00fv22v = 19.6 = 4.43 m s(b)Now we apply conservation of energy for the system of the 3.00 kgobject and the Earth during the time interval between the instantwhen the string goes slack and the instant at which the 3.00 kgobject reaches its highest position in its free fall.FIG. P8.13a f13.00 v 2 = mg y = 3.00 gy2y = 1.00 my max = 4.00 m + y = 5.00 mP8.14m1 > m 2(a)m1 gh =v=(b)bg1m1 + m 2 v 2 + m 2 gh2bbmgg2 m1 m 2 gh1 + m2Since m 2 has kinetic energy1m 2 v 2 , it will rise an additional height h determined from2m 2 g h =1m2 v 22or from (a),h =The total height m 2 reaches is h + h =P8.15bbggm1 m 2 hv2=m1 + m 22g2m1 h.m1 + m 2The force of tension and subsequent force of compression in therod do no work on the ball, since they are perpendicular to eachstep of displacement. Consider energy conservation of the ballEarth system between the instant just after you strike the ball andthe instant when it reaches the top. The speed at the top is zero ifyou hit it just hard enough to get it there.K i + U gi = K f + U gf :a fa fa f1mvi2 + 0 = 0 + mg 2L2vi = 4 gL = 4 9.80 0.770vi = 5.49 m sinitialfinalLviFIG. P8.15L224*P8.16Potential Energyefficiency =e=useful output energy useful output power=total input energytotal input powerm water gy tb1 2gm ev tjair2=bge v tj2 water v water t gy air r2bg2 w v w t gy=22 3 a r vwhere is the length of a cylinder of air passing through the mill and v w is the volume of waterpumped in time t. We need inject negligible kinetic energy into the water because it starts and endsat rest.eja fb gejejF 1 000 L IJ FG 60 s IJ = 160 L minsGH 1 m K H 1 min K23v w e a r 2 v 3 0.275 1.20 kg m 1.15 m 11 m s==2 w gyt2 1 000 kg m 3 9.80 m s 2 35 m= 2.66 10 3 m3P8.17(a)3K i + U gi = K f + U gf11mvi2 + 0 = mv 2 + mgy ff22111222mv xi + mv yi = mv xf + mgy f222But v xi = v xf , so for the first ballyf =2v yi2g=b1 000 sin 37.0g2a9.80f2= 1.85 10 4 mand for the secondb1 000g ==2a9.80f2yf(b)5.10 10 4 mThe total energy of each is constant with valuebgb120.0 kg 1 000 m s2g2= 1.00 10 7 J .3Chapter 8P8.18In the swing down to the breaking point, energy is conserved:mgr cos =1mv 22at the breaking point consider radial forces Fr = mar+Tmax mg cos = mEliminatev2rv2= 2 g cos rTmax mg cos = 2mg cos Tmax = 3mg cos = cos 1F T I = cosGH 3mg JKmax1FI44GG 3 2.00 kg .59.N m s JJge 80 j KHb2 = 40.8*P8.19(a)For a 5-m cord the spring constant is described by F = kx ,mg = k 1.5 m . For a longer cord of length L the stretch distanceis longer so the spring constant is smaller in inverse proportion:afk=5 m mg= 3.33 mg LL 1.5 meK + Ug+ Usj = eK + Uig+ Usjinitialf10 + mgyi + 0 = 0 + mgy f + kx 2f2mg 211mg yi y f = kx 2 = 3.33xff22LdFIG. P8.19(a)ihere yi y f = 55 m = L + x f1255.0 mL = 3.33 55.0 m L255.0 mL = 5.04 10 3 m 2 183 mL + 1.67 L2af0 = 1.67L2 238 L + 5.04 10 3 = 0L=a fe2a1.67f238 238 2 4 1.67 5.04 10 3j = 238 152 =3.33only the value of L less than 55 m is physical.(b)mg25.8 m F = mak = 3.33x max = x f = 55.0 m 25.8 m = 29.2 m+ kx max mg = mamg3.3329. 2 m mg = ma25.8 ma = 2.77 g = 27.1 m s 2final25.8 m225226*P8.20Potential EnergyWhen block B moves up by 1 cm, block A moves down by 2 cm and the separation becomes 3 cm.vhWe then choose the final point to be when B has moved up by and has speed A . Then A has322hmoved downand has speed v A :3eKA+ KB + Ug0+0+0=j = eKiA+ KB + UgFG IJH Kv112mv A + m A2222+jfmgh mg 2 h33mgh 52= mv A388 gh15vA =Section 8.3P8.21Conservative and Nonconservative ForcesbgejyFg = mg = 4.00 kg 9.80 m s 2 = 39.2 N(a)Ba fa f= a39.2 N fa5.00 mf + a39.2 N fa5.00 mfa 1fC(5.00, 5.00) mOWork along OAC = work along OA + work along AC= Fg OA cos 90.0+ Fg AC cos 180A= 196 J(b)W along OBC = W along OB + W along BC= 39.2 N 5.00 m cos 180+ 39.2 N 5.00 m cos 90.0afafafafFIG. P8.21= 196 J(c)a fF 1 IJ =2 mjG H 2KWork along OC = Fg OC cos135afe= 39.2 N 5.00 196 JThe results should all be the same, since gravitational forces are conservative.P8.22(a)zW = F dr and if the force is constant, this can be written aszdiW = F dr = F r f ri , which depends only on end points, not path.(b)zW = F dr =afzejej a f z dx + a4.00 Nf z dy+ a 4.00 N fy= 15.0 J + 20.0 J = 35.0 J3 i + 4 j dx i + dyj = 3.00 N5.00 mW = 3.00 N x 05.00 m0The same calculation applies for all paths.5.00 m5.00 m00xChapter 8P8.23zdx i 2 y i + x 2 j =zdyj 2 y i + x 2 j =5.00 mWOA =(a)0z2 ydxzx dyzx dyz2272 ydxj5.00 mj5 .00 m2ej5.00 m2ej5.00 me0WOA = 0and since along this path, y = 05 .00 mW AC =0eFor x = 5.00 m,W AC = 125 Jand0WOAC = 0 + 125 = 125 JWOB =(b)zdyj 2 y i + x 2 j =zdx i 2 yi + x 2 j =5 .00 m00WOB = 0since along this path, x = 0 ,WBC =5.00 m00WBC = 50.0 Jsince y = 5.00 m,WOBC = 0 + 50.0 = 50.0 JWOC =(c)WOC =Since x = y along OC,zezeje5.00 m2j2 x + x 2 dx = 66.7 J0(d)P8.24j z e2ydx + x dyjdx i + dyj 2 y i + x 2 j =F is nonconservative since the work done is path dependent.(a)a K fa= W = W g = mgh = mg 5.00 3.20112mv B mv 2 = m 9.80 1.80A22v B = 5.94 m sABfa fa faf2Similarly, vC = v A + 2 g 5.00 2.00 = 7.67 m s(b)WgA CAaf= mg 3.00 m = 147 JB5.00 m3.20 mFIG. P8.24C2.00 m228P8.25Potential Energy(a)ejF = 3.00 i + 5.00 j Nm = 4.00 kgejr = 2.00 i 3.00 j ma fafW = 3.00 2.00 + 5.00 3.00 = 9.00 JThe result does not depend on the path since the force is conservative.(b)W = KFa f IGHJK4.004.00 v 29.00 = 4.0022so v =(c)232.0 9.00= 3.39 m s2.00U = W = 9.00 JSection 8.4Changes in Mechanical Energy for Nonconservative ForcesP8.26U f = K i K f + Ui(a)U f = 30.0 18.0 + 10.0 = 22.0 JE = 40.0 J(b)P8.27Yes, Emech = K + U is not equal to zero. For conservative forces K + U = 0 .The distance traveled by the ball from the top of the arc to the bottom is R . The work done by thenon-conservative force, the force exerted by the pitcher,isa fE = Fr cos 0 = F R .We shall assign the gravitational energy of the ball-Earth system to be zero with the ball at thebottom of the arc.Thenbecomesor11mv 2 mvi2 + mgy f mgyif2211mv 2 = mvi2 + mgyi + F Rf22Emech =a fv f = vi2 + 2 gyi +a f a15.0f + 2a9.80fa1.20f + 2a30.0f a0.600f0.2502 F R=m2v f = 26.5 m s*P8.28The useful output energy isafdi120 Wh 1 0.60 = mg y f yi = Fg yy =bg FG J IJ FG N m IJ =H W sKH J K120 W 3 600 s 0.40890 N194 mChapter 8*P8.29229As the locomotive moves up the hill at constant speed, its output power goes into internal energyplus gravitational energy of the locomotive-Earth system:Pt = mgy + fr = mgr sin + frP = mgv f sin + fv fAs the locomotive moves on level track,F 746 W I = f b27 m sgf = 2.76 10 NGH 1 hp JKF 5 m IJ + e2.76 10 NjvThen also 746 000 W = b160 000 kg ge9.8 m s jv GH 100 m KP = fvi41 000 hp2vf =P8.30746 000 W1.06 10 5 N4ff= 7.04 m sWe shall take the zero level of gravitational potential energy to be at the lowest level reached by thediver under the water, and consider the energy change from when the diver started to fall until hecame to rest.11mv 2 mvi2 + mgy f mgyi = f k d cos 180f22E =di0 0 mg yi y f = f k dfk =P8.31dmg yi y fi = b70.0 kg ge9.80 m s ja10.0 m + 5.00 mf =25.00 mdm 2 gh fh =Ui + K i + Emech = U f + K f :2.06 kN11m1 v 2 + m 2 v 222f = n = m1 gm 2 gh m1 gh =v2 =v=P8.32di eEmech = K f K i + U gf U gibbg1m1 + m 2 v 22gb g2 m 2 m1 hgm1 + m 2eFIG. P8.31jafb2 9.80 m s 2 1.50 m 5.00 kg 0.400 3.00 kg8.00 kgg= 3.74 m sjBut Emech = Wapp fx , where Wapp is the work the boydid pushing forward on the wheels.di ej1= me v v j + mg a hf + fx21= a 47.0 f a6. 20f a1.40f a 47.0fa9.80fa 2.60 f + a 41.0 fa12.4f2Thus,Wapp = K f K i + U gf U gi + fxorWappWappWapp = 168 J2f2i22FIG. P8.32230Potential Energy11m v 2 vi2 = mvi2 = 160 Jf22ej(a)K =(b)U = mg 3.00 m sin 30.0 = 73.5 J(c)P8.33The mechanical energy converted due to friction is 86.5 Jaf=(d)FIG. P8.3386.5 J= 28.8 N3.00 mf = k n = k mg cos 30.0 = 28.8 N28.8 Nk == 0.6795.00 kg 9.80 m s 2 cos 30.0bP8.34fgejConsider the whole motion: K i + U i + Emech = K f + U f(a)0 + mgyi f1 x1 f 2 x 2 =1mv 2 + 0f2b80.0 kg ge9.80 m s j1 000 m a50.0 Nfa800 mf b3 600 Nga200 mf = 1 b80.0 kg gv21784 000 J 40 000 J 720 000 J = b80.0 kg gv22b 24 000 Jg= 24.5 m sv =22f2ff(b)(c)80.0 kgYes this is too fast for safety.Now in the same energy equation as in part (a), x 2 is unknown, and x1 = 1 000 m x 2 :afbg bg784 000 J 50.0 N 1 000 m x 2 3 600 N x 2 =bgbgb180.0 kg 5.00 m s2g2784 000 J 50 000 J 3 550 N x 2 = 1 000 Jx 2 =(d)733 000 J= 206 m3 550 NReally the air drag will depend on the skydivers speed. It will be larger than her 784 Nweight only after the chute is opened. It will be nearly equal to 784 N before she opens thechute and again before she touches down, whenever she moves near terminal speed.P8.35aK + Uf + E(a)a0+bf= K +U f :mechi1 21kx fx = mv 2 + 022ge18.00 N m 5.00 10 2 m2(b)j e3.20 10e2 5. 20 10 3 Jv=231Chapter 85.30 103j=kg22jaf 1 e5.30 102N 0.150 m =3jkg v 21.40 m sWhen the spring force just equals the friction force, the ball will stop speeding up. HereFs = kx ; the spring is compressed by3. 20 10 2 N= 0.400 cm8.00 N mand the ball has moved5.00 cm 0.400 cm = 4.60 cm from the start.(c)Between start and maximum speed points,1 211kxi fx = mv 2 + kx 2f22221118.00 5.00 10 2 3.20 10 2 4.60 10 2 = 5.30 10 3 v 2 + 8.00 4.00 10 3222v = 1.79 m seP8.36j ejej eje Fy = n mg cos 37.0 = 0 n = mg cos 37.0 = 400 Naff = n = 0.250 400 N = 100 N fx = Emecha100fa20.0f = U + U + K + KU = m g d h h i = a50.0fa9.80fa 20.0 sin 37.0f = 5.90 10U = m g d h h i = a100 fa9.80fa 20.0f = 1.96 101K = m e v v j2m1K = m e v v j =K = 2 K2mAAABfBABBABifAB4i2f2f32i2iBAAAAdding and solving, K A = 3.92 kJ .FIG. P8.36j2232P8.37Potential Energy(a)The object moved down distance 1.20 m + x. Choose y = 0 at its lower point.K i + U gi + U si + Emech = K f + U gf + U sf0 + mgyi + 0 + 0 = 0 + 0 +1 2kx2b1.50 kg ge9.80 m s ja1.20 m + xf = 1 b320 N mgx20 = b160 N mgx a14.7 N fx 17.6 J14.7 N a 14.7 N f 4b160 N mga17.6 N mfx=2b160 N mg2222x=14.7 N 107 N320 N mThe negative root tells how high the object will rebound if it is instantly glued to the spring.We wantx = 0.381 m(b)From the same equation,b1.50 kg ge1.63 m s ja1.20 m + xf = 1 b320 N mgx2220 = 160 x 2 2.44x 2.93The positive root is x = 0.143 m .(c)The equation expressing the energy version of the nonisolated system model has one moreterm:mgyi fx =1 2kx2b1.50 kg ge9.80 m s ja1.20 m + xf 0.700 Na1.20 m + xf = 1 b320 N mgx2217.6 J + 14.7 Nx 0.840 J 0.700 Nx = 160 N m x 2160 x 2 14.0 x 16.8 = 0x=14.0 a14.0f 4a160fa16.8fx = 0.371 m23202Chapter 8P8.38233The total mechanical energy of the skysurfer-Earth system is1mv 2 + mgh .2Emech = K + U g =Since the skysurfer has constant speed,a fdEmechdvdh= mv+ mg= 0 + mg v = mgv .dtdtdtThe rate the system is losing mechanical energy is thendEmech= mgv = 75.0 kg 9.80 m s 2 60.0 m s = 44.1 kW .dtb*P8.39(a)jbgLet m be the mass of the whole board. The portion on the rough surface has massnormal force supporting it isa=(b)ge mgxmxgand the frictional force is k= ma . ThenLL k gxopposite to the motion.LIn an incremental bit of forward motion dx, the kinetic energy converted into internal mgxenergy is f k dx = kdx . The whole energy converted isLzL k mgx mg x 21mv 2 =dx = k22LL0L=0 k mgL2v = k gLSection 8.5Relationship Between Conservative Forces and Potential EnergyP8.40U = Ax + Bx 2 dx =zex(a)j0(b)U = z3.00 mFdx =K =(a)Ax 2 Bx 323j a feA 3.00 2 2.002.00 mP8.41mx. TheLFG 5.00 A + 19.0 BIJH 23 KzW = Fx dx =2z a2x + 4fdx = FGH 2x25 .00 m21(b)K + U = 0(c)K = K f 2mv122+ 4xa f a2.00fB 3.0033IJK=5.0019.0AB235 .00 m= 25.0 + 20.0 1.00 4.00 = 40.0 J1U = K = W = 40.0 JK f = K +32mv1= 62.5 J2234P8.42Potential Energyeej ej e 3x 3 y 7xU== 9x 2 y 7 = 7 9x 2 yFx = xx 3x 3 y 7xU== 3 x 3 0 = 3 x 3Fy = yyjjb gThus, the force acting at the point x , y is F = Fx i + Fy j =P8.43afe7 9 x y ji 3 x j .23Ard AAUFr = == 2 . The positive value indicates a force of repulsion.dr rrrUr =Section 8.6FG IJH KEnergy Diagrams and the Equilibrium of a SystemP8.44stableunstableneutralFIG. P8.44P8.45(a)Fx is zero at points A, C and E; Fx is positive at point B and negative at point D.(b)A and E are unstable, and C is stable.(c)FxBACEDFIG. P8.45x (m)Chapter 8P8.46(a)235There is an equilibrium point wherever the graph of potential energy is horizontal:At r = 1.5 mm and 3.2 mm, the equilibrium is stable.At r = 2.3 mm , the equilibrium is unstable.A particle moving out toward r approaches neutral equilibrium.(b)(c)If the system energy is 3 J, its potential energy must be less than or equal to 3 J. Thus, theparticles position is limited to 0.6 mm r 3.6 mm .(d)K + U = E . Thus, K max = E U min = 3.0 J 5.6 J = 2.6 J .(e)Kinetic energy is a maximum when the potential energy is a minimum, at r = 1.5 mm .(f)P8.47The system energy E cannot be less than 5.6 J. The particle is bound if 5.6 J E < 1 J .3 J + W = 1 J . Hence, the binding energy is W = 4 J .(a)When the mass moves distance x, the length of each springachanges from L tokFHIKfx 2 + L2 , so each exerts forcex 2 + L2 L towards its fixed end. The y-componentscancel out and the x components add to:Fx = 2 kFHx 2 + L2 LIK FGHx22x +LI = 2kx +JK2 kLxx 2 + L2FIG. P8.47(a)Choose U = 0 at x = 0 . Then at any point the potential energy of the system isz z FGH 2kx + x2kLxL IJK dx = 2kz xdx 2kLz+U a x f = kx + 2 kLF L x + L IHKU a x f = 40.0 x + 96.0F 1.20 x + 1.44 IHKafxU x = Fx dx = 0x202(b)x2220x0x2x + L2dx22afFor negative x, U x has the same value as forpositive x. The only equilibrium point (i.e., whereFx = 0) is x = 0 .(c)K i + U i + Emech = K f + U f10 + 0.400 J + 0 = 1.18 kg v 2 + 0f2v f = 0.823 m sbgFIG. P8.47(b)236Potential EnergyAdditional ProblemsP8.48The potential energy of the block-Earth system is mgh.An amount of energy k mgd cos is converted into internal energy due to friction on the incline.Therefore the final height y max is found frommgy max = mgh k mgd cos wherey maxsin mgy max = mgh k mgy max cot d=hy maxSolving,y max =P8.49FIG. P8.48h1 + k cot .At a pace I could keep up for a half-hour exercise period, I climb two stories up, traversing fortysteps each 18 cm high, in 20 s. My output work becomes the final gravitational energy of the systemof the Earth and me,bjagefmgy = 85 kg 9.80 m s 2 40 0.18 m = 6 000 J6 000 J= ~ 10 2 W .20 smaking my sustainable powerP8.50v = 100 km h = 27.8 m sThe retarding force due to air resistance isR=afejb11DAv 2 = 0.330 1.20 kg m 3 2.50 m 2 27.8 m s22jeg2= 382 NComparing the energy of the car at two points along the hill,K i + U gi + E = K f + U gfora fK i + U gi + We R s = K f + U gfwhere We is the work input from the engine. Thus,a f di eWe = R s + K f K i + U gf U gijRecognizing that K f = K i and dividing by the travel time t gives the required power input fromthe engine asFG W IJ = RFG s IJ + mgFG y IJ = Rv + mgv sinH t K H t K H t KP = a382 N fb 27.8 m sg + b1 500 kg ge9.80 m s jb 27.8 m sg sin 3.20P=e2P = 33.4 kW = 44.8 hpChapter 8P8.51m = mass of pumpkinR = radius of silo top Fr = mar n mg cos = m237v2RWhen the pumpkin first loses contact with the surface, n = 0 .Thus, at the point where it leaves the surface: v 2 = Rg cos .FIG. P8.51Choose U g = 0 in the = 90.0 plane. Then applying conservation of energy for the pumpkin-Earthsystem between the starting point and the point where the pumpkin leaves the surface givesK f + U gf = K i + U gi1mv 2 + mgR cos = 0 + mgR2Using the result from the force analysis, this becomes1mRg cos + mgR cos = mgR , which reduces to2cos =b g2, and gives = cos 1 2 3 = 48.23as the angle at which the pumpkin will lose contact with the surface.bjagef(a)U A = mgR = 0. 200 kg 9.80 m s 2 0.300 m = 0.588 J(b)P8.52K A + U A = KB + UBK B = K A + U A U B = mgR = 0.588 J(c)(d)2K B=mvB =af2 0.588 J= 2.42 m s0.200 kgbgejafFIG. P8.52UC = mghC = 0.200 kg 9.80 m s 2 0.200 m = 0.392 JbK C = K A + U A U C = mg h A hCbgejagfK C = 0.200 kg 9.80 m s 2 0.300 0.200 m = 0.196 JP8.53bgb112mv B = 0.200 kg 1.50 m s22g2(a)KB == 0.225 J(b)Emech = K + U = K B K A + U B U Abg= 0.225 J + b0.200 kg ge9.80 m s ja0 0.300 mf= K B + mg hB hA2= 0.225 J 0.588 J = 0.363 J(c)Its possible to find an effective coefficient of friction, but not the actual value of since nand f vary with position.238P8.54Potential EnergyThe gain in internal energy due to friction represents a loss in mechanical energy that must be equalto the change in the kinetic energy plus the change in the potential energy.Therefore, k mgx cos = K +1 2kx mgx sin 2and since vi = v f = 0 , K = 0.Thus,a fa fafaf a100fa02.200f a2.00fa9.80fasin 37.0fa0.200f2 k 2.00 9.80 cos 37.0 0.200 =and we find k = 0.115 . Note that in the above we had a gain in elastic potential energy for thespring and a loss in gravitational potential energy.P8.55(a)Since no nonconservative work is done, E = 0k = 100 N/mAlso K = 02.00 kgtherefore, Ui = U fbgwhere Ui = mg sin xand U f =1 2kx2FIG. P8.55a fa fxa f 2 and solving we findSubstituting values yields 2.00 9.80 sin 37.0 = 100x = 0.236 m(b) F = ma . Only gravity and the spring force act on the block, so kx + mg sin = maFor x = 0.236 m ,a = 5.90 m s 2 . The negative sign indicates a is up the incline.The acceleration depends on position .(c)U(gravity) decreases monotonically as the height decreases.U(spring) increases monotonically as the spring is stretched.K initially increases, but then goes back to zero.Chapter 8k = 2.50 10 4 N m,m = 25.0 kgx A = 0.100 m,P8.56Ugx =0= Usx=0=0EmechEmech = K A + U gA + U sA1 2kx A2= 25.0 kg 9.80 m s 2 0.100 mEmech(a)12.50 10 42= 24.5 J + 125 J = 100 JEmech = 0 + mgx A +b(b)fjaN mja 0.100 mfgee+2Since only conservative forces are involved, the total energy of the child-pogo-stick-Earthsystem at point C is the same as that at point A.bgej0 + 25.0 kg 9.80 m s 2 xC + 0 = 0 24.5 J + 125 JK C + U gC + U sC = K A + U gA + U sA :x C = 0.410 mbagf1225.0 kg v B + 0 + 0 = 0 + 24.5 J + 125 J2v B = 2.84 m s(c)K B + U gB + U sB = K A + U gA + U sA :(d)K and v are at a maximum when a = F m = 0 (i.e., when the magnitude of the upwardspring force equals the magnitude of the downward gravitational force).This occurs at x < 0 wherek x = mgorx=Thus,K = K max at x = 9.80 mmb25.0 kg ge9.8 m s j = 9.80 102(e)e2.50 10 4 N mx =9.80 mmsAms x =9.80 mm2max24yielding3j + eU Uj1b25.0 kg gv = b25.0 kgge9.80 m s j a0.100 mf b0.009 8 mg21+ e 2.50 10 N mj a 0.100 mf b0.009 8 mg2K max = K A + U gA U gorP8.572392v max = 2.85 m sEmech = fxE f Ei = f d BC1 2kx mgh = mgd BC2mgh 1 kx 22= 0.328=mgd BCFIG. P8.572240Potential Energy(a)F=(b)P8.58d x3 + 2x 2 + 3x i =dxej e3 x2j 4x 3 iF=0when x = 1.87 and 0.535(c)The stable point is atafx = 0.535 point of minimum U x .The unstable point is atFIG. P8.58afx = 1.87 maximum in U x .P8.59aK + U f = aK + U f10 + b30.0 kg ge9.80 m s ja0.200 mf + b 250 N mga0.200 mf21= b50.0 kg gv + b 20.0 kg ge9.80 m s ja0.200 mf sin 40.0258.8 J + 5.00 J = b 25.0 kg gv + 25.2 Jif22222v = 1.24 m sP8.60(a)FIG. P8.59Between the second and the third picture, Emech = K + U mgd = b11mvi2 + kd 222gbgebge1150.0 N m d 2 + 0.250 1.00 kg 9.80 m s 2 d 1.00 kg 3.00 m s 2 = 0222.45 21.25 Nd== 0.378 m50.0 N m(b)jjBetween picture two and picture four, Emech = K + Ua f 1 mv 1 mv222v = b3.00 m sg b1.00 kg g a2.45 Nfa2fa0.378 mf2 f 2d =2i2= 2.30 m s(c)For the motion from picture two to picture five,Emech = K + Uafbgbg121.00 kg 3.00 m s29.00 JD= 2 0.378 m = 1.08 m2 0.250 1.00 kg 9.80 m s 2 f D + 2d = afbgejafFIG. P8.60Chapter 8P8.61(a)Initial compression of spring:ga fb1450 N m x2 x = 0. 400 m(b)21 2 1kx = mv 222bgb10.500 kg 12.0 m s2=g2Speed of block at top of track: Emech = fxFIG. P8.61FG mgh + 1 mv IJ FG mgh + 1 mv IJ = f aRfHK HK22b0.500 kgge9.80 m s ja2.00 mf + 1 b0.500 kg gv 1 b0.500 kggb12.0 m sg22= a7.00 N fa fa1.00 mfT2TB2B2T2220.250 vT = 4.21 vT = 4.10 m s(c)Does block fall off at or before top of track? Block falls if a c < gac =a f224.10vT== 16.8 m s 2R1.00Therefore a c > g and the block stays on the track .P8.62Let represent the mass of each one meter of the chain and Trepresent the tension in the chain at the table edge. We imagine theedge to act like a frictionless and massless pulley.(a)For the five meters on the table with motion impending, Fy = 0 :+n 5 g = 0n = 5 gb gfs sn = 0.6 5g = 3g Fx = 0 :+T f s = 0T = fsT 3 gFIG. P8.62The maximum value is barely enough to support the hanging segment according to Fy = 0 :+T 3 g = 0T = 3 gso it is at this point that the chain starts to slide.continued on next page241242Potential Energy(b)Let x represent the variable distance the chain has slipped since the start.a f+n a5 x fg = 0n = a5 xfgf = n = 0.4a5 xfg = 2g 0.4xgThen length 5 x remains on the table, with now Fy = 0 :kkConsider energies of the chain-Earth system at the initial moment when the chain starts toslip, and a final moment when x = 5 , when the last link goes over the brink. Measureheights above the final position of the leading end of the chain. At the moment the final linkslips off, the center of the chain is at y f = 4 meters.Originally, 5 meters of chain is at height 8 m and the middle of the dangling segment is at3height 8 = 6.5 m .2K i + U i + Emech = K f + U f :FG 1 mv + mgyIJH2Kb5g g8 + b3g g6.5 z b2g 0.4xg gdx = 1 b8 gv + b8g g42bzfg20 + m1 gy1 + m 2 gy 2 i f k dx =fi502zz5540.0 g + 19.5 g 2.00 g dx + 0.400 g x dx = 4.00 v 2 + 32.0 g00527.5 g 2.00 gx 0 + 0.400 ga f2 5x2= 4.00 v 20a f27.5 g 2.00 g 5.00 + 0.400 g 12.5 = 4.00 v 222.5 g = 4.00 v 2a22.5 mfe9.80 m s j =2v=P8.634.007.42 m sLaunch speed is found frommgFG 4 hIJ = 1 mv :H5 K 22v = 2gFG 4 IJ hH 5Kv y = v sin The height y above the water (by conservation of energyfor the child-Earth system) is found frommgy =1h2mv y + mg25FIG. P8.6312mv x is constant in projectile motion)21 2 h 1 2hy=vy + =v sin 2 +2g5 2g5(sincey=LM FG IJ OP sin + h =5N H KQ142g h2g524hh sin 2 +55Chapter 8*P8.64(a)The length of string between glider and pulley is given byNow(b)eKA22= x 2 + h0 . Then 2a243ddx= 2x+ 0.dtdtfddx= v y = v x = cos v x .is the rate at which string goes over the pulley:dtdt+ KB + U gj = eKib0 + 0 + m B g y 30 y 45Ag+ KB + U gjf1122= m A v x + mB v y22Now y 30 y 45 is the amount of string that has gone over the pulley, 30 45 . We havehhh0h0sin 30 = 0 and sin 45 = 0 , so 30 45 == 0. 40 m 2 2 = 0.234 m .sin 30 sin 453045From the energy equatione0.5 kg 9.8 m s 2 0.234 m =j11221.00 kg v x + 0.500 kg v x cos 2 45221.15 J= 1.35 m s0.625 kgvx =bg(c)(d)P8.65v y = v x cos = 1.35 m s cos 45 = 0.958 m sThe acceleration of neither glider is constant, so knowing distance and acceleration at onepoint is not sufficient to find speed at another point.bgThe geometry reveals D = L sin + L sin , 50.0 m = 40.0 m sin 50+ sin , = 28.9(a)From takeoff to alighting for the Jane-Earth systemeK + U j + W = eK + U j1mv + mg a L cos f + FDa 1f = 0 + mg b L cos g2150 kg v + 50 kg e9.8 m s ja 40 m cos 50f 110 Na50 mf = 50 kg e9.8 m s ja 40 m cos 28.9f2gwindigf2i2i22150 kg vi2 1.26 10 4 J 5.5 10 3 J = 1.72 10 4 J2vi =(b)af2 947 J= 6.15 m s50 kgFor the swing backbgea faff1mvi2 + mg L cos + FD +1 = 0 + mg L cos 21130 kg vi2 + 130 kg 9.8 m s 2 40 m cos 28.9 + 110 N 50 m2ejaja= 130 kg 9.8 m s 2 40 m cos 50f1130 kg vi2 4.46 10 4 J + 5 500 J = 3.28 10 4 J2vi =bg=2 6 340 J130 kg9.87 m saf244P8.66Potential Energy11mv 2 = kx 22225.00 kg 1.20 m smvk= 2 =x10 2 m 2Case I: Surface is frictionlessbejeeK + U j = eK + U jggA= 7. 20 10 2 N m(b)11.1 m sv2=ac =r6.3 m(c) Fy = ma yg22B120 + mgy A = mv B + 02b2j a0.300fb5.00 kg ge9.80 m s je105.00 kg 2 1v = 7.20 10 2 N m 10 1 m22v = 0.923 m s(a)g k = 0.30011mv 2 = kx 2 k mgx22Case II: Surface is rough,*P8.67gbe1jmjv B = 2 gy A = 2 9.8 m s 2 6.3 m = 11.1 m s2= 19.6 m s 2 up+n B mg = ma cejn B = 76 kg 9.8 m s 2 + 19.6 m s 2 = 2.23 10 3 N upaf(d)W = Fr cos = 2.23 10 3 N 0.450 m cos 0 = 1.01 10 3 J(e)eK + U jgBe+ W = K +UgjDbg1122mv B + 0 + 1.01 10 3 J = mv D + mg y D y B22112276 kg 11.1 m s + 1.01 10 3 J = 76 kg v D + 76 kg 9.8 m s 2 6.3 m22bge5.70 103ejJ 4.69 10 3 J 276 kg(f)eK + U j = eK + U j1mv + 0 = 0 + mg b y2ggD2D(g)Ej= v D = 5.14 m swhere E is the apex of his motionE yDgy E yD =bg225.14 m svD== 1.35 m2 g 2 9.8 m s 2ejConsider the motion with constant acceleration between takeoff and touchdown. The timeis the positive root of1y f = yi + v yi t + a y t 2212.34 m = 0 + 5.14 m s t + 9.8 m s 2 t 2224.9t 5.14t 2.34 = 0et=a fajf=5.14 5.14 2 4 4.9 2.349.81.39 sChapter 8*P8.68If the spring is just barely able to lift the lower block from the table, the spring lifts it through nonoticeable distance, but exerts on the block a force equal to its weight Mg. The extension of thespring, from Fs = kx , must be Mg k . Between an initial point at release and a final point when themoving block first comes to rest, we haveFGHIJKFGHIJKFG IJH K20 + mg c he4m j = m M=2c hOnly a positive mass is physical, so we take M = ma3 1f = 2m .m m 2 4221212(a)FG IJH K4mgMg1 Mg1 4mg+ k= 0 + mg+ k22kkkk2 22 222 2mMgM g4m g8m g+=+kkk2kM24m 2 = mM +2M2+ mM 4m 2 = 02K i + U gi + U si = K f + U gf + U sf :P8.692459m 2Take the original point where the ball isreleased and the final point where itsupward swing stops at height H andhorizontal displacementax = L2 L Hf2= 2LH H 2Since the wind force is purely horizontal, itdoes workzzWwind = F ds = F dx = F 2LH H 2FIG. P8.69The work-energy theorem can be written:K i + U gi + Wwind = K f + U gf , or0 + 0 + F 2LH H 2 = 0 + mgH giving F 2 2LH F 2 H 2 = m 2 g 2 H 2Here H = 0 represents the lower turning point of the balls oscillation, and the upper limit isat F 2 2L = F 2 + m 2 g 2 H . Solving for H yieldsa f ejH=2LF 22L=F + m2 g 21 + mg Fb2g2As F 0 , H 0 as is reasonable.As F , H 2L , which would be hard to approach experimentally.(b)H=af2 2.00 mbgej1 + 2.00 kg 9.80 m s 2 14.7 Ncontinued on next page2= 1.44 m246Potential Energy(c)Call the equilibrium angle with the vertical. Fx = 0 T sin = F , and Fy = 0 T cos = mgDividing: tan =F14.7 N== 0.750 , or = 36.9mg 19.6 Naf afafTherefore, H eq = L 1 cos = 2.00 m 1 cos 36.9 = 0.400 m(d)As F , tan , 90.0 and H eq LA very strong wind pulls the string out horizontal, parallel to the ground. Thus,eH jeqP8.70max=L .Call = 180 the angle between the upward vertical andthe radius to the release point. Call v r the speed here. Byconservation of energyvi = RgThe pathafter stringis cutK i + U i + E = K r + U r112mvi2 + mgR + 0 = mv r + mgR cos 222gR + 2 gR = v r + 2 gR cos CRv r = 3 gR 2 gR cos FIG. P8.70The components of velocity at release are v x = v r cos andv y = v r sin so for the projectile motion we havex = vxt1y = v y t gt 22R sin = v r cos t R cos = v r sin t 1 2gt2By substitution R cos = v r sin R sin g R 2 sin 2 2v r cos 2 v r cos 2 with sin 2 + cos 2 = 1 ,b2gR sin 2 = 2 v r cos = 2 cos 3 gR 2 gR cos gsin 2 = 6 cos 4 cos 2 = 1 cos 2 3 cos 2 6 cos + 1 = 0cos =6 36 126Only the sign gives a value for cos that is less than one:cos = 0.183 5 = 79.43so = 100.6247Chapter 8P8.71Applying Newtons second law at the bottom (b) and top (t) of thecircle givesTb mg =Adding these gives2mv band Tt mg = RTb = Tt + 2mg +me2vbvtmgmv t2Rv t2TbjRmgAlso, energy must be conserved and U + K = 0So,e2m v b v t22j + b0 2mgRg = 0 and mevTt2b v t2Rj = 4mgvbFIG. P8.71Substituting into the above equation gives Tb = Tt + 6mg .P8.72(a)(b)Energy is conserved in the swing of the pendulum, and thestationary peg does no work. So the balls speed does notchange when the string hits or leaves the peg, and the ballswings equally high on both sides.Relative to the point of suspension,aUi = 0, U f = mg d L dLdPegfFrom this we find thataf1 mg 2d L + mv 2 = 02Also for centripetal motion,mg =mv 2where R = L d .RUpon solving, we get d =3L.5FIG. P8.72248*P8.73Potential Energy(a)At the top of the loop the car and riders are in freefall: Fy = ma y :mg down =v = Rgmv 2downREnergy of the car-riders-Earth system is conservedbetween release and top of loop:K i + U gi = K f + U gf :0 + mgh =a f1mv 2 + mg 2 R2a f1Rg + g 2 R2h = 2.50 Rgh =(b)Let h now represent the height 2.5 R of the releasepoint. At the bottom of the loop we havemgh =12mv b2 Fy = ma y :or2v b = 2 ghn b mg =n b = mg +2mv bupRm 2 ghb gb gRa f1mv t2 + mg 2 R2v t2 = 2 gh 4 gRAt the top of the loop, mgh = Fy = ma y :n t mg = FIG. P8.73mv t2Rm2 gh 4 gRn t = mg +Rm 2 gh 5mgnt =Rbgb gThen the normal force at the bottom is larger byn b n t = mg +b g mb2 ghg + 5mg =m 2 ghRR6mg .Chapter 8*P8.74(a)249Conservation of energy for the sled-rider-Earth system,between A and C:K i + U gi = K f + U gfb1m 2.5 m s2vC =(b)g2jaef12mvC + 02+ m 9.80 m s 2 9.76 m =b2.5 m sg + 2e9.80 m s ja9.76 mf =22FIG. P8.74(a)14.1 m sIncorporating the loss of mechanical energy during the portion of the motion in the water,we have, for the entire motion between A and D (the riders stopping point),bgbg bjage7.90 10 3 J 7.90 10 3 N m== 158 Nx50 mThe water exerts a frictional forcefk =and also a normal force of(c)n = mg = 80 kg 9.80 m s 2 = 784 Nbgea158 Nf + a784 Nf2The magnitude of the water force is(d)f1280 kg 2.5 m s + 80 kg 9.80 m s 2 9.76 m f k x = 0 + 02 f k x = 7.90 10 3 JK i + U gi f k x = K f + U gf :j2= 800 NThe angle of the slide is = sin 19.76 m= 10.454.3 mFor forces perpendicular to the track at B, Fy = ma y :FIG. P8.74(d)n B mg cos = 0bgejn B = 80.0 kg 9.80 m s 2 cos 10.4 = 771 N(e) Fy = ma y :2mvCrnC = 80.0 kg 9.80 m s 2+nC mg =bgejb80.0 kggb14.1 m sg+220 mnC = 1.57 10 3 N upFIG. P8.74(e)The rider pays for the thrills of a giddy height at A, and a high speed and tremendous splashat C. As a bonus, he gets the quick change in direction and magnitude among the forces wefound in parts (d), (e), and (c).250Potential EnergyANSWERS TO EVEN PROBLEMSP8.61.84 msee the solution(a) r = 1.5 mm and 3.2 mm, stable; 2.3 mmand unstable; r neutral;(b) 5.6 J E < 1 J ; (c) 0.6 mm r 3.6 mm ;(d) 2.6 J; (e) 1.5 mm; (f) 4 JP8.48(a) 1.11 10 9 J ; (b) 0.2e7 9 x y ji 3 x jP8.44(a) 800 J; (b) 107 J; (c) 0P8.4P8.42P8.46P8.2see the solutionP8.5033.4 kWP8.52(a) 0.588 J; (b) 0.588 J; (c) 2.42 m s;(d) 0.196 J; 0.392 J6P8.8(a) 10.2 kW; (b) 10.6 kW; (c) 5.82 10 JP8.10d=P8.12(a) see the solution; (b) 60.0P8.14(a)kx 2x2mg sin bbmg+m g2 m1 m 2 gh12; (b)2m1 hm1 + m 223P8.16160 L minP8.540.115P8.1840.8P8.56P8.20FG 8 gh IJH 15 K(a) 100 J; (b) 0.410 m; (c) 2.84 m s ;(d) 9.80 mm ; (e) 2.85 m sP8.58(a) 3 x 2 4x 3 i ; (b) 1.87; -0.535;P8.22(a) see the solution; (b) 35.0 JP8.24(a) v B = 5.94 m s; vC = 7.67 m s ; (b) 147 JP8.26(a) U f = 22.0 J ; E = 40.0 J ; (b) Yes. The total12mechanical energy changes.ej(c) see the solutionP8.60(a) 0.378 m; (b) 2.30 m s ; (c) 1.08 mP8.62(a) see the solution; (b) 7.42 m sP8.64(a) see the solution; (b) 1.35 m s ;(c) 0.958 m s ; (d) see the solutionP8.28194 mP8.302.06 kN upP8.660.923 m sP8.32168 JP8.682mP8.34(a) 24.5 m s ; (b) yes; (c) 206 m; (d) Air dragdepends strongly on speed.P8.70100.6P8.363.92 kJP8.72see the solutionP8.3844.1 kWP8.74(a) 14.1 m s; (b) 7.90 J ; (c) 800 N;(d) 771 N; (e) 1.57 kN upP8.40(a)Ax 2 Bx 3;235 A 19B19B 5 A(b) U =; K =23329Linear Momentum and CollisionsCHAPTER OUTLINE9.1Linear Momentum and ItsConservationImpulse and MomentumCollisions in One DimensionTwo-Dimensional CollisionsThe Center of MassMotion of a System ofParticlesRocket PropulsionANSWERS TO QUESTIONSQ9.2The momentum doubles since it is proportional to the speed.The kinetic energy quadruples, since it is proportional to thespeed-squared.The momenta of two particles will only be the same if themasses of the particles of the same.(a)It does not carry force, for if it did, it could accelerateitself.(b)(c)No. Impulse, Ft , depends on the force and the time for whichit is applied.Q9.49.7Q9.1Q9.39.29.39.49.59.6It cannot deliver more kinetic energy than it possesses.This would violate the law of energy conservation.It can deliver more momentum in a collision than it possesses in its flight, by bouncing fromthe object it strikes.Q9.5Provided there is some form of potential energy in the system, the parts of an isolated system canmove if the system is initially at rest. Consider two air-track gliders on a horizontal track. If youcompress a spring between them and then tie them together with a string, it is possible for thesystem to start out at rest. If you then burn the string, the potential energy stored in the spring willbe converted into kinetic energy of the gliders.Q9.6No. Only in a precise head-on collision with momenta with equal magnitudes and oppositedirections can both objects wind up at rest. Yes. Assume that ball 2, originally at rest, is strucksquarely by an equal-mass ball 1. Then ball 2 will take off with the velocity of ball 1, leaving ball 1 atrest.Q9.7Interestingly, mutual gravitation brings the ball and the Earth together. As the ball movesdownward, the Earth moves upward, although with an acceleration 10 25 times smaller than that ofthe ball. The two objects meet, rebound, and separate. Momentum of the ball-Earth system isconserved.Q9.8(a)Linear momentum is conserved since there are no external forces acting on the system.(b)Kinetic energy is not conserved because the chemical potential energy initially in theexplosive is converted into kinetic energy of the pieces of the bomb.251252Linear Momentum and CollisionsQ9.9Momentum conservation is not violated if we make our system include the Earth along with theclay. When the clay receives an impulse backwards, the Earth receives the same size impulseforwards. The resulting acceleration of the Earth due to this impulse is significantly smaller than theacceleration of the clay, but the planet absorbs all of the momentum that the clay loses.Q9.10Momentum conservation is not violated if we choose as our system the planet along with you.When you receive an impulse forward, the Earth receives the same size impulse backwards. Theresulting acceleration of the Earth due to this impulse is significantly smaller than your accelerationforward, but the planets backward momentum is equal in magnitude to your forward momentum.Q9.11As a ball rolls down an incline, the Earth receives an impulse of the same size and in the oppositedirection as that of the ball. If you consider the Earth-ball system, momentum conservation is notviolated.Q9.12Suppose car and truck move along the same line. If one vehicle overtakes the other, the fastermoving one loses more energy than the slower one gains. In a head-on collision, if the speed of them + 3m ctimes the speed of the car, the car will lose more energy.truck is less than T3mT + m cQ9.13The rifle has a much lower speed than the bullet and much less kinetic energy. The butt distributesthe recoil force over an area much larger than that of the bullet.Q9.14His impact speed is determined by the acceleration of gravity and the distance of fall, inv 2 = vi2 2 g 0 yi . The force exerted by the pad depends also on the unknown stiffness of the pad.fQ9.15The product of the mass flow rate and velocity of the water determines the force the firefightersmust exert.Q9.16The sheet stretches and pulls the two students toward each other. These effects are larger for afaster-moving egg. The time over which the egg stops is extended so that the force stopping it isnever too large.Q9.17(c) In this case, the impulse on the Frisbee is largest. According to Newtons third law, the impulseon the skater and thus the final speed of the skater will also be largest.Q9.18Usually but not necessarily. In a one-dimensional collision between two identical particles with thesame initial speed, the kinetic energy of the particles will not change.Q9.19g downward.Q9.20As one finger slides towards the center, the normal force exerted by the sliding finger on the rulerincreases. At some point, this normal force will increase enough so that static friction between thesliding finger and the ruler will stop their relative motion. At this moment the other finger startssliding along the ruler towards the center. This process repeats until the fingers meet at the center ofthe ruler.Q9.21The planet is in motion around the sun, and thus has momentum and kinetic energy of its own. Thespacecraft is directed to cross the planets orbit behind it, so that the planets gravity has acomponent pulling forward on the spacecraft. Since this is an elastic collision, and the velocity of theplanet remains nearly unchanged, the probe must both increase speed and change direction for bothmomentum and kinetic energy to be conserved.bgChapter 9253Q9.22Noan external force of gravity acts on the moon. Yes, because its speed is constant.Q9.23The impulse given to the egg is the same regardless of how it stops. If you increase the impact timeby dropping the egg onto foam, you will decrease the impact force.Q9.24Yes. A boomerang, a kitchen stool.Q9.25The center of mass of the balls is in free fall, moving up and then down with the acceleration due togravity, during the 40% of the time when the jugglers hands are empty. During the 60% of the timewhen the juggler is engaged in catching and tossing, the center of mass must accelerate up with asomewhat smaller average acceleration. The center of mass moves around in a little circle, makingthree revolutions for every one revolution that one ball makes. Letting T represent the time for onecycle and Fg the weight of one ball, we have FJ 0.60T = 3 Fg T and FJ = 5 Fg . The average force exertedby the juggler is five times the weight of one ball.Q9.26In empty space, the center of mass of a rocket-plus-fuel system does not accelerate during a burn,because no outside force acts on this system. According to the texts basic expression for rocketpropulsion, the change in speed of the rocket body will be larger than the speed of the exhaustrelative to the rocket, if the final mass is less than 37% of the original mass.Q9.27The gun recoiled.Q9.28Inflate a balloon and release it. The air escaping from the balloon gives the balloon an impulse.Q9.29There was a time when the English favored position (a), the Germans position (b), and the Frenchposition (c). A Frenchman, Jean DAlembert, is most responsible for showing that each theory isconsistent with the others. All are equally correct. Each is useful for giving a mathematically simplesolution for some problems.SOLUTIONS TO PROBLEMSSection 9.1P9.1Linear Momentum and Its Conservatione(a)jv = 3.00 i 4.00 j m sm = 3.00 kg ,ejp = mv = 9.00 i 12.0 j kg m sThus,and(b)p x = 9.00 kg m sp y = 12.0 kg m sa9.00f + a12.0f = 15.0 kg m sF p I = tan a1.33f = 307GH p JK222p = px + py = = tan 1yx12254Linear Momentum and Collisions(a)At maximum height v = 0 , so p = 0 .(b)P9.2Its original kinetic energy is its constant total energy,Ki =af b11mvi2 = 0.100 kg 15.0 m s22g2= 11.2 J .At the top all of this energy is gravitational. Halfway up, one-half of it is gravitational andthe other half is kinetic:v=bgbbg10.100 kg v 222 5.62 J= 10.6 m s0.100 kgK = 5.62 J =gThen p = mv = 0.100 kg 10.6 m s jp = 1.06 kg m s j .P9.3I have mass 85.0 kg and can jump to raise my center of gravity 25.0 cm. I leave the ground withspeed given bydiv 2 vi2 = 2 a x f x i :fjae0 vi2 = 2 9.80 m s 2 0.250 mfvi = 2.20 m sTotal momentum of the system of the Earth and me is conserved as I push the earth down andmyself up:j begb0 = 5.98 10 24 kg v e + 85.0 kg 2.20 m sgv e ~ 10 23 m sP9.4(a)For the system of two blocks p = 0 ,orpi = p fTherefore,0 = Mv m + 3 M 2.00 m sSolving givesv m = 6.00 m s (motion toward thea fbgleft).(b)a f1 2 1122kx = Mv M + 3 M v 3 M = 8. 40 J222FIG. P9.4Chapter 9(a)The momentum is p = mv , so v =(b)P9.5K=Section 9.2*P9.61mv 2 implies v =2FG IJH Kpp11and the kinetic energy is K = mv 2 = mmm222K2K, so p = mv = m=mm2=Impulse and Momentuma fFrom the impulse-momentum theorem, F t = p = mv f mvi , the average force required to holdF=di = b12 kg gb0 60 mi hg F 1 m s I = 6.44 10GH 2.237 mi h JK0.050 s 0a t fm v f vi3N.Therefore, the magnitude of the needed retarding force is 6.44 10 3 N , or 1 400 lb. A personcannot exert a force of this magnitude and a safety device should be used.(a)zI = Fdt = area under curvejbgI=11.50 10 3 s 18 000 N = 13.5 N s2(b)F=13.5 N s= 9.00 kN1.50 10 3 s(c)*P9.8p2.2m2mK .onto the child isP9.7255From the graph, we see that Fmax = 18.0 kNeFIG. P9.71122mv1 = mgy1 . The rebound speed is given by mgy 2 = mv 2 . The22impulse of the floor is the change in momentum,The impact speed is given byb= megmv 2 up mv1 down = m v 2 + v1 upj2 gh2 + 2 gh1 upe= 0.15 kg 2 9.8 m s 2je= 1.39 kg m s upwardj0.960 m + 1.25 m up256P9.9Linear Momentum and Collisionsp = Ftfej a= ma v sin 60.0 v sin 60.0f = 2mv sin 60.0= 2b3.00 kg gb10.0 m sga0.866fp y = m v fy viy = m v cos 60.0 mv cos 60.0 = 0p x= 52.0 kg m sFave =P9.10p x 52.0 kg m s== 260 N0.200 stFIG. P9.9Assume the initial direction of the ball in the x direction.ga f bbga fe j(a)(b)P9.11Impulse, I = p = p f pi = 0.060 0 40.0 i 0.060 0 50.0 i = 5.40 i N sWork = K f K i =b10.060 02g a40.0f a50.0f22= 27.0 JTake x-axis toward the pitcher(a)b0.200 kg gb15.0 m sga cos 45.0f + I = b0.200 kggb40.0 m sg cos 30.0pix + I x = p fx :xI x = 9.05 N sb0.200 kg gb15.0 m sga sin 45.0f + I = b0.200 kg gb40.0 m sg sin 30.0I = e9.05 i + 6.12 jj N spiy + I y = p fy :(b)byfgaaeFm =P9.12fa110 + Fm 4.00 ms + Fm 20.0 ms + Fm 4.00 ms22Fm 24.0 10 3 s = 9.05 i + 6.12 j N sI=fje377 i + 255 jj NIf the diver starts from rest and drops vertically into the water, the velocity just before impact isfound fromK f + U gf = K i + U gi12mv impact + 0 = 0 + mgh v impact = 2 gh2With the diver at rest after an impact time of t , the average force during impact is given byF=em 0 v impacttj = m2 ghtor F =m 2 ght(directed upward).Assuming a mass of 55 kg and an impact time of 1.0 s , the magnitude of this average force isb55 kg g 2e9.8 m s ja10 mf = 770 N , orF =21.0 s~ 10 3 N .Chapter 9P9.13The force exerted on the water by the hose isF=bgbg0.600 kg 25.0 m s 0p water mv f mvi=== 15.0 N .1.00 sttAccording to Newton's third law, the water exerts a force of equal magnitude back on the hose.Thus, the gardener must apply a 15.0 N force (in the direction of the velocity of the exiting waterstream) to hold the hose stationary.*P9.14(a)Energy is conserved for the spring-mass system:K i + U si = K f + U sf :1 2 1kx = mv 2 + 022kv=xm0+klarger.m(b)From the equation, a smaller value of m makes v = x(c)I = p f p i = mv f = 0 = mx(d)From the equation, a larger value of m makes I = x km larger.(e)For the glider, W = K f K i =k= x kmm11mv 2 0 = kx 222The mass makes no difference to the work.Section 9.3P9.15Collisions in One Dimensionb200 g gb55.0 m sg = b46.0 g gv + b200 g gb40.0 m sgv = 65.2 m s*P9.16bm vg = bm v + m v g22.5 g b35 m sg + 300 g b2.5 m sg = 22.5 gv1 1v1 f =+ m2 v2i1 12 2 f1f+037.5 g m s= 1.67 m s22.5 gFIG. P9.16257258P9.17Linear Momentum and CollisionsMomentum is conserved10.0 10 3 kg v = 5.01 kg 0.600 m sj begbgv = 301 m sP9.18(a)mv1i + 3mv 2 i = 4mv f where m = 2.50 10 4 kgvf =a f4.00 + 3 2.00= 2.50 m s4a fLMNa f OPQ eP9.19jaf111224m v 2 mv1i + 3m v 2i = 2.50 10 4 12.5 8.00 6.00 = 3.75 10 4 Jf222(b)K f Ki =(a)The internal forces exerted by the actor donot change the total momentum of thesystem of the four cars and the movie actora4mfv = a3mfb2.00 m sg + mb4.00 m sgi6.00 m s + 4.00 m s= 2.50 m svi =4(b)(c)gbgfb ga fb11223m 2.00 m s + m 4.00 m s 4 m 2.50 m s222.50 10 4 kg2=12.0 + 16.0 25.0 m s = 37.5 kJ2Wactor = K f K i =Wactora fbjaFIG. P9.19eg2The event considered here is the time reversal of the perfectly inelastic collision in theprevious problem. The same momentum conservation equation describes both processes.P9.20v1 , speed of m1 at B before collision.12m1 v1 = m1 gh2a fa fv1 = 2 9.80 5.00 = 9.90 m sv1 f , speed of m1 at B just after collision.m m21v1 f = 1v1 = 9.90 m s = 3.30 m sm1 + m 23At the highest point (after collision)a fm1 ghmax =a1m1 3.302fFIG. P9.20b3.30 m sg =2e9.80 m s j22hmax =20.556 mChapter 9P9.21259(a), (b) Let v g and v p be the velocity of the girl and the plankrelative to the ice surface. Then we may say that v g v p isthe velocity of the girl relative to the plank, so thatv g v p = 1.50(1)But also we must have m g v g + m p v p = 0 , since totalmomentum of the girl-plank system is zero relative to theice surface. Therefore45.0 v g + 150 v p = 0 , or v g = 3.33 v pPutting this into the equation (1) above givesFIG. P9.213.33 v p v p = 1.50 or v p = 0.346 m safThen v g = 3.33 0.346 = 1.15 m s*P9.22For the car-truck-driver-driver system, momentum is conserved:p 1i + p 2 i = p 1 f + p 2 f :b4 000 kg gb8 m sgi + b800 kg gb8 m sge ij = b4 800 kg gv ifvf =25 600 kg m s= 5.33 m s4 800 kgFor the driver of the truck, the impulse-momentum theorem isFt = p f pi :af bgbg bgbgF 0.120 s = 80 kg 5.33 m s i 80 kg 8 m s ie jF = 1.78 10 3 N i on the truck driverFor the driver of the car,af bgbg bgbge jF 0.120 s = 80 kg 5.33 m s i 80 kg 8 m s iF = 8.89 10 3 Ni on the car driver , 5 times larger.P9.23(a)According to the Example in the chapter text, the fraction of total kinetic energy transferredto the moderator isf2 =4m1 m 2bm1+ m2g2where m 2 is the moderator nucleus and in this case, m 2 = 12m1f2 =b4m1 12m1b13m g12g = 48 =1690. 284 or 28.4%of the neutron energy is transferred to the carbon nucleus.(b)a fejJj == a0.716 fe1.6 10K C = 0.284 1.6 10 13 J = 4.54 10 14 JKn131.15 10 13 J260P9.24Linear Momentum and CollisionsEnergy is conserved for the bob-Earth system between bottom andtop of swing. At the top the stiff rod is in compression and the bobnearly at rest.12Mv b + 0 = 0 + Mg 222v b = g 4 so v b = 2 gK i + Ui = K f + U f :FIG. P9.24Momentum of the bob-bullet system is conserved in the collision:mv = mP9.25v+M 2 g2ejv=4MmgAt impact, momentum of the clay-block system is conserved, so:bgmv1 = m1 + m 2 v 2After impact, the change in kinetic energy of the clay-block-surfacesystem is equal to the increase in internal energy:bbggbgge12m1 + m 2 v 2 = f f d = m1 + m 2 gd2120.112 kg v 2 = 0.650 0.112 kg 9.80 m s 2 7.50 m22v 2 = 95.6 m 2 s 2v 2 = 9.77 m se12.0 10P9.263bj bgbkg v1 = 0.112 kg 9.77 m sjagfFIG. P9.25v1 = 91.2 m sWe assume equal firing speeds v and equal forces F required for the two bullets to push wood fibersapart. These equal forces act backward on the two bullets.For the first,For the second,17.00 10 3 kg v 2 F 8.00 10 2 m = 027.00 10 3 kg v = 1.014 kg v feK i + Emech = K fpi = p fejj be7.00 10 jv=ejg3vf1.014bejK i + Emech = K f :Substituting for v f ,117.00 10 3 v7.00 10 3 kg v 2 Fd = 1.014 kg221.014ebje311 7.00 1032Fd = 7.00 10 v 221.014eSubstituting for v,eg117.00 10 3 kg v 2 Fd = 1.014 kg v 2f22Again,jjFGHFd = F 8.00 10 2 m 1 7.00 10 31.014IJKjgFGHIJK22v2d = 7.94 cmChapter 9*P9.27(a)261c ph = c ph , givesa4.0 + 10 + 3.0f kg v = b4.0 kggb5.0 m sg + b10 kg gb3.0 m sg + b3.0 kggb4.0 m sg .Using conservation of momentum,afterbeforeTherefore, v = +2.24 m s , or 2. 24 m s toward the right .(b)No . For example, if the 10-kg and 3.0-kg mass were to stick together first, they wouldmove with a speed given by solvingb13 kggv = b10 kg gb3.0 m sg + b3.0 kg gb4.0 m sg , or v11= +1.38 m s .Then when this 13 kg combined mass collides with the 4.0 kg mass, we haveb17 kggv = b13 kggb1.38 m sg + b4.0 kg gb5.0 m sg , and v = +2.24 m sjust as in part (a). Coupling order makes no difference.Section 9.4P9.28(a)Two-Dimensional CollisionsFirst, we conserve momentum for the system of two football players in the x direction (thedirection of travel of the fullback).b90.0 kg gb5.00 m sg + 0 = b185 kg gV coswhere is the angle between the direction of the final velocity V and the x axis. We findV cos = 2.43 m s(1)Now consider conservation of momentum of the system in the y direction (the direction oftravel of the opponent).b95.0 kg gb3.00 m sg + 0 = b185 kggaV sin fwhich gives,V sin = 1.54 m sDivide equation (2) by (1)tan =V = 2.88 m sThen, either (1) or (2) givesbbgbgb190.0 kg 5.00 m s21K f = 185 kg 2.88 m s2Ki =1.54= 0.6332.43 = 32.3From which(b)(2)ggbgb195.0 kg 3.00 m s22+2g2= 7.67 10 2 J= 1.55 10 3 JThus, the kinetic energy lost is 783 J into internal energy.262P9.29Linear Momentum and Collisionsp xf = p xibmvO cos 37.0+ mv Y cos 53.0 = m 5.00 m s0.799 vO + 0.602 v Y = 5.00 m sg(1)p yf = p yimvO sin 37.0 mv Y sin 53.0 = 00.602 vO = 0.799 v Y(2)Solving (1) and (2) simultaneously,vO = 3.99 m s and v Y = 3.01 m s .P9.30p xf = p xi :afmvO cos + mv Y cos 90.0 = mvivO cos + v Y sin = vip yf = p yi :FIG. P9.29a(1)fmvO sin mv Y sin 90.0 = 0vO sin = v Y cos (2)From equation (2),vO = v YFG cos IJH sin K(3)FIG. P9.30Substituting into equation (1),vYsoF cos I + vGH sin JK2eYsin = vijv Y cos 2 + sin 2 = vi sin , and v Y = vi sin .Then, from equation (3), vO = vi cos .We did not need to write down an equation expressing conservation of mechanical energy. In theproblem situation, the requirement of perpendicular final velocities is equivalent to the condition ofelasticity.Chapter 9P9.31The initial momentum of the system is 0. Thus,a1.20mfvandBib= m 10.0 m sgv Bi = 8.33 m sbb gagafbgFG eH11122m 10.0 m s + 1. 20m 8.33 m s = m 183 m 2 s 2222111 122K f = m vG + 1.20m v B =m 183 m 2 s 2222 2Ki =orfb g22vG + 1.20 v B = 91.7 m 2 s 2ejIJKj(1)From conservation of momentum,afmvG = 1.20m v BorvG = 1.20 v B(2)Solving (1) and (2) simultaneously, we findvG = 7.07 m s (speed of green puck after collision)andP9.32v B = 5.89 m s (speed of blue puck after collision)We use conservation of momentum for the system of two vehiclesfor both northward and eastward components.For the eastward direction:bgM 13.0 m s = 2 MV f cos 55.0For the northward direction:Mv 2i = 2 MV f sin 55.0Divide the northward equation by the eastward equation to find:bgv 2 i = 13.0 m s tan 55.0 = 18.6 m s = 41.5 mi hThus, the driver of the north bound car was untruthful.FIG. P9.32263264P9.33Linear Momentum and CollisionsBy conservation of momentum for the system of the two billiardballs (with all masses equal),bg5.00 m s + 0 = 4.33 m s cos 30.0+ v 2 fxv 2 fx = 1.25 m sbg0 = 4.33 m s sin 30.0+ v 2 fyv 2 fy = 2.16 m sv 2 f = 2.50 m s at 60.0FIG. P9.33Note that we did not need to use the fact that the collision is perfectly elastic.(a)sopi = p fp xi = p xfandP9.34p yi = p yfmvi = mv cos + mv cos 0 = mv sin + mv sin From (2),(1)(2)sin = sin = Furthermore, energy conservation for the systemof two protons requires111mvi2 = mv 2 + mv 2222viv=so2so(b)Hence, (1) gives vi =b2 vi cos gFIG. P9.34 = 45.02 = 45.0a fv = e3.00 i 1.20 jj m sP9.35m1 v 1i + m 2 v 2i = m1 + m 2 v f :3.00 5.00 i 6.00 j = 5.00 vP9.36x-component of momentum for the system of the two objects:p1ix + p 2ix = p1 fx + p 2 fx : mvi + 3mvi = 0 + 3mv 2 xy-component of momentum of the system:0 + 0 = mv1 y + 3 mv 2 yby conservation of energy of the system:+also1111222mvi2 + 3mvi2 = mv1 y + 3m v 2 x + v 2 y22222vv2x = i3v1 y = 3 v 2 ySo the energy equation becomes24vi2 = 9 v 2 y +we haveorcontinued on next page8 vi22= 12 v 2 y32 viv2y =3e4vi22+ 3v2 y3jChapter 9(a)v1 y = 3 v 2 y =The object of mass m has final speed2 vi22v2x + v2y =and the object of mass 3 m moves at22v2x + v2y =(b) = tan 1Fv IGH v JK2y = tan 12xFGHm 0 = 17.0 10 27 kgIJK2 vi 3= 35.33 2 viv 1 = 6.00 10 6 j m sm 2 = 8.40 10 27 kg2vi3v i = 0 (the parent nucleus)m1 = 5.00 10 27 kgP9.374vi2 2 vi2+99v 2 = 4.00 10 6 i m s(a)m1 v 1 + m 2 v 2 + m 3 v 3 = 0where m 3 = m 0 m1 m 2 = 3.60 10 27 kgev3 =(b)jej ejej ejFIG. P9.375.00 10 27 6.00 10 6 j + 8.40 10 27 4.00 10 6 i + 3.60 10 27 v 3 = 0e9.33 10 i 8.33 10 jj m s6111222m 1 v1 + m 2 v 2 + m 3 v 322215.00 10 27 6.00 10 6E=26E=LMeNjej + e8.40 10 je4.00 10 j + e3.60 10 je12.5 10 j OQP2276 2276 2E = 4.39 10 13 JSection 9.5P9.38The Center of MassThe x-coordinate of the center of mass isx CM = m i xi mi=b0+0+0+02.00 kg + 3.00 kg + 2.50 kg + 4.00 kggxCM = 0and the y-coordinate of the center of mass isyCM = m i yi miyCM = 1.00 m=b2.00 kgga3.00 mf + b3.00 kg ga2.50 mf + b2.50 kg ga0f + b4.00 kg ga0.500 mf2.00 kg + 3.00 kg + 2.50 kg + 4.00 kg265266P9.39Linear Momentum and CollisionsTake x-axis starting from the oxygen nucleus and pointing toward themiddle of the V.yCM = 0Thenx CM =andx CM = mi x i mia=faf0 + 1.008 u 0.100 nm cos 53.0+1.008 u 0.100 nm cos 53.015.999 u + 1.008 u + 1.008 uFIG. P9.39xCM = 0.006 73 nm from the oxygen nucleus*P9.40Let the x axis start at the Earths center and point toward the Moon.ej24228m1 x1 + m 2 x 2 5.98 10 kg 0 + 7.36 10 kg 3.84 10 m=m1 + m 26.05 10 24 kgx CM == 4.67 10 6 m from the Earths centerThe center of mass is within the Earth, which has radius 6.37 10 6 m.P9.41Let A1 represent the area of the bottom row of squares, A 2the middle square, and A3 the top pair.A = A1 + A 2 + A 3M = M1 + M 2 + M 3M1 M=A1AA1 = 300 cm 2 , A 2 = 100 cm 2 , A3 = 200 cm 2 , A = 600 cm 2A300 cm 2MM1 = M 1 =M=2A2600 cmM2M3x CMx CMFG IJH KF A IJ = 100 cm= MGH A K 600 cmF A IJ = 200 cm= MGH A K 600 cm2222M=M6FIG. P9.41M3x M + x 2 M 2 + x 3 M 3 15.0 cm= 1 1=M= 11.7 cmyCM =312af2M=c M h + 5.00 cmc Mh + 10.0 cmc MhayCM = 13.3 cm16Mf c Mha25.0 cmf = 13.3 cmM 5.00 cm + 1 M 15.0 cm +6M121313Chapter 9*P9.42(a)267Represent the height of a particle of mass dm within the object as y. Its contribution to thegravitational energy of the object-Earth system is dm gy . The total gravitational energy is1Ug =gy dm = g y dm . For the center of mass we have yCM =y dm , so U g = gMyCM .Mall massz(b)a fzaefaf13.6 m 15.7 m 64.8 m = 1.83 10 3 m 3 . Its mass is21.83 10 3 m 3 = 6.96 10 6 kg . Its center of mass is above its base by one-The volume of the ramp isV = 3 800 kg m3fazjej1third of its height, yCM = 15.7 m = 5.23 m . Then3U g = MgyCM = 6.96 10 6 kg 9.8 m s 2 5.23 m = 3.57 10 8 J .eM=(a)zz0.300 m0 .300 m0P9.43j0dx =50.0 g m + 20.0 x g m 2 dxM = 50.0 x g m + 10.0 x 2 g m 2x CM =(b)x CM*P9.44zxdmall massM=LMNM1Mz0.300 m xdx =00.300 m01Mz= 15.9 g0.300 m50.0 x g m + 20.0 x 2 g m 2 dx0320 x g m 2125.0 x 2 g m +=15.9 g3OPQP0.300 m= 0.153 m0Take the origin at the center of curvature. We have L =r=2L12r ,4. An incremental bit of the rod at angle from the x axis hasdm MMr=, dm =d where we have used therd LLdefinition of radian measure. Nowmass given byyCM ==zz135 11Mrr2y dm =r sin d =M all massM = 45 LLFG 2L IJ 1 a cos fHK L213545 =4L22L4 2L2=F1 2 2 I L =GH JK2zx135 sin dFIG. P9.4445 FG 1 + 1 IJ = 4 2LH 2 2K The top of the bar is above the origin by r =byy0.063 5L .22L, so the center of mass is below the middle of the bar268Linear Momentum and CollisionsSection 9.6P9.45Motion of a System of Particles(a)v CM ==v CM = mi v iM=m1 v 1 + m 2 v 2Mb2.00 kg ge2.00i m s 3.00 j m sj + b3.00 kg ge1.00i m s + 6.00 j m sj5.00 kge1.40i + 2.40 jj m sbgeje7.00i + 12.0 jj kg m s(b)(a)See figure to the right.(b)P9.46p = Mv CM = 5.00 kg 1. 40 i + 2.40 j m s =Using the definition of the position vector at the center of mass,rCM =rCM =b2.00 kg ga1.00 m, 2.00 mf + b3.00 kg ga4.00 m, 3.00 mfrCM =(c)2.00 kg + 3.00 kge2.00i 1.00 jj mFIG. P9.46The velocity of the center of mass isv CM =v CM =(d)m1 r1 + m 2 r2m1 + m 2bgbgbgbe3.00i 1.00 jj m sThe total linear momentum of the system can be calculated as P = Mv CMor asP = m1 v 1 + m 2 v 2Either givesP9.47g b2.00 kg 3.00 m s , 0.50 m s + 3.00 kg 3.00 m s , 2.00 m sP m1 v 1 + m 2 v 2==Mm1 + m 22.00 kg + 3.00 kgP=e15.0 i 5.00 jj kg m sLet x = distance from shore to center of boat= length of boatx = distance boat moves as Juliet moves toward RomeoThe center of mass stays fixed.Before:h + M cx + hdM + M + M iM ax x f + M cx + x h + M c x + x h=xdM + M + M iFG 55.0 + 77.0 IJ = x a80.0 55.0 77.0f + a55.0 + 77.0fH 2 2K255.0a 2.70f55.0x === 0.700 mx CM =cMb x + M J x BAfter:B2RJRJ2R2CMB212212JR2FIG. P9.47gChapter 9P9.48(a)269Conservation of momentum for the two-ball system gives us:bgbg0.200 kg 1.50 m s + 0.300 kg 0.400 m s = 0.200 kg v1 f + 0.300 kg v 2 fRelative velocity equation:v 2 f v1 f = 1.90 m sd0.300 0.120 = 0.200 v1 f + 0.300 1.90 + v1 fThenv1 f = 0.780 m sv 2 f = 1.12 m sv 1 f = 0.780 i m s(b)v CM =Before,iv 2 f = 1.12 i m sb0.200 kg gb1.50 m sgi + b0.300 kg gb0.400 m sgibg0.500 kgv CM = 0.360 m s iAfterwards, the center of mass must move at the same velocity, as momentum of the systemis conserved.Section 9.7Rocket PropulsiondMdtejejThrust = 2.60 10 3 m s 1.50 10 4 kg s = 3.90 10 7 N(a)Thrust = v e(b)P9.49 Fy = Thrust Mg = Ma :ja f eej3.90 10 7 3.00 10 6 9.80 = 3.00 10 6 aa = 3.20 m s 2*P9.50(a)dM 12.7 g== 6.68 10 3 kg sdt1.90 sThe fuel burns at a rateThrust = v edM:dte5.26 N = v e 6.68 10 3 kg sjv e = 787 m s(b)F M I:GH M JKv f vi = v e lnggg FGH 53.553.+5 25.+5 25.512.7 g IJKggbiv f 0 = 797 m s lnfv f = 138 m sP9.51v = v e lnMiMf(a)Mi = e v v e M fMi = e 5 3.00 10 3 kg = 4.45 10 5 kgThe mass of fuel and oxidizer isM = Mi M f = 445 3.00 10 3 kg = 442 metric tons(b)aeajffM = e 2 3.00 metric tons 3.00 metric tons = 19.2 metric tonsBecause of the exponential, a relatively small increase in fuel and/or engine efficiency causesa large change in the amount of fuel and oxidizer required.F M I = v lnF M IGH M JKGH M JKF M kt IJ = v lnFG 1 k tIJ= M kt , so v = v lnGH M KH M KFrom Equation 9.41, v 0 = v e lniNow, M ffiiefeieiiMWith the definition, Tp i , this becomeskFGHIJKF t IJ= 144 s , v = b1 500 m sg lnG 1 H 144 s Kafv t = v e ln 1 With v e = 1 500 m s, and Tpa f vbm sgv (m/s)ts02244888081323730FHdtpIK OP F 1Q = v GGH 1 etTpIF 1 I F v IF 1JJ GH T JK = GH T JK GG 1 KHepptTpIJJ , orKveTp t1 500 m s144 s ta f aem s j210.42012.14014.46017.98023.410034.112062.5132125continued on next page140120100806040200t (s)FIG. P9.52(d)14002a (m/s )120ts100With v e = 1 500 m s, and Tp = 144 s , a =80(d)LMNd v e ln 1 Tt60afat =FIG. P9.52(b)40afdvat ==dtt (s)1402690120120(c)122017804080100100602500200015001000500080406002040003500300020(b)tTp20(a)0P9.52Linear Momentum and Collisions0270Chapter 9(e)LM F t I OPL t OF dt I v lnG 1 J dt = v T z ln M1 PG Jz z MN H T K PQMN T PQH T KLF t I F t I F t I Oxat f = v T MG 1 J lnG 1 J G 1 J PMNH T K H T K H T K PQF tIxat f = v eT t j lnG 1 J + v tH TKtafx t = 0 + vdt =0ttee pp00ppte peppp0epWith v e = 1 500 m s = 1.50 km s , and Tp = 144 s ,f FGHIJKt+ 1.50t144a f xakmfts02.199.2322.142.271.7120115132153t (s)FIG. P9.52(f)*P9.53The thrust acting on the spacecraft is F = b3 500 kg ge2.50 10 6 je9.80 F = ma :thrust =FG dM IJ v :H dt Ke8.58 10 2 N =M = 4.41 kgF M I b70 m sgGH 3 600 s JKjm s 2 = 8.58 10 2 N14010012080100806040200100608040140120602016000x (km)40ax = 1.50 144 t ln 1 20(f)p271272Linear Momentum and CollisionsAdditional ProblemsP9.54(a)When the spring is fully compressed, each cart moves with same velocity v. Applyconservation of momentum for the system of two glidersbpi = p f :(b)gv=m1 v 1 + m 2 v 2 = m1 + m 2 vm1 v 1 + m 2 v 2m1 + m 2bg1111 222m1 v1 + m 2 v 2 = m1 + m 2 v 2 + kx m2222Only conservative forces act, therefore E = 0 .Substitute for v from (a) and solve for x m .2xmbm=xm =(c)1gbgb22+ m 2 m 1 v1 + m 1 + m 2 m 2 v 2 m 1 v1bk m1 + m 2e22m 1 m 2 v1 + v 2 2 v 1 v 2bk m1 + m 2gj = bv1 v2gg bm v g22 22 2m 1 m 2 v1 v 2mg kbm mm g+1212m1 v 1 + m 2 v 2 = m1 v 1 f + m 2 v 2 fdidiConservation of momentum:m1 v 1 v 1 f = m 2 v 2 f v 2Conservation of energy:11112222m 1 v1 + m 2 v 2 = m 1 v1 f + m 2 v 2 f2222which simplifies to:Factoring givesem dvj eji dv + v i = m dv(1)2222m 1 v 1 v1 f = m 2 v 2 f v 211 v1 f11f22fid v2 v2 f + v2iand with the use of the momentum equation (equation (1)),this reduces todvorv1 f = v 2 f + v 2 v11i d+ v1 f = v 2 f + v 2i(2)Substituting equation (2) into equation (1) and simplifying yields:v2 f =FG 2m IJ v + FG mHm +m K Hm1112IJK m1v2+ m212Upon substitution of this expression for v 2 f into equation 2, one findsv1 f =FG mHmIJKFGHIJK m22m 2v1 +v2m1 + m 21 + m21Observe that these results are the same as Equations 9.20 and 9.21, which should have beenexpected since this is a perfectly elastic collision in one dimension.P9.55(a)b60.0 kg g4.00 m s = a120 + 60.0f kgvChapter 9273fv f = 1.33 m s i(b)b Fy = 0 :gn 60.0 kg 9.80 m s 2 = 0aff k = k n = 0.400 588 N = 235 NFIG. P9.55fk = 235 N i(c)For the person, pi + I = p fmvi + Ft = mv fb60.0 kg g4.00 m s a235 Nft = b60.0 kg g1.33 m st = 0.680 safperson:mv f mv i = 60.0 kg 1.33 4.00 m s = 160 N sicart:(d)120 kg 1.33 m s 0 = +160 N sibgaf(e)x f xi =114.00 + 1.33 m s 0.680 s = 1.81 mvi + v f t =22(f)x f xi =11vi + v f t = 0 + 1.33 m s 0.680 s = 0.454 m22(g)111mv 2 mvi2 = 60.0 kg 1.33 m sf222(h)111mv 2 mvi2 = 120.0 kg 1.33 m sf222(i)ddibigbbg2g2b160.0 kg 4.00 m s2g2= 427 J 0 = 107 JThe force exerted by the person on the cart must equal in magnitude and opposite indirection to the force exerted by the cart on the person. The changes in momentum ofthe two objects must be equal in magnitude and must add to zero. Their changes inkinetic energy are different in magnitude and do not add to zero. The followingrepresent two ways of thinking about why. The distance the cart moves is differentfrom the distance moved by the point of application of the friction force to the cart.The total change in mechanical energy for both objects together, 320 J, becomes+320 J of additional internal energy in this perfectly inelastic collision.P9.56The equation for the horizontal range of a projectile is R =vi2 sin 2. Thus, with = 45.0 , the initialgvelocity isvi = Rg =a fa200 mfe9.80 m s j = 44.3 m s2I = F t = p = mvi 0Therefore, the magnitude of the average force acting on the ball during the impact is:F=ejbg46.0 10 3 kg 44.3 m smvi== 291 N .t7.00 10 3 s274P9.57Linear Momentum and CollisionsWe hope the momentum of the wrench provides enough recoil so that the astronaut can reach theship before he loses life support! We might expect the elapsed time to be on the order of severalminutes based on the description of the situation.No external force acts on the system (astronaut plus wrench), so the total momentum is constant.Since the final momentum (wrench plus astronaut) must be zero, we have final momentum = initialmomentum = 0.m wrench v wrench + m astronaut v astronaut = 0bgbg0.500 kg 20.0 m sm wrench v wrench== 0.125 m sm astronaut80.0 kgAt this speed, the time to travel to the ship isThus v astronaut = t=30.0 m= 240 s = 4.00 minutes0.125 m sThe astronaut is fortunate that the wrench gave him sufficient momentum to return to the ship in areasonable amount of time! In this problem, we were told that the astronaut was not drifting awayfrom the ship when he threw the wrench. However, this is not quite possible since he did notencounter an external force that would reduce his velocity away from the ship (there is no airfriction beyond earths atmosphere). If this were a real-life situation, the astronaut would have tothrow the wrench hard enough to overcome his momentum caused by his original push away fromthe ship.P9.58Using conservation of momentum from just before to justafter the impact of the bullet with the block:afmmvi = M + m v forvi =FG M + m IJ vH m Kfvi(1)MhThe speed of the block and embedded bullet just afterimpact may be found using kinematic equations:d = v f t and h =Thus, t =d1 2gt2g2hd=and v f = = dgt2hFIG. P9.58gd 22hSubstituting into (1) from above gives vi =FG M + m IJH m Kgd 2.2hChapter 9*P9.59(a)275Conservation of momentum:ejej0.5 kg 2 i 3 j + 1k m s + 1.5 kg 1i + 2 j 3k m sej= 0.5 kg 1i + 3 j 8k m s + 1.5 kg v 2 fv2 f =e0.5 i + 1.5 j 4kj kg m s + e0.5i 1.5 j + 4kj kg m s =1.5 kg0The original kinetic energy is110.5 kg 2 2 + 3 2 + 1 2 m 2 s 2 + 1.5 kg 1 2 + 2 2 + 3 2 m 2 s 2 = 14.0 J22ejej10.5 kg 1 2 + 3 2 + 8 2 m 2 s 2 + 0 = 18.5 J different from the original2energy so the collision is inelastic .eThe final kinetic energy is(b)jWe follow the same steps as in part (a):e0.5 i + 1.5 j 4kj kg m s = 0.5 kge0.25i + 0.75 j 2kj m s + 1.5 kg ve0.5i + 1.5 j 4kj kg m s + e0.125 i 0.375 j + 1kj kg m sv =2f2f1.5 kge0.250i + 0.750 j 2.00kj m s=We see v 2 f = v 1 f , so the collision is perfectly inelastic .(c)Conservation of momentum:e0.5 i + 1.5 j 4kj kg m s = 0.5 kge1i + 3 j + akj m s + 1.5 kg ve0.5 i + 1.5 j 4kj kg m s + e0.5i 1.5 j 0.5akj kg m sv =2f2f1.5 kg=a2.67 0.333 afk m sConservation of energy:a110.5 kg 1 2 + 3 2 + a 2 m 2 s 2 + 1.5 kg 2.67 + 0.333 a222= 2.5 J + 0. 25 a + 5.33 J + 1.33 a + 0.083 3 a 214.0 J =ejf2m2 s 20 = 0.333 a 2 + 1.33 a 6.167a=afa1.33 1.33 2 4 0.333 6.167f0.667a = 2.74 or 6.74. Either value is possible. a = 2.74 ,a fhc= c2.67 0.333a 6.74fhk m s =v 2 f = 2.67 0.333 2.74 k m s = 3.58k m s a = 6.74 , v 2 f0.419 k m s276P9.60Linear Momentum and Collisions(a)The initial momentum of the system is zero, whichremains constant throughout the motion.Therefore, when m1 leaves the wedge, we musthavem 2 v wedge + m1 v block = 0orso(b)b3.00 kg gvv wedge = 0.667 m swedgebgbg+ 0.500 kg +4.00 m s = 0v wedge+xUsing conservation of energy for the block-wedgeEarth system as the block slides down the smooth(frictionless) wedge, we haveFIG. P9.60K block + U system + K wedge = K block + U systemior*P9.61(a)0 + m1 gh + 0 =ifLM 1 m a4.00f + 0OP + 1 m a0.667fN2Q 21222+ K wedgefwhich gives h = 0.952 m .Conservation of the x component of momentum for the cart-bucket-water system:bgmvi + 0 = m + V v(b)v block = 4.00 m/svi =m + VvmRaindrops with zero x-component of momentum stop in the bucket and slow its horizontalmotion. When they drip out, they carry with them horizontal momentum. Thus the cartslows with constant acceleration.Chapter 9P9.62Consider the motion of the firefighter during the threeintervals:(1) before, (2) during, and (3) after collision with theplatform.(a)v1While falling a height of 4.00 m, his speed changesfrom vi = 0 to v1 as found fromi bdv2gE = K f + U f K i U i , orK f = E U f + K i + U iWhen the initial position of the platform is taken asthe zero level of gravitational potential, we havea f12mv1 = fh cos 180 0 + 0 + mgh2FIG. P9.62Solving for v1 givesv1 =(b)b2 fh + mghmg = 2c300a4.00f + 75.0a9.80f4.00h =6.81 m s75.0During the inelastic collision, momentum is conserved; and if v 2 is the speed of thefirefighter and platform just after collision, we have mv1 = m + M v 2 orav2 =fa f75.0 6.81m 1 v1== 5.38 m sm + M 75.0 + 20.0Following the collision and again solving for the work done by non-conservative forces,using the distances as labeled in the figure, we have (with the zero level of gravitationalpotential at the initial position of the platform):E = K f + U fg + U fs K i U ig U is , or11 fs = 0 + m + M g s + ks 2 m + M v 2 0 022afa fafThis results in a quadratic equation in s:a f2 000s 2 931 s + 300s 1 375 = 0 or s = 1.00 m277278*P9.63Linear Momentum and Collisions(a)Each object swings down according tomgR =12mv12MgR =12Mv12av1 = 2 gRfThe collision: mv1 + Mv1 = + m + M v 2Mmv2 =v1M+mSwinging up:af afaf2 gRa1 cos 35fa M + mf = a M m f 2 gRfa12M + m v 2 = M + m gR 1 cos 352v 2 = 2 gR 1 cos 350.425 M + 0.425m = M m1.425m = 0.575 Mm= 0.403M(b)P9.64No change is required if the force is different. The nature of the forces within the system ofcolliding objects does not affect the total momentum of the system. With strong magneticattraction, the heavier object will be moving somewhat faster and the lighter object fasterstill. Their extra kinetic energy will all be immediately converted into extra internal energywhen the objects latch together. Momentum conservation guarantees that none of the extrakinetic energy remains after the objects join to make them swing higher.(a)Use conservation of the horizontal component ofmomentum for the system of the shell, the cannon,and the carriage, from just before to just after thecannon firing.p xf = p xi :or(b)m shell v shell cos 45.0+ m cannon v recoil = 0a200fa125f cos 45.0+b5 000gvrecoil=0v recoil = 3.54 m sFIG. P9.64Use conservation of energy for the system of the cannon, the carriage, and the spring fromright after the cannon is fired to the instant when the cannon comes to rest.K f + U gf + U sf = K i + U gi + U si :0+0+x max =1 212kx max = mv recoil + 0 + 0222mv recoil=keb5 000ga3.54f22.00 10 4jam = 1.77 mfFs, max = 2.00 10 4 N m 1.77 m = 3.54 10 4 N(c)Fs, max = kx max(d)No. The rail exerts a vertical external force (the normal force) on the cannon and prevents itfrom recoiling vertically. Momentum is not conserved in the vertical direction. The springdoes not have time to stretch during the cannon firing. Thus, no external horizontal force isexerted on the system (cannon, carriage, and shell) from just before to just after firing.Momentum of this system is conserved in the horizontal direction during this interval.Chapter 9P9.65(a)Utilizing conservation ofmomentum,bv1igm + m2= 1m12792 ghm 1 v1 A = m 1 + m 2 v Bv1 Ayv1 A 6.29 m s(b)xUtilizing the two equations,FIG. P9.651 2gt = y and x = v1 A t2we combine them to findv1 A =x2ygFrom the data, v1 A = 6.16 m sMost of the 2% difference between the values for speed is accounted for by the uncertainty0.01 0.1110.1in the data, estimated as++++= 1.1% .8.68 68.8 263 257 85.3*P9.66The ice cubes leave the track with speed determined by mgyi =e1mv 2 ;2jv = 2 9.8 m s 2 1.5 m = 5.42 m s .Its speed at the apex of its trajectory is 5.42 m s cos 40 = 4.15 m s . For its collision with the wall wehavemvi + Ft = mv fFGH0.005 kg 4.15 m s + Ft = 0.005 kg IJK14.15 m s2Ft = 3.12 10 2 kg m sThe impulse exerted by the cube on the wall is to the right, +3.12 10 2 kg m s. Here F could referto a large force over a short contact time. It can also refer to the average force if we interpret t as1s, the time between one cubes tap and the nexts.10Fav =3.12 10 2 kg m s= 0.312 N to the right0.1 s280P9.67Linear Momentum and Collisions(a)Find the speed when the bullet emerges from theblock by using momentum conservation:400 m/smvi = MVi + mvThe block moves a distance of 5.00 cm. Assume foran approximation that the block quickly reaches itsmaximum velocity, Vi , and the bullet kept goingwith a constant velocity, v. The block thencompresses the spring and stops.FIG. P9.6711MVi2 = kx 222Vi =v=v5.00 cmb900 N mge5.00 10j2m2= 1.50 m s1.00 kgjbeg bgb5.00 10 3 kg 400 m s 1.00 kg 1.50 m smvi MVi=m5.00 10 3 kggv = 100 m s(b)jbmj15.00 10 3 kg 100 m s2e1+ b900 N mge5.00 102 E = K + U =2g2jb15.00 10 3 kg 400 m s2eg22E = 374 J , or there is an energy loss of 374 J .*P9.68The orbital speed of the Earth isvE =SCM2r 2 1.496 10 11 m== 2.98 10 4 m sT3.156 10 7 sEIn six months the Earth reverses its direction, to undergomomentum changeejeFIG. P9.68jm E v E = 2m E v E = 2 5.98 10 24 kg 2.98 10 4 m s = 3.56 10 25 kg m s .Relative to the center of mass, the sun always has momentum of the same magnitude in theopposite direction. Its 6-month momentum change is the same size, mS v S = 3.56 10 25 kg m s .Then v S =3.56 10 25 kg m s1.991 10 30 kg= 0.179 m s .Chapter 9P9.69(a)b3.00 kg gb7.00 m sg j + e12.0 Nija5.00 sf = b3.00 kg gvv = e 20.0 i + 7.00 jj m sp i + Ft = p f :ffv f vi(b)a=(c)a=(d)r = v i t +ta=:F :a=m1 2at :2e20.0 i + 7.00 j 7.00 jj m s =5.00 s12.0 N i= 4.00 i m s 23.00 kg(f)(g)P9.70jaef 1 e4.00 m s ija5.00 sf22r = 7.00 m s j 5.00 s +e50.0i + 35.0 jj mr =(e)4.00 i m s 2eje2jW = 12.0 N i 50.0 m i + 35.0 m j = 600 JW = F r :bge11mv 2 = 3.00 kg 20.0 i + 7.00 j 20.0 i + 7.00 j m 2 s 2f221mv 2 = 1.50 kg 449 m 2 s 2 = 674 Jf2bjebgb11mvi2 + W = 3.00 kg 7.00 m s22g2jgej+ 600 J = 674 JbgM = 360 kg 2.50 kg s t .We find the mass frombgbg1 500 m s 2.50 kg s3 750 NThrust v e dM dt===MMMMWe find the velocity and position according to Euler,v new = v old + a tfromx new = x old + v tandIf we take t = 0.132 s , a portion of the output looks like this:We find the acceleration froma=a fa fTimet(s)Total mass(kg)Accelerationa m s2Speed, v(m/s)Positionx(m)0.0000.1320.264...65.86866.00066.132...131.736131.868132.000360.00359.67359.3410.416710.426210.43580.00001.37502.75130.00000.18150.54467195.330195.000194.67019.198319.230819.2634916.54919.08921.6127191273122743330.66030.33030.000122.3092123.6400125.00003687.33703.53719.8152382152871153362(a)The final speed is(b)The rocket travelsejv f = 3.7 km s153 km281282P9.71Linear Momentum and CollisionsThe force exerted by the table is equal to the change in momentumof each of the links in the chain.By the calculus chain rule of derivatives,F1 =a fdp d mvdmdv==v+m .dtdtdtdtWe choose to account for the change in momentum of each link byhaving it pass from our area of interest just before it hits the table,so thatvFIG. P9.71dmdv 0 and m= 0.dtdtSince the mass per unit length is uniform, we can express each link of length dx as having a mass dm:dm =Mdx .LThe magnitude of the force on the falling chain is the force that will be necessary to stop each of theelements dm.F1 = vFG IJH KFG IJH KdmM dxM 2=v=vdtL dtLAfter falling a distance x, the square of the velocity of each link v 2 = 2 gx (from kinematics), henceF1 =2 Mgx.LThe links already on the table have a total length x, and their weight is supported by a force F2 :F2 =Mgx.LHence, the total force on the chain isFtotal = F1 + F2 =3 Mgx.LThat is, the total force is three times the weight of the chain on the table at that instant.Chapter 9P9.72A picture one second later differs by showing five extra kilograms of sand moving on the belt.bgbg(a)5.00 kg 0.750 m sp x== 3.75 Nt1.00 s(b)The only horizontal force on the sand is belt friction,p xi + ft = p xfso from(c)p x= 3.75 Ntthis isf=andFext = 3.75 NThe belt is in equilibrium: Fx = ma x :+ Fext f = 0af(d)W = Fr cos = 3.75 N 0.750 m cos 0 = 2.81 J(e)11m v 2 = 5.00 kg 0.750 m s22a f(f)*P9.73283x CM =bg2= 1.41 JFriction between sand and belt converts half of the input work into extra internal energy. m i xi mi=cm1 R +2h + m a0 f =2cm1 R +m1 + m 22hym1 + m 2xR2FIG. P9.73ANSWERS TO EVEN PROBLEMSP9.2(a) 0; (b) 1.06 kg m s ; upwardP9.200.556 mP9.4(a) 6.00 m s to the left; (b) 8.40 JP9.221.78 kN on the truck driver; 8.89 kN in theopposite direction on the car driverP9.6The force is 6.44 kNP9.24v=4MmgP9.81.39 kg m s upwardP9.10(a) 5.40 N s toward the net; (b) 27.0 JP9.267.94 cmP9.12~ 10 3 N upwardP9.28(a) 2.88 m s at 32.3; (b) 783 J becomesinternal energyP9.14(a) and (c) see the solution; (b) small;(d) large; (e) no differenceP9.30v Y = vi sin ; vO = vi cos P9.161.67 m sP9.32No; his speed was 41.5 mi hP9.18(a) 2.50 m s ; (b) 3.75 10 4 JP9.34(a) v =vi2; (b) 45.0 and 45.0284Linear Momentum and Collisions2vi ; (b) 35.33P9.36(a)P9.38P9.404.67 10 6 m from the Earths centerP9.42(c) v 1 f =a0, 1.00 mf(a) see the solution; (b) 3.57 10 8 J2vi ;v2 f =FG mHmIJKFGH1 m1v21 + m2FG 2m IJ v + FG mHm +m K Hm11IJK m22m 2v1 +v2 ;m1 + m 21 + m2122IJKP9.60(a) 0.667 m s; (b) 0.952 mP9.62(a) 6.81 m s; (b) 1.00 mP9.64(a) 3.54 m s ; (b) 1.77 m; (c) 35.4 kN;(d) No. The rails exert a vertical force tochange the momentumP9.660.312 N to the rightP9.680.179 m sP9.70(a) 3.7 km s ; (b) 153 kmP9.72(a) see the solution;(b) 2.00 m, 1.00 m ;FG M + m IJH m K(a) 3.75 N to the right; (b) 3.75 N to theright; (c) 3.75 N; (d) 2.81 J; (e) 1.41 J;(f) Friction between sand and belt convertshalf of the input work into extra internalenergy.0.063 5LP9.46291 NP9.58P9.44P9.56af(c) e3.00 i 1.00 jj m s ;(d) e15.0 i 5.00 jj kg m sP9.48(a) 0.780 i m s ; 1.12 i m s; (b) 0.360 i m sP9.50(a) 787 m s; (b) 138 m sP9.52see the solutionP9.54(a)m1 v 1 + m 2 v 2;m1 + m 2b(b) v1 v 2mg kbm mm g ;+1122gd 22h10Rotation of a Rigid ObjectAbout a Fixed AxisANSWERS TO QUESTIONSCHAPTER OUTLINE10.110.210.310.410.510.610.710.810.9Angular Position, Velocity,and AccelerationRotational Kinematics:Rotational Motion withConstant AngularAccelerationAngular and LinearQuantitiesRotational EnergyCalculation of Moments ofInertiaTorqueRelationship BetweenTorque and AngularAccelerationWork, Power, and Energyin Rotational MotionRolling Motion of a RigidObjectQ10.11 rev/min, orrad/s. Into the wall (clockwise rotation). = 0.30FIG. Q10.1Q10.2+k , kQ10.3Yes, they are valid provided that is measured in degrees persecond and is measured in degrees per second-squared.Q10.4The speedometer will be inaccurate. The speedometer measures the number of revolutions persecond of the tires. A larger tire will travel more distance in one full revolution as 2r .Q10.5Smallest I is about x axis and largest I is about y axis.Q10.6ML2if the mass was nonuniformly distributed, nor12could it be calculated if the mass distribution was not known.The moment of inertia would no longer beQ10.7The object will start to rotate if the two forces act along different lines. Then the torques of the forceswill not be equal in magnitude and opposite in direction.Q10.8No horizontal force acts on the pencil, so its center of mass moves straight down.Q10.9You could measure the time that it takes the hanging object, m, to fall a measured distance afterbeing released from rest. Using this information, the linear acceleration of the mass can becalculated, and then the torque on the rotating object and its angular acceleration.Q10.10You could use = t and v = at . The equation v = R is valid in this situation since a = R .Q10.11The angular speed would decrease. The center of mass is farther from the pivot, but the momentof inertia increases also.285286Rotation of a Rigid Object About a Fixed AxisQ10.12The moment of inertia depends on the distribution of mass with respect to a given axis. If the axis ischanged, then each bit of mass that makes up the object is a different distance from the axis. Inexample 10.6 in the text, the moment of inertia of a uniform rigid rod about an axis perpendicular tothe rod and passing through the center of mass is derived. If you spin a pencil back and forth aboutthis axis, you will get a feeling for its stubbornness against changing rotation. Now change the axisabout which you rotate it by spinning it back and forth about the axis that goes down the middle ofthe graphite. Easier, isnt it? The moment of inertia about the graphite is much smaller, as the massof the pencil is concentrated near this axis.Q10.13Compared to an axis through the center of mass, any other parallel axis will have larger averagesquared distance from the axis to the particles of which the object is composed.Q10.14A quick flip will set the hardboiled egg spinning faster and more smoothly. The raw egg losesmechanical energy to internal fluid friction.Q10.15I CM = MR 2 , I CM = MR 2 , I CM =Q10.16Yes. If you drop an object, it will gain translational kinetic energy from decreasing gravitationalpotential energy.Q10.17No, just as an object need not be moving to have mass.Q10.18No, only if its angular momentum changes.Q10.19Yes. Consider a pendulum at its greatest excursion from equilibrium. It is momentarily at rest, butmust have an angular acceleration or it would not oscillate.Q10.20Since the source reel stops almost instantly when the tape stops playing, the friction on the sourcereel axle must be fairly large. Since the source reel appears to us to rotate at almost constant angularvelocity, the angular acceleration must be very small. Therefore, the torque on the source reel due tothe tension in the tape must almost exactly balance the frictional torque. In turn, the frictional torqueis nearly constant because kinetic friction forces dont depend on velocity, and the radius of the axlewhere the friction is applied is constant. Thus we conclude that the torque exerted by the tape onthe source reel is essentially constant in time as the tape plays.vmust increase to keep theAs the source reel radius R shrinks, the reels angular speed =Rtape speed v constant. But the biggest change is to the reels moment of inertia. We model the reel asa roll of tape, ignoring any spool or platter carrying the tape. If we think of the roll of tape as a1uniform disk, then its moment of inertia is I = MR 2 . But the rolls mass is proportional to its base2area R 2 . Thus, on the whole the moment of inertia is proportional to R 4 . The moment of inertiadecreases very rapidly as the reel shrinks!The tension in the tape coming into the read-and-write heads is normally dominated bybalancing frictional torque on the source reel, according to TR friction . Therefore, as the tape playsthe tension is largest when the reel is smallest. However, in the case of a sudden jerk on the tape, therotational dynamics of the source reel becomes important. If the source reel is full, then the momentof inertia, proportional to R 4 , will be so large that higher tension in the tape will be required to givethe source reel its angular acceleration. If the reel is nearly empty, then the same tape accelerationwill require a smaller tension. Thus, the tape will be more likely to break when the source reel isnearly full. One sees the same effect in the case of paper towels; it is easier to snap a towel free whenthe roll is new than when it is nearly empty.11MR 2 , I CM = MR 232Chapter 10287Q10.21The moment of inertia would decrease. This would result in a higher angular speed of the earth,shorter days, and more days in the year!Q10.22There is very little resistance to motion that can reduce the kinetic energy of the rolling ball. Eventhough there is static friction between the ball and the floor (if there were none, then no rotationwould occur and the ball would slide), there is no relative motion of the two surfacesby thedefinition of rollingand so no force of kinetic friction acts to reduce K. Air resistance and frictionassociated with deformation of the ball eventually stop the ball.Q10.23In the frame of reference of the ground, no. Every pointmoves perpendicular to the line joining it to theinstantaneous contact point. The contact point is notmoving at all. The leading and trailing edges of thecylinder have velocities at 45 to the vertical as shown.vvCMCMvPFIG. Q10.23Q10.24The sphere would reach the bottom first; the hoop would reach the bottom last. If each object hasthe same mass and the same radius, they all have the same torque due to gravity acting on them.The one with the smallest moment of inertia will thus have the largest angular acceleration andreach the bottom of the plane first.Q10.25To win the race, you want to decrease the moment of inertia of the wheels as much as possible.Small, light, solid disk-like wheels would be best!SOLUTIONS TO PROBLEMSSection 10.1P10.1(a)Angular Position, Velocity, and Acceleration t= 0 = 5.00 rad t =0 = t=0 =(b)ddtddtt=0= 10.0 + 4.00t t = 0 = 10.0 rad s= 4.00 rad s 2t=0 t= 3.00 s = 5.00 + 30.0 + 18.0 = 53.0 rad t = 3.00 s =ddt t = 3.00 s =ddtt = 3 .00 s= 10.0 + 4.00t t = 3.00 s = 22.0 rad s= 4.00 rad s 2t = 3 .00 s288Rotation of a Rigid Object About a Fixed AxisSection 10.2*P10.2Rotational Kinematics: Rotational Motion with Constant Angular Acceleration f = 2.51 10 4 rev min = 2.63 10 3 rad s f i2.63 10 3 rad s 0= 8.22 10 2 rad s 23.2 s(a)(b)P10.411 f = i t + t 2 = 0 + 8.22 10 2 rad s 2 3.2 s22(a)=(b)P10.3=11 = i t + t 2 = 4.00 rad s 2 3.00 s22=tja fejaef2= 18.0 rad i = 2 000 rad s , = 80.0 rad s 2a fa f f = i + t = 2 000 80.0 10.0 = 1 200 rad s(b)0 = i + tt=i =i=2 000= 25.0 s80.0FGHIJ FG 2 rad IJ = 10K H 1.00 rev K 3100 rev 1 min1.00 min 60.0 s f i(a)t=(b)P10.6= 4.21 10 3 rad i 12.0 rad s== 4.00 rad s 2t3.00 s(a)P10.52 f = t ==FG Hf0 1032.00+i2rad s , f = 0s = 5.24 sIJ t = FG 10K H6rad sIJ FG 10 sIJ =KH 6 K27.4 rad i = 3 600 rev min = 3.77 10 2 rad s = 50.0 rev = 3.14 10 2 rad and f = 0 2 = i2 + 2fe0 = 3.77 10 2 rad sj2e+ 2 3.14 10 2 radj = 2.26 10 2 rad s 2P10.7 = 5.00 rev s = 10.0 rad s . We will break the motion into two stages: (1) a period during which thetub speeds up and (2) a period during which it slows down.0 + 10.0 rad s8.00 s = 40.0 rad2a f10.0 rad s + 0= t =a12.0 sf = 60.0 rad2While speeding up, 1 = t =While slowing down,2So, total = 1 + 2 = 100 rad = 50.0 revChapter 10P10.81 f i = i t + t 2 and f = i + t are two equations in two unknowns i and 211 f i = f t t + t 2 = f t t 2222 rad137.0 rev= 98.0 rad s 3.00 s 3.00 s1 rev2d i = f t :iFGHej232 rad = 294 rad 4.50 s 2 :*P10.10(a)=(b)P10.9289t ==IJKafaf261.5 rad= 13.7 rad s 224.50 s 1 rev2 rad=== 7.27 10 5 rad st 1 day 86 400 s=FGHIJK= b0.750 rad sgt . For the bone,1072 rad= 2.57 10 4 s or 428 min57.27 10 rad s 360The location of the dog is described by db =112 rad + 0.015 rad s 2 t 2 .32We look for a solution to2+ 0.007 5t 230 = 0.007 5t 2 0.75t + 2.09 = 00.75t =t=bg0.75 0.75 2 4 0.007 5 2.090.015= 2.88 s or 97.1 s22 2 + 0.007 5t 2 or if 0.75t =+ 2 + 0.007 5t 2 that is, if33either the dog or the turntable gains a lap on the other. The first equation hasThe dog and bone will also pass if 0.75t =t=bga0.75 0.75 2 4 0.007 5 4.190.015f = 105 s or 5.30 sonly one positive root representing a physical answer. The second equation hast=bg0.75 0.75 2 4 0.007 5 8.380.015= 12.8 s or 87.2 s .In order, the dog passes the bone at 2.88 s after the merry-go-round starts to turn, and again at12.8 s and 26.6 s, after gaining laps on the bone. The bone passes the dog at 73.4 s, 87.2 s, 97.1 s,105 s, and so on, after the start.290Rotation of a Rigid Object About a Fixed AxisSection 10.3P10.11Angular and Linear QuantitiesEstimate the tires radius at 0.250 m and miles driven as 10 000 per year.= = 6.44 10 7P10.137rev yr or ~ 10 7 rev yrv 45.0 m s== 0.180 rad sr250 m(a)v = r ; =(b)P10.12FGIJHKF 1 rev IJ = 1.02 10rad yr GH 2 rad Ks 1.00 10 4 mi 1 609 m== 6.44 10 7 rad yr0.250 m1 mir45.0 m sv2=ar =r250 mbg2= 8.10 m s 2 toward the center of trackGiven r = 1.00 m, = 4.00 rad s 2 , i = 0 and i = 57.3 = 1.00 rad(a) f = i + t = 0 + tafAt t = 2.00 s , f = 4.00 rad s 2 2.00 s = 8.00 rad s(b)bgv = r = 1.00 m 8.00 rad s = 8.00 m sba r = a c = r 2 = 1.00 m 8.00 rad seg2= 64.0 m s 2ja t = r = 1.00 m 4.00 rad s 2 = 4.00 m s 2The magnitude of the total acceleration is:2a = a r + a t2 =e64.0 m s j + e4.00 m s j2 22 2= 64.1 m s 2The direction of the total acceleration vector makes an angle with respect to the radius topoint P: = tan 1FG a IJ = tan FG 4.00 IJ =H 64.0 KHa Kt13.58c(c)af eja11 f = i + i t + t 2 = 1.00 rad + 4.00 rad s 2 2.00 s22f2= 9.00 radChapter 10*P10.14(a)Consider a tooth on the front sprocket. It gives this speed, relative to the frame, to the link ofthe chain it engages:v = r =(b)FG 0.152 m IJ 76 rev min FG 2 rad IJ FG 1 min IJ =H 2 KH 1 rev K H 60 s Kv 0.605 m s= 0.07 m = 17.3 rad sr2chConsider the wheel tread and the road. A thread could be unwinding from the tire with thisspeed relative to the frame:v = r =(d)FG 0.673 m IJ 17.3 rad s =H 2 KFG 1 IJ = 1.39 m sH 1 rad Kv 25.0 m s== 25.0 rad sr1.00 m(a)=(b) 2 = i2 + 2 fa f=b25.0 rad sg 0 ==2a f2 a1.25 revfb 2 rad rev g2 2 i2ft =(a)39.8 rad s 2s = vt = 11.0 m s 9.00 s = 99.0 mb=(b)=25.0 rad s(c)P10.165.82 m sWe did not need to know the length of the pedal cranks, but we could use that informationto find the linear speed of the pedals:v = r = 0.175 m 7.96 rad sP10.150.605 m sConsider the chain link engaging a tooth on the rear sprocket:=(c)29139.8 rad s 2ga= 0.628 sfs 99.0 m== 341 rad = 54.3 revr 0.290 mf =vfr=22.0 m s= 75.9 rad s = 12.1 rev s0.290 m292Rotation of a Rigid Object About a Fixed AxisFGHIJK2 rad 1 200 rev= 126 rad s1 rev60.0 s = 2f =v = r = 126 rad s 3.00 10 2 m = 3.77 m s(c)a c = 2 r = 126(d)P10.18(a)(b)P10.17s = r = rt = 126 rad s 8.00 10 2 m 2.00 s = 20.1 mbgeja f e8.00 10 j = 1 260 m s22b2so a r = 1.26 km s 2 toward the centerjagefThe force of static friction must act forward and then more and more inward on the tires, to produceboth tangential and centripetal acceleration. Its tangential component is m 1.70 m s 2 . Its radiallyeinward component ismv. This takes the maximum valuerejFGHm 2 r = mr i2 + 2 = mr 0 + 2fWith skidding impending we haveIJK= mr = ma t = m 1.70 m s 2 .2es =(a)ej Fy = ma y , + n mg = 0, n = mgfs = sn = s mg = m 2 1.70 m s 2*P10.19j21.70 m s 2gj2e+ m 2 2 1.70 m s 2j21 + 2 = 0.572Let RE represent the radius of the Earth. The base of the building moves east at v1 = REwhere is one revolution per day. The top of the building moves east at v 2 = RE + h . Itseastward speed relative to the ground is v 2 v1 = h . The objects time of fall is given byby = 0 +1 2gt , t =2g2h. During its fall the objects eastward motion is unimpeded so itsgbgdeflection distance is x = v 2 v1 t = hf FGH 92.8sm IJK2F IGH JK2h2= h3 2gg12.12(b)2 rad50 m86 400 s(c)The deflection is only 0.02% of the original height, so it is negligible in many practical cases.a32= 1.16 cmChapter 10Section 10.4P10.20293Rotational Energym1 = 4.00 kg , r1 = y1 = 3.00 m ;m 2 = 2.00 kg , r2 = y 2 = 2.00 m;m 3 = 3.00 kg , r3 = y 3 = 4.00 m ; = 2.00 rad s about the x-axis(a)I x = m1 r12 + m 2 r22 + m 3 r32a fa f + 3.00a4.00f1= a92.0fa 2.00f = 184 J2I x = 4.00 3.00KR =(b)1I x 222+ 2.00 2.0022= 92.0 kg m 22a f= r = 2.00a 2.00f == r = 4.00a 2.00f =FIG. P10.20v2v3K1 =24.00 m s38.00 m sK = K 1 + K 2 + K 3 = 72.0 + 16.0 + 96.0 = 184 J =P10.21(a)a fa fa fa fa fa f1122m1 v1 = 4.00 6.00 = 72.0 J221122K 2 = m 2 v 2 = 2.00 4.00 = 16.0 J221122K 3 = m 3 v 3 = 3.00 8.00 = 96.0 J22v1 = r1 = 3.00 2.00 = 6.00 m s1I x 22I = m j r j2y (m)j4In this case,3.00 kgr1 = r2 = r3 = r4r=I=213.0 m2.00 kg2a3.00 mf + a2.00 mf232= 13.0 m13.00 + 2.00 + 2.00 + 4.00 kgx (m)012= 143 kg m 2(b)KR =jb1 2 1I = 143 kg m 2 6.00 rad s22eg22.00 kg4.00 kg= 2.57 10 3 JFIG. P10.213294P10.22Rotation of a Rigid Object About a Fixed AxisaI = Mx 2 + m L xf2axfdI= 2 Mx 2m L x = 0 (for an extremum)dxmLx =M+md2I= 2m + 2 M ; therefore I is minimum when the axis ofdx 2mLwhich is also the centerrotation passes through x =M+mof mass of the system. The moment of inertia about an axispassing through x isI CM = MLM mL OPNM + mQ2LMN+m 1mM+mOP LQ22=LmMxMm 2L = L2M+mLxFIG. P10.22Mm.where =M+mSection 10.5P10.23Calculation of Moments of InertiaWe assume the rods are thin, with radius much less than L.Call the junction of the rods the origin of coordinates, andthe axis of rotation the z-axis.For the rod along the y-axis, I =y1mL2 from the table.3xFor the rod parallel to the z-axis, the parallel-axis theoremgivesI=FG IJH K1Lmr 2 + m222axis of rotationz1mL24FIG. P10.23In the rod along the x-axis, the bit of material between x and x + dx has massdistance r = x 2 +FG L IJH 2KFG m IJ dx and is atH LK2from the axis of rotation. The total rotational inertia is:z FGHL2I total =11L2mL2 + mL2 +x2 +344L 2FG IJH K7m x3=mL2 +12L 3=L2+L 2I FG m IJ dxJK H L KL2mLx4 L 22711mL2mLmL2+=mL2 +1212412Note: The moment of inertia of the rod along the x axis can also be calculated from the parallel-axis21Ltheorem as.mL2 + m122FG IJH KChapter 10P10.24295Treat the tire as consisting of three parts. The two sidewalls are each treated as a hollow cylinder ofinner radius 16.5 cm, outer radius 30.5 cm, and height 0.635 cm. The tread region is treated as ahollow cylinder of inner radius 30.5 cm, outer radius 33.0 cm, and height 20.0 cm.Use I =122m R1 + R 2 for the moment of inertia of a hollow cylinder.2ejSidewall:af a0.165 mf e6.35 10 mje1.10 101= b1.44 kg g a0.165 mf + a0.305 mf = 8.68 102m = 0.305 mI side2232232jkg m 3 = 1.44 kgkg m 2Tread:af a0.305 mf a0.200 mfe1.10 10 kg m j = 11.0 kg1= b11.0 kg g a0.330 mf + a0.305 mf = 1.11 kg m2m = 0.330 mI tread2233222Entire Tire:ejI total = 2 I side + I tread = 2 8.68 10 2 kg m 2 + 1.11 kg m 2 = 1.28 kg m 2P10.25Every particle in the door could be slid straight down into a high-density rod across its bottom,without changing the particles distance from the rotation axis of the door. Thus, a rod 0.870 m longwith mass 23.0 kg, pivoted about one end, has the same rotational inertia as the door:I=bgaf11ML2 = 23.0 kg 0.870 m332= 5.80 kg m 2 .The height of the door is unnecessary data.P10.26Model your body as a cylinder of mass 60.0 kg and circumference 75.0 cm. Then its radius is0.750 m= 0.120 m2and its moment of inertia isbgaf11MR 2 = 60.0 kg 0.120 m222= 0.432 kg m 2 ~ 10 0 kg m 2 = 1 kg m 2 .296P10.27Rotation of a Rigid Object About a Fixed AxisFor a spherical shell dI =224r 2 dr r 2dmr 2 =33ez zzeje j af2rIF4r jG 14. 2 11.6 J e10 kg m jdrI=H3RKRRF 2IF 2I= G J 4 e14. 2 10 j G J 4 e11.6 10 jH 3KH 3K56814.2 11.6 II=e10 jR FGH 5 6 JK3rIFM = z dm = z 4r G 14.2 11.6 J 10 drHRKF 14.2 11.6 IJ R= 4 10 GH3 4Kb8 3ge10 jR b14.2 5 11.6 6g = 2 FG .907 IJ = 0.330I=4 10 R R b14.2 3 11.6 4g 3 H 1.83 KMR24r 2 r 2 r dr3I = dI =R4330335355R23033323532 I = 0.330 MR 2*P10.28(a)y hhx. The area of the front face= , y=Lx L11is hL. The volume of the plate is hLw . Its density is222MMM. The mass of the ribbon is== 1=VhLw2 hLwBy similar triangles,dm = dV = ywdx =yhxL2 Mywdx 2 Mhx2 Mxdx.=dx =hLwhLLL2FIG. P10.28The moment of inertia isI=zr 2 dm =x2x=0all mass(b)zLzL2 Mxdx 2 M 32 M L4ML2= 2 x dx = 2=.22LL 0L 4From the parallel axis theorem I = I CM + MI h = I CM + MFG L IJH 3Kinertia I CM +2FG 2L IJH3K2= I CM +4ML2and9ML2. The two triangles constitute a rectangle with moment of91ML2 1+= 2 M L2 . Then 2 I CM = ML2939= I CM +4ML2+ I CM9I = I CM +a f4ML2181=ML2 +ML2 =ML2 .918182Chapter 10*P10.29297We consider the cam as the superposition of the original solid disk and a disk of negative mass cutfrom it. With half the radius, the cut-away part has one-quarter the face area and one-quarter thevolume and one-quarter the mass M 0 of the original solid cylinder:M0 1M0 = M4M0 =4M.3By the parallel-axis theorem, the original cylinder had moment of inertiaFG R IJ = 1 M R + M R = 3 M R .H 2K 2441F 1I F R I M R . The whole cam hasThe negative-mass portion has I = G M J G J = H 4 KH 2 K2322I CM + M 02020I=00022M R 2 23323 42311 2323M0 R 2 0=M0 R 2 =MR 2 =MR 2 and K = I 2 =MR 2 2 =MR 2 2 .4323232 32422 2448Section 10.6P10.302TorqueResolve the 100 N force into components perpendicularto and parallel to the rod, asafFpar = 100 N cos 57.0 = 54.5 NandafFperp = 100 N sin 57.0 = 83.9 NThe torque of Fpar is zero since its line of action passesthrough the pivot point.aFIG. P10.30fThe torque of Fperp is = 83.9 N 2.00 m = 168 N m (clockwise)P10.31 = 0.100 ma12.0 N f 0.250 ma9.00 N f 0.250 ma10.0 N f =3.55 N mThe thirty-degree angle is unnecessary information.FIG. P10.31P10.32The normal force exerted by the ground on each wheel isbgej1 500 kg 9.80 m s 2mgn=== 3 680 N44The torque of friction can be as large as max = f max r = sn r = 0.800 3 680 N 0.300 m = 882 N mb g afbgafThe torque of the axle on the wheel can be equally as large as the light wheel starts to turn withoutslipping.298P10.33Rotation of a Rigid Object About a Fixed AxisIn the previous problem we calculated the maximum torque that can be applied without skidding tobe 882 N m. This same torque is to be applied by the frictional force, f, between the brake pad andthe rotor for this wheel. Since the wheel is slipping against the brake pad, we use the coefficient ofkinetic friction to calculate the normal force.b g = fr = k n r , so n =Section 10.7P10.34(a)882 N m== 8.02 10 3 N = 8.02 kN0.500 0.220 m krafafRelationship Between Torque and Angular Accelerationbge11MR 2 = 2.00 kg 7.00 10 2 m220.600= == 122 rad s 2I 4.90 10 3=t2 1 200 60t === 1.03 s122I=j2= 4.90 10 3 kg m 2c h(b)P10.35 =bga1 2 1t = 122 rad s 1.03 s22f2= 64.7 rad = 10.3 revm = 0.750 kg , F = 0.800 Naf(a)(b)=(c)P10.36 = rF = 30.0 m 0.800 N = 24.0 N mat24.0rF==I mr 2 0.750 30.0= 0.035 6 rad s 2a f= r = 0.035 6a30.0 f = 1.07 m s f = i + t :2FIG. P10.352af10.0 rad s = 0 + 6.00 s10.00rad s 2 = 1.67 rad s 2=6.0036.0 N m= 21.6 kg m 221.67 rad s(a) = 36.0 N m = I :I=(b) f = i + t :0 = 10.0 + 60.0=a f = 0.167 rad s 2ejej = I = 21.6 kg m 2 0.167 rad s 2 = 3.60 N m(c)Number of revolutions f = i + i t +During first 6.00 sDuring next 60.0 s1 2t2a fa fa f a fa fFGIJHK121.67 6.00 = 30.1 rad212 f = 10.0 60.0 0.167 60.0 = 299 rad21 rev total = 329 rad= 52.4 rev2 radf =Chapter 10P10.37For m1 , Fy = ma y :+n m 1 g = 0n1 = m1 g = 19.6 Nf k 1 = k n1 = 7.06 Nb Fx = ma x :g7.06 N + T1 = 2.00 kg aFor the pulley, = I :FG IJH K1= b10.0 kg ga2= b5.00 kg gaT1 R + T2 R =T1 + T2T1 + T21aMR 22R+n 2 m 2 g cos = 0For m 2 ,(1)(2)jaefn 2 = 6.00 kg 9.80 m s 2 cos 30.0= 50.9 NFIG. P10.37fk 2 = kn 2= 18.3 N : 18.3 N T2 + m 2 sin = m 2 a18.3 N T2 + 29.4 N = 6.00 kg a (3)b(a)gAdd equations (1), (2), and (3):bg7.06 N 18.3 N + 29.4 N = 13.0 kg aa=ej4.01 N= 0.309 m s 213.0 kgT1 = 2.00 kg 0.309 m s 2 + 7.06 N = 7.67 N(b)ejT2 = 7.67 N + 5.00 kg 0.309 m s 2 = 9.22 NP10.38bgaf11mR 2 = 100 kg 0.500 m22 i = 50.0 rev min = 5.24 rad sI== f it=2= 12.5 kg m 20 5.24 rad s= 0.873 rad s 26.00 sej = I = 12.5 kg m 2 0.873 rad s 2 = 10.9 N mThe magnitude of the torque is given by fR = 10.9 N m, where f isthe force of friction.10.9 N m0.500 mTherefore,f=yieldsk =f 21.8 N== 0.312n 70.0 Nandf = knFIG. P10.38299300*P10.39Rotation of a Rigid Object About a Fixed Axis1 = I = 2 MR 2a25f 1 b80 kggFGH 1.2 mIJK e1.67 rad s j2f a2135 N 0.230 m + T 0.230 m =2T = 21.5 NSection 10.8P10.40Work, Power, and Energy in Rotational MotionThe moment of inertia of a thin rod about an axis through one end is I =kinetic energy is given asKR =withIh =afm h L2 60.0 kg 2.70 mh=3321ML2 . The total rotational3112I h 2 + I m mh22= 146 kg m 2a fFIGHJK2 rad F 1 h I=1 h G 3 600 s JHK = 1.75 10 rad s11= a146 fe1.45 10 j + a675 fe1.75 10 j222andIn addition,100 kg 4.50 mm L2= 675 kg m 2Im = m m =332 rad1hh == 1.45 10 4 rad s12 h 3 600 swhileTherefore,*P10.41mKR34 23 2= 1.04 10 3 J111Ewhere E = I 2 =MR 2 2 is the stored energy and222t1Pxxis the time it can roll. Then MR 2 2 = Pt =andt =4vvThe power output of the bus is P =afc2h22MR 2 2 v 1 600 kg 0.65 m 4 000 60 s 11.1 m sx === 24.5 km .4P4 18 746 WP10.42afaaffWork done = Fr = 5.57 N 0.800 m = 4.46 J11and Work = K = I 2 I i2f22(The last term is zero because the top starts from rest.)Thus, 4.46 J =14.00 10 4 kg m 2 2f2eAFjAand from this, f = 149 rad s .FIG. P10.42Chapter 10*P10.43(a)g a f a f = 2.28 10 kg mj bbgej.1b0.850 kg gb0.82 m sg + 1 b0.42 kggb0.82 m sg + 1 e2.28 10 kg m jFGH 0082 m s IJK222.03 m+0.42 kg e9.8 m s ja0.7 mf 0.25b0.85 kg ge9.8 m s ja0.7 mfF v IJ111= b0.85 kg gv + b0.42 kg gv + e 2.28 10 kg m jG222H 0.03 m K0.512 J + 2.88 J 1.46 J = b0.761 kg gv11222M R1 + R 2 = 0.35 kg 0.02 m + 0.03 m22K 1 + K 2 + K rot + U g 2 f k x = K 1 + K 2 + K rot feI=242i22422222f42ff222fvf =(b)P10.44=1.94 J= 1.59 m s0.761 kgv 1.59 m s== 53.1 rad s0.03 mrWe assume the rod is thin. For the compound objectLMNOPQ12M rod L2 + m ball R 2 + M ball D 235122I = 1.20 kg 0.240 m + 2.00 kg 4.00 10 2 m35I = 0.181 kg m 2I=a(a)fej2af+ 2.00 kg 0.280 m2K f + U f = K i + U i + EFG IJH Kafja1 2L+ M ball g L + R + 0I + 0 = 0 + M rod g2210.181 kg m 2 2 = 1.20 kg 9.80 m s 2 0.120 m + 2.00 kg 9.80 m s 2 0.280 m210.181 kg m 2 2 = 6.90 J2eejje(b)v = r = 0.280 m 8.73 rad s = 2.44 m s(d)v 2 = vi2 + 2 a y f yife = 8.73 rad s(c)fafdeijafv f = 0 + 2 9.80 m s 2 0.280 m = 2.34 m sThe speed it attains in swinging is greater by2.44= 1.043 2 times2.34jaf301302P10.45Rotation of a Rigid Object About a Fixed Axis(a)For the counterweight, Fy = ma y becomes:For the reel = I50.0 T =FG 50.0 IJ aH 9.80 Kreads TR = I = II=whereaR1MR 2 = 0.093 8 kg m 22We substitute to eliminate the acceleration:50.0 T = 5.10F TR IGH I JK2T = 11.4 Na=dv f = 2 7.57 6.00 = 9.53 m sUse conservation of energy for the system of the object, the reel, and the Earth:aK + U f = aK + U f :i11mv 2 + I 222v2I2mgh = mv 2 + I 2 = v 2 m + 2RRmgh =f2mghm + I2v=F I FGIJGH JK HK2a50.0 N fa6.00 mf==5.10 kg +RP10.46FIG. P10.4550.0 11.4= 7.57 m s 25.10a fiv 2 = vi2 + 2 a x f xi :f(b)and0.093 8a0. 250 f9.53 m s2Choose the zero gravitational potential energy at the level where the masses pass.K f + U gf = K i + U gi + E111m1 v 2 + m 2 v 2 + I 2 = 0 + m1 gh1i + m 2 gh2 i + 0222abLM a f OPFG IJNQH Kf2a fa fa fa11 1v15.0 + 10.0 v 2 +3.00 R 2= 15.0 9.80 1.50 + 10.0 9.80 1.5022 2R126.5 kg v 2 = 73.5 J v = 2.36 m s2P10.47gfFrom conservation of energy for the object-turntable-cylinder-Earthsystem,FG IJH K1 vI2 r2+1mv 2 = mgh2v2= 2mgh mv 2r22 ghI = mr 21v2IFGHIJKFIG. P10.47Chapter 10P10.48The moment of inertia of the cylinder isI=bfga11mr 2 = 81.6 kg 1.50 m222= 91.8 kg m 2and the angular acceleration of the merry-go-round is found as=a f afaf50.0 N 1.50 mFr=== 0.817 rad s 2 .2II91.8 kg mejAt t = 3.00 s, we find the angular velocity = i + tjaef = 0 + 0.817 rad s 2 3.00 s = 2.45 rad sjband K =P10.491 2 1I = 91.8 kg m 2 2. 45 rad s22(a)gFind the velocity of the CMeaK + U f = aK + U fi2= 276 J .PivotfR10 + mgR = I 222mgR2mgR= 3=2I2 mRg4gRg= 23R3vCM = RFIG. P10.49Rg3(b)(c)*P10.50v L = 2 vCM = 4vCM =(a)The moment of inertia of the cord on the spool is2mgR=2mRgeaf + a0.09 mf j = 4.16 101122M R1 + R 2 = 0.1 kg 0.015 m22eje224kg m 2 .jThe protruding strand has mass 10 2 kg m 0.16 m = 1.6 10 3 kg andI = I CM + Md 2 =FG aHf + a0.09 m + 0.08 mf IJK110.16 mML2 + Md 2 = 1.6 10 3 kg121222= 4.97 10 5 kg m 2For the whole cord, I = 4.66 10 4 kg m 2 . In speeding up, the average power isP=(b)afaE=t12I 2t=affFGH 2 000 s2 IJK =60P = = 7.65 N 0.16 m + 0.09 mFGH4.66 10 4 kg m 2 2 500 22 0.215 s60 s401 WIJK2= 74.3 W303304Rotation of a Rigid Object About a Fixed AxisSection 10.9Rolling Motion of a Rigid Objectbgb11mv 2 = 10.0 kg 10.0 m s22K trans =K rot =(c)P10.52(a)(b)P10.512K total = K trans + K rot = 750 JFGH1 2 1 1I =mr 222 2bW = K f K i = K trans + K rotW=orP10.53gIJ FG v IJ = 1 b10.0 kg gb10.0 m sgKH r K 4g bKf222trans+ K rotFGHgi1111 2Mv 2 + I 2 0 0 = Mv 2 +MR 222 5227W=Mv 210(a)= 500 JFG IJH KIJ FG v IJKH RK= 250 J2 = ImgR sin = I CM + mR 2 ea=jmgR 2 sin I CM + mRR2mgR 2 sin a hoop =2mR 2f1=g sin 2mgR 2 sin (b)FIG. P10.53Rf = If = n = mg cos IfR===mg cos mg cos P10.54mgn2=a disk = 3g sin 3mR 224The disk moves with the acceleration of the hoop.3LMNOPQc111Imv 2 + I 2 =m + 2 v2222RAlso,Ui = mgh , U f = 0 ,K=23g sin he12mR 22R mg cos whereandj=1tan 3vRvi = 0=since no slipping.LMNOPQ1Im + 2 v 2 = mgh2R2 ghv2 =1+ I 2Therefore,Thus,e jmR1I = mR 22For a disk,SoFor a ring,Since v diskv2 =2 gh1+ 12orv disk =4 gh32 ghorv ring = gh2> v ring , the disk reaches the bottom first.I = mR 2 so v 2 =Chapter 10P10.553051x 3.00 m== 2.00 m s = 0 + v f1.50 s2tvf4.00 m s8.00rad sv f = 4.00 m s and f ===2r6.38 10 26.38 10 m 2dv=iejWe ignore internal friction and suppose the can rolls without slipping.e K + K + U j + Eb0 + 0 + mgy g + 0 = FGH 1 mv2transrotgmechii2f+e= K trans + K rot + U g1 2I f + 02IJKjfej a3.00 mf sin 25.0 = 1 b0.215 kg gb4.00 m sg22.67 J = 1.72 J + e7 860 s jt0.215 kg 9.80 m s 22+FGHIJK18.00rad sI2 6.38 10 222I=P10.56(a)0.951 kg m 2 s 2= 1.21 10 4 kg m 27 860 s 2The height of the can is unnecessary data.Energy conservation for the system of the ball and theEarth between the horizontal section and top of loop:1111 222mv 2 + I 2 + mgy 2 = mv1 + I 122222FGHIJ FG v IJ + mgyKH r K1F2IF v I+ G mr J G JKH r K2H311 22mv 2 +mr 222 312mv125 25 2v 2 + gy 2 = v166=2v 2 = v1 22226gy 2 =51b4.03 m sg22FIG. P10.56fejab2.38 m sg = 12.6 m s=69.80 m s 2 0.900 m = 2.38 m s52v22The centripetal acceleration is 2>g0.450 mrThus, the ball must be in contact with the track, with the track pushing downward on it.(b)FGH2v 3 = v1 (c)IJ FG v IJ + mgy = 1 mv + 1 FG 2 mr IJ FG v IJKH r KKH r K22H36= b 4.03 m sg e9.80 m s ja 0.200 mf =511 22mv 3 +mr 222 36gy 3532321221224.31 m s1122mv 2 + mgy 2 = mv1222v 2 = v1 2 gy 2 =b4.03 m sg 2e9.80 m s ja0.900 mf =221.40 m 2 s 2This result is imaginary. In the case where the ball does not roll, the ball starts with lessenergy than in part (a) and never makes it to the top of the loop.306Rotation of a Rigid Object About a Fixed AxisAdditional ProblemsP10.57sin =mg23 gsin =23 gsin rat =23 gThenr > g sin 22for r >31 Aboutthe length of the chimney will have a3tangential acceleration greater than g sin .FGHP10.581m 23FG IJH KIJKttg sin gFIG. P10.57The resistive force on each ball is R = DAv 2 . Here v = r , where r is the radius of each balls path.The resistive torque on each ball is = rR , so the total resistive torque on the three ball system is total = 3rR .The power required to maintain a constant rotation rate is P = total = 3rR . This required powermay be written asa f = e3r DA j2 rad F 10 rev I F 1 min I 1 000=1 rev G 1 min J H 60.0 s KK G J = 30.0 rad sHF 1 000 IJ P = 3a0.100 mf a0.600 fe 4.00 10 m jGH 30.0 s KP = e0.827 m s j , where is the density of the resisting medium.P = total = 3r DA rWith23or(a)33354233In air, = 1.20 kg m3 ,ejand P = 0.827 m 5 s 3 1. 20 kg m 3 = 0.992 N m s = 0.992 W(b)P10.59atIn water, = 1 000 kg m3 and P = 827 W .(a)W = K =(b)(c)1 2 1 2 11I f I i = I 2 i2 where I = mR 2f22221221.00 kg 0.500 m 8.00 rad s 0 = 4.00 J2ejF 1IF I= G JG Jbf bgagH 2KH K 0 r b8.00 rad sga0.500 mf=== 1.60 st=f2.50 m s 2a1 f = i + i t + t 2 ; i = 0 ; i = 021 2 1 2.50 m s 221.60 s = 6.40 rad f = t =22 0.500 mFGHIaJKs = r = a0.500 mfa6. 40 radf =f3.20 m < 4.00 m YesChapter 10*P10.60StartThe quantity of tape is constant. Then the area of the rings yousee it fill is constant. This is expressed byrt rt2 rs2 = r 2 rs2 + r22 rs2 or r2 = rt2 + rs2 r 2 is theouter radius of spool 2.(a)P10.61(a)jLaterr2rvvand 2 = . Thertrstakeup reel must spin at maximum speed. At the end,vvand 1 = . The angularr = rs and r2 = rt so 2 =rtrsspeeds are just reversed.At the start, r = rt and r2 = rs so 1 =vFIG. P10.60Since only conservative forces act within the system of therod and the Earth,E = 0where I =K f + U f = K i + UisoFG IJH K1 2LI + 0 = 0 + Mg221ML23FIG. P10.613gL=Therefore,(b)vvWhere the tape comes off spool 1, 1 = . Where ther1 2v22.tape joins spool 2, 2 = = v rs + rt r 2r2e(b)rs = I , so that in the horizontal orientation,MgFG L IJ = MLH 2K 3=FG L IJ H 2Ka x = a r = r 2 = (d)= a y = a t = r = 3 Mg2R y Mg = Ma y = 3 Mg43g2LUsing Newtons second law, we haveR x = Ma x = 23g2(c)2Ry =Mg4FG L IJ =H 2K3g4307308P10.62Rotation of a Rigid Object About a Fixed Axisej = 10.0 rad s 2 5.00 rad s 3 t =z zd =65.0tddt10.0 5.00t dt = 10.0t 2.50t 2 = 65.0 rad s0d== 65.0 rad s 10.0 rad s 2 t 2.50 rad s 3 t 2dte(a)j ejAt t = 3.00 s,jaef ejej = 65.0 rad s 10.0 rad s 2 3.00 s 2.50 rad s 3 9.00 s 2 = 12.5 rad sz z(b)0td = dt =b0ztej ej65.0 rad s 10.0 rad s 2 t 2.50 rad s 3 t 2 dt0g ej ej = 65.0 rad s t 5.00 rad s 2 t 2 0.833 rad s 3 t 3At t = 3.00 s,bgaf ejej = 65.0 rad s 3.00 s 5.00 rad s 2 9.00 s 2 0.833 rad s 3 27.0 s 3 = 128 radP10.63The first drop has a velocity leaving the wheel given by1mvi2 = mgh1 , so2ejafejafv1 = 2 gh1 = 2 9.80 m s 2 0.540 m = 3.25 m sThe second drop has a velocity given byv 2 = 2 gh2 = 2 9.80 m s 2 0.510 m = 3.16 m sFrom =v, we findr1 =3.16 m sv1 3.25 m sv== 8.53 rad s and 2 = 2 == 8.29 rad s0.381 m0.381 mrror=bg b228.29 rad s 8.53 rad s 2 12=24g2= 0.322 rad s 2Chapter 10P10.64At the instant it comes off the wheel, the first drop has a velocity v1 , directed upward. Themagnitude of this velocity is found fromK i + U gi = K f + U gf12mv1 + 0 = 0 + mgh1 or v1 = 2 gh12and the angular velocity of the wheel at the instant the first drop leaves is1 =2 gh1v1=RR2Similarly for the second drop: v 2 = 2 gh2 and 2 =v2=R.2 gh2R2.The angular acceleration of the wheel is then2 gh222 1=a= 22P10.65R22 gh1R2a f2 2=bg h2 h12R1111Mv 2 + I 2 : U f = Mgh f = 0 ; K i = Mvi2 + I i2 = 0ff22221vUi = Mgh i : f = N = Mg cos ; = ; h = d sin and I = mr 22rKf =b g(a)E = E f Ei or fd = K f + U f K i U i11 fd = Mv 2 + I 2 Mghff22FGHI Mgd sinbgJK 21LmMNM + 2 OPQv = Mgd sin bMg cos gd or2bsin cos gv = 2 Mgd1mr 2 Mg cos d = Mv 2 +22v2r222m2vd(b)+MLM= M4 gdbsin cos gOPN am + 2 M fQ122v 2 = vi2 + 2 ax , v d = 2 adfa=FGHIJ bK2vdMsin cos = 2g2dm + 2Mg2g.309310P10.66Rotation of a Rigid Object About a Fixed Axis(a)E=FGHIJ e jK1 2MR 2 22 51 2E = 5.98 10 24 6.37 10 62 5e(b)jeLM FGMN Hj FGH 862400 IJK2IJ FG 2 IJ OPK H T K PQ1dT= MR a 2 f e 2T j5dt1F 2 I F 2 I dT= MR G J G JH T K H T K dt5F 2 I F 10 10= e 2.57 10 JjGH 86 400 s JK GH 3.16 10= 2.57 10 29 J2dE d 1 2MR 2=dt dt 2 52223226297I b86 400 s daygJsKsdE= 1.63 10 17 J daydt*P10.67(a) f = i + t= f it=2Tfte2Ti=d2 Ti T fiTi T f tj F 1 d I FG 1 yr IJ =~1 d 1 d 100 yr G 86 400 s J H 3.156 10 s KHK2 10 3 s27(b)10 22 s 2The Earth, assumed uniform, has moment of inertiaI=22MR 2 = 5.98 10 24 kg 6.37 10 6 m55e = I ~ 9.71 10 37jej = 9.71 10kg m e2.67 10s j = 10222223716kg m 2NmThe negative sign indicates clockwise, to slow the planets counterclockwise rotation.(c) = Fd . Suppose the person can exert a 900-N force.d= 2.59 10 16 N m~ 10 13 m=900 NFThis is the order of magnitude of the size of the planetary system.Chapter 10P10.68 = tt== = 31c h rev = 0.005 74 sv31.0 360900 rev60 s0.800 m= 139 m sv=0.005 74 sdFIG. P10.68P10.69 f will oppose the torque due to the hanging object: = I = TR f : f = TR I(1)Now find T, I and in given or known terms and substitute intoequation (1).b Fy = T mg = ma :T=m gaat 22also y = vi t +a=g(2)2y(3)t2=2ya= 2:R RtI=and1RM R2 +22LMNMFIG. P10.69(4)FG IJ OP = 5 MRH K QP 822(5)Substituting (2), (3), (4), and (5) into (1),FGH f =m gwe findP10.70(a)2yt2IJ R 5 MR b2yg = RLMmFG g 2 y IJ 5 My OPK 8 RtN H t K 4t Q2222W = K + UW = K f K i + U f Ui111mv 2 + I 2 mgd sin kd 22221 21 2 I + mR 2 = mgd sin + kd220=e=j2mgd sin + kd 2I + mR 2FIG. P10.70(b)==bgejafafaf2 0.500 kg 9.80 m s 2 0.200 m sin 37.0 + 50.0 N m 0.200 maf1.00 kg m 2 + 0.500 kg 0.300 m1.18 + 2.00= 3.04 = 1.74 rad s1.0522311312P10.71Rotation of a Rigid Object About a Fixed Axis(a)m 2 g T2 = m 2 abgejT2 = m 2 g a = 20.0 kg 9.80 m s 2 2.00 m s 2 = 156 NT1 m1 g sin 37.0 = m1 abfgaT1 = 15.0 kg 9.80 sin 37.0+2.00 m s 2 = 118 N(b)FG a IJH RKbT T gR = a156 N 118 Nfa0.250 mfI=bT2g T1 R = I = I2212.00 m s 2aP10.722FIG. P10.71= 1.17 kg m 2For the board just starting to move,FG IJ cos = FG 1 m IJ H 2KH3 K3 F gI = G J cos 2H K = I :2mg3g cos 23The vertical component isa y = a t cos = g cos 2 2If this is greater than g, the board will pull ahead of the ball falling:The tangential acceleration of the end isat = =23FIG. P10.72(a)32g cos 2 g gives cos 2 so23(b)When = 35.3 , the cup will land underneath the release-point of the ball ifcos When = 1.00 m, and = 35.3rc = 1.00 maf 35.3and2= 0.816 m3so the cup should be 1.00 m 0.816 m = 0.184 m from the moving endP10.73At t = 0 , = 3.50 rad s = 0 e 0 . Thus, 0 = 3.50 rad sAt t = 9.30 s, = 2.00 rad s = 0 e a 9.30 s f , yielding = 6.02 10 2 s 1(a)ed d 0 e=dtdtAt t = 3.00 s,=b tj = a fe0getj = 3.50 rad s 6.02 10 2 s 1 ezt(b) = 0 e t dt =0e3.00 6 .02 10 2j=0.176 rad s 2 0 te 1 = 0 1 e tAt t = 2.50 s ,3.50 rad s 6 .02 10 2 ja 2.50 f1e e== 8.12 rad = 1.29 rev26.02 10 1 se(c)As t , LMj NOPQ3.50 rad s01 e == 58.2 rad = 9.26 rev6.02 10 2 s 1ejrc = cos Chapter 10P10.74313Consider the total weight of each hand to act at the center of gravity (mid-point) of that hand. Thenthe total torque (taking CCW as positive) of these hands about the center of the clock is given by = m h gFG L IJ sinH 2Khh mm gFG L IJ sinH2Kmm=gm h L h sin h + m m Lm sin m2bgIf we take t = 0 at 12 oclock, then the angular positions of the hands at time t are h = ht ,rad h6whereh =and m = mt ,where m = 2 rad hTherefore, = 4.90 m s 2 60.0 kg 2.70 m sinf FGH 6t IJK + 100 kga4.50 mf sin 2tOPQL F t IO = 794 N mMsinG J + 2.78 sin 2t P , where t is in hours.H6KNQor(a)(i)At 3:00, t = 3.00 h ,so(ii)LMNaLM FG IJ + 2.78 sin 6 OP =N H 2KQ = 794 N m sinAt 5:15, t = 5 h +794 N m15h = 5.25 h , and substitution gives:60 = 2 510 N m(iii) = 0 N m(iv)At 8:20, = 1 160 N m(v)(b)At 6:00,At 9:45, = 2 940 N mThe total torque is zero at those times whenFG t IJ + 2.78 sin 2t = 0H6KsinWe proceed numerically, to find 0, 0.515 295 5, ..., corresponding to the times12:00:002:33:254:58:147:27:3610:02:5912:30:552:56:295:30:528:03:0510:27:2912:58:193:33:226:00:008:26:3811:01:411:32:313:56:556:29:089:03:3111:29:051:57:014:32:247:01:469:26:35314*P10.75Rotation of a Rigid Object About a Fixed Axis(a)As the bicycle frame moves forward at speed v, the center of each wheel moves forward atvthe same speed and the wheels turn at angular speed = . The total kinetic energy of theRbicycle isK = K trans + K rotorK=bIJ bKFGHgg FGH1111m frame + 2m wheel v 2 + 2 I wheel 2 = m frame + 2m wheel v 2 +m wheel R 22222IJ FG v IJ .KH R K22This yieldsK=(b)bgb11m frame + 3m wheel v 2 = 8.44 kg + 3 0.820 kg22g b3.35 m sg2= 61.2 J .As the block moves forward with speed v, the top of each trunk moves forward at the samevspeed and the center of each trunk moves forward at speed . The angular speed of each2vroller is =. As in part (a), we have one object undergoing pure translation and two2Ridentical objects rolling without slipping. The total kinetic energy of the system of the stoneand the trees isK = K trans + K rotorK=FG IJH K11vm stone v 2 + 2 m tree2222+2FG 1 IH2tree2IJ = 1 FG mK 2Hstone+IJKFGH11m tree v 2 +m tree R 222IJ FG v IJ .K H 4R KThis givesK=P10.76FGHIJKb131m stone + m tree v 2 = 844 kg + 0.75 82.0 kg242g b0.335 m sg2= 50.8 J .Energy is conserved so U + K rot + K trans = 0af LMN 1 mv2famg R r cos 1 +2OP 1 LM 2 mr OPQ 2 N5 Q0 +22=0RSince r = v , this gives=or=afaf10 R r 1 cos g7r2a10 Rg 1 cos 7r2fsince R >> r .FIG. P10.7622Chapter 10P10.77F aI1 F = T Mg = Ma: = TR = I = 2 MR 2 G R JH K(a)Combining the above two equations we findbT =M gaganda=2TMFIG. P10.77T=thusMg3FG IJH K2T2 Mg2==g3M M 3(b)a=(c)v 2 = vi2 + 2 a x f xifdiv2 = 0 + 2fvf =FG 2 gIJ ah 0fH3 K4 gh3For comparison, from conservation of energy for the system of the disk and the Earth wehaveU gi + K rot i + K trans i = U gf + K rot f + K trans f :Mgh + 0 + 0 = 0 +vf =P10.78(a)4 gh3 Fx = F f = Ma: = fR = IUsing I =(b)FGH1 1MR 22 21a2FMR 2 and = , we find a =23MRWhen there is no slipping, f = Mg .Substituting this into the torque equation of part (a), we have MgR =1F.MRa and =23 MgIJ FG v IJKH R Kf2+1Mv f 22315316P10.79Rotation of a Rigid Object About a Fixed AxisK rot + K trans + U = 0mNote that initially the center of mass of the sphere is adistance h + r above the bottom of the loop; and as themass reaches the top of the loop, this distance abovethe reference level is 2R r . The conservation ofenergy requirement gives(a)ha farf11mg h + r = mg 2 R r + mv 2 + I 222For the sphere I =gh + 2 gr = 2 gR +RPFIG. P10.792mr 2 and v = r so that the expression becomes57 2v10(1)Note that h = hmin when the speed of the sphere at the top of the loop satisfies the conditionmv 2 F = mg = aR r faor v 2 = g R rfSubstituting this into Equation (1) givesafafafhmin = 2 R r + 0.700 R r or hmin = 2.70 R r = 2.70 R(b)When the sphere is initially at h = 3 R and finally at point P, the conservation of energyequation givesaf11mg 3 R + r = mgR + mv 2 + mv 2 , or251022R + r gv =7afTurning clockwise as it rolls without slipping past point P, the sphere is slowing down withcounterclockwise angular acceleration caused by the torque of an upward force f of static2friction. We have Fy = ma y and = I becoming f mg = m r and fr =mr 2 .5FG IJH KEliminating f by substitution yields =mv 2 Fx = n = R r = c h( 2R + r ) mg =107Rr5gso that7r20mg7 Fy =5 mg7(since R >> r )317Chapter 10P10.80Consider the free-body diagram shown. The sum of torquesabout the chosen pivot is = I F(a)=FG 1 ml IJ FG a IJ = FG 2 mlIJ aH 3 KH K H 3 K2CMl2pivotHx(1)CM= l = 1.24 m : In this case, Equation (1) becomes3 14.7 N3F== 35.0 m s 2aCM =2m 2 0.630 kgbafCMmglg Fx = maCM F + H x = maCM or H x = maCM FbHygeF = 14.7 NjThus, H x = 0.630 kg 35.0 m s 2 14.7 N = +7.35 N orFIG. P10.80H x = 7.35 i N .(b)=1= 0.620 m : For this situation, Equation (1) yields2aCM =Again,af3 14.7 N3F== 17.5 m s 2 .4m 4 0.630 kgbg Fx = maCM H x = maCM F , sobgejH x = 0.630 kg 17.5 m s 2 14.7 N = 3.68 N or H x = 3.68 i N .(c)If H x = 0, thenF Fx = maCM F = maCM , or aCM = m .Thus, Equation (1) becomesF =P10.81FG 2 mlIJ FG F IJ soH 3 KH mKafbg22= l = 1.24 m = 0.827 m from the top .332Let the ball have mass m and radius r. Then I = mr 2 . If the ball takes four seconds to go down5twenty-meter alley, then v = 5 m s . The translational speed of the ball will decrease somewhat asthe ball loses energy to sliding friction and some translational kinetic energy is converted torotational kinetic energy; but its speed will always be on the order of 5.00 m s , including at thestarting point.As the ball slides, the kinetic friction force exerts a torque on the ball to increase the angular speed.vWhen = , the ball has achieved pure rolling motion, and kinetic friction ceases. To determine therelapsed time before pure rolling motion is achieved, consider:F 2 I L b5.00 m sg r OP which givesMN t PQ = I b k mg gr = G 5 mr 2 J MH Kt=2( 5.00 m s) 2.00 m s=5 k gkgNote that the mass and radius of the ball have canceled. If k = 0.100 for the polished alley, thesliding distance will be given bybx = vt = 5.00 m sL 2 ms OgMM a0.100f.00.80 m s PP = 10.2 m or x ~jQN e9210 1 m .318P10.82Rotation of a Rigid Object About a Fixed AxisConservation of energy between apex and the point wherethe grape leaves the surface:mgy =11mv 2 + I 2ff22afFGHIJ FG v IJKH R KfR2fnF IGH JK27 vfwhich gives g 1 cos =10 Rfy = RR cos11 2mgR 1 cos = mv 2 +mR 2f22 5ai(1)mg cosmg sinConsider the radial forces acting on the grape:mg cos n =mv 2fRFIG. P10.82.At the point where the grape leaves the surface, n 0 .Thus, mg cos =mv 2fRorv2fR= g cos .Substituting this into Equation (1) givesg g cos =P10.83(a)107and = 54.0 .g cos or cos =1710There are not any horizontal forces acting on the rod, so the center of mass will not movehorizontally. Rather, the center of mass drops straight downward (distance h/2) with the rodrotating about the center of mass as it falls. From conservation of energy:K f + U gf = K i + U giFG IJH KI = MgFG h IJ which reduces toJKH 2K11h2orMvCM + I 2 + 0 = 0 + Mg222FGH11 12MvCM +Mh 222 12vCM =(b)IJ FG vKH2CMh23 gh4In this case, the motion is a pure rotation about a fixed pivot point (the lower end of the rod)with the center of mass moving in a circular path of radius h/2. From conservation of energy:K f + U gf = K i + U giFG IJH K1 2horI + 0 = 0 + Mg221 1Mh 22 3IJ FG v IJKH KvCM =3 gh4FGHCMh22= MgFG h IJ which reduces toH 2KChapter 10P10.84(a)319Mr 2where M is the initial mass ofR2mr 2,+ K rot = 0 . Thus, when I =2The mass of the roll decreases as it unrolls. We have m =the roll. Since E = 0 , we then have U g + K transbmgr MgRg + mv + LM mr2N22Since r = v , this becomes v =e4g R3 r 33r2OPQ2=02j2(b)Using the given data, we find v = 5.31 10 4 m s(c)We have assumed that E = 0 . When the roll gets to the end, we will have an inelasticcollision with the surface. The energy goes into internal energy . With the assumption wemade, there are problems with this question. It would take an infinite time to unwrap thetissue since dr 0 . Also, as r approaches zero, the velocity of the center of mass approachesinfinity, which is physically impossible.P10.85(a) Fx = F + f = MaCM = FR fR = IbgFR MaCM F R =f = MaCM F = M(c)v 2 = vi2 + 2 a x f xifvf =8 Fd3MMgIaCMRFG 4F IJ F =H 3M K(b)dFiaCM =1F34F3MnfFIG. P10.85320P10.86Rotation of a Rigid Object About a Fixed AxisCall ft the frictional force exerted by each rollerbackward on the plank. Name as fb the rollingresistance exerted backward by the ground oneach roller. Suppose the rollers are equally farfrom the ends of the plank.MmRmFor the plank,b Fx = ma xFRFIG. P10.86g6.00 N 2 f t = 6.00 kg a pThe center of each roller moves forward only half as far as the plank. Each roller has accelerationand angular accelerationap 2ap2ap=a5.00 cmf a0.100 mfThen for each,bg a21f a5.00 cmf + f a5.00 cmf = b 2.00 kg ga5.00 cmf = I2F1 If + f = G kg J aH2 K Fx = ma xtSotp+ f t f b = 2.00 kgbbap210.0 cmpAdd to eliminate fb :bg2 f t = 1.50 kg a p(a)bg bgAnd 6.00 N 1.50 kg a p = 6.00 kg a pap =For each roller, a =(b)ap2a6.00 Nf =b7.50 kg g0.800 m s 2= 0.400 m s 2bgSubstituting back, 2 f t = 1.50 kg 0.800 m s 2Mg6.00 Nft = 0.600 N0.600 N + f b =fb = 0.200 N1kg 0.800 m s 22ejftftntntntntftftThe negative sign means that the horizontal forceof ground on each roller is 0.200 N forwardmgmgrather than backward as we assumed.fbnbfbnbFIG. P10.86(b)Chapter 10P10.87Rolling is instantaneous rotation about the contact point P. Theweight and normal force produce no torque about this point.Now F1 produces a clockwise torque about P and makes the321F3F2F4spool roll forward.Counterclockwise torques result from F3 and F4 , making thecspool roll to the left.The force F2 produces zero torque about point P and doesPnot cause the spool to roll. If F2 were strong enough, it wouldcause the spool to slide to the right, but not roll.P10.88FIG. P10.87F2The force applied at the critical angle exerts zero torque aboutthe spools contact point with the ground and so will not makethe spool roll.From the right triangle shown in the sketch, observe that c = 90 = 90 90 = .bgRrThus, cos c = cos =.RrcPFIG. P10.88P10.89(a)Consider motion starting from rest over distance x along the incline:bKtransgb+ K rot + U i + E = K trans + K rot + U0 + 0 + Mgx sin + 0 =aFGH11Mv 2 + 2 mR 222fgfIJ FG v IJKH RK2+02 Mgx sin = M + 2m v 2Since acceleration is constant,v 2 = vi2 + 2 ax = 0 + 2 ax , soaf2 Mgx sin = M + 2m 2 axa=Mg sin M + 2mafyxxFIG. P10.88continued on next pageF1322Rotation of a Rigid Object About a Fixed Axis(c)Suppose the ball is fired from a cart at rest. It moves with acceleration g sin = a x down theincline and a y = g cos perpendicular to the incline. For its range along the ramp, we havey yi = v yi t t=2 v yig cos x xi = v xi t +d=0+d=(b)1g cos t 2 = 0 021axt 22FGHIJK24v yi1g sin 22g cos 2 22 v yi sin g cos 2 In the same time the cart movesx xi = v xi t +12dc = 0 +dc =a1axt 22F g sinM I FG 4v IJGH aM + 2mf JK H g cos K2yi2222 v yi sin Mfg M + 2m cos 2 So the ball overshoots the cart byx = d d c =x =x =22 v yi22 v yi sin g cos 2 sin M +24v yi222 v yi sin Msin m aag cos 2 M + 2m22 v yig cos M + 2m24mv yisin aM + 2mfg cos2ffsin MChapter 10P10.90 Fx = ma x reads f + T = ma . If we take torques around the center of mass,we can use = I , which reads + fR 2 TR1 = I . For rolling withoutslipping, =a. By substitution,R2bIaI=T fR2 R 2 m2fR 2 m TR1 R 2 m = IT Ifej b2f I + mR 2 = T I + mR1 R 2f=mgTffR 2 TR1 =F I + mR R I TGH I + mR JK122323gnFIG. P10.90g2Since the answer is positive, the friction force is confirmed to be to the left.ANSWERS TO EVEN PROBLEMSP10.281ML22P10.30168 N m clockwise226 rad s 2P10.32882 N mP10.813.7 rad s 2P10.34(a) 1.03 s; (b) 10.3 revP10.10(a) 2.88 s; (b) 12.8 sP10.36(a) 21.6 kg m 2 ; (b) 3.60 N m ; (c) 52.4 revP10.12(a) 0.180 rad s;(b) 8.10 m s 2 toward the center of thetrackP10.380.312P10.401.04 10 3 JP10.14(a) 0.605 m s ; (b) 17.3 rad s ; (c) 5.82 m s ;(d) The crank length is unnecessaryP10.42149 rad sP10.16(a) 54.3 rev; (b) 12.1 rev sP10.44(a) 6.90 J; (b) 8.73 rad s ; (c) 2.44 m s ;(d) 1.043 2 times largerP10.180.572P10.462.36 m sP10.20(a) 92.0 kg m 2 ; 184 J ;(b) 6.00 m s ; 4.00 m s ; 8.00 m s ; 184 JP10.48276 JP10.50(a) 74.3 W; (b) 401 WP10.22see the solutionP10.527 Mv 210P10.54The disk;P10.2(a) 822 rad s 2 ; (b) 4.21 10 3 radP10.4(a) 1.20 10 2 rad s ; (b) 25.0 sP10.62P10.241.28 kg mP10.26~ 10 0 kg m 24 ghversus3gh324P10.56Rotation of a Rigid Object About a Fixed Axisaf(a) 2.38 m s ; (b) 4.31 m s;(c) It will not reach the top of the loop.P10.76P10.58(a) 0.992 W; (b) 827 WP10.78see the solutionP10.60see the solutionP10.80(a) 35.0 m s 2 ; 7.35 i N ;P10.62(a) 12.5 rad s ; (b) 128 radP10.64P10.66bg h2 h12Rg(a) 2.57 10 29 J ; (b) 1.63 10 17 J dayP10.68(a)P10.72P10.74(a) 794 N m ; 2 510 N m; 0;1 160 N m; 2 940 N m;(b) see the solutionI + mR 2; (b) 1.74 rad se4g R3 r 3j ; (b) 5.31 104m s;(a)P10.86(a) 0.800 m s 2 ; 0.400 m s 2 ;(b) 0.600 N between each cylinder and theplank; 0.200 N forward on each cylinderby the groundP10.88see the solutionP10.90see the solution; to the leftsee the solution2mgd sin + kd 254.0P10.84139 m sP10.707r 2(b) 17.5 m s 2 ; 3.68 i N ;(c) At 0.827 m from the top.P10.82210 Rg 1 cos 23r(c) It becomes internal energy.11Angular MomentumCHAPTER OUTLINE11.111.211.311.411.511.6The Vector Product andTorqueAngular MomentumAngular Momentum of aRotating Rigid ObjectConservation of AngularMomentumThe Motion of Gyroscopesand TopsAngular Momentum as aFundamental QuantityANSWERS TO QUESTIONSQ11.1No to both questions. An axis of rotation must be defined tocalculate the torque acting on an object. The moment arm ofeach force is measured from the axis.Q11.2A B C is a scalar quantity, since B C is a vector. SinceA B is a scalar, and the cross product between a scalar and avector is not defined, A B C is undefined.afafa fe j(a)Downcrossleft is away from you: j i = k(b)Q11.3Leftcrossdown is toward you: i j = ke jFIG. Q11.3Q11.4The torque about the point of application of the force is zero.Q11.5You cannot conclude anything about the magnitude of the angular momentum vector without firstdefining your axis of rotation. Its direction will be perpendicular to its velocity, but you cannot tellits direction in three-dimensional space until an axis is specified.Q11.6Yes. If the particles are moving in a straight line, then the angular momentum of the particles aboutany point on the path is zero.Q11.7Its angular momentum about that axis is constant in time. You cannot conclude anything about themagnitude of the angular momentum.Q11.8No. The angular momentum about any axis that does not lie along the instantaneous line of motionof the ball is nonzero.325326Angular MomentumQ11.9There must be two rotors to balance the torques on the body of the helicopter. If it had only onerotor, the engine would cause the body of the helicopter to swing around rapidly with angularmomentum opposite to the rotor.Q11.10The angular momentum of the particle about the center of rotation is constant. The angularmomentum about any point that does not lie along the axis through the center of rotation andperpendicular to the plane of motion of the particle is not constant in time.Q11.11The long pole has a large moment of inertia about an axis along the rope. An unbalanced torque willthen produce only a small angular acceleration of the performer-pole system, to extend the timeavailable for getting back in balance. To keep the center of mass above the rope, the performer canshift the pole left or right, instead of having to bend his body around. The pole sags down at theends to lower the system center of gravity.Q11.12The diver leaves the platform with some angular momentum about a horizontal axis through hercenter of mass. When she draws up her legs, her moment of inertia decreases and her angular speedincreases for conservation of angular momentum. Straightening out again slows her rotation.Q11.13Suppose we look at the motorcycle moving to the right. Its drive wheel is turning clockwise. Thewheel speeds up when it leaves the ground. No outside torque about its center of mass acts on theairborne cycle, so its angular momentum is conserved. As the drive wheels clockwise angularmomentum increases, the frame of the cycle acquires counterclockwise angular momentum. Thecycles front end moves up and its back end moves down.Q11.14The angular speed must increase. Since gravity does not exert a torque on the system, its angularmomentum remains constant as the gas contracts.Q11.15Mass moves away from axis of rotation, so moment of inertia increases, angular speed decreases,and period increases.Q11.16The turntable will rotate counterclockwise. Since the angular momentum of the mouse-turntablesystem is initially zero, as both are at rest, the turntable must rotate in the direction opposite to themotion of the mouse, for the angular momentum of the system to remain zero.Q11.17Since the cat cannot apply an external torque to itself while falling, its angular momentum cannotchange. Twisting in this manner changes the orientation of the cat to feet-down without changingthe total angular momentum of the cat. Unfortunately, humans arent flexible enough to accomplishthis feat.Q11.18The angular speed of the ball must increase. Since the angular momentum of the ball is constant, asthe radius decreases, the angular speed must increase.Q11.19Rotating the book about the axis that runs across the middle pages perpendicular to thebindingmost likely where you put the rubber bandis the one that has the intermediate momentof inertia and gives unstable rotation.Q11.20The suitcase might contain a spinning gyroscope. If the gyroscope is spinning about an axis that isoriented horizontally passing through the bellhop, the force he applies to turn the corner results in atorque that could make the suitcase swing away. If the bellhop turns quickly enough, anything at allcould be in the suitcase and need not be rotating. Since the suitcase is massive, it will want to followan inertial path. This could be perceived as the suitcase swinging away by the bellhop.Chapter 11SOLUTIONS TO PROBLEMSSection 11.1The Vector Product and TorqueiP11.1jkMN= 621 = 7.00 i + 16.0 j 10.0k2 1 3afaf af(a)area = A B = AB sin = 42.0 cm 23.0 cm sin 65.015.0 = 740 cm 2(b)P11.2A + B = 42.0 cm cos 15.0+ 23.0 cm cos 65.0 i + 42.0 cm sin 15.0+ 23.0 cm sin 65.0 jafafA + B = a50.3 cmfi + a31.7 cmf jlength = A + B = a50.3 cmf + a31.7 cmf2iP11.3(a)2faf= 59.5 cmj kA B = 3 4 0 = 17.0k2(b)a30A B = A B sin 17 = 5 13 sin FG 17 IJ = 70.6H 5 13 KA B = 3.00a6.00f + 7.00a 10.0f + a 4.00fa9.00 f = 124AB = a3.00f + a7.00f + a 4.00f a6.00f + a 10.0 f + a9.00 fF A B IJ = cos a0.979f = 168(a)cos GH AB K = arcsinP11.42221jA B = 3.00= 127k9.00a23.0f + a3.00f + a12.0f = 26.1FG A B IJ = sin a0.206f = 11.9 or 168H AB KAB =(c)27.00 4.00 = 23.0 i + 3.00 j 12.0k6.00 10.0sin 121i(b)22221Only the first method gives the angle between the vectors unambiguously.327328*P11.5Angular Momentumaf af = r F = 0.450 m 0.785 N sin 9014 up east= 0.343 N m northFIG. P11.5P11.6The cross-product vector must be perpendicular to both of the factors, so its dot product with eitherfactor must be zero:ejejDoes 2 i 3 j + 4k 4i + 3 j k = 0 ?8 9 4 = 5 0No . The cross product could not work out that way.P11.7A B = A B AB sin = AB cos tan = 1 or = 45.0ij ka f a f a f a7.00 N mfk(a) =rF= 1 3 0 = i 00 j 00 +k 29 =3 2 0(b)P11.8The particles position vector relative to the new axis is 1 i + 3 j 6 j = 1 i 3 j .ijkaf = 1 3 0 = 11.0 N m k32 0P11.9BF3 = F1 + F2The torque produced by F3 depends on theperpendicular distance OD, therefore translating thepoint of application of F3 to any other point alongF3DOBC will not change the net torque .ACF1FIG. P11.9F2Chapter 11*P11.10i i = 1 1 sin 0 = 0i j=kkk j = iki= jij i = kjk = ijj j and k k are zero similarly since thevectors being multiplied are parallel.329i k = ji j = 1 1 sin 90 = 1FIG. P11.10Section 11.2P11.11Angular MomentumL = mi vi ribygbf bgagbgaf= 4.00 kg 5.00 m s 0.500 m + 3.00 kg 5.00 m s 0.500 m3.00 kg2L = 17.5 kg m s , andejL = 17.5 kg m 2 s kx1.00 m4.00 kgFIG. P11.11P11.12L=rpgeej bjL = e 8.10k 13.9k j kg m s = e 22.0 kg m sjkL = 1.50 i + 2.20 j m 1.50 kg 4.20 i 3.60 j m s2P11.13ejv=r = 6.00 i + 5.00t j msoedr= 5.00 j m sdtjp = mv = 2.00 kg 5.00 j m s = 10.0 j kg m siand2L = r p = 6.000j5.00t10.0k0 =0e60.0 kg m sjk2330Angular Momentummv 2r Fx = ma xT sin = Fy = ma yP11.14T cos = mgSo2sin v=cos rgv = rglsin cos L = rmv sin 90.0L = rm rgmsin cos FIG. P11.14sin cos r = sin , soL = m 2 gr 3L=P11.15m2 g3sin 4 cos The angular displacement of the particle around the circle is = t =vt.RyvThe vector from the center of the circle to the mass is thenR cos i + R sin j .RmQPThe vector from point P to the mass isr = R i + R cos i + R sin jLMFGNHr = R 1 + cosFG vt IJ IJ i + sinFG vt IJ jOPH R KK H R K QFIG. P11.15The velocity isv=SoFG IJH KFG IJH Kdrvtvt= v sini + v cosjdtRRL = r mvafL F vt I OmvRk McosG J + 1PN H RK QL = mvR 1 + cos t i + sin t j sin t i + cos t jL=P11.16(a)The net torque on the counterweight-cord-spool system is:bgej = r F = 8.00 10 2 m 4.00 kg 9.80 m s 2 = 3.14 N m .(b)L = r mv + I(c)=bL = Rmv +gdL= 0.400 kg m adta=FG IJ FGH K HIJK1vMMR 2=R m+v=2R23.14 N m= 7.85 m s 20.400 kg mb0.400 kg mgvxChapter 11P11.17(a)(b)vi = vxi izeroAt the highest point of the trajectory,viv 2 sin 21x= R= iand22gy = hmax =bvi sin Lv=MMN=2igbv2R2FIG. P11.17bvi sin sin 2i+2g2g m vi sin O2gL 1 = r1 mv 1(c)331g jOP mvPQ2xi ig v cos k2i2gL 2 = R i mv 2 , where R =vi2 sin 2gej= mRvi sin k = mvi3 sin 2 sin kg= mR i vi cos i vi sin j(d)P11.18The downward force of gravity exerts a torque in the z direction.Whether we think of the Earths surface as curved or flat, we interpret the problem to mean that theplanes line of flight extended is precisely tangent to the mountain at its peak, and nearly parallel tothe wheat field. Let the positive x direction be eastward, positive y be northward, and positive z bevertically upward.(a)af ejp = mv = 12 000 kg e 175 i m sj = 2.10 10 i kg m sL = r p = e 4.30 10 k mj e 2.10 10 i kg m sj = e 9.03 10r = 4.30 km k = 4.30 10 3 m k63(b)6a9jkg m 2 s jfNo . L = r p sin = mv r sin , and r sin is the altitude of the plane. Therefore, L =constant as the plane moves in level flight with constant velocity.(c)Zero . The position vector from Pikes Peak to the plane is anti-parallel to the velocity ofthe plane. That is, it is directed along the same line and opposite in direction.Thus, L = mvr sin180 = 0 .332P11.19Angular MomentumThe vector from P to the falling ball is1r = ri + v i t + at 22r=e cos i +jsin j + 0 mFG 1 gt IJ jH2 Kl2The velocity of the ball isPv = v i + at = 0 gt jL = r mvSoL=mLMe cos i +Njsin j + 0 FG 1 gt IJ jOP e gtjjH2 K Q2FIG. P11.19L = m gt cos kP11.20In the vertical section of the hose, the water has zero angularmomentum about our origin (point O between the firemans feet).As it leaves the nozzle, a parcel of mass m has angular momentum:afbL = r mv = mrv sin 90.0 = m 1.30 m 12.5 m se2jvfgL = 16.3 m s m1.30 mThe torque on the hose is the rate of change in angular momentum.Thus,=jbOgdLdm= 16.3 m 2 s= 16.3 m 2 s 6.31 kg s = 103 N mdtdtejeviFIG. P11.20Section 11.3*P11.21P11.22K=Angular Momentum of a Rotating Rigid Object1 2 1 I 2 2 L2I ==22 I2IThe moment of inertia of the sphere about an axis through its center isI=bgaf22MR 2 = 15.0 kg 0.500 m552= 1.50 kg m 2Therefore, the magnitude of the angular momentum isejbgL = I = 1.50 kg m 2 3.00 rad s = 4.50 kg m 2 sSince the sphere rotates counterclockwise about the vertical axis, the angular momentum vector isdirected upward in the +z direction.ejThus, L = 4.50 kg m 2 s k .Chapter 11P11.23(a)(b)FG 1 MR IJ = 1 b3.00 kg ga0.200 mf b6.00 rad sg =H2 K 2L1F RI OL = I = M MR + M G J PH 2 K PQMN 23= b3.00 kg ga0.200 mf b6.00 rad sg = 0.540 kg m s422L = I =222P11.242The total angular momentum about the center point is given by L = I h h + I m mIh =andI m3 =In addition,h =whilef2a= 146 kg m 2f100 kg 4.50 mm m L2m=332= 675 kg m 2FIGHJK2 rad F 1 h I =1 h G 3 600 s JHK = 1.75 10 rad sL = 146 kg m e1.45 10 rad sj + 675 kg m e1.75 102 rad1h= 1.45 10 4 rad s12 h 3 600 s3m42Thus,23rad sjL = 1.20 kg m 2 sor(a)am h L2 60.0 kg 2.70 mh=33withP11.250.360 kg m 2 sI=a1m1 L2 + m 2 0.50012fa f2=afa f10.100 1.00122a+ 0.400 0.500f2= 0.108 3 kg m 2L = I = 0.108 3 4.00 = 0.433 kg m 2 s(b)I=afa f11m1 L2 + m 2 R 2 = 0.100 1.00332a f+ 0.400 1.002= 0.433a fL = I = 0.433 4.00 = 1.73 kg m 2 s*P11.26 Fx = ma x :+ fs = ma xWe must use the center of mass as the axis inaf af anf = I :Fg 0 n 77.5 cm + fs 88 cm = 0 Fy = ma y :Fg88 cm+n Fg = 0fs155 cm2FIG. P11.26We combine the equations by substitution:af a fe9.80 m s j77.5 cm = 8.63 m s= mg 77.5 cm + ma x 88 cm = 02ax88 cm2333334*P11.27Angular Momentumv2= 2rrWe require a c = g =e9.80 m s j = 0.313 rad s2g=r=100 ma2fI = Mr = 5 10 4 kg 100 m2= 5 10 8 kg m 2(a)L = I = 5 10 8 kg m 2 0.313 s = 1.57 10 8 kg m 2 s(c) = I =dI f iit t = I f I i = L f LiThis is the angular impulse-angular momentum theorem.(b)Section 11.4P11.28(a)t =1.57 10 8 kg m 2 s= 6.26 10 3 s = 1.74 h2 125 N 100 mafafFrom conservation of angular momentum for the system of two cylinders:1g+ I 2 f = I 1 iKf =soP11.29=Conservation of Angular MomentumbI(b)Lf 0Ii i = I f f :borandg1I1 + I 2 2f2KfKi=12bI1+ I2212 I 1 if =Ki =g FG I IJHI +I K112i2= 2 = 7.14 rev min1I 1 i22I1which is less than 1 .I1 + I 2e250 kg m jb10.0 rev ming = 250 kg m2I1iI1 + I 22af+ 25.0 kg 2.00 m22Chapter 11P11.30(a)335The total angular momentum of the system of the student, the stool, and the weights aboutthe axis of rotation is given bye jI total = I weights + I student = 2 mr 2 + 3.00 kg m 2Before:r = 1.00 m .Thus,I i = 2 3.00 kg 1.00 mAfter:r = 0.300 mThus,I f = 2 3.00 kg 0.300 mbfgab2+ 3.00 kg m 2 = 9.00 kg m 2fga2+ 3.00 kg m 2 = 3.54 kg m 2We now use conservation of angular momentum.I f f = I i if =orF I I = FG 9.00 IJ b0.750 rad sg =GH I JK H 3.54 KifijbgjbgKi =P11.3111I i i2 = 9.00 kg m 2 0.750 rad s22Kf =(b)11I f 2 = 3.54 kg m 2 1.91 rad sf22ee1.91 rad s2= 2.53 J2= 6.44 JLet M = mass of rod and m = mass of each bead. From Ii i = I f f , we have(a)LM 1 MN 122OPQ+ 2mr12 i =LM 1 MN 122OPQ+ 2mr22 fWhen = 0.500 m , r1 = 0.100 m , r2 = 0.250 m , and with other values as stated in theproblem, we find f = 9.20 rad s .(b)Since there is no external torque on the rod,L = constant and is unchanged .*P11.32Let M represent the mass of all the ribs together and L the length of each. The original moment of1inertia is ML2 . The final effective length of each rib is L sin 22.5 and the final moment of inertia is312M L sin 22.5 angular momentum of the umbrella is conserved:3af11ML2 i = ML2 sin 2 22.5 f331.25 rad sf == 8.54 rad ssin 2 22.5336P11.33Angular Momentum(a)The table turns opposite to the way the woman walks, so its angular momentum cancelsthat of the woman. From conservation of angular momentum for the system of the womanand the turntable, we have L f = Li = 0so,L f = I woman woman + I table table = 0and table = tablewomanwoman rv womanI table500 kg m 2af11 22m woman v woman + I table22work done = K = K f 0 =W=P11.34I womanmr2 woman = womanI tableI table table = 0.360 rad s counterclockwiseor(b)FI FG v IJ = mIJKGHJK H r K60.0 kg a 2.00 mfb1.50 m sg== 0.360 rad sFGHbgb160 kg 1.50 m s2g2+jb1500 kg m 2 0.360 rad s2eg2= 99.9 JWhen they touch, the center of mass is distant from the center of the larger puck byyCM =af0 + 80.0 g 4.00 cm + 6.00 cm= 4.00 cm120 g + 80.0 gejbjeg(a)L = r1 m1 v1 + r2 m 2 v 2 = 0 + 6.00 10 2 m 80.0 10 3 kg 1.50 m s = 7.20 10 3 kg m 2 s(b)The moment of inertia about the CM isFG 1 m r + m d IJ + FG 1 m r + m d IJH2K H2K1I = b0.120 kg ge6.00 10 mj + b0.120 kg ge 4.00 10 j21+ e80.0 10 kg je 4.00 10 mj + e80.0 10 kg je6.00 102I=21 121 122 22322 2222 223I = 7.60 10 4 kg m 2Angular momentum of the two-puck system is conserved: L = I=L 7.20 10 3 kg m 2 s== 9. 47 rad sI7.60 10 4 kg m 22mj2Chapter 11P11.35(a)Li = mv ext = 0 , so L f337= Li = mva fF m IJ v=GH m + MKLf = m + M vfvf(b)fFGHIJKFraction of K lost =12m2M +m2mv 2 1212v2mv=FIG. P11.35MM+mFor one of the crew,mv 2= m i2 rr Fr = mar :n=We requiren = mg , so i =Now,Ii i = I f fgraf5.00 10 8 kg m 2 + 150 65.0 kg 100 mF 5.98 10 IGH 5.32 10 JK88(a)afg= 5.00 10 8 kg m 2 + 50 65.0 kg 100 mr2fa r = 2 r = 1.26 g = 12.3 m s 2fConsider the system to consist of the wad of clayand the cylinder. No external forces acting on thissystem have a torque about the center of thecylinder. Thus, angular momentum of the systemis conserved about the axis of the cylinder.L f = Li :I = mvi dorLM 1 MRN2Thus,(b)2gg= f = 1.12rrNow,P11.37v1mv 221K f = M + m v2f2mvf =v velocity of the bulletM+mand blockKi =aP11.36lM=2OPQ+ mR 2 = mvi d2mvi da M + 2 m fR2FIG. P11.37.No . Some mechanical energy changes to internal energy in this perfectly inelastic collision.338*P11.38Angular Momentum(a)Let be the angular speed of the signboard when it is vertical.1 2I = Mgh21 11ML2 2 = Mg L 1 cos 2 32FGHaIJKa3 g 1 cos L =ffMgmjaevf3 9.80 m s 2 1 cos 25.0=FIG. P11.380.50 m= 2.35 rad s(b)I f f = Ii i mvL represents angular momentum conservationFG 1 MLH32 f ==(c)1313IJK+ mL2 f =1ML2 i mvL3ML i mvc13hM+m Lb2.40 kg ga0.5 mfb2.347 rad sg b0.4 kg gb1.6 m sg =b2.40 kgg + 0.4 kg a0.5 mf130. 498 rad sLet hCM = distance of center of mass from the axis of rotation.hCM =b2.40 kg ga0.25 mf + b0.4 kgga0.50 mf = 0.285 7 m .2.40 kg + 0.4 kgApply conservation of mechanical energy:aM + mfgh a1 cos f = 1 FGH 1 ML + mL IJK2 3L c M + mhL OP = cos M1 MN 2aM + mfgh PQR b2.40 kg g + 0.4 kg a0.50 mf b0.498 rad sg|= cos S1 | 2b2.40 kg + 0.4 kg ge9.80 m s jb0.285 7 mgT2CM1312222CM11322= 5.581L22U|V|WChapter 11P11.39339The meteor will slow the rotation of the Earth by the largest amount if its line of motion passesfarthest from the Earths axis. The meteor should be headed west and strike a point on the equatortangentially.Let the z axis coincide with the axis of the Earth with +z pointing northward. Then, conservingangular momentum about this axis, L f = Li I f = I i + mv rorThus,22MR 2 f k = MR 2 i k mvRk55mvR5mvor=i f = 222 MR5 MRi f =ejej = 5.91 10kg je6.37 10 mj5 3.00 10 13 kg 30.0 10 3 m se2 5.98 1024146rad s max ~ 10 13 rad sSection 11.5*P11.40The Motion of Gyroscopes and TopsAngular momentum of the system of the spacecraft and the gyroscope is conserved. The gyroscopeand spacecraft turn in opposite directions.0 = I 1 1 + I 2 2 :t I 1 1 = I 2bg20 kg m 2 100 rad s = 5 10 5 kg m 2t=*P11.41I=2.62 10 5 s= 131 s2 00022MR 2 = 5.98 10 24 kg 6.37 10 6 m55ejej2= 9.71 10 37 kg m 2F 2 rad I = 7.06 10 kg m sGH 86 400 s JKF 2 rad I FG 1 yr IJ F 1 d I =kg m sjGH 2.58 10 yr JK H 365.25 d K GH 86 400 s JKL = I = 9.71 10 37 kg m 2e = L p = 7.06 10 33Section 11.6P11.42FG 30 IJ FG rad IJH t K H 180 K3322245.45 10 22 N mAngular Momentum as a Fundamental Quantity6.626 1 10 34 J s(a)L=hh= mvr so v =2mr2(b)K=11mv 2 = 9.11 10 31 kg 2.19 10 6 m s22(c)=L1.055 10 34 J s==I mr 29.11 10 31 kg 0.529 10 10 meev=jejeejej2 9.11 10 -31 kg 0.529 10 10 mj2j= 2.18 10 18 J2= 4.13 10 16 rad s= 2.19 10 6 m s340Angular MomentumAdditional Problems*P11.43First, we define the following symbols:I P = moment of inertia due to mass of people on the equatorI E = moment of inertia of the Earth alone (without people) = angular velocity of the Earth (due to rotation on its axis)2T== rotational period of the Earth (length of the day)R = radius of the EarthThe initial angular momentum of the system (before people start running) isbgLi = I P i + I E i = I P + I E iWhen the Earth has angular speed , the tangential speed of a point on the equator is v t = R .Thus, when the people run eastward along the equator at speed v relative to the surface of the Earth,vpv= + .their tangential speed is v p = v t + v = R + v and their angular speed is P =RRThe angular momentum of the system after the people begin to run isFGHL f = I P p + I E = I P +IJKbgI vv+ I E = I P + I E + P .RRdiSince no external torques have acted on the system, angular momentum is conserved L f = Li ,bgbgIPv= I P + I E i . Thus, the final angular velocity of the Earth isRIPvIPv = i = i 1 x = , where x .I P + I E R iIP + IE Rgiving I P + I E +ba fgbThe new length of the day is T =day is T = T Ti Ti x = TiLMMN bIP2=gT2= i Ti 1 + x , so the increase in the length of thei 1 x 1 xOPg PQa fa fIPvTi2 I P v2, this may be written as T . Since i =.Ti+ I E R i2 I P + I E RbTo obtain a numeric answer, we computejbeI P = m p R 2 = 5.5 10 9 70 kgg e6.37 10 mj62= 1.56 10 25 kg m 22= 9.71 10 37 kg m 2 .andIE =22m E R 2 = 5.98 10 24 kg 6.37 10 6 m55ejeje8.64 10 sj e1.56 10 kg m jb2.5 m sg =Thus, T 2 e1.56 10 + 9.71 10 j kg m e6.37 10 mj425225372267.50 10 11 s .gChapter 11*P11.44(a)341bK + U g = bK + U gs As B120 + mgy A = mv B + 02ejv B = 2 gy A = 2 9.8 m s 2 6.30 m = 11.1 m s(b)L = mvr = 76 kg 11.1 m s 6.3 m = 5.32 10 3 kg m 2 s toward you along the axis of thechannel.(c)The wheels on his skateboard prevent any tangential force from acting on him. Then notorque about the axis of the channel acts on him and his angular momentum is constant. Hislegs convert chemical into mechanical energy. They do work to increase his kinetic energy.The normal force acts forward on his body on its rising trajectory, to increase his linearmomentum.(d)L = mvr(e)v=5.32 10 3 kg m 2 s= 12.0 m s76 kg 5.85 meK + U j + W = e K + U j1176 kg b11.1 m sg + 0 + W = 76 kg b12.0 m sg22ggBC22+ 76 kg 9.8 m s 2 0.45 mW = 5.44 kJ 4.69 kJ + 335 J = 1.08 kJ(f)eK + U j = e K + U j1176 kg b12.0 m sg + 0 = 76 kgv22gCgD22D+ 76 kg 9.8 m s 2 5.85 mv D = 5.34 m s(g)Let point E be the apex of his flight:eK + U j = e K + U j176 kg b5.34 m sg + 0 = 0 + 76 kg e9.8 m s jb y2by y g = 1.46 mgDgE2E(h)E yDgDFor the motion between takeoff and touchdown1y f = yi + v yi t + a y t 222.34 m = 0 + 5.34 m s t 4.9 m s 2 t 2t=(i)2a fa f =5.34 5.34 2 + 4 4.9 2.349.81. 43 sThis solution is more accurate. In chapter 8 we modeled the normal force as constant whilethe skateboarder stands up. Really it increases as the process goes on.342P11.45Angular Momentum(a)I = mi ri2=mFG 4d IJH3K2+mFG d IJH 3K2+mFG 2d IJH3K2mm1d2= 7m32d32dThink of the whole weight, 3mg, acting at the center of gravity. =rF=FG d IJ e ij 3mge jj = bmgdgkH 3K3g 3mgd==counterclockwise2I 7md7d(c)=(d)a = r =FG 3 g IJ FG 2d IJ =H 7d K H 3 K2gup7The angular acceleration is not constant, but energy is.a K + U f + E = a K + U fF dI 10 + a3m f g G J + 0 = IH 3K 2i(e)maximum kinetic energy = mgd(f)f =(g)L f = I f =(h)vf = fr =6g7d7md 236g=7d6g d=7d 3FG 14 g IJH3K2 gd2112md 3 2f2f3dFIG. P11.45(b)mP+0Chapter 11P11.46(a)af b343gThe radial coordinate of the sliding mass is r t = 0.012 5 m s t . Its angular momentum isbgbgbgbg2L = mr 2 = 1.20 kg 2.50 rev s 2 rad rev 0.012 5 m s t 2ejL = 2.95 10 3 kg m 2 s 3 t 2orThe drive motor must supply torque equal to the rate of change of this angular momentum:=bja f b0.005 89 WgtdL= 2.95 10 3 kg m 2 s 3 2t =dtefga(b) f = 0.005 89 W 440 s = 2.59 N m(c)P = = 0.005 89 W t 5 rad s =(d)P f = 0.092 5 W s 440 s = 40.7 W(e)T =m(f)W=bgbbfgav2= mr 2 = 1.20 kg 0.012 5 m s t 5 rad srbzzb440 sgb440 s0(g)g b0.092 5 W sgtgb0Pdt =g0.092 5 W s tdt =jag210.092 5 J s 2 440 s2eb3.70 N sgt=f2= 8.96 kJThe power the brake injects into the sliding block through the string isbgbg bgPb = F v = Tv cos 180 = 3.70 N s t 0.012 5 m s = 0.046 3 W s t =zzb440 s440 s0Wb =0Pb dt = g0.046 3 W s tdtbga10.046 3 W s 440 s2=(h)dWbdt W = W + Wb = 8.96 kJ 4.48 kJ =f2= 4.48 kJ4.48 kJJust half of the work required to increase the angular momentum goes into rotational kineticenergy. The other half becomes internal energy in the brake.P11.47Using conservation of angular momentum, we havee je jLaphelion = Lperihelion or mra2 a = mrp2 p .vpe j v = emr j rrThus, mra2aara v a = rp v p or v a =2prpragivingpvp =bg0.590 AU54.0 km s = 0.910 km s .35.0 AU344Angular Momentum(a) = MgR MgR =(b)P11.48 =0dL, and sincedt = 0 , L =constant.Since the total angular momentum of the system is zero, themonkey and bananas move upward with the same speedat any instant, and he will not reach the bananas (until theyget tangled in the pulley). Also, since the tension in the rope isthe same on both sides, Newtons second law applied to themonkey and bananas give the same acceleration upwards.FIG. P11.48P11.49(a) = r F = r F sin180 = 0Angular momentum is conserved.L f = Limrv = mri viv=b gm ri vimv 2=rr3ri vir2(b)T=(c)The work is done by the centripetal force in thenegative direction.Method 1:b gW = z F d = z Tdr = zdr =2a r fIF Imbr v g F 1=GH r r1 JK = 1 mv GH rr 1JK22rrii ib gar fFIG. P11.49m ri vi2iMethod 2:(d)m ri vi322W = K =22i2 r2ri2i2FGHIJKr2111mv 2 mvi2 =mvi2 i2 1222rUsing the data given, we findv = 4.50 m sT = 10.1 NW = 0.450 JChapter 11P11.50(a)Angular momentum is conserved:FGHFGHmvi d1d=Md 2 + m2122=(b)mIJ IJKK2vi6mviMd + 3mddOO(a)1The original energy is mvi2 .2(b)FIG. P11.50The final energy isFGHI 36m vJK aMd + 3mdf2 2i1 2 1 1md 2I =Md 2 +22 1242=3m 2 vi2 d.2 Md + 3mdafThe loss of energy is3m 2 vi2 dmMvi2 d1=mvi2 22 Md + 3md2 Md + 3mdaf afand the fractional loss of energy isamMvi2 d 22 Md + 3mdP11.51(a)fmvi2=M.M + 3mFG d IJH 2KL = 2b75.0 kg gb5.00 m sga5.00 mfLi = m1 v1i r1i + m 2 v 2i r2i = 2mviLi = 3 750 kg m 2 s(b)3451122m1 v1i + m 2 v 2i22175.0 kg 5.00 m sKi = 22Ki =FG IJ bH Kgbg2= 1.88 kJ(c)Angular momentum is conserved: L f = Li = 3 750 kg m 2 s(d)vf =(e)Kf = 2(f)W = K f K i = 5.62 kJLf=3 750 kg m 2 s= 10.0 m s2 75.0 kg 2.50 md i b2 mr ffgaFG 1 IJ b75.0 kggb10.0 m sgH 2K2= 7.50 kJFIG. P11.51346P11.52Angular Momentum(a)(b)(c)LM FG d IJ OP = MvdN H 2K QF1 IK = 2G Mv J = MvH2 KLi = 2 Mv22L f = Li = MvdLf(d)(e)(f)*P11.53Kf = 2FIG. P11.52Mvd== 2vvf =2 Mr f 2 M d4W = K f K i = 3 Mv 2chFG 1 Mv IJ = Ma2 vfH2 K2f2= 4Mv 2The moment of inertia of the rest of the Earth isI=22MR 2 = 5.98 10 24 kg 6.37 10 6 m55ej2= 9.71 10 37 kg m 2 .For the original ice disks,I=11Mr 2 = 2.30 10 19 kg 6 10 5 m22ej2= 4.14 10 30 kg m 2 .For the final thin shell of water,I=22Mr 2 = 2.30 10 19 kg 6.37 10 6 m33ej2= 6.22 10 32 kg m 2 .Conservation of angular momentum for the spinning planet is expressed by Ii i = I f f2e4.14 10 + 9.71 10 j 86 2 s = e6.22 10 + 9.71 10 j b86 400 s + g400F1 + I F 1 + 4.14 10 I = F 1 + 6.22 10 IGH 86 400 s JK GH 9.71 10 JK GH 9.71 10 JK303732303237376.22 10 32 4.14 10 30=86 400 s 9.71 10 37 9.71 10 37 = 0.550 s37Chapter 11P11.54For the cube to tip over, the center of mass (CM) must rise so that itis over the axis of rotation AB. To do this, the CM must be raised aedistance of ajCM2 1 .e1I cube 22j Mga2 1 =DAFrom conservation of angular momentum,FGH4a8 Mamv =33mv=2 MaFGH2P11.55CIJKDB4a/3AIJK1 8 Ma 2 m 2 v 2= Mga234M 2 a 2M3 gamv=347e2 1e2 1jFIG. P11.54jAngular momentum is conserved during theinelastic collision.Mva = I=Mva 3 v=I8aThe condition, that the box falls off the table, is thatthe center of mass must reach its maximum heightas the box rotates, hmax = a 2 . Using conservationof energy:1 2I = Mg a 2 a2eFGHjI FG 3v IJ = MgeaJK H 8 a K16v =gae 2 1j3OL gav = 4M e 2 1jP3QN1 8 Ma 22322 ajFIG. P11.55212P11.56(a)The net torque is zero at the point of contact, so the angular momentum before and after thecollision must be equal.FG 1 MR IJ = FG 1 MR IJ + eMR jH2 K H2 K2(b)E=Ee1 12 2MR 2i 23je jR i 23e je MR j+1M21 12 22i22ie1 12 22jMR 2 i2= =23i3348Angular MomentumpR iMvMR===Mg Mg3 gf(a)t =(b)P11.57W = K =Mgx =1 2 1I =MR 2 i22181MR 2 i218x=(See Problem 11.56)R 2 i218 gANSWERS TO EVEN PROBLEMSP11.2(a) 740 cm 2 ; (b) 59.5 cmP11.328.54 rad sP11.4(a) 168; (b) 11.9 principal value;(c) Only the first is unambiguous.P11.34(a) 7.20 10 3 kg m 2 s ; (b) 9.47 rad sP11.3612.3 m s 2P11.38(a) 2.35 rad s; (b) 0.498 rad s ; (c) 5.58P11.40131 sP11.42(a) 2.19 10 6 m s ; (b) 2.18 10 18 J ;P11.6No; see the solutionP11.8(a) 7.00 N m k ; (b) 11.0 N m kP11.10see the solutionP11.12e22.0 kg m sjkP11.14see the solutionP11.16(a) 3.14 N m ; (b) 0. 400 kg m v ;af(c) 4.13 10 16 rad s2P11.44b(c) 7.85 m sP11.18afg2ej(a) +9.03 10 9 kg m 2 s south; (b) No;P11.46bgbg(e) b3.70 N sgt ; (f) 8.96 kJ; (g) 4.48 kJ(a) 0.005 89 W t ; (b) 2.59 N m ;(c) 0.092 5 W s t ; (d) 40.7 W ;(c) 0(h) +4.48 kJP11.20103 N mP11.22e4.50 kg m sj upP11.24(a) 11.1 m s; (b) 5.32 10 3 kg m 2 s ;(c) see the solution; (d) 12.0 m s;(e) 1.08 kJ ; (f) 5.34 m s; (g) 1.46 m;(h) 1.43 s; (i) see the solution221.20 kg m s perpendicularly into theclock faceP11.26P11.28P11.30(a) 1.91 rad s ; (b) 2.53 J; 6.44 J(a) 0; (b) 0; noP11.50(a)P11.52(a) Mvd ; (b) Mv 2 ; (c) Mvd ; (d) 2v;(e) 4Mv 2 ; (f) 3 Mv 28.63 m s 2KfI1I=(a) 1 i ; (b)Ki I1 + I 2I1 + I 2P11.486mviM; (b)M + 3mMd + 3mdP11.54M3 gamP11.56(a)e2 1ji2E=; (b)E3312Static Equilibrium and ElasticityCHAPTER OUTLINE12.1The Conditions forEquilibriumMore on the Center ofGravityExamples of Rigid Objectsin Static EquilibriumElastic Properties of SolidsANSWERS TO QUESTIONS12.312.4Q12.1When you bend over, your center of gravity shifts forward.Once your CG is no longer over your feet, gravity contributesto a nonzero net torque on your body and you begin to rotate.Q12.2Yes, it can. Consider an object on a spring oscillating back andforth. In the center of the motion both the sum of the torquesand the sum of the forces acting on the object are (separately)zero. Again, a meteoroid flying freely through interstellar spacefeels essentially no forces and keeps moving with constantvelocity.Q12.312.2Noone condition for equilibrium is that F = 0 . For this tobe true with only a single force acting on an object, that forcewould have to be of zero magnitude; so really no forces act onthat object.Q12.4(a)Consider pushing up with one hand on one side of a steering wheel and pulling downequally hard with the other hand on the other side. A pair of equal-magnitude oppositelydirected forces applied at different points is called a couple.(b)An object in free fall has a non-zero net force acting on it, but a net torque of zero about itscenter of mass.Q12.5No. If the torques are all in the same direction, then the net torque cannot be zero.Q12.6(a)Yes, provided that its angular momentum is constant.(b)Yes, provided that its linear momentum is constant.Q12.7A V-shaped boomerang, a barstool, an empty coffee cup, a satellite dish, and a curving plastic slideat the edge of a swimming pool each have a center of mass that is not within the bulk of the object.Q12.8Suspend the plywood from the nail, and hang the plumb bob from the nail. Trace on the plywoodalong the string of the plumb bob. Now suspend the plywood with the nail through a different pointon the plywood, not along the first line you drew. Again hang the plumb bob from the nail and tracealong the string. The center of gravity is located halfway through the thickness of the plywoodunder the intersection of the two lines you drew.349350Static Equilibrium and ElasticityQ12.9The center of gravity must be directly over the point where the chair leg contacts the floor. Thatway, no torque is applied to the chair by gravity. The equilibrium is unstable.Q12.10She can be correct. If the dog stands on a relatively thick scale, the dogs legs on the ground mightsupport more of its weight than its legs on the scale. She can check for and if necessary correct forthis error by having the dog stand like a bridge with two legs on the scale and two on a book ofequal thicknessa physics textbook is a good choice.Q12.11If their base areas are equal, the tall crate will topple first. Its center of gravity is higher off the inclinethan that of the shorter crate. The taller crate can be rotated only through a smaller angle before itscenter of gravity is no longer over its base.Q12.12The free body diagram demonstrates that it is necessary to havefriction on the ground to counterbalance the normal force of thewall and to keep the base of the ladder from sliding. Interestinglyenough, if there is friction on the floor and on the wall, it is notpossible to determine whether the ladder will slip from theequilibrium conditions alone.FIG. Q12.12Q12.13When you lift a load with your back, your back muscles must supply the torque not only to rotateyour upper body to a vertical position, but also to lift the load. Since the distance from thepivotyour hipsto the loadessentially your shouldersis great, the force required to supplythe lifting torque is very large. When lifting from your knees, your back muscles need only keepyour back straight. The force required to do that is much smaller than when lifting with your back,as the torque required is small, because the moment arm of the load is smallthe line of action ofthe load passes close to your hips. When you lift from your knees, your much stronger leg and hipmuscles do the work.Q12.14Shear deformation.Q12.15The vertical columns experience simple compression due to gravity acting upon their mass. Thehorizontal slabs, however, suffer significant shear stress due to gravity. The bottom surface of asagging lintel is under tension. Stone is much stronger under compression than under tension, sohorizontal slabs are more likely to fail.351Chapter 12SOLUTIONS TO PROBLEMSSection 12.1P12.1The Conditions for EquilibriumTo hold the bat in equilibrium, the player must exert both aforce and a torque on the bat to make Fx = Fy = 0andF0.600 m = 0 Fy = 0 F 10.0 N = 0 , or the player must exert a netOupward force of F = 10.0 N10.0 NTo satisfy the second condition of equilibrium, the player mustexert an applied torque a to makeFIG. P12.1 = a a0.600 mfa10.0 N f = 0 . Thus, the required torque is a = +6.00 N m or 6.00 N m counterclockwiseP12.2Use distances, angles, and forces as shown. The conditions ofequilibrium are: Fy = 0 Fx = 0 = 0 FyFx R x = 0Fy cos FgFxlFy + R y Fg = 0RyFG IJ cos FH 2Ksin = 0xFgRxOFIG. P12.2P12.3Take torques about P.LOLmb gm1 gO p = n0 M 2 + d P + m1 g M 2 + d P + m b gd m 2 gx = 0N Q N Qdm1We want to find x for which n 0 = 0 .x=bm g + m ggd + m g1bm2 g12=bm1g+ m b d + m1m2m2 gO2m2PCGx2nOnPFIG. P12.3352Static Equilibrium and ElasticitySection 12.2P12.4More on the Center of GravityThe hole we can count as negative massxCG =m 1 x1 m 2 x 2m1 m 2Call the mass of each unit of pizza area.xCG =xCG =P12.5c h c h c hR 2 0 R 2R834=R 22R2R 22R6The coordinates of the center of gravity of piece 1 are4.00 cmx1 = 2.00 cm and y1 = 9.00 cm .The coordinates for piece 2 are18.0 cm1x 2 = 8.00 cm and y 2 = 2.00 cm .2The area of each piece is12.0 cmA1 = 72.0 cm 2 and A 2 = 32.0 cm 2 .FIG. P12.5And the mass of each piece is proportional to the area. Thus, m i xixCG = mie72.0 cm ja2.00 cmf + e32.0 cm ja8.00 cmf ==2272.0 cm 2 + 32.0 cm 23.85 cmandyCG = m i yi mie72.0 cm ja9.00 cmf + e32.0 cm ja2.00 cmf =2=2104 cm 26.85 cm .4.00 cmChapter 12P12.6Let represent the mass-per-face area. Avertical strip at position x, with width dx andheightax 3.00fy21.00 mhas mass9afy = (x 3.00)2/92 x 3.00 dxdm =.9The total mass isxza fz3.002 x 3 dxM = dm =9x =0FG IJ z ex 6 x + 9jdxH 9KF I L x 6x + 9xOPM = G JM H 9 KN 3 2 Q3.00M=353x02FIG. P12.6033.00 mdx3.002=0The x-coordinate of the center of gravity isxCG =P12.7zxdmM=19z3.00a f2x x 3 dx =09ze3.00jx 3 6 x 2 + 9 x dx =0LMN1 x 4 6x3 9x 2+9 432OPQ3.00=06.75 m= 0.750 m9.00Let the fourth mass (8.00 kg) be placed at (x, y), thenxCG = 0 =x=Similarly,yCG = 0 =a3.00fa4.00f + m axf412.0 + m 412.0= 1.50 m8.00a3.00fa4.00f + 8.00byg12.0 + 8.00y = 1.50 mP12.8In a uniform gravitational field, the center of mass and center of gravity of an object coincide. Thus,the center of gravity of the triangle is located at x = 6.67 m , y = 2.33 m (see the Example on thecenter of mass of a triangle in Chapter 9).The coordinates of the center of gravity of the three-object system are then:b6.00 kg ga5.50 mf + b3.00 kgga6.67 mf + b5.00 kg ga3.50 mfa6.00 + 3.00 + 5.00f kgxCG = m i xi mixCG =35.5 kg m= 2.54 m and14.0 kgyCG = m i yi miyCG =66.5 kg m= 4.75 m14.0 kg==b6.00 kg ga7.00 mf + b3.00 kgga2.33 mf + b5.00 kgga+3.50 mf14.0 kg354Static Equilibrium and ElasticitySection 12.3P12.9Examples of Rigid Objects in Static Equilibrium = 0 = mga3r f Tr2T Mg sin 45.0 = 03rbgMg sin 45.0 1 500 kg g sin 45.0=22= 530 9.80 NT=a fa fm=T 530 g== 177 kg3g3gm1 500 kg = 45FIG. P12.9*P12.10(a)For rotational equilibrium of the lowest rod about its point of support,m1 = 9.00 g+12.0 g g 3 cm m1 g 4 cm(b)For the middle rod,+ m 2 2 cm 12.0 g + 9.0 g 5 cm = 0b(c)P12.11gm 2 = 52.5 gFor the top rod,52.5 g + 12.0 g + 9.0 g 4 cm m3 6 cm = 0b = 0 .gm3 = 49.0 gFg standard weight24.0 cm26.0 cmFg weight of goods soldaf a fF 13 IF = F G JH 12 KF F F I 100 = F 13 1I 100 =GH F JK GH 12 JKFg 0.240 = Fg 0.260gFgFggggFIG. P12.118.33%g*P12.12(a)Consider the torques about an axis perpendicularto the page and through the left end of thehorizontal beam.T = +aT sin 30.0fd a196 N fd = 0 ,Vgiving T = 392 N .(b)30.0H196 NdFIG. P12.12afFrom Fx = 0 ,From Fy = 0 , V + T sin 30.0200 N = 0 , or V = 196 N a392 N f sin 30.0 =H T cos 30.0 = 0 , or H = 392 N cos 30.0 = 339 N to the right .0 .355Chapter 12P12.13(a) Fx = f n w = 0 Fy = n g 800 N 500 N = 0nwTaking torques about an axis at the foot of the ladder,a800 Nfa4.00 mf sin 30.0+a500 Nfa7.50 mf sin 30.0n a15.0 cmf cos 30.0 = 0500 NngwSolving the torque equation,nwa4.00 mfa800 Nf + a7.50 mfa500 Nf tan 30.0 = 268 N .=800 NfA15.0 mNext substitute this value into the Fx equation to findf = n w = 268 NSolving the equationFIG. P12.13in the positive x direction. Fy = 0 ,n g = 1 300 N in the positive y direction.(b)In this case, the torque equation A = 0gives:a9.00 mfa800 Nf sin 30.0+a7.50 mfa500 Nf sin 30.0a15.0 mfbn g sin 60.0 = 0worn w = 421 N .Since f = n w = 421 N and f = fmax = n g , we findfmax421 N== 0.324 .1 300 Nng=P12.14(a) Fx = f n w = 0 Fy = n g m1 g m 2 g = 0nw(1)m2 g(2)F LI A = m1 g G 2 J cos m 2 gx cos + n w L sin = 0H KFrom the torque equation,nw =LM 1 m g + FG x IJ m g OP cot N2 H LK Q1Then, from equation (1):and from equation (2):(b)m1 g2L1 F xI Of = n = M m g + G J m g P cot N2 H LK Qn = bm + m g gwg1122If the ladder is on the verge of slipping when x = d ,then=fx=dng=em12+m2dLj cot m1 + m 2.fAngFIG. P12.14356P12.15Static Equilibrium and Elasticity(a)Taking moments about P,aR sin 30.0f0 + aR cos 30.0fa5.00 cmf a150 Nfa30.0 cmf = 0R = 1 039.2 N = 1.04 kNThe force exerted by the hammer on the nail is equal in magnitudeand opposite in direction:1.04 kN at 60 upward and to the right.(b)FIG. P12.15f = R sin 30.0150 N = 370 Nn = R cos 30.0 = 900 Naf afFsurface = 370 N i + 900 N jP12.16See the free-body diagram at the right.When the plank is on the verge of tipping about point P, thenormal force n1 goes to zero. Then, summing torques aboutpoint P gives p = mgd + Mgx = 0orF mIx = G Jd .H MKFrom the dimensions given on the free-body diagram, observethat d = 1.50 m Thus, when the plank is about to tip,x=P12.17F 30.0 kg I a1.50 mf =GH 70.0 kg JKMg3.00 mxPn1mgd6.00 mFIG. P12.160.643 m .Torque about the front wheel is zero.afb g afb g0 = 1.20 m mg 3.00 m 2 FrThus, the force at each rear wheel isFr = 0.200mg = 2.94 kN .The force at each front wheel is thenFf =mg 2 Fr= 4. 41 kN .2FIG. P12.17n21.50 mChapter 12P12.18 Fx = Fb Ft + 5.50 N = 0(1)3575.50 N Fy = n mg = 0Summing torques about point O,10.0 m O = Ft a1.50 mf a5.50 mfa10.0 mf = 0mgFtwhich yields Ft = 36.7 N to the left1.50 mThen, from Equation (1),FbOFb = 36.7 N 5.50 N = 31.2 N to the rightnFIG. P12.18P12.20(a)Te sin 42.0 = 20.0 NTe = 29.9 N(b)P12.19Te cos 42.0 = TmTm = 22.2 NRelative to the hinge end of the bridge, the cable is attachedhorizontally out a distance x = 5.00 m cos 20.0 = 4.70 m anda fvertically down a distance y = a5.00 mf sin 20.0 = 1.71 m . Thecable then makes the following angle with the horizontal: = tan 1(a)LM a12.0 + 1.71f m OP = 71.1 .N 4.70 m Qafafffawhich yields T = 35.5 kN Fx = 0 Rx T cos 71.1 = 0or(c)afbR x = 35.5 kN cos 71.1 = 11.5 kN rightg Fy = 0 R y 19.6 kN + T sin 71.19.80 kN = 0Thus,afR y = 29. 4 kN 35.5 kN sin 71.1 = 4.19 kN= 4.19 kN downy19.6 kN9.80 kNf9.80 kN 7.00 m cos 20.0 = 0(b)20.0Rx7.00 mT cos 71.1 1.71 m + T sin 71.1 4.70 max5.00 mR x 0 + R y 0 19.6 kN 4.00 m cos 20.0aT4.00 mTake torques about the hinge end of the bridge:afRyFIG. P12.20358*P12.21Static Equilibrium and Elasticity(a)We model the horse as a particle. The drawbridge will fallout from under the horse. = mg=(b)ecos 01213m=2jaf1 2I = mgh21 111 sin 0 m 2 2 = mg 2 32b =(c)3g= 1.73 rad sFIG. P12.21(a)gb1 sin g = e 8.00 m j a1 sin 20f =01.56 rad sThe linear acceleration of the bridge is:feRy011a= = 8.0 m 1.73 rad s 2 = 6.907 m s 222jThe force at the hinge + the force of gravity produce theacceleration a = 6.907 m s 2 at right angles to the bridge.bRxmg23 9.80 m s 2a03gcos 023 9.80 m s 2 cos 20.02 8.00 mRygeaRxmgFIG. P12.21(c)jR x = ma x = 2 000 kg 6.907 m s 2 cos 250 = 4.72 kNRy mg = ma yj begej R y = m g + a y = 2 000 kg 9.80 m s 2 + 6.907 m s 2 sin 250 = 6.62 kNejThus: R = 4.72 i + 6.62 j kN .(d)Rx = 0a=2FG 1 IJ = b1.56 rad sg a4.0 mf = 9.67 m sH2 K2R y mg = mabgejRy2 R y = 2 000 kg 9.8 m s 2 + 9.67 m s 2 = 38.9 kNThus: R y = 38.9 j kNamgFIG. P12.21(d)RxChapter 12P12.22Call the required force F, withcomponents Fx = F cos 15.0 andFxcenter of the wheel by the handles.RF cos 15.0n xbbnx8.00 cmJust as the wheel leaves the ground, theground exerts no force on it. Fx = 0 : Fy = 0 :400 NFyFy = F sin 15.0 , transmitted to thenyaadistances(1)forces F sin 15.0400 N + n y = 0 (2)FIG. P12.22Take torques about its contact point with the brick. The needed distances are seen to be:afb = R 8.00 cm = 20.0 8.00 cm = 12.0 cm22a = R b = 16.0 cm(a)a = 0 :f Fx b + Fy a + 400 N a = 0 , orafafafafF 12.0 cm cos 15.0+ 16.0 cm sin 15.0 + 400 N 16.0 cm = 0F=so(b)6 400 N cm= 859 N7.45 cmThen, using Equations (1) and (2),afn x = 859 N cos 15.0 = 830 N andafn y = 400 N + 859 N sin 15.0 = 622 N22n = n x + n y = 1.04 kN = tan 1F n I = tan a0.749f =GH n JKy136.9 to the left and upwardx*P12.23When x = x min , the rod is on the verge of slipping, sob gf = fsFrommax= sn = 0.50n . Fx = 0 , n T cos 37 = 0 , or n = 0.799T .af Fy = 0 ,Using = 0a2.0 m37nfxFgFg2.0 mFIG. P12.23Thus, f = 0.50 0.799T = 0.399TFromf + T sin 372 Fg = 0 , or 0.399T 0.602T 2 Fg = 0 , giving T = 2.00 Fg .for an axis perpendicular to the page and through the left end of the beam givesf e j359af Fg x min Fg 2.0 m + 2 Fg sin 37 4.0 m = 0 , which reduces to x min = 2.82 m .360P12.24Static Equilibrium and Elasticityx=3L4LIf the CM of the two bricks does not lie over the edge, thenthe bricks balance.If the lower brick is placedLover the edge, then the4second brick may be placed so that its end protrudesover the edge.P12.25x3L4FIG. P12.24To find U, measure distances and forces from point A. Then, balancing torques,a0.750fU = 29.4a2.25fU = 88.2 NTo find D, measure distances and forces from point B. Then, balancing torques,a0.750fD = a1.50fa29.4fAlso, notice that U = D + Fg , so*P12.26D = 58.8 N Fy = 0 .Consider forces and torques on the beam. Fx = 0 : Fy = 0 : = 0 :R cos T cos 53 = 0R sin + T sin 53800 N = 0aT sin 53f8 m a600 Nfx a200 Nf4 m = 0600 Nx + 800 N m= 93.9 N m x + 125 N . As x increases from 2 m, this expression8 m sin 53grows larger.b(a)Then T =(b)gFrom substituting back,R cos = 93.9 x + 125 cos 53R sin = 800 N 93.9 x + 125 sin 53Dividing, tan =800 NR sin = tan 53+93.9 x +125 cos 53R cos afF 32 1IJtan = tan 53 GH 3x + 4 KAs x increases the fraction decreases and decreases .continued on next pageChapter 12(c)To find R we can work out R 2 cos 2 + R 2 sin 2 = R 2 . From the expressions above forR cos and R sin ,aR 2 = T 2 cos 2 53+T 2 sin 2 531 600 NT sin 53+ 800 NR 2 = T 2 1 600T sin 53+640 000aR 2 = 93.9 x + 125f2af2f 1 278 93.9 x + 125 + 640 000eR = 8 819 x 2 96 482 x + 495 678j12At x = 0 this gives R = 704 N . At x = 2 m , R = 581 N . At x = 8 m , R = 537 N . Over therange of possible values for x, the negative term 96 482x dominates the positive term8 819 x 2 , and R decreases as x increases.Section 12.4P12.27LF=YALiL =P12.28Elastic Properties of Solids(a)a fa fa fejej200 9.80 4.00FLi== 4.90 mmAY0.200 10 4 8.00 10 10stress =FF= 2A raf FGH d IJK2e8F = stress F = 1.50 102F 2.50 10 m IN m j GH 2 JK-222F = 73.6 kN(b)aL =*P12.29fstress = Y strain =YLLiastressfL = e1.50 10iYThe definition of Y =8jaf=N m 2 0.250 m1.50 1010N m22.50 mmstressmeans that Y is the slope of the graph:strainY=300 10 6 N m 2= 1.0 10 11 N m 2 .0.003361362P12.30Static Equilibrium and ElasticityCount the wires. If they are wrapped together so that all support nearly equal stress, the numbershould be20.0 kN= 100 .0.200 kNSince cross-sectional area is proportional to diameter squared, the diameter of the cable will bea1 mmfP12.31100 ~ 1 cm .From the defining equation for the shear modulus, we find x asejaf5.00 10 3 m 20.0 Nhfx === 2.38 10 5 m4622SA3.0 10 N m 14.0 10 mejejor x = 2.38 10 2 mm .P12.32The force acting on the hammer changes its momentum according toa fmvi + F t = mv f so F =Hence, F =30.0 kg 10.0 m s 20.0 m s0.110 sm v f vit.= 8.18 10 3 N .By Newtons third law, this is also the magnitude of the average force exerted on the spike by thehammer during the blow. Thus, the stress in the spike is:stress =and the strain is: strain =P12.33(a)a fa f= e5.00 10F 8.18 10 3 N== 1.97 10 7 N m 2b0.023 0 mg2A4stress 1.97 10 7 N m 2== 9.85 10 5 .Y20.0 10 10 N m 2F = A stressF3mj e4.00 1028N m2j3.0 ft= 3.14 10 4 N(b)tAThe area over which the shear occurs is equal tothe circumference of the hole times its thickness.Thus,a fejejA = 2r t = 2 5.00 10 3 m 5.00 10 3 m= 1.57 10af4emFIG. P12.332jejSo, F = A Stress = 1.57 10 4 m 2 4.00 10 8 N m 2 = 6.28 10 4 N .Chapter 12P12.34363Let the 3.00 kg mass be mass #1, with the 5.00 kg mass, mass # 2. Applying Newtons second law toeach mass gives:m1 a = T m1 g(1)m2 a = m2 g Tand(2)where T is the tension in the wire.Solving equation (1) for the acceleration gives: a =and substituting this into equation (2) yields:T g,m1m2T m2 g = m2 g T .m1Solving for the tension T givesbgbgej22m1 m 2 g 2 3.00 kg 5.00 kg 9.80 m s== 36.8 N .8.00 kgm 2 + m1T=From the definition of Youngs modulus, Y =FLi, the elongation of the wire is:A La fa36.8 Nfa2.00 mfTLL ==YAe2.00 10 N m j e2.00 10i11P12.3523jm2= 0.029 3 mm .Consider recompressing the ice, which has a volume 1.09V0 .F V IJ = e2.00 10 N m ja0.090f =P = BG1.09HV K921.65 10 8 N m 2i*P12.36B=PVVi=PViVej1.13 10 8 N m 2 1 m3PVi== 0.053 8 m 3B0.21 10 10 N m 2(a)V = (b)The quantity of water with mass 1.03 10 3 kg occupies volume at the bottom1 m 3 0.053 8 m 3 = 0.946 m 3 . So its density is(c)*P12.371.03 10 3 kg0.946 m3= 1.09 10 3 kg m3 .With only a 5% volume change in this extreme case, liquid water is indeed nearlyincompressible.Part of the load force extends the cable and part compresses the column by the same distance :F=YA A A +Ys As A =sFYA A AA+Ys As=s= 8.60 10e8 500 N7 10 10 0.162 4 2 0 .161 4 2a f4 3. 254mj + 20 10 b0.012 7 g4a 5.75 f102364Static Equilibrium and ElasticityAdditional Problems(a)The beam is perpendicular to the wall, since 3 2 + 4 2 = 5 2 . Then sin =(b)*P12.38 hinge = 0 :a ff+T sin 3 m 250 N 10 m = 0T=(c)a4m; = 53.1 .5m2 500 Nm= 1.04 10 3 N3 m sin 53.11.04 10 3 NT== 0.126 mk 8.25 10 3 N mThe cable is 5.126 m long. From the law of cosines,x=a fa5.126 m4mf4 2 = 5.126 2 + 3 2 2 3 5.126 cos = cos 1(d)223m23 + 5.126 4= 51.22 3 5.126a fafFIG. P12.38From the law of sines, the angle the hinge makes with the wall satisfiessin sin 51.2=5.126 m4msin = 0.998 58 hinge = 0a fafaf+T 3 m sin 51.2250 N 10 m 0.998 58 = 0T = 1.07 10 3 Nx=(e)1.07 10 3 N= 0.129 m8.25 10 3 N m = cos 1(f)3 2 + 5.129 2 4 2= 51.12 3 5.129a fafNow the answers are self-consistent:sin 51.1= 0.998 514mT 3 m sin 51.1250 N 10 m 0.998 51 = 0sin = 5.129 ma fafaf3T = 1.07 10 Nx = 0.129 5 m = 51.1P12.39Let n A and n B be the normal forces at the points ofsupport.Choosing the origin at point A with Fy = 0 and = 0,ABwe find:ej eje3.00 10 jb g g15.0 e8.00 10 jb g g25.0 + n a50.0 f = 0n A + n B 8.00 10 4 g 3.00 10 4 g = 0 and44B15.0 m50.0 mFIG. P12.39The equations combine to give n A = 5.98 10 5 N and b B = 4.80 10 5 N .365Chapter 12P12.40When the concrete has cured and the pre-stressing tension has been released, the rod presses in onthe concrete and with equal force, T2 , the concrete produces tension in the rod.(a)Thus, L =or(b)f FGH L IJKLaIn the concrete: stress = 8.00 10 6 N m 2 = Y strain = YastressfL = eija8.00 10 6 N m 2 1.50 m9Y30.0 10 N mif2L = 4.00 10 4 m = 0.400 mm .In the concrete: stress =T2= 8.00 10 6 N m 2 , soAcejejT2 = 8.00 10 6 N m 2 50.0 10 4 m 2 = 40.0 kN(c)FG IJH KT2 LiT2L=Ysteel so L =ARLiA R YsteelFor the rod:e4.00 10 Nja1.50 mfL == 2.00 10e1.50 10 m je20.0 10 N m j4410232m = 2.00 mm(d)The rod in the finished concrete is 2.00 mm longer than its unstretched length. To removestress from the concrete, one must stretch the rod 0.400 mm farther, by a total of 2. 40 mm .(e)For the stretched rod around which the concrete is poured:FGHIJKFG L IJ A YH L KF 2.40 10 m I e1.50 10 m je20.0 10T =GH 1.50 m JKLtotalT1=YsteelARLior T1 =341*P12.41totali2With as large as possible, n1 and n 2 will both be large. Theequality sign in f 2 sn 2 will be true, but the less-than signin f1 < sn1 . Take torques about the lower end of the pole.n 2 cos + FgFG 1 IJ cos fH2 K2R steelafn2sin = 01Fg = 02Since n 2 > 0 , it is necessary that1 0.576 tan < 01= 1.7360.576 > 60.1 tan > =jN m 2 = 48.0 kNSetting f 2 = 0.576n 2 , the torque equation becomesn 2 1 0.576 tan +107.80 ftd<= 9.00 ftsin sin 60.1f2n1Fgf1FIG. P12.41d366P12.42Static Equilibrium and ElasticityCall the normal forces A and B. They make angles and with the vertical. Fx = 0: Fy = 0:MgA sin B sin = 0AA cos Mg + B cos = 0A sin sin Substitute B =A cos + A cos bsin = Mgsin gMgA cos sin + sin cos = Mg sin sin A = Mgsin + bsin B = Mgsin + bP12.43B(a)gB cos A cos gFIG. P12.42See the diagram.(b)B sin A sin If x = 1.00 m , thenTRy60.0RxO O = a700 N fa1.00 mf a200 N fa3.00 mfa fa f+aT sin 60.0fa6.00 mf = 0x700 N 80.0 N 6.00 m3.00 m200 N80.0 N3.00 mFIG. P12.43Solving for the tension gives: T = 343 N .FromFrom(c) Fx = 0 , R x = T cos 60.0 = Fy = 0 , R y = 980 N T sin 60.0 =171 N .683 N .If T = 900 N : O = a700 N fx a 200 N fa3.00 mf a80.0 N fa6.00 mf +Solving for x gives: x = 5.13 m .a900 Nf sin 60.0 a6.00 mf = 0 .367Chapter 12P12.44(a)Sum the torques about top hinge: = 0:T cos 30.0Caf afafC 0 + D 0 + 200 N cos 30.0 0af+200 N sin 30.0 3.00 maf a1.80 mf392 N 1.50 m + A 1.80 maf160 N bright gGiving A =392 NA+B 0 = 01.50 m1.50 mB.FIG. P12.44 Fx = 0 :(b)T sin 30.0DC 200 N cos 30.0+ A = 0C = 160 N 173 N = 13.2 NIn our diagram, this means 13.2 N to the right . Fy = 0 : +B + D 392 N + 200 N sin 30.0 = 0(c)b gB + D = 392 N 100 N = 292 N upGiven C = 0: Take torques about bottom hinge to obtain(d)af af af afafafafA 0 + B 0 + 0 1.80 m + D 0 392 N 1.50 m + T sin 30.0 3.00 m + T cos 30.0 1.80 m = 0so T =P12.45Using588 N m= 192 N .1.50 m + 1.56 maf Fx = Fy = = 0, choosing the origin at the left endof the beam, we have (neglecting the weight of the beam) Fx = Rx T cos = 0 , Fy = Ry + T sin Fg = 0 ,and = Fg aL + d f + T sin a2L + d f = 0.Solving these equations, we find:(a)T=(b)Rxa fsin a 2L + d fF aL + d f cot =Fg L + dg2L + dRy =Fg L2L + dFIG. P12.45368P12.46Static Equilibrium and Elasticity point 0 = 0 givesT sin 25.0aT cos 25.0fFGH 34 sin 65.0IJK + aT sin 25.0fFGH 34 cos 65.0IJKFI= b 2 000 N ga cos 65.0f + b1 200 N gG cos 65.0JH2Kl3l41 200 NFrom which, T = 1 465 N = 1.46 kNFrom65.0H Fx = 0 ,bgH = T cos 25.0 = 1 328 N toward right = 1.33 kNFrom2 000 NT cos 25.0VFIG. P12.46 Fy = 0 ,bgV = 3 200 N T sin 25.0 = 2 581 N upward = 2.58 kNP12.47We interpret the problem to mean that the support atpoint B is frictionless. Then the support exerts a force inthe x direction andFBy = 0 Fx = FBx FAx = 0bgand = b3 000 g ga 2.00f b10 000 g ga6.00f + F a1.00 f = 0 .FAy 3 000 + 10 000 g = 0BxThese equations combine to giveFIG. P12.475FAx = FBx = 6.47 10 NFAy = 1.27 10 5 NP12.48afn= M+m g H = faHfH max = f max = s m + M gmgLcos 60.0+ Mgx cos 60.0 HL sin 60.02 m + M tan 60.0 mx H tan 60.0 m== s2M2MLMgM A = 0 =a31= s tan 60.0 = 0.78924xfMgmgn60.0AfFIG. P12.48Chapter 12P12.49From the free-body diagram, the angle T makes with the rod isT20 = 60.0+20.0 = 80.0and the perpendicular component of T is T sin 80.0.Summing torques around the base of the rod,a = 0:fbagf10 000 N 4.00 m 10 000 N cos 60+T 4.00 m sin 80 = 0T= Fx = 0 :b10 000 Ng cos 60.0 =sin 80.0FV5.08 10 3 N60FH T cos 20.0 = 0FHFH = T cos 20.0 = 4.77 10 3 N Fy = 0 :FIG. P12.49FV + T sin 20.010 000 N = 0bgand FV = 10 000 N T sin 20.0 = 8.26 10 3 NP12.50Choosing the origin at R,(1)(2)(3)R Fx = + R sin 15.0T sin = 0 Fy = 700 R cos 15.0+T cos = 0 = 700 cos a0.180f + T b0.070 0g = 0Solve the equations for from (3), T = 1 800 cos from (1), R =T9015.01 800 sin cos sin 15.01 800 sin cos cos 15.0Then (2) gives 700 + 1 800 cos 2 = 0sin 15.0orcos 2 + 0.388 9 3.732 sin cos = 018.0 cm25.0 cmnFIG. P12.50Squaring, cos 4 0.880 9 cos 2 + 0.010 13 = 0Letu = cos 2 then using the quadratic equation,u = 0.011 65 or 0.869 3Only the second root is physically possible, = cos 1 0.869 3 = 21.2 T = 1.68 10 3 NP12.51andChoosing torques about R, withaf aR = 2.34 10 3 N = 0fFGH IJK af2LL350 N + T sin 12.0 200 N L = 0 .23From which, T = 2.71 kN .Let R x = compression force along spine, and fromR x = Tx = T cos 12.0 = 2.65 kN . Fx = 0FIG. P12.51369370P12.52Static Equilibrium and Elasticity(a)(b)Just three forces act on the rod: forces perpendicular to thesides of the trough at A and B, and its weight. The lines ofaction of A and B will intersect at a point above the rod.They will have no torque about this point. The rods weightwill cause a torque about the point of intersection as inFigure 12.52(a), and the rod will not be in equilibriumunless the center of the rod lies vertically below theintersection point, as in Figure 12.52(b). All three forcesmust be concurrent. Then the line of action of the weight isa diagonal of the rectangle formed by the trough and thenormal forces, and the rods center of gravity is verticallyabove the bottom of the trough.2AO =So cos =(a)22L1+cos 2 30.0 cos 2 60.0 2F cosGH cos230.0260.0FgOFIG. P12.52(a)IJKBFgA30.0L=260.0OFIG. P12.52(b)AO 1= and = 60.0 .2LLocate the origin at the bottom left corner of the cabinetand let x = distance between the resultant normal force andthe front of the cabinet. Then we have Fx = 200 cos 37.0 n = 0 Fy = 200 sin 37.0+n 400 = 0 = na0.600 xf 400a0.300f + 200 sin 37.0 a0.600 f(2)200 cos 37.0 0.400 = 0(3)aFrom (2),From (3),From (1),(b)AIn Figure (b), AO cos 30.0 = BO cos 60.0 andL2 = AO + BO = AO + AOP12.53B(1)fn = 400 200 sin 37.0 = 280 Naf72.2 120 + 280 0.600 64.0280x = 20.1 cm to the left of the front edgex=k =200 cos 37.0= 0.571280In this case, locate the origin x = 0 at the bottom rightcorner of the cabinet. Since the cabinet is about to tip, wecan use = 0 to find h: = 400a0.300f a300 cos 37.0fh = 0h=FIG. P12.53120= 0.501 m300 cos 37.0Chapter 12P12.54(a), (b) Use the first diagram and sum the torques about the lowerfront corner of the cabinet. = 0 F 1.00 m + 400 N 0.300 m = 0af a faa400 Nfa0.300 mf = 120 Nyielding F =1.00 mFx = 0 f + 120 N = 0 , Fy = 0 400 N + n = 0 ,0.300 mFf400 N1.00 mf = 120 Nn = 400 Norsoff 120 NThus, s = == 0.300 .n 400 N(c)nApply F at the upper rear corner and directed so + = 90.0 to obtain the largest possible lever arm. = tan 1FFG 1.00 m IJ = 59.0H 0.600 m K400 Nfa1.00 mf + a0.600 mf + a400 Nfa0.300 mf = 021.00 mThus, = 90.059.0 = 31.0 .Sum the torques about the lower front corner of thecabinet:F3712n0.600 m120 N m= 103 N .soF =1.17 mTherefore, the minimum force required to tip the cabinet isFIG. P12.54103 N applied at 31.0 above the horizontal at the upper left corner .P12.55(a)We can use Fx = Fy = 0 and = 0 with pivot point atthe contact on the floor.ThenPT Fx = T sn = 0 ,L/2 Fy = n Mg mg = 0, andFLMgI = MgaL cos f + mg G 2 cos J T aL sin f = 0HKL/2mgSolving the above equations givesM=FGHm 2 s sin cos 2 cos s sin nIJKfFIG. P12.55This answer is the maximum vaue for M if s < cot . If s cot , the mass M can increasewithout limit. It has no maximum value, and part (b) cannot be answered as stated either. Inthe case s < cot , we proceed.(b)At the floor, we have the normal force in the y-direction and frictional force in the xdirection. The reaction force then isb gR = n 2 + sn2=a M + mf g21 + s .At point P, the force of the beam on the rope isb gF = T 2 + Mg2a2= g M2 + s M + mf2.372P12.56Static Equilibrium and Elasticity(a)The height of pin B is1000 Na10.0 mf sin 30.0 = 5.00 m .B10.0 mThe length of bar BC is thennAnC45.030.0BC =5.00 m= 7.07 m.sin 45.0CAFIG. P12.56(a)Consider the entire truss: Fy = n A 1 000 N + nC = 0 A = b1 000 N g10.0 cos 30.0+nC 10.0 cos 30.0+7.07 cos 45.0=0Which gives nC = 634 N .Then, n A = 1 000 N nC = 366 N .(b)(c)Suppose that a bar exerts on a pin a force not along thelength of the bar. Then, the pin exerts on the bar aforce with a component perpendicular to the bar. Theonly other force on the bar is the pin force on the otherend. For F = 0 , this force must also have acomponent perpendicular to the bar. Then, the totaltorque on the bar is not zero. The contradiction provesthat the bar can only exert forces along its length.FIG. P12.56(b)Joint A:CAB Fy = 0 : C AB sin 30.0+366 N = 0 ,soC AB = 732 NATACnA = 366 N Fx = 0 : C AB cos 30.0+TAC = 0a1000 NfTAC = 732 N cos 30.0 = 634 N30.0Joint B:af Fx = 0 : 732 N cos 30.0CBC cos 45.0 = 0C BC =a732 Nf cos 30.0 =cos 45.0897 NCAB = 732 NB45.0CBCFIG. P12.56(c)Chapter 12P12.57From geometry, observe thatcos =14 = 75.5andFor the left half of the ladder, we have Fx = T R x = 0 Fy = R y + n A 686 N = 0 top = 686 Na1.00 cos 75.5f + T a2.00 sin 75.5f(1)n A 4.00 cos 75.5 = 0(3)af(2)FIG. P12.57For the right half of the ladder we have Fx = Rx T = 0 Fy = nB Ry = 0 top = nB a 4.00 cos 75.5f T a 2.00 sin 75.5f = 0(4)(5)Solving equations 1 through 5 simultaneously yields:(a)T = 133 N(b)n A = 429 Nandn B = 257 N(c)R x = 133 NandR y = 257 NThe force exerted by the left half of the ladder on the right half is to the right anddownward.P12.58 (a)x CG ==yCG = m i xi mib1 000 kg g10.0 m + b125 kgg0 + b125 kg g0 + b125 kg g20.0 m =1 375 kg9.09 mb1 000 kg g10.0 m + b125 kgg20.0 m + b125 kgg20.0 m + b125 kg g01 375 kg= 10.9 m(b)By symmetry, x CG = 10.0 mThere is no change in yCG = 10.9 m(c)P12.59vCG =FG 10.0 m 9.09 m IJ =H 8.00 s K0.114 m sConsidering the torques about the point at the bottom of the bracket yields:b0.050 0 mga80.0 Nf Fb0.060 0 mg = 0 soF = 66.7 N .373374P12.60Static Equilibrium and ElasticityWhen it is on the verge of slipping, the cylinder is in equilibrium.andf1 = n 2 = sn1f 2 = sn 2 Fx = 0 : Fy = 0 : = 0:P + n1 + f 2 = FgP = f1 + f 2As P grows so do f1 and f 2n1Therefore, since s = , f1 = 122nthen(1)P + n1 + 1 = Fg45SoP + n1 = Fg43Therefore, P =Fg8P12.61(a)(a)=(2)or8P = Fg3FLiA La fkLiYAand k =ALizzaL0P12.62FALLiFIG. P12.60FG IJH Kbecomesa fL(b)andF = k L , Youngs modulus is Y =Thus, Y =n 2 n1=24nn3P = 1 + 1 = n124 45 4P+P = Fg4 3f2 =and0W = Fdx = f kx dx =YALizLxdx = YAa L f022 LiTake both balls together. Their weight is 3.33 Nand their CG is at their contact point.P1 Fx = 0 : + P3 P1 = 0P2 = 3.33 N Fy = 0 : + P2 3.33 N = 0 A = 0: P3 R + P2 R 3.33 NaR + R cos 45.0faf+ P1 R + 2 R cos 45.0 = 0Substituting,af afa+ P Ra1 + 2 cos 45.0f = 0 P1 R + 3.33 N R 3.33 N R 1 + cos 45.03.33 NP3Fgf1P2a3.33 Nf cos 45.0 = 2 P cos 45.01P1 = 1.67 N so P3 = 1.67 N(b)FIG. P12.62(a)Take the upper ball. The lines of action of its weight, of P1 ,and of the normal force n exerted by the lower ball all gothrough its center, so for rotational equilibrium there can beno frictional force. Fx = 0 : n cos 45.0 P1 = 01.67 N= 2.36 Ncos 45.0 Fy = 0 : n sin 45.01.67 N = 0 gives the same resultn=1.67 Nn cos 45.0n sin 45.0FIG. P12.62(b)P1Chapter 12P12.63 Fy = 0 :375+380 N Fg + 320 N = 0Fg = 700 NTake torques about her feet:a = 0:f af af380 N 2.00 m + 700 N x + 320 N 0 = 0x = 1.09 mP12.64FIG. P12.63The tension in this cable is not uniform, so this becomes a fairly difficult problem.dLF=L YAAt any point in the cable, F is the weight of cable below that point. Thus, F = gy where is the massper unit length of the cable.Then, y =z FGHLi0IJKzL2g i1 gLidLdy =ydy =2 YALYA 0a fa fa fejeja10.0 1.00f m s =F v IF = mG J = b1.00 kg gH t K0.002 s2y =P12.65(a)(b)(c)2. 40 9.80 5001= 0.049 0 m = 4.90 cm2 2.00 10 11 3.00 10 4stress =4 500 N4 500 NF== 4.50 10 6 N m 20.010 m 0.100 mAafafYes . This is more than sufficient to break the board.376P12.66Static Equilibrium and ElasticityThe CG lies above the center of the bottom. Consider a disk of water at height y above the bottom.Its radius isyfFGH 30.0y cm IJK = 25.0 cm + 3yIyIyIFFFIts area is G 25.0 cm + J . Its volume is G 25.0 cm + J dy and its mass is G 25.0 cm + JHKHKH333Ka25.0 cm + 35.0 25.0 cm22whole mass of the water iszz30 .0 cm30 .0 cmy =0M=FGH0dm =IJK50.0 y y 2+dy39 625 +L50.0 yy O+ PM = M625 y +627 PMNQL50.0a30.0fa30.0f OP+M = M625a30.0f +627 PMNQM = e10 kg cm je 27 250 cm j = 85.6 kg2330.0023333The height of the center of gravity isyCG =z30 .0 cmy =0= ydmMz30 .0 cm0F 625y + 50.0 yGH32+IJKy 3 dy9 MLMOPMNPQ.0 L 625a30.0f=MM 2 + 50.0a30.0f + a3036 f9MN e10 kg cm j453 750 cm= 625 y 2 50.0 y 3 y 4=++2936M2330 .0 cm0334yCGM1.43 10 3 kg cm== 16.7 cm85.6 kg4OPPQ2dy . TheChapter 12P12.67Let represent the angle of the wire with the vertical. The radius ofthe circle of motion is r = 0.850 m sin .For the mass:afv2= mr 2rT sin = m 0.850 m sin 2aa ffa f af e3.90 10 mj e7.00 10 N m je1.00 10 jAY astrainf==ma0.850 mfb1.20 kg ga0.850 mfa2P12.6810mgFIG. P12.67TFurther, = Y strain or T = AY strainAThus, AY strain = m 0.850 m 2 , givingorrf4T Fr = mar = m37732 = 5.73 rad s .For the bridge as a whole:DB A = n A a0f a13.3 kNfa100 mf + nE a 200 mf = 0sonE =a13.3 kNfa100 mf =200 mA6.66 kNECnA Fy = n A 13.3 kN + n E = 0 gives100 m100 m13.3 kNn A = 13.3 kN n E = 6.66 kNAt Pin A: Fy = FAB sin 40.0+6.66 kN = 0 orb6.66 kN= 10.4 kN compressionsin 40.0 Fx = FAC 10.4 kN cos 40.0 = 0 soFAB =afafFABaFAC = 10.4 kN cos 40.0 = 7.94 kN tensionAt Pin B:40.0gFACf Fy = a10.4 kN f sin 40.0 FBC sin 40.0 = 0aThus, FBC = 10. 4 kN tensiona40.0fbbg= 10.4 kN atensionf= 7.94 kN atensionfBy symmetry: FDE = FAB = 10.4 kN compressionFEC = FACWe can check by analyzing Pin C: Fx = +7.94 kN 7.94 kN = 0 or 0 = 0 Fy = 2a10.4 kNf sin 40.013.3 kN = 0which yields 0 = 0 .40.0FBCFBD = 2 10.4 kN cos 40.0 = 15.9 kN compressionFDC = FBCFBDf Fx = FAB cos 40.0+ FBC cos 40.0 FBD = 0andnA = 6.66 kNFAB = 10.4 kNg10.4 kN10.4 kN40.040.07.94 kN7.94 kN13.3 kNFIG. P12.68nE378P12.69Static Equilibrium and ElasticityMember AC is not in pure compression or tension. Italso has shear forces present. It exerts a downwardforce S AC and a tension force FAC on Pin A and onPin C. Still, this member is in equilibrium.SAC25.0 mFACFACCA Fx = FAC FAC = 0 FAC = FAC A = 0: a14.7 kNfa25.0 mf + S AC a50.0 mf = 0or14.7 kNS = 7.35 kNAC Fy = S AC 14.7 kN + 7.35 kN = 0 S AC = 7.35 kNDBThen S AC = S and we have proved that the loading by the car AACis equivalent to one-half the weight of the car pulling down on nAeach of pins A and C, so far as the rest of the truss is concerned.For the Bridge as a whole:SAC25.0 mEC75.0 mnE25.0 m 14.7 kN A = 0:afaf af7.35 kN 14.7 kN 25.0 m + n E 100 m = 0n E = 3.67 kNFAB30.0 Fy = n A 14.7 kN + 3.67 kN = 0FACn A = 11.0 kNAt Pin A:nA = 11.0 kN Fy = 7.35 kN + 11.0 kN FAB sin 30.0 = 0bFAB = 7.35 kN compressionFBDg30.0 Fx = FAC a7.35 kN f cos 30.0 = 0aFAC = 6.37 kN tensionAt Pin B:7.35 kN60.0FBCf4.24 kN Fy = a7.35 kN f sin 30.0 FBC sin 60.0 = 060.0af F = a7.35 kNf cos 30.0+a 4.24 kN f cos 60.0 FF = 8.49 kN bcompressiongFBC = 4.24 kN tensionxBDFCD60.06.37 kNFCE=07.35 kNBDAt Pin C: Fy = a 4.24 kNf sin 60.0+ FCD sin 60.07.35 kN = 0aFCD = 4.24 kN tensionf30.0 Fx = 6.37 kN a4.24 kNf cos 60.0+a 4.24 kNf cos 60.0+ FCE = 0afFCE = 6.37 kN tensionAt Pin E:6.37 kN3.67 kNFIG. P12.69 Fy = FDE sin 30.0+3.67 kN = 0bFDE = 7.35 kN compressionor Fx = 6.37 kN FDE cos 30.0 = 0which gives FDE = 7.35 kN as before.FDEg379Chapter 12P12.70(1)(2)ph = Ipp = MvCMhvCMIf the ball rolls without slipping, R = vCMSo, h =P12.71(a)2III===R5pMvCM MRFIG. P12.70If the acceleration is a, we have Px = ma andPy + n Fg = 0 . Taking the origin at the center ofHLdgravity, the torque equation givesafCGPy L d + Px h nd = 0 .PhSolving these equations, we find(b)F d ah I .L GH g JKah e 2.00 m s ja1.50 mf==If P = 0 , then d =(c)Using the given data, Px = 306 N and Py = 553 N .Py =nFgFygFIG. P12.712y9.80 m s 2ge0.306 m .jThus, P = 306 i + 553 j N .*P12.72When the cyclist is on the point of tipping over forward,the normal force on the rear wheel is zero. Parallel to theplane we have f1 mg sin = ma . Perpendicular to theplane, n1 mg cos = 0 . Torque about the center of mass:af af amgfmg 0 f1 1.05 m + n1 0.65 m = 0 .f1Combining by substitution,FIG. P12.72ma = f1 mg sin =FGHn1 0.65 m0.65 m mg sin = mg cos mg sin 1.05 m1.05 mIJK0.65 sin 20 = 2.35 m s 21.05When the car is on the point of rolling over, the normalforce on its inside wheels is zero.a = g cos 20*P12.73 Fy = ma y :n mg = 0 Fx = ma x :f=n1mghmv 2RTake torque about the center of mass: fh nThen by substitution2mv maxRhmgd=02fd= 0.2v max =mgdgdR2hFIG. P12.73A wider wheelbase (larger d) and a lower center of mass (smaller h) will reduce the risk of rollover.380Static Equilibrium and ElasticityANSWERS TO EVEN PROBLEMSP12.2Fy + R y Fg = 0 ; Fx R x = 0 ;P12.42FG IJ cos FH 2K(a) 0.400 mm; (b) 40.0 kN; (c) 2.00 mm;(d) 2.40 mm; (e) 48.0 kNat A: MgP12.44Fy cos FgP12.40(a) 160 N to the right;(b) 13.2 N to the right; (c) 292 N up;(d) 192 Nsin = 0xsin sin ; at B: Mgsin + sin + bgbgP12.4see the solutionP12.60.750 mP12.8a2.54 m, 4.75 mfP12.10(a) 9.00 g; (b) 52.5 g; (c) 49.0 gP12.461.46 kN ; 1.33 i + 2.58 j kNP12.12(a) 392 N; (b) 339 i + 0 j NeP12.480.789P12.14(a) f =P12.50T = 1.68 kN ; R = 2.34 kN; = 21.2P12.52(a) see the solution; (b) 60.0P12.54(a) 120 N; (b) 0.300; (c) 103 N at 31.0 abovethe horizontal to the rightP12.56(a), (b) see the solution;(c) C AB = 732 N ; TAC = 634 N ; C BC = 897 NP12.58(a) 9.09 m, 10.9 m ; (b) 10.0 m, 10.9 m ;(c) 0.114 m s to the rightngjLM m g + m gx OP cot ;N2 L Qe + j cot = bm + m g g ; (b) =m +m12m1212m2dL12P12.16see the solution; 0.643 mP12.1836.7 N to the left ; 31.2 N to the rightP12.20(a) 35.5 kN; (b) 11.5 kN to the right;(c) 4.19 kN downP12.22(a) 859 N; (b) 104 kN at 36.9 above thehorizontal to the leftP12.243L4P12.60eajfaf3Fg8P12.62(a) P1 = 1.67 N ; P2 = 3.33 N ; P3 = 1.67 N ;(b) 2.36 NP12.26(a) see the solution; (b) decreases ;(c) R decreasesP12.644.90 cmP12.28(a) 73.6 kN; (b) 2.50 mmP12.6616.7 cm above the center of the bottomP12.30~ 1 cmP12.68P12.329.85 10 5P12.340.029 3 mmC AB = 10.4 kN ; TAC = 7.94 kN ;TBC = 10.4 kN ; C BD = 15.9 kN ;C DE = 10.4 kN ; TDC = 10. 4 kN ;TEC = 7.94 kNP12.36(a) 0.053 8 m3 ; (b) 1.09 10 3 kg m3 ;(c) Yes, in most practical circumstancesP12.38(a) 53.1; (b) 1.04 kN; (c) 0.126 m, 51.2;(d) 1.07 kN; (e) 0.129 m, 51.1; (f) 51.1P12.702R5P12.722.35 m s 213Universal GravitationCHAPTER OUTLINE13.113.213.313.413.513.613.7Newtons Law of UniversalGravitationMeasuring the GravitationalConstantFree-Fall Acceleration andthe Gravitational ForceKeplers Laws and theMotion of PlanetsThe Gravitational FieldGravitational PotentialEnergyEnergy Considerations inPlanetary and SatelliteMotionANSWERS TO QUESTIONSQ13.1Because g is the same for all objects near the Earths surface.The larger mass needs a larger force to give it just the sameacceleration.Q13.2To a good first approximation, your bathroom scale reading isunaffected because you, the Earth, and the scale are all in freefall in the Suns gravitational field, in orbit around the Sun. Toa precise second approximation, you weigh slightly less atnoon and at midnight than you do at sunrise or sunset. TheSuns gravitational field is a little weaker at the center of theEarth than at the surface subsolar point, and a little weaker stillon the far side of the planet. When the Sun is high in your sky,its gravity pulls up on you a little more strongly than on theEarth as a whole. At midnight the Sun pulls down on you alittle less strongly than it does on the Earth below you. So youcan have another doughnut with lunch, and your bedspringswill still last a little longer.Q13.3Keplers second law states that the angular momentum of the Earth is constant as the Earth orbitsthe sun. Since L = mr , as the orbital radius decreases from June to December, then the orbital speedmust increase accordingly.Q13.4Because both the Earth and Moon are moving in orbit about the Sun. As described byFgravitational = ma centripetal , the gravitational force of the Sun merely keeps the Moon (and Earth) in anearly circular orbit of radius 150 million kilometers. Because of its velocity, the Moon is kept in itsorbit about the Earth by the gravitational force of the Earth. There is no imbalance of these forces, atnew moon or full moon.Q13.5Air resistance causes a decrease in the energy of the satellite-Earth system. This reduces the diameterof the orbit, bringing the satellite closer to the surface of the Earth. A satellite in a smaller orbit,however, must travel faster. Thus, the effect of air resistance is to speed up the satellite!Q13.6Keplers third law, which applies to all planets, tells us that the period of a planet is proportional tor 3 2 . Because Saturn and Jupiter are farther from the Sun than Earth, they have longer periods. TheSuns gravitational field is much weaker at a distant Jovian planet. Thus, an outer planet experiencesmuch smaller centripetal acceleration than Earth and has a correspondingly longer period.381382Q13.7Universal GravitationTen terms are needed in the potential energy:U = U 12 + U 13 + U 14 + U 15 + U 23 + U 24 + U 25 + U 34 + U 35 + U 45 .With N particles, you needN ai 1f =i =1Q13.8N2 Nterms.2No, the escape speed does not depend on the mass of the rocket. If a rocket is launched at escapespeed, then the total energy of the rocket-Earth system will be zero. When the separation distanceGM E m1= 0 , the massbecomes infinite U = 0 the rocket will stop K = 0 . In the expression mv 2 2rm of the rocket divides out.afafQ13.9It takes 100 times more energy for the 10 5 kg spacecraft to reach the moon than the 10 3 kgspacecraft. Ideally, each spacecraft can reach the moon with zero velocity, so the only term that needbe analyzed is the change in gravitational potential energy. U is proportional to the mass of thespacecraft.Q13.10The escape speed from the Earth is 11.2 km/s and that from the Moon is 2.3 km/s, smaller by a factorof 5. The energy requiredand fuelwould be proportional to v 2 , or 25 times more fuel is requiredto leave the Earth versus leaving the Moon.Q13.11The satellites used for TV broadcast are in geosynchronous orbits. The centers of their orbits are thecenter of the Earth, and their orbital planes are the Earths equatorial plane extended. This is theplane of the celestial equator. The communication satellites are so far away that they appear quiteclose to the celestial equator, from any location on the Earths surface.Q13.12For a satellite in orbit, one focus of an elliptical orbit, or the center of a circular orbit, must be locatedat the center of the Earth. If the satellite is over the northern hemisphere for half of its orbit, it mustbe over the southern hemisphere for the other half. We could share with Easter Island a satellite thatwould look straight down on Arizona each morning and vertically down on Easter Island eachevening.Q13.13The absolute value of the gravitational potential energy of the Earth-Moon system is twice thekinetic energy of the moon relative to the Earth.Q13.14In a circular orbit each increment of displacement is perpendicular to the force applied. The dotproduct of force and displacement is zero. The work done by the gravitational force on a planet in anelliptical orbit speeds up the planet at closest approach, but negative work is done by gravity andthe planet slows as it sweeps out to its farthest distance from the Sun. Therefore, net work in onecomplete orbit is zero.Q13.15Every point q on the sphere that does not liealong the axis connecting the center of thesphere and the particle will have companionpoint q for which the components of thegravitational force perpendicular to the axiswill cancel. Point q can be found by rotatingthe sphere through 180 about the axis. Theforces will not necessarily cancel if the mass isnot uniformly distributed, unless the center ofmass of the non-uniform sphere still lies alongthe axis.qFpqFpqq (behind the sphere)FIG. Q13.15pChapter 13Q13.16Speed is maximum at closest approach. Speed is minimum at farthest distance.Q13.17Set the universal description of the gravitational force, Fg =383Fg = ma gravitational , where M X and R XGM X m, equal to the local description,2RXare the mass and radius of planet X, respectively, and m is themass of a test particle. Divide both sides by m.Q13.18The gravitational force of the Earth on an extra particle at its center must be zero, not infinite as oneinterpretation of Equation 13.1 would suggest. All the bits of matter that make up the Earth will pullin different outward directions on the extra particle.Q13.19Cavendish determined G. Then from g =Q13.20The gravitational force is conservative. An encounter with a stationary mass cannot permanentlyspeed up a spacecraft. Jupiter is moving. A spacecraft flying across its orbit just behind the planetwill gain kinetic energy as the planets gravity does net positive work on it.Q13.21Method one: Take measurements from an old kinescope of Apollo astronauts on the moon. From themotion of a freely falling object or from the period of a swinging pendulum you can find theacceleration of gravity on the moons surface and calculate its mass. Method two: One coulddetermine the approximate mass of the moon using an object hanging from an extremely sensitivebalance, with knowledge of the position and distance of the moon and the radius of the Earth. Firstweigh the object when the moon is directly overhead. Then weigh of the object when the moon isjust rising or setting. The slight difference between the measured weights reveals the cause of tidesin the Earths oceans, which is a difference in the strength of the moons gravity between differentpoints on the Earth. Method three: Much more precisely, from the motion of a spacecraft in orbitaround the moon, its mass can be determined from Keplers third law.Q13.22The spacecraft did not have enough fuel to stop dead in its high-speed course for the Moon.GM, one may determine the mass of the Earth.R2SOLUTIONS TO PROBLEMSSection 13.1P13.1Newtons Law of Universal GravitationFor two 70-kg persons, modeled as spheres,Fg =P13.2F = m1 g =g=Gm 2r2Gm1 m 2r2e6.67 10=11jbgbN m 2 kg 2 70 kg 70 kga2 mf2g~ 10 7 N .Gm1 m 2=r2e6.67 1011jea100 mfN m 2 kg 2 4.00 10 4 10 3 kg2j=2.67 10 7 m s 2384P13.3Universal Gravitation(a)At the midpoint between the two objects, the forces exerted by the 200-kg and 500-kg objectsare oppositely directed,Gm1 m 2and fromwe have(b)Fg =F =r2bgbG 50.0 kg 500 kg 200 kga0.200 mf2g=2.50 10 5 N toward the 500-kg object.At a point between the two objects at a distance d from the 500-kg objects, the net force onthe 50.0-kg object will be zero whenbgb g = Gb50.0 kggb500 kg gda0.400 m dfG 50.0 kg 200 kg2d = 0.245 morP13.42m1 + m 2 = 5.00 kgF =Gm 2 = 5.00 kg m1m1 m 2r2b5.00 kg gmgbm 3.00 kg gbme 1.00 10 8 N = 6.67 10 11 N m 2 kg 221 m1 =be1.00 1086.67 10jeN 0.040 0 m 2112N m kg2j m ba5..00 kg m g0 200 mf112j = 6.00 kg22Thus, m1 5.00 kg m1 + 6.00 kg = 0or11g 2.00 kg = 0giving m1 = 3.00 kg, so m 2 = 2.00 kg . The answer m1 = 2.00 kg and m 2 = 3.00 kg is physicallyequivalent.P13.5The force exerted on the 4.00-kg mass by the 2.00-kg mass isdirected upward and given byF24 = Gm4m22r24ej = 6.67 10 11 N m 2 kg 22j b4.00a3kg gbm.00 kg g j.00 f2= 5.93 10 11 j NThe force exerted on the 4.00-kg mass by the 6.00-kg mass isdirected to the leftF64 = Gm 4 m62r64e ij = e6.67 1011N m 2 kg 26j b4.00a4kg gbm.00 kg g i.00 f2FIG. P13.5= 10.0 10 11 iNTherefore, the resultant force on the 4.00-kg mass is F4 = F24 + F64 =e10.0i + 5.93 jj 1011N .Chapter 13P13.6(a)385The Sun-Earth distance is 1.496 10 11 m and the Earth-Moon distance is 3.84 10 8 m , so thedistance from the Sun to the Moon during a solar eclipse is1.496 10 11 m 3.84 10 8 m = 1.492 10 11 mM S = 1.99 10 30 kgThe mass of the Sun, Earth, and Moon areM E = 5.98 10 24 kgM M = 7.36 10 22 kgandWe have FSM =(b)(c)FEM =FSEe6.67 10e6.67 10=Gm1 m 2r211e6.67 10 je1.99 10 je7.36 10 j ==e1.492 10 j11302211 2jejej=jej=N m 2 kg 2 5.98 10 24 7.36 10 22e3.84 10 j8 211jeN m 2 kg 2 1.99 10 30 5.98 10 24e1.496 10 j11 24.39 10 20 N1.99 10 20 N3.55 10 22 NNote that the force exerted by the Sun on the Moon is much stronger than the force of theEarth on the Moon. In a sense, the Moon orbits the Sun more than it orbits the Earth. TheMoons path is everywhere concave toward the Sun. Only by subtracting out the solarorbital motion of the Earth-Moon system do we see the Moon orbiting the center of mass ofthis system.Section 13.2P13.7Measuring the Gravitational Constantbgej1.50 kg 15.0 10 3 kgGMm1122F== 6.67 10N m kg= 7.41 10 10 N22r24.50 10 mejej386P13.8Universal GravitationLet represent the angle each cable makes with the vertical, L thecable length, x the distance each ball scrunches in, and d = 1 m theoriginal distance between them. Then r = d 2 x is the separation ofthe balls. We have Fy = 0 :T cos mg = 0 Fx = 0 :T sin Thentan =Gmm=0r2FIG. P13.8Gmmr 2 mgx2L x2=aGmg d 2xfax d 2x2f2=Gm 2L x2 .gGmis numerically small. There are two possibilities: either x is small or else d 2 x isgThe factorsmall.Possibility one: We can ignore x in comparison to d and L, obtaininge6.67 10xa1 mf =211jbN m 2 kg 2 100 kge9.8 m s j2eg 45 mx = 3.06 10 8 m.jThe separation distance is r = 1 m 2 3.06 10 8 m = 1.000 m 61.3 nm .Possibility two: If d 2 x is small, x 0.5 m and the equation becomesma0.5 mfr = e6.67 10 b9N N kgkg jb100 kgg a45 mf a0.5 mf.8g1122222r = 2.74 10 4 m .For this answer to apply, the spheres would have to be compressed to a density like that of thenucleus of atom.Section 13.3P13.9a=Free-Fall Acceleration and the Gravitational ForceMGb 4R gE2=9.80 m s 2= 0.613 m s 216.0toward the Earth.e j = 4 GR4 R 3GM G 3g= 2 =RR2P13.10IfgM 1= =gE 6thengM= MgEE34G M R M34G E RE3F I F R I = FG 1 IJ a4f =GH JK GH R JK H 6 KEM2.3Chapter 13P13.11(a)At the zero-total field point,so(b)GmM E2rE387GmM M=2rMMMr7.36 10 22= rE= E24ME9.015.98 10r8rE + rM = 3.84 10 m = rE + E9.013.84 10 8 mrE == 3.46 10 8 m1.11rM = rEAt this distance the acceleration due to the Earths gravity isgE =GM E2rEe6.67 10=11jemjN m 2 kg 2 5.98 10 24 kge3.46 108j2g E = 3.34 10 3 m s 2 directed toward the EarthSection 13.4Keplers Laws and the Motion of Planetsbg32r 2 384 400 10 m== 1.02 10 3 m s .T27.3 86 400 s(a)v=(b)P13.12In one second, the Moon falls a distancebgeej a fj2311 v 2 2 1 1.02 10x = at 2 =t = 1.0022 r2 3.844 10 82= 1.35 10 3 m = 1.35 mm .The Moon only moves inward 1.35 mm for every 1020 meters it moves along a straight-linepath.P13.13Applying Newtons 2nd Law,GMMa 2r f2= F = ma yields Fg = ma c for each star:Mv 2rM=or4v 2 r.GWe can write r in terms of the period, T, by considering the time anddistance of one complete cycle. The distance traveled in one orbit is thecircumference of the stars common orbit, so 2r = vT . ThereforeM=eFG IJH K4v 2 r 4v 2 vT=GG 2ja3fbFIG. P13.13g32 v 3 T 2 220 10 m s 14.4 d 86 400 s d== 1.26 10 32 kg = 63.3 solar massesso, M =1122G 6.67 10N m kgej388P13.14Universal GravitationSince speed is constant, the distance traveled between t1 and t 2 is equal to the distance traveledbetween t 3 and t 4 . The area of a triangle is equal to one-half its (base) width across one side times its(height) dimension perpendicular to that side.Sobgb11bv t 2 t1 = bv t 4 t 322gstates that the particles radius vector sweeps out equal areas in equal times.P13.15T2 =M=4 2 a 3GM4 2 a 3GT 2(Keplers third law with m << M )=ej4 2 4.22 10 8 mejb36.67 10 11 N m 2 kg 2 1.77 86 400 sg2= 1.90 10 27 kg(Approximately 316 Earth masses)P13.16By conservation of angular momentum for the satellite,vprp v p = ra v ava=ra 2 289 km + 6.37 10 3 km 8 659 km=== 1.27 .6 829 kmrp459 km + 6.37 10 3 kmWe do not need to know the period.P13.17By Keplers Third Law, T 2 = ka 3(a = semi-major axis)For any object orbiting the Sun, with T in years and a in A.U.,k = 1.00 . Therefore, for Comet Halleya75.6f = a1.00fFGH 0.570 + y IJK223The farthest distance the comet gets from the Sun isa fy = 2 75.6P13.18 F = ma :23FIG. P13.17 0.570 = 35.2 A. U. (out around the orbit of Pluto)Gm planet M starr2=m planet v 2rGM star= v 2 = r 2 2r33GM star = r 3 3 = rx 2 = ry 2xyy =xFr IGH r JKxy3 2y =F 90.0 I 3GH 5.00 yr JK3 2=4685.00 yrSo planet Y has turned through 1.30 revolutions .FIG. P13.18Chapter 13P13.19GM JdR + di=2d4 2 R J + dTJd11i2GM J T 2 = 4 2 R J + de6.67 10389i3jbjeN m 2 kg 2 1.90 10 27 kg 9.84 3 600g2e= 4 2 6.99 10 7 + dj3d = 8.92 10 7 m = 89 200 km above the planetP13.20The gravitational force on a small parcel of material at the stars equator supplies the necessarycentripetal force:GM s mRs2so==mv 2= mRs 2RsGM s3Rs=e6.67 1011j emjN m 2 kg 2 2 1.99 10 30 kge10.0 103j3 = 1.63 10 4 rad s*P13.21The speed of a planet in a circular orbit is given byGM sun m F = ma :r2=mv 2rGM sun.rv=For Mercury the speed isvM =e6.67 10 je1.99 10 j me5.79 10 j sand for Pluto,vP =e6.67 10 je1.99 10 j me5.91 10 j s113010211301222= 4.79 10 4 m s2= 4.74 10 3 m s .With greater speed, Mercury will eventually move farther from the Sun than Pluto. With originaldistances rP and rM perpendicular to their lines of motion, they will be equally far from the Sunafter time t where2222rP + v P t 2 = rM + v M t 2ej2222rP rM = v M v P t 2t=e5.91 10e4.79 104j e5.79 10m sj e 4.74 1012m22103jm2msj2=3.49 10 25 m 2= 1.24 10 8 s = 393 yr .2.27 10 9 m 2 s 2390*P13.22Universal GravitationGM s m F = ma :For the Earth,r2Also the angular momentumWe eliminateFG LT IJH 2m KFG IJH Kmv 2 m 2r=rr T2.GM s T 2 = 4 2 r 3 .2rL = mvr = mr is a constant for the Earth.TLTr=between the equations:2mThenGM s T 2 = 4 2=32GM sT 1 2 = 4 2FG L IJH 2m K3 2.Now the rate of change is described byGM sFG 1 TH2T t1 2dM F T I TdTIJIJ FG=G2 J K HKdtdt H M K TFG 2 T IJ = 5 000 yrF 3.16 10 s I e3.64 10 kg sjF 2 1 yrGH 1 yr JKGH 1.991 10H MKdM s 1 2dT+G 1=0TdtdtdM sdtss79s30IJkg KT = 1.82 10 2 sSection 13.5P13.23g=sog=The Gravitational FieldGmGmGmi + 2 j + 2 cos 45.0 i + sin 45.0 j2ll2leg=Gml2FGHFGHjGM11+2l2 22+ymIJ ei + jj orKlmlIJK1toward the opposite corner2mOFIG. P13.23P13.24(a)(b)ej ejej6.67 10 11 N m 2 kg 2 100 1.99 10 30 kg 10 3 kgGMmF=== 1.31 10 17 N24r21.00 10 m + 50.0 meF =GMm GMm 22rfrontrbackg =22F GM rback rfront=22mrfront rbackjeje6.67 10 j 100e1.99 10g =e1.00 1011g = 2.62 10 12 N kgj LNMe1.01 10 mj e1.00 10 mj OQPmj e1.01 10 mj3042424422FIG. P13.24xChapter 13P13.25g1 = g 2 =MGr 2 + a2g1y = g 2yg y = g1y + g 2 yrcos =2a + r2g 1 x = g 2 x = g 2 cos ee jj39112g = 2 g2x iorSection 13.6P13.26g=2 MGre2r + a2j3 2toward the center of massFIG. P13.25Gravitational Potential Energy(a)ejbjefg6.67 10 11 N m 2 kg 2 5.98 10 24 kg 100 kgGM E m== 4.77 10 9 J .U=6r6.37 + 2.00 10 ma(b), (c) Planet and satellite exert forces of equal magnitude on each other, directed downward onthe satellite and upward on the planet.F=P13.27U = Gso thatMmrGM E mr2and=e6.67 1011jbjeN m 2 kg 2 5.98 10 24 kg 100 kge8.37 10 mj6g=2569 NGM E2REFG 1 1 IJ = 2 mgRH 3R R K 32U = b1 000 kg ge9.80 m s je6.37 10 mj =3U = GMmEEE2P13.28g=64.17 10 10 J .The height attained is not small compared to the radius of the Earth, so U = mgy does not apply;GM 1 M 2U=does. From launch to apogee at height h,rGM E M pGM E M p1+0=0K i + Ui + Emch = K f + U f :M p vi2 2RERE + h110.0 10 3 m s2ej e2e= 6.67 10 11 N m 2 kg 2e5.00 107.98jFGH 56.3710105.98 10jFGH 6.37 10 mkgh IJK+ 6.67 10 11 N m 2 kg 2j eh = 2.52 10 7 m3.99 10 14 m3 s 21. 26 10 7 m 2 s 26IJmKkg246jm 2 s 2 6.26 10 7 m 2 s 2 =6.37 10 6 m + h =243.99 10 14 m3 s 26.37 10 6 m + h= 3.16 10 7 m392Universal GravitationMS=g=(c)P13.30(a)(b)P13.29Ug = GM SW = U = (a)jmje4 6.37 10e6.67 103611= 1.84 10 9 kg m 3jemjN m 2 kg 2 1.99 10 30 kge6.37 1062ej=3.27 10 6 m s 2jbjeFG Gm mH r11120IJKjejeN m 2 kg 2 7.36 10 22 kg 1.00 10 3 kg61.74 10 mFGHU Tot = U 12 + U 13 + U 23 = 3U 12 = 3 (b)(a)eg6.67 10 11 N m 2 kg 2 1.99 10 30 kg 1.00 kgGM S m== 2.08 10 13 JrE6.37 10 6 mU Tot = *P13.32=2rEe+6.67 10W=P13.31=243 rE3 1.99 10 30 kgeGm 1 m 2r12je2.82 10 9 JIJK3 6.67 10 11 N m 2 kg 2 5.00 10 3 kg0.300 mj=j2= 1.67 10 14 JAt the center of the equilateral triangleEnergy conservation of the object-Earth system from release to radius r:eK + U jgaltitude he= K +Ugjradius rGM E mGM E m 1= mv 2 0RE + h 2rFGHv = 2GM Efdt = it =iRE + h=drdtdr i dr=. The time of fall isv f v2GM EREt =12Ez z zz FGH FGHz LMNf(b)FG 1 1 IJ IH r R + h K JK6.87 10 6 m11r RE + hIJ IK JK1 2dr2 6.67 10 11 5.98 10 246.37 10 6 mFG 1 1 IJ OPH r 6.87 10 m K Q61 2drWe can enter this expression directly into a mathematical calculation program.rAlternatively, to save typing we can change variables to u = 6 . Then101 26. 871 210 6 6.87 111110 6 du = 3.541 10 8t = 7.977 10 141 266u 6.876.87 106. 37 10 u6. 3710 6ejz FGHIJKe jz FGHIJK1 2duA mathematics program returns the value 9.596 for this integral, giving for the time offall t = 3.541 10 8 10 9 9.596 = 339.8 = 340 s .Chapter 13Section 13.7Energy Considerations in Planetary and Satellite MotionFGHIJKFGHIJK11 11= mv 2mvi2 + GM E mf22r f ri1 211= v2vi + GM E 0 f22REorP13.332v 2 = v1 fandvfvf2GM EREF 2GM IJ= Gv H R K= Le 2.00 10 j 1.25 10 OMNPQ2112EE4 2812= 1.66 10 4 m s2 M SunG= 42.1 km sRESun(a)v solar escape =(b)P13.34Let r = RES x represent variable distance from the Sun, with x in astronomical units.v=2 M SunG 42.1=RES xxIf v =125 000 km, then x = 1.47 A.U. = 2.20 10 11 m3 600 s(at or beyond the orbit of Mars, 125 000 km/h is sufficient for escape).To obtain the orbital velocity, we useF =orP13.35v=orvi2GM E=RE + hRE + hbgMGR1mMG2mv esc =2R2 MG=v esc =RWe can obtain the escape velocity fromP13.36mMG mv 2=RR22v2FGHIJK11 GM E m1=K i = mvi2 =+h22 RE2LM e6.67 10 N m kg je5.98 10 kg jb500 kgg OPMNPQ = 1.45 10e6.37 10 mj + e0.500 10 mj1122426610JThe change in gravitational potential energy of the satellite-Earth system isFIGHJK= e6.67 10N m kg je5.98 10 kg jb500 kg ge 1.14 1011= mv = b500 kg ge 2.00 10 m sj = 1.00 10 J .22U =GM E m GM E m11= GM E mRiRfRi R f11Also,Kf222f24329The energy transformed due to friction isafEint = K i K f U = 14.5 1.00 + 2.27 10 9 J = 1.58 10 10 J .8jm 1 = 2.27 10 9 J393394P13.37Universal GravitationFc = FG givesmv 2 GmM E=rr2which reduces to v =2rr= 2r.vGM Eand period =(a)GM Err = RE + 200 km = 6 370 km + 200 km = 6 570 kmThus,e6period = 2 6.57 10 mj e6.67 10e6.57 10 mjN m kg je5.98 10611223T = 5.30 10 s = 88.3 min = 1.47 hGM E=re6.67 1011jeN m 2 kg 2 5.98 10 24 kg(b)v=(c)j=K f + U f = K i + Ui + energy input, givesinput =e6.57 10 mj6FGHI FJK GHGM E mGM E m11mv 2 mvi2 +f22rfri7.79 km sIJK(1)ri = RE = 6.37 10 6 mvi =2RE= 4.63 10 2 m s86 400 sSubstituting the appropriate values into (1) yields theminimum energy input = 6.43 10 9 J24kgjChapter 13P13.38The gravitational force supplies the needed centripetal acceleration.GM E mThus,=b R + hg bE2bmv 2RE + h2r 2 RE + h=T=GM Ev(a)v2 =orggGM ERE + hbR + hgET = 23GM Eb R + hgEGM ERE + h(b)v=(c)Minimum energy input isej ejEmin = K f + U gf K i U gi .It is simplest tolaunch the satellite from a location on the equator, and launch it toward the east.1mvi22GM E m.U gi = REKi =This choice has the object starting with energy2RE2RE=1.00 day 86 400 swithvi =Thus,Emin =Etot = E =GMm2rFGHLMMN bIJK2GM EGM E m 14 2 RE1m m2RE + hRE + h 286 400 sEmin = GM E morP13.39FGHandI eJKLM R + 2 h OP 2 R mNM 2 R bR + hg QP b86 400 sgje2EEEj2EOP GM m+g PQ RE2E2FGH6.67 10 11 5.98 10 24 10 3 kg11GMm 1 1=ri r f2210 3 m 6 370 + 100 6 370 + 200IJKE = 4.69 10 8 J = 469 MJgE =P13.40Gm EGmUgU =2rEFG IJH K(a)2gU mU rE1== 14.023.70g E m E rU(b)v esc ,E =22rUa fegU = 1.02 9.80 m s 2 = 10.0 m s 22GmU:rUv esc ,E2Gm E; v esc ,U =rEv esc ,U=mU rE14.0== 1.953.70m E rUFor the Earth, from the texts table of escape speeds, v esc ,E = 11.2 km sa fbj= 1.02g v esc ,U = 1.95 11.2 km s = 21.8 km s395396P13.41Universal GravitationThe rocket is in a potential well at Ganymedes surface with energye6.67 10 11 N m 2 m 2 1.495 10 23 kgGm 1 m 2=U1 = rkg 2 2.64 10 6 me26U1 = 3.78 10 m 2 m sjj2The potential well from Jupiter at the distance of Ganymede isU2 = e6.67 10 11 N m 2 m 2 1.90 10 27 kgGm1 m 2=rkg 2 1.071 10 9 me28U 2 = 1.18 10 m 2 m sTo escape from both requiresjj212m 2 v esc = + 3.78 10 6 + 1.18 10 8 m 2 m 2 s 22ejv esc = 2 1.22 10 8 m 2 s 2 = 15.6 km sP13.42We interpret lunar escape speed to be the escape speed from the surface of a stationary moonalone in the Universe:GM m m12mv esc =2Rmv esc =2GM mRm2GM mRmv launch = 2Now for the flight from moon to EarthaK + U f = a K + U fifGmM m GmM E 1GmM m GmM E122= mv impact mv launch 22Rmrelrm 2RE4GM m GM m GM E 1 2GM m GM E= v impact 2RmRmrelrm 2REL F 3 M + M + M M I OP= M 2G GMN H R r R r JK PQL F 3 7.36 10 kg + 7.36 10= M 2G GNM H 1.74 10 m 3.84 1012v impactmmmEm2EEel22226kg8m+5.98 10 24 kg6.37 10 6 mej3.84 10 8= 2G 1.27 10 17 + 1.92 10 14 + 9.39 10 17 1.56 10 16 kg mej= 2 6.67 10 11 N m 2 kg 2 10.5 10 17 kg m12I OPm JQKP5.98 10 24 kg12= 11.8 km s12Chapter 13*P13.43(a)397Energy conservation for the object-Earth system from firing to apex:eK + U j = eK + U jggifGmM EGmM E1=0mvi2 RERE + h2whereGmM E12. Thenmv esc =RE2RE1 2 1 21 2vi v esc = v esc222RE + h2v esc vi2 =12v esch=h=a fa11.2f a8.76f6.37 10 6 m 8.762 vi2=2v esc RERE + hRE + h2v esc RE2v esc RE2v esc vi2 RE =22v esc RE v esc RE + vi2 RE2v esc vi2RE vi22v esc vi2= 1.00 10 7 m(b)h=(c)The fall of the meteorite is the time-reversal of the upward flight of the projectile, so it isdescribed by the same energy equation22FGH2vi2 = v esc 1 FGHIJKIJ eKREh2= v esc= 11.2 10 3 m sRE + hRE + h2mj FGH 6.37 10.51 +10.51 10 m IJKm 27267= 1.00 10 8 m 2 s 2vi = 1.00 10 4 m s(d)With vi << v esc , h RE vi22v escb ga f0 2 = vi2 + 2 g h 0 .P13.44=RE vi2 REGM Ev2. But g =, so h = i , in agreement with22GM E2gREFor a satellite in an orbit of radius r around the Earth, the total energy of the satellite-Earth system isGM EE=. Thus, in changing from a circular orbit of radius r = 2 RE to one of radius r = 3 RE , the2rrequired work isW = E = LMNOPQGM E m GM E mGM E m11+= GM E m=.2r f2ri4 RE 6 R E12 RE398*P13.45Universal Gravitation(a)The major axis of the orbit isFurther, in Figure 13.5,Then(b)2 a = 50.5 AUsoa + c = 50 AUsoc 24.75= 0.980e= =a 25.25a = 25.25 AUc = 24.75 AUIn T 2 = K s a 3 for objects in solar orbit, the Earth gives usb1 yrg2a= K s 1 AUfb1 yrg=a1 AU f23Ks3b1 yrg a25.25 AU f T = 127 yrT =a1 AU fN m kg je1.991 10 kg je1.2 10 kg j= 2.13 1050e1.496 10 mj2Thene(c)*P13.466.67 10 11GMmU==r(a)For the satellitev0 =(b)32FG GM IJH r KE F = ma232301711GmM Er2=J2mv 0r12Conservation of momentum in the forward direction for the exploding satellite:c mvh = c mvhif5mv 0 = 4mvi + m0vi =(c)10FGH55 GM Ev0 =r44IJK12With velocity perpendicular to radius, the orbiting fragment is at perigee. Its apogeedistance and speed are related to r and vi by 4mrvi = 4mr f v f andGM E 4m 1vrGM E 4m1= 4mv 2 . Substituting v f = i we have4mvi2 f22rrfrf1 2 GM E 1 vi2 r 2 GM E25 GM Egives=. Further, substituting vi2 =vi 222 rf16 rrrf25 GM E GM E 25 GM E r GM E=32 r32 r f2rrf725r1=32r 32r f2 r fClearing of fractions, 7r f2 = 25r 2 32rr f or 7rf=14f2 32FG r IJ + 25 = 0 givingHrKfa fa f = 50 or 14 . The latter root describes the starting point. The outer+32 32 2 4 7 2514142525r=; rf =.end of the orbit has77rrFG r IJHrKrf399Chapter 13Additional ProblemsP13.47Let m represent the mass of the spacecraft, rE the radius of the Earths orbit, and x the distance fromEarth to the spacecraft.Fs =The Sun exerts on the spacecraft a radial inward force ofGM s mbr x gEFE =while the Earth exerts on it a radial outward force of2GM E mx2The net force on the spacecraft must produce the correct centripetal acceleration for it to have anorbital period of 1.000 year.FS FE =Thus,GM S mbr x gEwhich reduces toGM Sbr x gE2GM Ex22=GM E mbbr x g .x24 2mv 2m==rE xrE xg bET2LM 2 br xg OPg NM T QP2E(1)Cleared of fractions, this equation would contain powers of x ranging from the fifth to the zeroth.We do not solve it algebraically. We may test the assertion that x is between 1.47 10 9 m and1.48 10 9 m by substituting both of these as trial solutions, along with the following data:M S = 1.991 10 30 kg , M E = 5.983 10 24 kg , rE = 1.496 10 11 m, and T = 1.000 yr = 3.156 10 7 s .With x = 1. 47 10 9 m substituted into equation (1), we obtain6.052 10 3 m s 2 1.85 10 3 m s 2 5.871 10 3 m s 2or5.868 10 3 m s 2 5.871 10 3 m s 2With x = 1. 48 10 9 m substituted into the same equation, the result is6.053 10 3 m s 2 1.82 10 3 m s 2 5.870 8 10 3 m s 2or5.870 9 10 3 m s 2 5.870 8 10 3 m s 2 .Since the first trial solution makes the left-hand side of equation (1) slightly less than the right handside, and the second trial solution does the opposite, the true solution is determined as between thetrial values. To three-digit precision, it is 1.48 10 9 m .As an equation of fifth degree, equation (1) has five roots. The Sun-Earth system has five Lagrangepoints, all revolving around the Sun synchronously with the Earth. The SOHO and ACE satellitesare at one. Another is beyond the far side of the Sun. Another is beyond the night side of the Earth.Two more are on the Earths orbit, ahead of the planet and behind it by 60. Plans are under way togain perspective on the Sun by placing a spacecraft at one of these two co-orbital Lagrange points.The Greek and Trojan asteroids are at the co-orbital Lagrange points of the Jupiter-Sun system.400P13.48Universal GravitationThe acceleration of an object at the center of the Earth dueto the gravitational force of the Moon is given byMa = G Moond2MMAt the point A nearest the Moon,a+ = G2dra fAt the point B farthest from the Moon, a = GMMad + r fFIG. P13.482a = a + a = GM M2GM M rLM 1MN ad r f21d2OPPQ= 1.11 10 6 m s 2For d >> r ,Across the planet,*P13.49a =g 2 a 2. 22 10 6 m s 2=== 2. 26 10 7gg9.80 m s 2Energy conservation for the two-sphere system from release to contact:GmmGmm 11=+ mv 2 + mv 2R2r22Gm(a)(b)FG 1 1 IJ = vH 2r R KFG L 1 1 OIJH MN 2r R PQK212v = GmThe injected impulse is the final momentum of each sphere,FG L 1 1 OIJH MN 2r R PQKmv = m 2 2 Gm12LMN= Gm 3FG 1 1 IJ OPH 2r R K Q12.If they now collide elastically each sphere reverses its velocity to receive impulseLMNa fmv mv = 2mv = 2 Gm 3P13.50d3FG 1 1 IJ OPH 2r R K Q12Momentum is conserved:m1 v 1i + m 2 v 2i = m1 v 1 f + m 2 v 2 f0 = Mv 1 f + 2 Mv 2 fv2 f = 1v1 f2Energy is conserved:aK + U f + E = aK + U fi0fGm 1 m 2Gm1 m 21122+ 0 = m 1 v1 f + m 2 v 2 f rfri22a fa fFGHGM 2 M1112v1 f= Mv1 f + 2 M12 R222v1 f =2 GMR3v2 f =IJK2a fGM 2 M4R11 GMv1 f =R23Chapter 13e1.25 10=(a)(b)P13.51v2ac =rdiff = 10.2 9.90 = 0.312 m s 2 =ac6.67 10P13.52(a)11112ms1.53 10N m kg11j2= 10.2 m s 2mGMr2e0.312 m s je1.53 10 mjM=264012= 1.10 10 32 kg2FIG. P13.51GM EThe free-fall acceleration produced by the Earth is g =r2= GM E r 2 (directed downward)a fdg= GM E 2 r 3 = 2GM E r 3 .drIts rate of change isThe minus sign indicates that g decreases with increasing height.dg2GM E=.3drREAt the Earths surface,(b)For small differences,gr=gh=2GM Eg =Thus,3REejaje2GM E hf=2 6.67 10 11 N m 2 kg 2 5.98 10 24 kg 6.00 m3RE1.85 10 5 m s 2(c)*P13.53g =(a)Each bit of mass dm in the ring is at the same distance from the object at A. The separateGmM ringGmdmto the system energy add up to . When the object is at A,contributions rrthis ise6.37 10 mj636.67 10 11 N m 2 1 000 kg 2.36 10 20 kgkg 2(b)882= 7.04 10 4 J .When the object is at the center of the ring, the potential energy is(c)e1 10 mj + e2 10 mj26.67 10 11 N m 2 1 000 kg 2.36 10 20 kgkg 2 1 10 8 m= 1.57 10 5 J .Total energy of the object-ring system is conserved:e K + U j = eK + U jggAB120 7.04 10 J = 1 000 kgv B 1.57 10 5 J24vB =F 2 8.70 10 J IGH 1 000 kg JK412= 13.2 m s402P13.54Universal GravitationTo approximate the height of the sulfur, setmv 2= mg Io h2h = 70 000 mGM= 1.79 m s 2r2g Io =a fbgbv = 2 1.79 70 000 500 m s over 1 000 mi hv = 2 g Io hgA more precise answer is given by1GMmGMm=mv 2 2r1r21 2v = 6.67 10 11 8.90 10 222eP13.55je611.89 106IJKFrom the walk, 2r = 25 000 m. Thus, the radius of the planet is r =From the drop:y =so,*P13.56jFGH 1.82 1 10g=a1 2 1gt = g 29. 2 s22aff = 3.28 102 1.40 ma29.2 sf22v = 492 m s25 000 m= 3.98 10 3 m2= 1.40 m3m s2 =MGr2 M = 7.79 10 14 kgThe distance between the orbiting stars is d = 2r cos 30 = 3 r since3. The net inward force on one orbiting star iscos 30 =2GmmGMm Gmmmv 2cos 30+ 2 + 2 cos 30 =rd2rdGm 2 cos 30 GM 4 2 r 2+ 2 =rrT 23r 2m4 2 r 3+M =GT23FGHT2 =rd60FF4 2 r 3G M+30rIJKejF rT = 2 GGH GeM +m33m3P13.5730FIJJjK12FIG. P13.56For a 6.00 km diameter cylinder, r = 3 000 m and to simulate 1 g = 9.80 m s 2v2= 2rrg= 0.057 2 rad s=rg=The required rotation rate of the cylinder is1 rev110 s(For a description of proposed cities in space, see Gerard K. ONeill in Physics Today, Sept. 1974.)Chapter 13P13.58(a)G has unitsN m 2 kg m m 2m3= 2= 2kg 2s kg 2s kgand dimensions G =L3.T2 MThe speed of light has dimensions of c =as angular momentum or h =M L2.TL, and Plancks constant has the same dimensionsTWe require G p c q h r = L , or L3 p T 2 p M p Lq T q M r L2 r T r = L1 M 0 T 0 .Thus, 3 p + q + 2r = 12 p q r = 0p + r = 0which reduces (using r = p ) to3p + q + 2p = 12 p q p = 0These equations simplify toThen, 5 p 3 p = 1 , yielding p =5 p + q = 1 and q = 3 p .131, q = , and r = .222Therefore, Planck length = G 1 2 c 3 2 h1 2 .(b)P13.59e6.67 10 j e3 10 j e6.63 10 j11 1 28 3 234 1 2e= 1.64 10 69Gm p m 012m 0 v esc =2Rv esc =2Gm pR4With m p = R 3 , we have3v esc ==So, v esc R .40342G 3 R 3R8GR3j12= 4.05 10 35 m ~ 10 34 m404*P13.60Universal GravitationFor both circular orbits,GM E m F = ma :=r2v=mv 2rGM ErFIG. P13.60e6.67 10 N m kg je5.98 10e6.37 10 m + 2 10 mjkgj=7.79 10 3 m s .N m 2 kg 2 5.98 10 24 kgj=7.85 10 3 m s .11(a)(b)The original speed is vi =The final speed isvi =226e6.67 1011e245je6j6.47 10 mThe energy of the satellite-Earth system isK +Ug =GM E mGM E m 1 GM E GM E1= m=mv 2 222rrrrg=3.04 10 9 J .jbg=3.08 10 9 J .Originally(d)FinallyEf = (e)Thus the object speeds up as it spirals down to the planet. The loss of gravitational energy isso large that the total energy decreases by11N m 2 kg 2 5.98 10 24 kg 100 kgeie6.67 10jejb(c)e6.67 10E =6j2 6.57 10 m11jeN m 2 kg 2 5.98 10 24 kg 100 kge6j2 6. 47 10 mejEi E f = 3.04 10 9 J 3.08 10 9 J = 4.69 10 7 J .(f)The only forces on the object are the backward force of air resistance R, comparatively verysmall in magnitude, and the force of gravity. Because the spiral path of the satellite is notperpendicular to the gravitational force, one component of the gravitational force pullsforward on the satellite to do positive work and make its speed increase.405Chapter 13P13.61(a)At infinite separation U = 0 and at rest K = 0 . Since energy of the two-planet system isconserved we have,0=Gm1 m 21122m 1 v1 + m 2 v 2 d22(1)The initial momentum of the system is zero and momentum is conserved.0 = m1 v1 m 2 v 2Therefore,bTherefore,(a)K1 =bggdandv 2 = 2.58 10 3 m sand12m1 v1 = 1.07 10 32 J2K2 =12m 2 v 2 = 2.67 10 31 J2The net torque exerted on the Earth is zero. Therefore, the angular momentum of the Earthis conserved;mra v a = mrp v p and v a = v p(b)b2Gd m1 + m 2Substitute given numerical values into the equation found for v1 and v 2 in part (a) to findv1 = 1.03 10 4 m sP13.62v 2 = m1andg2G m 1 + m 2b gv r = v1 v 2 =Relative velocity(b)2Gd m1 + m 2v1 = m 2Combine equations (1) and (2):(2)Kp =F r I = e3.027 10GH r JKpjeejemsa112mv p = 5.98 10 24 3.027 10 422e4j2471jFGH 1..521 IJK =12.93 10 4 m s= 2.74 10 33 Jjej6.673 10 11 5.98 10 24 1.99 10 30GmM== 5.40 10 33 JUp = 11rp1.471 10(c)Using the same form as in part (b), K a = 2.57 10 33 J and U a = 5.22 10 33 J .Compare to find that K p + U p = 2.66 10 33 J and K a + U a = 2.65 10 33 J . They agree.406P13.63Universal Gravitation(a)The work must provide the increase in gravitational energyW = U g = U gf U giGM E M p=rfGM E M p=RE + y= GM E M p=++F1GH RF 6.67 10GHkgGM E M priGM E M pRE1RE + yE11N m22IJKI 5.98 10JK e24jbkg 100 kggFGH 6.37 1106mW = 850 MJ(b)In a circular orbit, gravity supplies the centripetal force:GM E M pbRThen,E+yMpv2=g bR2E+yg11 GM E M pMp v2 =22 RE + ybgSo, additional work = kinetic energy required=ejejb11N m 2 5.98 10 24 kg 100 kg1 6.67 102kg 2 7.37 10 6 me jejW = 2.71 10 9 JP13.64Centripetal acceleration comes from gravitational acceleration.v 2 M c G 4 2 r 2= 2 =rrT 2rGM c T 2 = 4 2 r 3e6.67 10 ja20fe1.99 10 je5.00 10 j113 230= 4 2 r 3rorbit = 119 kmP13.65ej152r 2 30 000 9.46 10 m== 7 10 15 s = 2 10 8 yrv2.50 10 5 m s(a)T=(b)4 2 30 000 9.46 10 15 m4 2 a 3M==GT 26.67 10 11 N m 2 kg 2 7.13 10 15 seej3jeM = 1.34 10 11 solar masses ~ 10 11 solar massesThe number of stars is on the order of 10 11 .j2= 2.66 10 41 kgg17.37 10 6 mIJKChapter 13P13.66(a)From the data about perigee, the energy of the satellite-Earth system isE=or(b)4076.67 10 je5.98 10 ja1.60fj e7.02 10a feGM E m 112= 1.60 8.23 10 3mv p rp22211246E = 3.67 10 7 JbgejejL = mvr sin = mv p rp sin 90.0 = 1.60 kg 8.23 10 3 m s 7.02 10 6 m= 9.24 10 10 kg m 2 s(c)Since both the energy of the satellite-Earth system and the angular momentum of the Earthare conserved,at apogee we must have1GMm2=Emv a 2raandmv a ra sin 90.0 = L .Thus,6.67 10 11 5.98 10 24 1.6012= 3.67 10 7 J1.60 v a 2rsandb1.60 kg gv rea fa aja fje= 9.24 10 10 kg m 2 s .eja fa fje6.67 10 11 5.98 10 24 1.60 1.60 v a12Solving simultaneously,= 3.67 10 71.60 v a 29.24 10 10a f20.800 v a 11 046 v a + 3.672 3 10 7 = 0which reduces tova =so11 046 b11 046g 4a0.800fe3.672 3 10 j .2a0.800 f27This gives v a = 8 230 m s or 5 580 m s . The smaller answer refers to the velocity at theapogee while the larger refers to perigee.ra =Thus,(d)bgejThe major axis is 2a = rp + ra , so the semi-major axis isa=(e)9.24 10 10 kg m 2 sL== 1.04 10 7 m .mv a1.60 kg 5.58 10 3 m sT=4 2 a 3=GM E17.02 10 6 m + 1.04 10 7 m = 8.69 10 6 m2ejej4 2 8.69 10 6 me6.67 10T = 8 060 s = 134 min11je3N m 2 kg 2 5.98 10 24 kgj408*P13.67Universal GravitationLet m represent the mass of the meteoroid and vi its speed when far away.No torque acts on the meteoroid, so its angular momentum is conserved asit moves between the distant point and the point where it grazes the Earth,moving perpendicular to the radius:Li = L f :FIG. P13.67mri v i = mr f v fbgm 3 RE vi = mRE v fv f = 3 viNow energy of the meteoroid-Earth system is also conserved:GM E m11mvi2 + 0 = mv 2 f22REeK + U j = eK + U j :ggifGM E1 2 1vi = 9 vi2 22REe jGM E= 4vi2 :RE*P13.68vi =GM E4 REFrom Keplers third law, minimum period means minimum orbit size. The treetop satellite inFigure P13.35 has minimum period. The radius of the satellites circular orbit is essentially equal tothe radius R of the planet.FGHRR e 4 R jGV =GMm F = ma :2=mv 2 m 2R=RR T22IJK22RT 244 2 R 3G R 3 =3T2FGHIJKThe radius divides out: T 2 G = 3P13.69T=3GIf we choose the coordinate of the center of mass at the origin, then0=bMr2 mr1gM+mandMr2 = mr1(Note: this is equivalent to saying that the net torque must be zero andthe two experience no angular acceleration.) For each mass F = ma so2mr1 1 =MGmd2and2Mr2 2 =MGmd2FIG. P13.69bgCombining these two equations and using d = r1 + r2 gives r1 + r2 2 =with 1 = 2 = 2and T =we find T 2 =4 2 d 3G M+maf.aM + mfGd2409Chapter 13P13.70The gravitational force exerted on m 2 by the Earth (mass m1 ) accelerates m 2 according to:Gm1 m 2. The equal magnitude force exerted on the Earth by m 2 produces negligiblem2 g 2 =r2acceleration of the Earth. The acceleration of relative approach is then(a)g2 =Gm1r2=e6.67 1011jemjN m 2 kg 2 5.98 10 24 kge1.20 1072j=2.77 m s 2 .Again, m 2 accelerates toward the center of mass with g 2 = 2.77 m s 2 . Now the Earthaccelerates toward m 2 with an acceleration given as(b)m1 g 1 =g1 =Gm1 m 2r2Gm 2r2e6.67 10=11jemjN m 2 kg 2 2.00 10 24 kge1.20 1072j = 0.926 m s2The distance between the masses closes with relative acceleration ofg rel = g 1 + g 2 = 0.926 m s 2 + 2.77 m s 2 = 3.70 m s 2 .P13.71Initial Conditions and Constants:Mass of planet:5.98 10 24 kgRadius of planet:Initial x:Initial y:Initial v x :6.37 10 6 m0.0 planet radii2.0 planet radii+5 000 m/sTime interval:10.9 sInitialvy :0.0 m/sFIG. P13.71t (s)0.010.921.732.65 431.65 442.45 453.35 464.110 841.310 852.210 863.1x (m)y (m)r (m)vx(m/s)vyax(m/s)em s j2ayem s j20.054 315.3108 629.4162 941.112 740 000.012 740 000.012 739 710.012 739 130.012 740 000.012 740 115.812 740 173.112 740 172.15 000.04 999.94 999.74 999.30.026.753.480.10.000 00.010 00.021 00.031 02.457 52.457 42.457 32.457 2112 843.831 121.450 603.4132 324.38 466 816.08 467 249.78 467 026.98 466 147.78 467 567.98 467 306.98 467 178.28 467 181.77 523.07 523.27 522.87 521.939.920.580.9141.40.074 00.020 00.033 00.087 05.562 55.563 35.563 45.562 8108 629.054 314.90.412 739 134.412 739 713.412 740 002.412 739 597.512 739 829.212 740 002.44 999.95 000.05 000.053.326.60.10.021 00.010 00.000 02.457 52.457 52.457 5The object does not hit the Earth ; its minimum radius is 1.33 RE .Its period is 1.09 10 4 s . A circular orbit would require a speed of 5.60 km s .410Universal GravitationANSWERS TO EVEN PROBLEMSP13.22.67 10 7 m s 2P13.40(a) 10.0 m s 2 ; (b) 21.8 km sP13.43.00 kg and 2.00 kgP13.4211.8 km sP13.6(a) 4.39 10 20 N toward the Sun;(b) 1.99 10 20 N toward the Earth;(c) 3.55 10 22 N toward the SunP13.44GM E m12 REP13.8see the solution; either 1 m 61.3 nm or2.74 10 4 mP13.46(a) v 0F GM IJ=GH r K(c) r f =P13.1023P13.12(a) 1.02 km s ; (b) 1.35 mmP13.14see the solutionP13.161.27P13.18Planet Y has turned through1.30 revolutionsP13.201.63 10 4 rad sP13.2218.2 msP13.24(a) 1.31 10 17 N toward the center;(b) 2.62 10 12 N kgE12; (b) vi =5GM E 1 2re j25r7P13.48;2.26 10 7P13.5042 GM 1 GM;3R3RP13.52(a), (b) see the solution;(c) 1.85 10 5 m s 2P13.54492 m sP13.56see the solutionP13.58(a) G 1 2 c 3 2 h1 2 ; (b) ~ 10 34 mP13.60(a) 7.79 km s; (b) 7.85 km s;(c) 3.04 GJ ;(d) 3.08 GJ ; (e) loss = 46.9 MJ ;(f) A component of the Earths gravitypulls forward on the satellite in itsdownward banking trajectory.P13.62(a) 29.3 km s ; (b) K p = 2.74 10 33 J ;P13.26(a) 4.77 10 9 J ; (b) 569 N down;(c) 569 N upP13.282.52 10 7 mP13.302.82 10 9 JP13.32(a) see the solution; (b) 340 sP13.34(a) 42.1 km s ; (b) 2.20 10 11 mP13.64119 kmP13.361.58 10 10 JP13.66(a) 36.7 MJ ; (b) 9.24 10 10 kg m 2 s ;(c) 5.58 km s; 10.4 Mm; (d) 8.69 Mm;(e) 134 minP13.68see the solutionP13.70(a) 2.77 m s 2 ; (b) 3.70 m s 2P13.38bU p = 5.40 10 33 J ;(c) K a = 2.57 10 33 J;U a = 5.22 10 33 J; yesg bGM g ;(b) bGM g b R + hg;L R + 2 h OP 2 R m(c) GM m MMN 2 R bR + hg PQ b86 400 sg(a) 2 RE + hEE1 2E1 23 212E2EEE2E2The satellite should be launched from theEarths equator toward the east.14Fluid MechanicsCHAPTER OUTLINE14.114.214.314.414.514.614.7PressureVariation of Pressure withDepthPressure MeasurementsBuoyant Forces andArchimedes PrincipleFluid DynamicsBernoullis EquationOther Applications of FluidDynamicsANSWERS TO QUESTIONSQ14.1The weight depends upon the total volume of glass. Thepressure depends only on the depth.Q14.2Both must be built the same. The force on the back of each damis the average pressure of the water times the area of the dam.If both reservoirs are equally deep, the force is the same.FIG. Q14.2Q14.3If the tube were to fill up to the height of several stories of the building, the pressure at the bottom ofthe depth of the tube of fluid would be very large according to Equation 14.4. This pressure is muchlarger than that originally exerted by inward elastic forces of the rubber on the water. As a result,water is pushed into the bottle from the tube. As more water is added to the tube, more watercontinues to enter the bottle, stretching it thin. For a typical bottle, the pressure at the bottom of thetube can become greater than the pressure at which the rubber material will rupture, so the bottlewill simply fill with water and expand until it bursts. Blaise Pascal splintered strong barrels by thismethod.Q14.4About 1 000 N: thats about 250 pounds.Q14.5The submarine would stop if the density of the surrounding water became the same as the averagedensity of the submarine. Unfortunately, because the water is almost incompressible, this will bemuch deeper than the crush depth of the submarine.Q14.6Yes. The propulsive force of the fish on the water causes the scale reading to fluctuate. Its averagevalue will still be equal to the total weight of bucket, water, and fish.Q14.7The boat floats higher in the ocean than in the inland lake. According to Archimedess principle, themagnitude of buoyant force on the ship is equal to the weight of the water displaced by the ship.Because the density of salty ocean water is greater than fresh lake water, less ocean water needs tobe displaced to enable the ship to float.411412Fluid MechanicsQ14.8In the ocean, the ship floats due to the buoyant force from salt water. Salt water is denser than freshwater. As the ship is pulled up the river, the buoyant force from the fresh water in the river is notsufficient to support the weight of the ship, and it sinks.Q14.9Exactly the same. Buoyancy equals density of water times volume displaced.Q14.10At lower elevation the water pressure is greater because pressure increases with increasing depthbelow the water surface in the reservoir (or water tower). The penthouse apartment is not so farbelow the water surface. The pressure behind a closed faucet is weaker there and the flow weakerfrom an open faucet. Your fire department likely has a record of the precise elevation of every firehydrant.Q14.11As the wind blows over the chimney, it creates a lower pressure at the top of the chimney. Thesmoke flows from the relatively higher pressure in front of the fireplace to the low pressure outside.Science doesnt suck; the smoke is pushed from below.Q14.12The rapidly moving air above the ball exerts less pressure than the atmospheric pressure below theball. This can give substantial lift to balance the weight of the ball.Q14.13The skijumper gives her body the shape of an airfoil. Shedeflects downward the air stream as it rushes past and itdeflects her upward by Newtons third law. The air exertson her a lift force, giving her a higher and longer trajectory.To say it in different words, the pressure on her back is lessthan the pressure on her front.FIG. Q14.13Q14.14The horizontal force exerted by the outside fluid, on an area element of the objects side wall, hasequal magnitude and opposite direction to the horizontal force the fluid exerts on another elementdiametrically opposite the first.Q14.15The glass may have higher density than the liquid, but the air inside has lower density. The totalweight of the bottle can be less than the weight of an equal volume of the liquid.Q14.16Breathing in makes your volume greater and increases the buoyant force on you. You instinctivelytake a deep breath if you fall into the lake.Q14.17No. The somewhat lighter barge will float higher in the water.Q14.18The level of the pond falls. This is because the anchor displaces more water while in the boat. Afloating object displaces a volume of water whose weight is equal to the weight of the object. Asubmerged object displaces a volume of water equal to the volume of the object. Because the densityof the anchor is greater than that of water, a volume of water that weighs the same as the anchor willbe greater than the volume of the anchor.Q14.19The metal is more dense than water. If the metal is sufficiently thin, it can float like a ship, with thelip of the dish above the water line. Most of the volume below the water line is filled with air. Themass of the dish divided by the volume of the part below the water line is just equal to the density ofwater. Placing a bar of soap into this space to replace the air raises the average density of thecompound object and the density can become greater than that of water. The dish sinks with itscargo.Chapter 14413Q14.20The excess pressure is transmitted undiminished throughout the container. It will compress airinside the wood. The water driven into the wood raises its average density and makes if float lowerin the water. Add some thumbtacks to reach neutral buoyancy and you can make the wood sink orrise at will by subtly squeezing a large clearplastic softdrink bottle. Bored with graph paper andproving his own existence, Ren Descartes invented this toy or trick.Q14.21The plate must be horizontal. Since the pressure of a fluid increases with increasing depth, otherorientations of the plate will give a non-uniform pressure on the flat faces.Q14.22The air in your lungs, the blood in your arteries and veins, and the protoplasm in each cell exertnearly the same pressure, so that the wall of your chest can be in equilibrium.Q14.23Use a balance to determine its mass. Then partially fill a graduated cylinder with water. Immerse therock in the water and determine the volume of water displaced. Divide the mass by the volume andyou have the density.Q14.24When taking off into the wind, the increased airspeed over the wings gives a larger lifting force,enabling the pilot to take off in a shorter length of runway.Q14.25Like the ball, the balloon will remain in front of you. It will not bob up to the ceiling. Air pressurewill be no higher at the floor of the sealed car than at the ceiling. The balloon will experience nobuoyant force. You might equally well switch off gravity.Q14.26Styrofoam is a little more dense than air, so the first ship floats lower in the water.Q14.27We suppose the compound object floats. In both orientations it displaces its own weight of water, soit displaces equal volumes of water. The water level in the tub will be unchanged when the object isturned over. Now the steel is underwater and the water exerts on the steel a buoyant force that wasnot present when the steel was on top surrounded by air. Thus, slightly less wood will be below thewater line on the block. It will appear to float higher.Q14.28A breeze from any direction speeds up to go over the mound and the air pressure drops. Air thenflows through the burrow from the lower entrance to the upper entrance.Q14.29Regular cola contains a considerable mass of dissolved sugar. Its density is higher than that of water.Diet cola contains a very small mass of artificial sweetener and has nearly the same density as water.The lowdensity air in the can has a bigger effect than the thin aluminum shell, so the can of dietcola floats.Q14.30(a)Lowest density: oil; highest density: mercury(b)The density must increase from top to bottom.(a)Since the velocity of the air in the right-hand section of the pipe is lower than that in themiddle, the pressure is higher.(b)The equation that predicts the same pressure in the far right and left-hand sections of thetube assumes laminar flow without viscosity. Internal friction will cause some loss ofmechanical energy and turbulence will also progressively reduce the pressure. If thepressure at the left were not higher than at the right, the flow would stop.Q14.31414Q14.32Fluid MechanicsClap your shoe or wallet over the hole, or a seat cushion, or your hand. Anything that can sustain aforce on the order of 100 N is strong enough to cover the hole and greatly slow down the escape ofthe cabin air. You need not worry about the air rushing out instantly, or about your body beingsucked through the hole, or about your blood boiling or your body exploding. If the cabin pressuredrops a lot, your ears will pop and the saliva in your mouth may boilat body temperaturebutyou will still have a couple of minutes to plug the hole and put on your emergency oxygen mask.Passengers who have been drinking carbonated beverages may find that the carbon dioxidesuddenly comes out of solution in their stomachs, distending their vests, making them belch, and allbut frothing from their ears; so you might warn them of this effect.SOLUTIONS TO PROBLEMSSection 14.1P14.1PressureeM = ironV = 7 860 kg m3M = 0.111 kgP14.24jLMN 3 b0.015 0 mg OPQ3The density of the nucleus is of the same order of magnitude as that of one proton, according to theassumption of close packing:=m 1.67 10 27 kg~~ 10 18 kg m3 .315V410m3ejWith vastly smaller average density, a macroscopic chunk of matter or an atom must be mostlyempty space.a f50.0 9.80F=A 0.500 10 2= 6.24 10 6 N m 2P14.3P=P14.4Let Fg be its weight. Then each tire supportssoyieldingP14.5ej2Fg4,F Fg=A 4AFg = 4 AP = 4 0.024 0 m 2 200 10 3 N m 2 = 1.92 10 4 NP=ejejThe Earths surface area is 4R 2 . The force pushing inward over this area amounts toejejF = P0 A = P0 4R 2 .This force is the weight of the air:Fg = mg = P0 4R 2so the mass of the air ism=eP0 4R 2gj = e1.013 105jLMN ej OPQN m 2 4 6.37 10 6 m9.80 m s 22= 5.27 10 18 kg .Chapter 14Section 14.2P14.6415Variation of Pressure with DepthejbjegP = P0 + gh = 1.013 10 5 Pa + 1 024 kg m3 9.80 m s 2 1 000 m(a)P = 1.01 10 7 Pa(b)The gauge pressure is the difference in pressure between the water outside and the airinside the submarine, which we suppose is at 1.00 atmosphere.Pgauge = P P0 = gh = 1.00 10 7 PaThe resultant inward force on the porthole is thenaF = Pgauge A = 1.00 10 7 Pa 0.150 mP14.7Fel = Ffluidandh== 7.09 10 5 N .kxgA323322OP =jQm215 000F= 22003.00e1.62 mFIG. P14.7F1F= 2A1 A 2Since the pressure is the same on both sides,In this case,P14.92kx = ghAore1 000 N m je5.00 10 mjh=e10 kg m je9.80 m s jLMN e1.00 10P14.8fF2 = 225 NorjFg = 80.0 kg 9.80 m s 2 = 784 NWhen the cup barely supports the student, the normal force of theceiling is zero and the cup is in equilibrium.ejFg = F = PA = 1.013 10 5 Pa AA=FgP=784= 7.74 10 3 m 251.013 10FIG. P14.9P14.10(a)Suppose the vacuum cleaner functions as a highvacuum pump. The air below the brickwill exert on it a lifting forceLM eNj OPQ =F = PA = 1.013 10 5 Pa 1.43 10 2 m(b)265.1 N .The octopus can pull the bottom away from the top shell with a force that could be no largerthanbgejejaf LNM e1.43 10F = PA = P0 + gh A = 1.013 10 5 Pa + 1 030 kg m3 9.80 m s 2 32.3 mF = 275 N2j OQPm2416P14.11Fluid MechanicsThe excess water pressure (over air pressure) halfway down isejajefPgauge = gh = 1 000 kg m3 9.80 m s 2 1.20 m = 1.18 10 4 Pa .The force on the wall due to the water isjaefafF = Pgauge A = 1.18 10 4 Pa 2.40 m 9.60 m = 2.71 10 5 Nhorizontally toward the back of the hole.P14.12The pressure on the bottom due to the water is Pb = gz = 1.96 10 4 PaSo,On each end,F = PA = 9.80 10 3 Pa 20.0 m 2 = 196 kNOn the side,P14.13Fb = Pb A = 5.88 10 6 NF = PA = 9.80 10 32588 kNIn the reference frame of the fluid, the carts acceleration causes a fictitious force to act backward, as ifathe acceleration of gravity were g 2 + a 2 directed downward and backward at = tan 1from thegdvertical. The center of the spherical shell is at depth below the air bubble and the pressure there is2F IGH JKP = P0 + g eff h = P0 +P14.14ejPae60.0 m j =1d g 2 + a 2 .2The air outside and water inside both exert atmospheric pressure,so only the excess water pressure gh counts for the net force. Takea strip of hatch between depth h and h + dh . It feels forceafdF = PdA = gh 2.00 m dh .(a)1.00 mThe total force iszF = dF =z2.00 m2.00 mafgh 2.00 m dhh = 1.00 mFIG. P14.14a2.00 mf a2.00 mf a1.00 mf= e1 000 kg m je9.80 m s jf2F = 29.4 kN bto the right gThe lever arm of dF is the distance a h 1.00 mf from hinge to strip: = z d = z gha 2.00 mfa h 1.00 mfdhLhh O = g a 2.00 mfM a1.00 mf P2 QN3F 7.00 m 3.00 m I = e1 000 kg m je9.80 m s ja 2.00 mfGJK2H 3h2F = g 2.00 m2a(b)2.00 m2.00 m3221.00 m2.00 mh = 1.00 m322.00 m1.00 m32 = 16.3 kN m counterclockwise332Chapter 14P14.15417The bell is uniformly compressed, so we can model it with any shape. We choose a sphere ofdiameter 3.00 m.The pressure on the ball is given by: P = Patm + w gh so the change in pressure on the ball fromwhen it is on the surface of the ocean to when it is at the bottom of the ocean is P = w gh .In addition:V = ghV4 w ghr 3VP= w=, where B is the Bulk Modulus.3BBBV = jejba3fe14.0 10 Pajegaf4 1 030 kg m3 9.80 m s 2 10 000 m 1.50 m103= 0.010 2 m3Therefore, the volume of the ball at the bottom of the ocean isV V =af4 1.50 m33 0.010 2 m 3 = 14.137 m3 0.010 2 m 3 = 14.127 m 3 .This gives a radius of 1.499 64 m and a new diameter of 2.999 3 m. Therefore the diameter decreasesby 0.722 mm .Section 14.3P14.16(a)Pressure MeasurementsWe imagine the superhero to produce a perfect vacuum in the straw. Take point 1 at thewater surface in the basin and point 2 at the water surface in the straw:P1 + gy1 = P2 + gy 2ejej1.013 10 5 N m 2 + 0 = 0 + 1 000 kg m 3 9.80 m s 2 y 2(b)P14.17y 2 = 10.3 mNo atmosphere can lift the water in the straw through zero height difference.P0 = ghh=P010.13 10 5 Pa== 10.5 mg 0.984 10 3 kg m3 9.80 m s 2ejejNo. Some alcohol and water will evaporate. The equilibriumvapor pressures of alcohol and water are higher than the vaporpressure of mercury.FIG. P14.17418P14.18Fluid Mechanics(a)Using the definition of density, we havehw =(b)100 gm water== 20.0 cm2A 2 water 5.00 cm 1.00 g cm 3ejSketch (b) at the right represents the situation afterthe water is added. A volume A 2 h2 of mercuryhas been displaced by water in the right tube. Theadditional volume of mercury now in the left tubeis A1 h . Since the total volume of mercury has notchanged,bA 2 h2 = A1 hgFIG. P14.18h2 =orA1hA2(1)At the level of the mercurywater interface in the right tube, we may write the absolutepressure as:P = P0 + water ghwThe pressure at this same level in the left tube is given bybgP = P0 + Hg g h + h2 = P0 + water ghwwhich, using equation (1) above, reduces toLMN Hg h 1 +or h = water hwe Hg 1 +A1A2jOPQA1= water hwA2.e1.00 g cm ja20.0 cmf =Thus, the level of mercury has risen a distance of h =e13.6 g cm jc1 + h3310 .05.000.490 cmabove the original level.bgP14.19P0 = gh = 2.66 10 3 Pa :P = P0 + P0 = 1.013 0.026 6 10 5 Pa = 0.986 10 5 PaP14.20Let h be the height of the water column added to the rightside of the Utube. Then when equilibrium is reached, thesituation is as shown in the sketch at right. Now considertwo points, A and B shown in the sketch, at the level of thewatermercury interface. By Pascals Principle, the absolutepressure at B is the same as that at A. But,PA = P0 + w gh + Hg gh2 andbh1waterhBh2MercuryAgPB = P0 + w g h1 + h + h2 .Thus, from PA = PB , w h1 + w h + w h2 = w h + Hg h2 , orh1 =LM NHgwOPQafaf 1 h2 = 13.6 1 1.00 cm = 12.6 cm .FIG. P14.20Chapter 14*P14.21P = P0 + ghThe gauge pressure is(a)jaefP P0 = gh = 1 000 kg 9.8 m s 2 0.160 m = 1.57 kPa = 1.57 10 3 Pa419FG 1 atm IJH 1.013 10 Pa K5= 0.015 5 atm .It would lift a mercury column to height1 568 PaP P0h=== 11.8 mm .g13 600 kg m 3 9.8 m s 2e(b)jejIncreased pressure of the cerebrospinal fluid will raise the level of the fluid in thespinal tap.(c)Blockage of the fluid within the spinal column or between the skull and the spinalcolumn would prevent the fluid level from rising.Section 14.4P14.22Buoyant Forces and Archimedes Principle(a)The balloon is nearly in equilibrium: Fg Fy = ma y B Fge jheliume jpayload=0 air gV helium gV m payload g = 0This reduces tom payload = air helium V = 1.29 kg m3 0.179 kg m 3 400 m 3orbg ejejm payload = 444 kg(b)Similarly,m payload = air hydrogen V = 1.29 kg m3 0.089 9 kg m3 400 m 3ej ejejm payload = 480 kgThe air does the lifting, nearly the same for the two balloons.P14.23At equilibrium F = 0 orwhereThe applied force,B is the buoyant force.Fapp = B mgbgB = Vol water gwherea fSo,F = aVolf g b gb4F = e1.90 10 mj e9.80 m s je10 kg m3F = bm + V g g must be equal to F = Vgandappwater2appP14.24Fapp + mg = Bg3m = Vol ball .4 ball = r 3 g water ball33bs23gFIG. P14.23j 84.0 kg m 3 = 0.258 NwSince V = Ah , m + s Ah = w Ahand A =bmw s hgFIG. P14.24420Fluid MechanicsP14.25(a)Before the metal is immersed: Fy = T1 Mg = 0 orbgeT1 = Mg = 1.00 kg 9.80 m s 2scalej= 9.80 NBT1(b)After the metal is immersed: Fy = T2 + B Mg = 0orbMggMgT2 = Mg B = Mg w V gV=M=1.00 kg2 700 kg m3aeT2 = Mg B = 9.80 N 1 000 kg m3(a)bFIG. P14.25Thus,*P14.26T2Fg Fy = 0 :(b)F 1. kgjGH 2 70000kg m3I 9.80 m s =jJK e26.17 N .15 N 10 N + B = 0B = 25.0 NTBFIG. P14.26(a)(c)The oil pushes horizontally inward on each side of the block.(d)String tension increases . The oil causes the water below to beunder greater pressure, and the water pushes up more stronglyon the bottom of the block.(e)Consider the equilibrium just before the string breaks:15 N15 N 60 N + 25 N+ Boil = 0Boil = 50 N60 NFor the buoyant force of the water we haveB = Vgjbeg25 N = 1 000 kg m3 0.25Vblock 9.8 m s 2Vblock = 1.02 10 2 m325 NBoilFIG. P14.26(e)For the buoyant force of the oilej ej50 N = 800 kg m3 f e 1.02 10 2 m 3 9.8 m s 2f e = 0.625 = 62.5%(f)ej ej15 N + 800 kg m3 f f 1.02 10 2 m 3 9.8 m s 2 = 015 Nf f = 0.187 = 18.7%BoilFIG. P14.26(f)Chapter 14P14.27421P = P0 + ghTaking P0 = 1.013 10 5 N m 2 and h = 5.00 cm(a)we findPtop = 1.017 9 10 5 N m 2For h = 17.0 cm, we getPbot = 1.029 7 10 5 N m 2Since the areas of the top and bottom areA = 0.100 mwe findFtop = Ptop A = 1.017 9 10 3 NandFbot = 1.029 7 10 3 NT + B Mg = 0(b)eafje2= 10 2 m 2jeFIG. P14.27jwhereB = w Vg = 10 3 kg m3 1.20 10 3 m3 9.80 m s 2 = 11.8 NandMg = 10.0 9.80 = 98.0 NTherefore,T = Mg B = 98.0 11.8 = 86.2 Na fbgFbot Ftop = 1.029 7 1.017 9 10 3 N = 11.8 N(c)which is equal to B found in part (b).P14.28Consider spherical balloons of radius 12.5 cm containing helium at STP and immersed in air at 0Cand 1 atm. If the rubber envelope has mass 5.00 g, the upward force on each isB Fg ,He Fg , env = air Vg HeVg m env g4bgFGH 3 r IJK g m gL4O= a1. 29 0.179 f kg m M a0.125 mf Pe9.80 m s j 5.00 103NQ3Fup = air HeFupenv3323ejkg 9.80 m s 2 = 0.040 1 NIf your weight (including harness, strings, and submarine sandwich) isej70.0 kg 9.80 m s 2 = 686 N686 N= 17 000 ~ 10 4 .0.040 1 Nyou need this many balloons:P14.29(a)eajejaf3But B = Weight of block = mg = woodVwood g = 0.650 g cm 3 20.0 cm ga fa fa faf20.0 h = 20.0a0.650 f so h = 20.0a1 0.650f = 7.00 cm30.650 20.0 g = 1.00 20.0 20.0 20.0 h g(b)fAccording to Archimedes, B = water Vwater g = 1.00 g cm 3 20.0 20.0 20.0 h gB = Fg + Mg where M = mass of leada fa fM = a1.00 0.650fa 20.0f = 0.350a 20.0 f331.00 20.0 g = 0.650 20.0 g + Mg33= 2 800 g = 2.80 kg422*P14.30Fluid Mechanics(a)The weight of the ball must be equal to the buoyant force of the water:4 3router g31.26 kgg = waterrouter(b)F 3 1.26 kg I=GH 4 1 000 kg m JK133= 6.70 cmThe mass of the ball is determined by the density of aluminum:FG 4 rH3IJKFG IJ eaH K4 ri3341.26 kg = 2 700 kg m3 0.067 m3m = Al V = Al30f3 ri3j1.11 10 4 m 3 = 3.01 10 4 m 3 ri3eri = 1.89 10 4 m 3*P14.31j13= 5.74 cmLet A represent the horizontal cross-sectional area of the rod, which we presume to be constant. Therod is in equilibrium: Fy = 0 : mg + B = 0 = 0 Vwhole rod g + fluidVimmersed gaf 0 ALg = A L h gThe density of the liquid is*P14.32= 0L.LhWe use the result of Problem 14.31. For the rod floating in a liquid of density 0.98 g cm 3 , = 0LLh0.98 g cm3 =a3 0LL 0.2 cmfej0.98 g cm L 0.98 g cm 3 0.2 cm = 0 LFor floating in the dense liquid, 0L1.14 g cm 3 =1.14 g cm 3(a)By substitution,aL 1.8 cmf e1.14 g cm j1.8 cm = L3a0fa f1.14L 1.14 1.8 cm = 0.98L 0.2 0.980.16L = 1.856 cmL = 11.6 cm(b)Substituting back,a 0 = 0.963 g cm(c)f0.98 g cm3 11.6 cm 0.2 cm = 0 11.6 cm3 0Lis not of the form = a + bh , equal-sizeLhsteps of do not correspond to equal-size steps of h.The marks are not equally spaced. Because =Chapter 14P14.34The balloon stops rising whenbTherefore,P14.33V=airg He gV = Mgand400M= air He 1.25 e 1 0.180bair423g He V = M ,V = 1 430 m3Since the frog floats, the buoyant force = the weight of the frog. Also, the weight of the displacedwater = weight of the frog, so oozeVg = m frog gorm frog = oozeV = oozeFGHIJ eK1 4 32r = 1.35 10 3 kg m36.00 10 2 m2 33j ej3Hence, m frog = 0.611 kg .P14.35B = FgV= sphere gV21 sphere = H 2O = 500 kg m324 glycerin gV sphere gV = 01010 glycerin =500 kg m3 = 1 250 kg m34 H 2O gFGHIJKeP14.36FIG. P14.35jConstant velocity implies zero acceleration, which means that the submersible is in equilibriumunder the gravitational force, the upward buoyant force, and the upward resistance force: Fy = ma y = 0ej 1. 20 10 4 kg + m g + w gV + 1 100 N = 0where m is the mass of the added water and V is the spheres volume.1.20 10 4 kg + m = 1.03 10 3soP14.37LM 4 a1.50f OP + 1 100 NN3Q 9.8 m s32m = 2.67 10 3 kgBy Archimedess principle, the weight of the fifty planes is equal to the weight of a horizontal slice ofwater 11.0 cm thick and circumscribed by the water line:a f50e 2.90 10 kg j g = e1 030 kg m j g a0.110 mf AB = water g V43giving A = 1.28 10 4 m 2 . The acceleration of gravity does not affect the answer.424Fluid MechanicsSection 14.5Fluid DynamicsSection 14.6Bernoullis EquationP14.38By Bernoullis equation,bbggdm= Av = 1 000 5.00 10 2dteP14.39bg111 000 v 2 = 6.00 10 4 N m 2 + 1 000 16 v 22212.00 10 4 N m 2 = 1 000 15 v 22v = 1.63 m s8.00 10 4 N m 2 +jb2FIG. P14.38g1.63 m s = 12.8 kg sAssuming the top is open to the atmosphere, thenP1 = P0 .NoteP2 = P0 .Flow rate = 2.50 10 3 m3 min = 4.17 10 5 m3 s .(a)soA1 >> A 2Assuming v1 = 0 ,v1 << v 2P1 +2v 1v 2+ gy1 = P2 + 2 + gy 222bv 2 = 2 gy1(b)Flow rate = A 2 v 2 =F dGH 42g12a fa f= 2 9.80 16.0I a17.7f = 4.17 10JK512= 17.7 m sm3 sd = 1.73 10 3 m = 1.73 mm*P14.40Take point 1 at the free surface of the water in the tank and 2 inside the nozzle. FFairwater1 21 2(a)With the cork in place P1 + gy1 + v1 = P2 + gy 2 + v 2 becomes22fP0 + 1 000 kg m3 9.8 m s 2 7.5 m + 0 = P2 + 0 + 0 ; P2 P0 = 7.35 10 4 Pa .For the stopper Fx = 0FIG. P14.40Fwater Fair f = 0P2 A P0 A = faff = 7.35 10 4 Pa 0.011 m(b)2= 27.9 NNow Bernoullis equation givesP0 + 7.35 10 4 Pa + 0 = P0 + 0 +v 2 = 12.1 m s121 000 kg m 3 v 22ejThe quantity leaving the nozzle in 2 h isejaf b12.1 m sg7 200 s =V = Av 2 t = 1 000 kg m3 0.011 mcontinued on next page23.32 10 4 kg .Chapter 14(c)Take point 1 in the wide hose and 2 just outside the nozzle. Continuity:A1 v 1 = A 2 v 2FG 6.6 cm IJH 2 K2v1 = FG 2.2 cm IJH 2 K212.1 m s12.1 m s= 1.35 m s91 21 2P1 + gy1 + v1 = P2 + gy 2 + v 222112P1 + 0 + 1 000 kg m3 1.35 m s = P0 + 0 + 1 000 kg m 3 12.1 m s22v1 =jbegjbeP1 P0 = 7.35 10 4 Pa 9.07 10 2 Pa = 7.26 10 4 PaP14.41Flow rate Q = 0.012 0 m3 s = v1 A1 = v 2 A 2v2 =*P14.43(a)P=(b)*P14.42PELFG IJH K= 0.85e8.5 10 ja9.8 fa87f =Q 0.012 0== 31.6 m sA2A2E mghm==gh = Rghttt5616 MWThe volume flow rate isFGH125 cm30.96 cm= Av1 = 16.3 s2IJK2v1 .The speed at the top of the falling column isv1 =7.67 cm 3 s0.724 cm 2= 10.6 cm s .Take point 2 at 13 cm below:P1 + gy1 +1 21 2v1 = P2 + gy 2 + v 222eje= P0 + 0 +ej121 000 kg m3 v 22eejbv 2 = 2 9.8 m s 2 0.13 m + 0.106 m sg2= 1.60 m sThe volume flow rate is constant:7.67 cm 3 s = d = 0. 247 cmFG d IJH 2Kjb11 000 kg m 3 0.106 m s2jP0 + 1 000 kg m 3 9.8 m s 2 0.13 m +2160 cm sg2g2425426*P14.44Fluid Mechanics(a)P1 +Between sea surface and clogged hole:eja fje1 atm + 0 + 1 030 kg m3 9.8 m s 2 2 m = P2 + 0 + 01 21 2v1 + gy1 = P2 + v 2 + gy 222P2 = 1 atm + 20.2 kPaThe air on the back of his hand pushes opposite the water, so the net force on his hand iseF = PA = 20. 2 10 3 N m 2(b)jFGH IJK e1.2 1042jm2F = 2.28 NNow, Bernoullis theorem is1 atm + 0 + 20. 2 kPa = 1 atm +121 030 kg m 3 v 2 + 02ejv 2 = 6. 26 m s1.2 10 2 m4j b6.26 m sg = 7.08 10e2The volume rate of flow isA2 v 2 =One acrefoot is(a)1 234 m 37.08 1043m s= 1.74 10 6 s = 20.2 daysFG P + 1 v + gyIJ = FG P + 1 v + gyIJK H 2KH 2P + 0 + g a564 mf = 1 atm + 0 + g b 2 096 mgP = 1 atm + e1 000 kg m je9.8 m s jb1 532 mg = 1 atm + 15.0 MPa2Suppose the flow is very slow:2river3(b)erim24 500 m3 d = Av =The volume flow rate isv = 4 500 m3 d(c)m3 s4 047 m 2 0.304 8 m = 1 234 m3RequiringP14.454d 2 v4F 4 IdjFGH 861400 s IJK GH a0.150 mf JK =22.95 m sImagine the pressure as applied to stationary water at the bottom of the pipe:FG P + 1 vH 2IJ = FG P + 1 v + gyIJKH 2K1P + 0 = 1 atm + e1 000 kg m jb 2.95 m sg + 1 000 kg e9.8 m s jb1 532 mg222+ gybottomtop3P = 1 atm + 15.0 MPa + 4.34 kPaThe additional pressure is 4.34 kPa .22Chapter 14P14.46(a)427For upward flight of a water-drop projectile from geyser vent to fountaintop,22v yf = v yi + 2 a y yjaefThen 0 = vi2 + 2 9.80 m s 2 +40.0 m andBetween geyser vent and fountaintop:P1 +Air is so low in density that very nearlyP1 = P2 = 1 atmThen,(b)vi = 28.0 m s1 2vi + 0 = 0 + 9.80 m s 2 40.0 m21 21 2v1 + gy1 = P2 + v 2 + gy 222jaefv1 = 28.0 m s(c)P1 +Between the chamber and the fountain-top:fjejaP P = e1 000 kg m je9.80 m s ja 215 mf =e1 21 2v1 + gy1 = P2 + v 2 + gy 222ejajefP1 + 0 + 1 000 kg m 3 9.80 m s 2 175 m = P0 + 0 + 1 000 kg m 3 9.80 m s 2 +40.0 m1P14.47P1 +3022.11 MPa2v 12A= P2 + 2 (Bernoulli equation), v1 A1 = v 2 A 2 where 1 = 422A2P = P1 P2 =FGHIJK2 2 2 A1v 22 1 and P = 1 15 = 21 000 Pav 2 v1 = v 12222A2ejv1 = 2.00 m s ; v 2 = 4v1 = 8.00 m s :The volume flow rate isSection 14.7v1 A1 = 2.51 10 3 m3 sOther Applications of Fluid Dynamicsa fP14.49Mg = P1 P2 Afor a balanced condition16 000 9.80= 7.00 10 4 P2AwhereP14.48A = 80.0 m 2 P2 = 7.0 10 4 0.196 10 4 = 6.80 10 4 Pab airv=gv2= P = Hg gh22 Hg gh air= 103 m sv airAhMercuryFIG. P14.49428P14.50Fluid MechanicsThe assumption of incompressibility is surely unrealistic, but allows an estimate of the speed:1 21 2v1 = P2 + gy 2 + v 222121.00 atm + 0 + 0 = 0.287 atm + 0 + 1.20 kg m 3 v 22P1 + gy1 +ev2 =P14.51(a)P0 + gh + 0 = P0 + 0 +afe2 1.00 0.287 1.013 10 5 N m 21.20 kg m1 2v 32P + gy +j=347 m sv 3 = 4.43 m s1 21 2v 2 = P0 + 0 + v 322P = P0 gySince v 2 = v 3 ,ySince P 0*P14.523v 3 = 2 ghIf h = 1.00 m ,(b)jFIG. P14.515P01.013 10 Pa== 10.3 m3g 10 kg m3 9.8 m s 2ejejTake points 1 and 2 in the air just inside and outside the window pane.1 21 2v1 + gy1 = P2 + v 2 + gy 2221P0 + 0 = P2 + 1.30 kg m3 11.2 m s2P1 +jbe(a)g2P2 = P0 81.5 PaThe total force exerted by the air is outward,ja fafjbeg a4 mfa1.5 mf =P1 A P2 A = P0 A P0 A + 81.5 N m 2 4 m 1.5 m = 489 N outward(b)P14.53P1 A P2 A =1 21v 2 A = 1.30 kg m 3 22.4 m s22e21.96 kN outward2.00 N= 8.00 10 4 Pa2.50 10 5 m 2In the reservoir, the gauge pressure isP =From the equation of continuity:A1 v1 = A 2 v 2e2.50 105j ejm 2 v1 = 1.00 10 8 m 2 v 2ejv1 = 4.00 10 4 v 222Thus, v1 is negligible in comparison to v 2 .Then, from Bernoullis equation:bP P g + 1 v21221+ gy1 =8.00 10 4 Pa + 0 + 0 = 0 +v2 =e2 8.00 10 4 Pa1 000 kg m3j=1 2v 2 + gy 22121 000 kg m3 v 22e12.6 m sjChapter 14Additional ProblemsP14.54Consider the diagram and apply BernoullisAequation to points A and B, taking y = 0 atthe level of point B, and recognizing that v Ais approximately zero. This gives:afaaf12 w 0 + w g h L sin 212= PB + w v B + w g 02PA +hfLValveBNow, recognize that PA = PB = Patmospheresince both points are open to the atmosphere(neglecting variation of atmosphericpressure with altitude). Thus, we obtainafFIG. P14.54eajfv B = 2 g h L sin = 2 9.80 m s 2 10.0 m 2.00 m sin 30.0v B = 13.3 m sNow the problem reduces to one of projectile motion with v yi = v B sin 30.0 = 6.64 m s . Then,b g22v yf = v yi + 2 a y gives at the top of the arc (where y = y max and v yf = 0 )bg + 2e9.80 m s jby2.25 m babove the level where the water emergesg .0 = 6.64 m sor y max =P14.5522max0gWhen the balloon comes into equilibrium, we must have Fy = B Fg , balloon Fg , He Fg , string = 0HeFg , string is the weight of the string above the ground, and Bis the buoyant force. NowFg , balloon = m balloon ghFg , He = HeVgB = air VgandFg , string = m stringhgLFIG. P14.55Therefore, we have air Vg m balloon g HeVg m stringorh=bairg He V m balloonm stringhg=0LLgiving,a1.29 0.179fekg m jFH3h=a4 0. 400 m30.050 0 kgf3I 0.250 kgKa2.00 mf =1.91 m .429430P14.56Fluid MechanicsAssume v inside 0PgaugeP14.57a faga fbf121 000 30.0 + 1 000 9.80 0.5002= P 1 atm = 4.50 10 5 + 4.90 10 3 = 455 kPaP + 0 + 0 = 1 atm +The balanced condition is one in which the apparent weight of thebody equals the apparent weight of the weights. This condition can bewritten as:Fg B = Fg B where B and B are the buoyant forces on the body and weightsrespectively. The buoyant force experienced by an object of volume Vin air equals:bFIG. P14.57gBuoyant force = Volume of object air gso we haveTherefore,P14.58B = V air gFg = Fg + V FGHB =andIg JKFgF F I GH g JKgair g .air g .The crosssectional area above water isaf a0.200 cmfa0.566 cmf = 0.330 cm2.46 rad 0.600 cm2aAall = 0.6000.400 cmf2220.80 cm= 1.13 cm 2 water gAunder = wood Aall g1.13 0.330= 0.709 g cm 3 = 709 kg m3 wood =1.13P14.59At equilibrium, Fy = 0 :FIG. P14.58B Fspring Fg , He Fg , balloon = 0bggivingButB = weight of displaced air = air Vgandm He = HeV .Therefore, we have:kL = air Vg HeVg m balloon gorL=From the data given,L=Thus, this givesP14.60Fspring = kL = B m He + m balloon g .L = 0.604 m .bairg He V m balloonke1.29 kg m3FIG. P14.59g.j 0.180 kg m3 5.00 m3 2.00 10 3 kg90.0 N ma fP = gh1.013 10 5 = 1.29 9.80 hh = 8.01 kmFor Mt. Everest,29 300 ft = 8.88 kmYese9.80 m s j .2Chapter 14z zz bThe torque is = d = rdFFrom the figureP14.61 = y g H y wdy =431Hg01gwH 361gwH 2The total force is given as2If this were applied at a height y eff such that the torque remainsunchanged, we haveLMN11gwH 3 = y eff gwH 262P14.62(a)OPQy eff =and1H .3FIG. P14.61The pressure on the surface of the two hemispheres is constantat all points, and the force on each element of surface area isdirected along the radius of the hemispheres. The applied forcealong the axis must balance the force on the effective area,which is the projection of the actual surface onto a planeperpendicular to the x axis,A = R 2bP PgR2Therefore,(b)P14.63F=For the values givenF = P0 0.100 P0 0.300 m0g abf2FIG. P14.62= 0.254P0 = 2.58 10 4 NLooking first at the top scale and the iron block, we have:T1 + B = Fg , ironwhere T1 is the tension in the spring scale, B is the buoyant force, and Fg , iron is the weight of the ironblock. Now if m iron is the mass of the iron block, we havem iron = ironVV=som iron iron= Vdisplaced oilThen, B = oil Viron gTherefore, T1 = Fg , iron oilViron g = m iron g oilorFGHT1 = 1 IJKFGHm iron irongI a fa fJK oil916m iron g = 1 2.00 9.80 = 17.3 N7 860 ironNext, we look at the bottom scale which reads T2 (i.e., exerts an upward force T2 on the system).Consider the external vertical forces acting on the beakeroiliron combination. Fy = 0 givesT1 + T2 Fg , beaker Fg , oil Fg , iron = 0orbgbgejT2 = m beaker + m oil + m iron g T1 = 5.00 kg 9.80 m s 2 17.3 NThus, T2 = 31.7 N is the lower scale reading.432P14.64Fluid MechanicsLooking at the top scale and the iron block:FG m IJ gH Kis the buoyant force exerted on the iron block by the oil.F m IJ gB=m g GThus,T =FH KF IJ m g is the reading on the top scale.orT = G1 H KT1 + B = Fg , Fe1FeB = 0 VFe g = 0whereg , FeFe01FeFe0FeFeFeNow, consider the bottom scale, which exerts an upward force of T2 on the beakeroilironcombination.T1 + T2 Fg , beaker Fg , oil Fg , Fe = 0 Fy = 0 :g FGHbT2 = Fg , beaker + Fg , oil + Fg , Fe T1 = m b + m 0 + m Fe g 1 P14.65LMmMNT2 =orb+ m0 +FG IJ m OP gH K PQ0is the reading on the bottom scale.FeFeIJK0m Fe g Fe CuV = 3.083 g Zn xV + Cu 1 x V = 2.517 ga f a fF 3.083 IJ x + 3.083a1 xf = 2.517 GH KFG 1 7.133 IJ x = FG 1 2.517 IJH 8.960 K H 3.083 KZnCux = 0.900 4%Zn = 90.04%P14.66(a)From F = mabgB m shell g m He g = m total a = m shell + m He aWhereB = water Vgandd 34Also,V = r 3 =36Putting these into equation (1) above,FmGHshell+ HeI FJK GHba=(b)t=m He = HeVIJKd 3d 3d 3 m shell Hea = waterg666which givesor(1)water Hegd 36 m shell3m shell + He d6gb1 000 0.180gekg m j a f 4.00 kg 9.80 m sa=af4.00 kg + e0.180 kg m j2a h d f2a 4.00 m 0.200 mf2x=== 4.06 s3 0. 200 m633 0. 200 m6aa0.461 m s 232= 0.461 m s 2Chapter 14P14.67Inertia of the disk: I =bfga11MR 2 = 10.0 kg 0.250 m222= 0.312 kg m 2Angular acceleration: f = i + t=Braking torque:FG 0 300 rev min IJ FG 2 rad IJ FG 1 min IJ = 0.524 rad sH 60.0 s K H 1 rev K H 60.0 s K = I fd = I , sof=2 Ide0.312 kg m je0.524 rad s j = 0.744 NFriction force: f =220.220 mNormal force: f = k n n =gauge pressure: P =P14.68f0.744 N== 1.49 Nk0.5001.49 Nn=A 2.50 10 2 mej2= 758 PaThe incremental version of P P0 = gy isdP = gdyWe assume that the density of air is proportional to pressure, orP=P00dP = PCombining these two equations we havezPP00gdyP0zhdP= g 0 dyPP0 0FG P IJ = ghHP K Pand integrating gives0ln0so where =P14.690 g,P0P = P0 e hEnergy for the fluid-Earth system is conserved.aK + Uf + Eimechaf= K +U f :0+mgL1+ 0 = mv 2 + 022ejv = gL = 2.00 m 9.8 m s 2 = 4.43 m s0433434P14.70Fluid MechanicsLet s stand for the edge of the cube, h for the depth of immersion, ice stand for the density of theice, w stand for density of water, and a stand for density of the alcohol.(a)According to Archimedess principle, at equilibrium we have ice gs 3 = w ghs 2 h = sWith icew ice = 0.917 10 3 kg m3 w = 1.00 10 3 kg m3andwe get(b)s = 20.0 mmh = 20.0 0.917 = 18.34 mm 18.3 mmafWe assume that the top of the cube is still above the alcohol surface. Letting ha stand for thethickness of the alcohol layer, we have a gs 2 ha + w gs 2 hw = ice gs 3sohw =With(c)wwhw = 18.34 0.806 5.00 = 14.31 mm 14.3 mmHereaha = 5.00 mmwe obtainice a = 0.806 10 3 kg m3andFG IJ s FG IJ hH K H Khw = s ha , so Archimedess principle givesa fbgbg a gs 2 ha + w gs 2 s ha = ice gs 3 a ha + w s ha = ice sha = sbbwwg = 20.0 a1.000 0.917f = 8.557 ga1.000 0.806f icea8.56 mmaChapter 14P14.71435Note: Variation of atmospheric pressure with altitude is included inthis solution. Because of the small distances involved, this effect isunimportant in the final answers.(a)Consider the pressure at points A and B in part (b) of thefigure:afUsing the left tube: PA = Patm + a gh + w g L h where thesecond term is due to the variation of air pressure withaltitude.Using the right tube: PB = Patm + 0 gLBut Pascals principle says that PA = PB .afTherefore,Patm + 0 gL = Patm + a gh + w g L horbh=(b)FG Hwg bg a h = w 0 L , givingIJ FK GHIJK1 000 750 0L=5.00 cm = 1.25 cm a1 000 1.29wwConsider part (c) of the diagram showing the situationwhen the air flow over the left tube equalizes the fluidlevels in the two tubes. First, apply Bernoullis equation topoints A and B y A = y B , v A = v , and v B = 0bThis gives: PA +1 a v 2 + a gy A2g1= P + a0 f2Baand since y A = y B , this reduces to: PB PA =2+ a gy B1av22FIG. P14.71(1)Now consider points C and D, both at the level of theoilwater interface in the right tube. Using the variation ofpressure with depth in static fluids, we have:PC = PA + a gH + w gLandPD = PB + a gH + 0 gLBut Pascals principle says that PC = PD . Equating these two gives:PB + a gH + 0 gL = PA + a gH + w gLorSubstitute equation (1) for PB PA into (2) to obtainorv=b2 gL w 0av = 13.8 m sbbg1 a v 2 = w 0 gL2g = 2e9.80 m s jb0.050 0 mgFG 1 000 750 IJH 1.29 K2gPB PA = w 0 gL(2)436P14.72Fluid Mechanics(a)The flow rate, Av, as given may be expressed as follows:25.0 liters= 0.833 liters s = 833 cm3 s .30.0 sThe area of the faucet tap is cm 2 , so we can find the velocity asv=(b)flow rate 833 cm 3 s== 265 cm s = 2.65 m s .A cm 2We choose point 1 to be in the entrance pipe and point 2 to be at the faucet tap. A1 v1 = A 2 v 2gives v1 = 0.295 m s . Bernoullis equation is:P1 P2 =j b122 v 2 v 1 + g y 2 y 12egand givesP1 P2 =orP14.73(a)110 3 kg m32ej b2.65 m sg b0.295 m sg + e10223jejafkg m3 9.80 m s 2 2.00 mPgauge = P1 P2 = 2.31 10 4 Pa .Since the upward buoyant force is balanced by the weight of the sphere,m1 g = Vg = FG 4 R IJ g .H3 K3In this problem, = 0.789 45 g cm 3 at 20.0C, and R = 1.00 cm so we find:m1 = (b)FG 4 R IJ = e0.789 45 g cm jLM 4 a1.00 cmf OP =H3 KN3Q333.307 g .Following the same procedure as in part (a), with = 0.780 97 g cm 3 at 30.0C, we find:m2 = (c)3FG 4 R IJ = e0.780 97 g cm jLM 4 a1.00 cmf OP =H3 KN3Q3333.271 g .When the first sphere is resting on the bottom of the tube,n + B = Fg 1 = m1 g , where n is the normal force.Since B = Vgejafn = m1 g Vg = 3.307 g 0.780 97 g cm 3 1.00 cmn = 34.8 g cm s 2 = 3.48 10 4 N3980 cm s 2Chapter 14*P14.74(a)437Take point 1 at the free water surface in the tank and point 2 at the bottom end of the tube:1 21 2v1 = P2 + gy 2 + v 2221 2P0 + gd + 0 = P0 + 0 + v 22v 2 = 2 gdP1 + gy1 +The volume flow rate isV AhAhAh.== v 2 A . Then t ==ttv 2 A A 2 gda0.5 mf 0.5 m=m 2e9.8 m s j10 m2(b)*P14.75t=(a)For diverging stream lines that pass just above and just below the hydrofoil we have2 10 42244.6 sPt + gy t +1 21 2v t = Pb + gy b + v b .22Ignoring the buoyant force means taking y t y bb g11 22 nv b = Pb + v b221 2 2Pb Pt = v b n 12Pt +ebgThe lift force is Pb Pt A =(b)j1 2 2v b n 1 A .2ejFor liftoff,1 2 2v b n 1 A = Mg2evbjF 2Mg I=GGH en 1jA JJK122The speed of the boat relative to the shore must be nearly equal to this speed of the waterbelow the hydrofoil relative to the boat.(c)ejv 2 n 2 1 A = 2 MgA=b gb9.5 m sg e1.05 1j1 000 kg m2 800 kg 9.8 m s 2223= 1.70 m 2438Fluid MechanicsANSWERS TO EVEN PROBLEMSP14.2~ 10 18 kg m3 ; matter is mostly emptyspaceP14.4P14.6(a) 1.01 10 7 Pa ;(b)7.09 10 5 N outward12.8 kg sP14.401.92 10 4 NP14.38(a) 27.9 N; (b) 3.32 10 4 kg ;(c) 7.26 10 4 PaP14.10P14.44(a) 2.28 N toward Holland; (b) 1.74 10 6 s(a), (b) 28.0 m s ; (c) 2.11 MPaP14.486.80 10 4 PaP14.50347 m sP14.52(a) 489 N outward; (b) 1.96 kN outwardP14.54255 N(a) see the solution; (b) 616 MWP14.46P14.8P14.422.25 m above the level where the wateremerges(a) 65.1 N; (b) 275 NP14.1265.88 10 N down; 196 kN outward;588 kN outwardP14.14(a) 29.4 kN to the right;(b) 16.3 kN m counterclockwiseP14.16(a) 10.3 m; (b) zeroP14.18(a) 20.0 cm; (b) 0.490 cmP14.56455 kPaP14.2012.6 cmP14.58709 kg m3P14.22(a) 444 kg; (b) 480 kgP14.608.01 km; yesP14.62(a) see the solution; (b) 2.58 10 4 NP14.64top scale: 1 mw s hP14.24bP14.26(a) see the solution; (b) 25.0 N up;(c) horizontally inward;(d) tension increases; see the solution;(e) 62.5%; (f) 18.7%gP14.28~ 10 4 balloons of 25-cm diameterP14.30(a) 6.70 cm; (b) 5.74 cmFGHIJK0m Fe g ; FeFGHbottom scale: m b + m 0 +P14.66(a) 0.461 m s 2 ; (b) 4.06 sP14.68IJK 0 m Feg Fesee the solution3P14.32(a) 11.6 cm; (b) 0.963 g cm ;(c) no; see the solutionP14.70(a) 18.3 mm; (b) 14.3 mm; (c) 8.56 mmP14.340.611 kgP14.72(a) 2.65 m s ; (b) 2.31 10 4 PaP14.362.67 10 3 kgP14.74(a) see the solution; (b) 44.6 s15Oscillatory MotionCHAPTER OUTLINE15.1Motion of an ObjectAttached to a SpringMathematical Representationof Simple Harmonic MotionEnergy of the SimpleHarmonic OscillatorComparing Simple HarmonicMotion with Uniform CircularMotionThe PendulumDamped OscillationsForced OscillationsANSWERS TO QUESTIONS15.315.415.515.615.7Q15.4Q15.1Neither are examples of simple harmonic motion, althoughthey are both periodic motion. In neither case is the accelerationproportional to the position. Neither motion is so smooth asSHM. The balls acceleration is very large when it is in contactwith the floor, and the students when the dismissal bell rings.Q15.2You can take = , or equally well, = . At t = 0 , the particleis at its turning point on the negative side of equilibrium, atx = A .Q15.315.2The two will be equal if and only if the position of the particleat time zero is its equilibrium position, which we choose as theorigin of coordinates.(a)In simple harmonic motion, one-half of the time, the velocity is in the same direction as thedisplacement away from equilibrium.(b)Velocity and acceleration are in the same direction half the time.(c)Acceleration is always opposite to the position vector, and never in the same direction.Q15.5No. It is necessary to know both the position and velocity at time zero.Q15.6The motion will still be simple harmonic motion, but the period of oscillation will be a bit larger. TheF kIeffective mass of the system in = GH m JK12will need to include a certain fraction of the mass of theeffspring.439440Q15.7Oscillatory MotionWe assume that the coils of the spring do not hit one another. The frequency will be higher than f bythe factor 2 . When the spring with two blocks is set into oscillation in space, the coil in the centerof the spring does not move. We can imagine clamping the center coil in place without affecting themotion. We can effectively duplicate the motion of each individual block in space by hanging asingle block on a half-spring here on Earth. The half-spring with its center coil clampedor its otherhalf cut offhas twice the spring constant as the original uncut spring, because an applied force ofthe same size would produce only one-half the extension distance. Thus the oscillation frequency inspace isFG 1 IJ FG 2 k IJH 2 K H m K12= 2 f . The absence of a force required to support the vibrating system inorbital free fall has no effect on the frequency of its vibration.Q15.8No; Kinetic, Yes; Potential, No. For constant amplitude, the total energy1 2kA stays constant. The21mv 2 would increase for larger mass if the speed were constant, but here the greater2mass causes a decrease in frequency and in the average and maximum speed, so that the kinetic andpotential energies at every point are unchanged.kinetic energyQ15.9Since the acceleration is not constant in simple harmonic motion, none of the equations in Table 2.2are valid.Equationx t = A cos t + Information given by equationposition as a function of timev t = A sin t + velocity as a function of timeafb gafb gva x f = e A x jaat f = A cosbt + gaat f = xat f22 12velocity as a function of position2acceleration as a function of time2acceleration as a function of positionThe angular frequency appears in every equation. It is a good idea to figure out the value of angularfrequency early in the solution to a problem about vibration, and to store it in calculator memory.Q15.10LfLi2 Liand T f === 2Ti . The period gets larger bygggmass has no effect on the period of a simple pendulum.We have Ti =2 times. Changing theQ15.11(a)Q15.12No, the equilibrium position of the pendulum will be shifted (angularly) towards the back of the car.The period of oscillation will increase slightly, since the restoring force (in the reference frame of theaccelerating car) is reduced.Q15.13The motion will be periodicthat is, it will repeat. The period is nearly constant as the angularamplitude increases through small values; then the period becomes noticeably larger as increasesfarther.Q15.14Shorten the pendulum to decrease the period between ticks.Q15.15No. If the resistive force is greater than the restoring force of the spring (in particular, if b 2 > 4mk ),the system will be overdamped and will not oscillate.Period decreases.(b)Period increases.(c)No change.Chapter 15441Q15.16Yes. An oscillator with damping can vibrate at resonance with amplitude that remains constant intime. Without damping, the amplitude would increase without limit at resonance.Q15.17The phase constant must be rad .Q15.18Higher frequency. When it supports your weight, the center of the diving board flexes down lessthan the end does when it supports your weight. Thus the stiffness constant describing the center of1kis greaterthe board is greater than the stiffness constant describing the end. And then f =2 mfor you bouncing on the center of the board.FG IJH KQ15.19The release of air from one side of the parachute can make the parachute turn in the oppositedirection, causing it to release air from the opposite side. This behavior will result in a periodic drivingforce that can set the parachute into side-to-side oscillation. If the amplitude becomes large enough,the parachute will not supply the needed air resistance to slow the fall of the unfortunate skydiver.Q15.20An imperceptibly slight breeze may be blowing past the leaves in tiny puffs. As a leaf twists in thewind, the fibers in its stem provide a restoring torque. If the frequency of the breeze matches thenatural frequency of vibration of one particular leaf as a torsional pendulum, that leaf can be driveninto a large-amplitude resonance vibration. Note that it is not the size of the driving force that setsthe leaf into resonance, but the frequency of the driving force. If the frequency changes, another leafwill be set into resonant oscillation.Q15.21We assume the diameter of the bob is not very small compared to the length of the cord supportingit. As the water leaks out, the center of mass of the bob moves down, increasing the effective lengthof the pendulum and slightly lowering its frequency. As the last drops of water dribble out, thecenter of mass of the bob hops back up to the center of the sphere, and the pendulum frequencyquickly increases to its original value.SOLUTIONS TO PROBLEMSSection 15.1P15.1Motion of an Object Attached to a Spring(a)Since the collision is perfectly elastic, the ball will rebound to the height of 4.00 m and thenrepeat the motion over and over again. Thus, the motion is periodic .(b)To determine the period, we use: x =1 2gt .2The time for the ball to hit the ground is t =aaf2 4.00 m2x== 0.909 sg9.80 m s 2fThis equals one-half the period, so T = 2 0.909 s = 1.82 s .(c)No . The net force acting on the ball is a constant given by F = mg (except when it is incontact with the ground), which is not in the form of Hookes law.442Oscillatory MotionSection 15.2Mathematical Representation of Simple Harmonic Motionf FGHa6IJKx = 5.00 cm cos 2t +v=dx= 10.0 cm s sin 2t +dt6(c)a=dv= 20.0 cm s 2 cos 2t +dt6(d)P15.3(a)(b)P15.2A = 5.00 cmaf aIJKv = 5.00 cm sAt t = 0 ,IJKj FGHeAt t = 0 ,a = 17.3 cm s 2andg FGHbx = 5.00 cm cosT=fb2=4.33 cm2= 3.14 s2gx = 4.00 m cos 3.00t + Compare this with x = A cos t + to find(a) = 2 f = 3.00orT=f = 1.50 Hz1= 0.667 sf(b)A = 4.00 m(c) = rad(d)*P15.4f FGH IJK =6aAt t = 0 ,x t = 0.250 s = 4.00 m cos 1.75 = 2.83 m(a)The spring constant of this spring isaf af afk=F 0.45 kg 9.8 m s 2== 12.6 N mx0.35 mwe take the x-axis pointing downward, so = 0x = A cos t = 18.0 cm cos(d)12.6 kg0.45 kg s 284. 4 s = 18.0 cm cos 446.6 rad = 15.8 cma fNow 446.6 rad = 71 2 + 0.497 rad . In each cycle the object moves 4 18 = 72 cm , so it hasaf afmoved 71 72 cm + 18 15.8 cm = 51.1 m .(b)By the same steps, k =x = A cos(e)a f0. 44 kg 9.8 m s 2= 12.1 N m0.355 mk12.1t = 18.0 cm cos84.4 = 18.0 cm cos 443.5 rad = 15.9 cmm0.44443.5 rad = 70 2 + 3.62 radafDistance moved = 70 72 cm + 18 + 15.9 cm = 50.7 m(c)The answers to (d) and (e) are not very different given the difference in the data about thetwo vibrating systems. But when we ask about details of the future, the imprecision in ourknowledge about the present makes it impossible to make precise predictions. The twooscillations start out in phase but get completely out of phase.Chapter 15P15.5(a)At t = 0 , x = 0 and v is positive (to the right). Therefore, this situation corresponds tox = A sin tandv = vi cos tSince f = 1.50 Hz , = 2 f = 3.00aafv max = vi = A = 2.00 3.00 = 6.00 cm s = 18.8 cm sThe particle has this speed at t = 0 and next at(c)aa max = A 2 = 2.00 3.00f2t=T1=s23t=3T = 0.500 s4= 18.0 2 cm s 2 = 178 cm s 2This positive value of acceleration first occurs at(d)fx = 2.00 cm sin 3.00 tAlso, A = 2.00 cm, so that(b)4432s and A = 2.00 cm, the particle will travel 8.00 cm in this time.33Hence, in 1.00 s = T , the particle will travel8.00 cm + 4.00 cm = 12.0 cm .2Since T =FGHIJKafFG v IJ sin tHKiThe proposed solutionx t = xi cos t +implies velocityv=dx= x i sin t + vi cos tdtand accelerationP15.6a=vdv= x i 2 cos t vi sin t = 2 x i cos t + i sin t = 2 xdtFGHFG IJH KIJK(a)The acceleration being a negative constant times position means we do have SHM, and itsangular frequency is . At t = 0 the equations reduce to x = xi and v = vi so they satisfy allthe requirements.(b)v 2 ax = x i sin t + vi cos tbg e x 2i2jFGHcos t vi sin t xi cos t +FG v IJ sin tIJHK Kiv 2 ax = xi2 2 sin 2 t 2 xi vi sin t cos t + vi2 cos 2 t+ x i2 2 cos 2 t + x i vi cos t sin t + xi vi sin t cos t + vi2 sin 2 t = x i2 2 + vi2So this expression is constant in time. On one hand, it must keep its original value vi2 ai xi .On the other hand, if we evaluate it at a turning point where v = 0 and x = A , it isA 2 2 + 0 2 = A 2 2 . Thus it is proved.P15.7(a)T=12.0 s= 2.40 s5(b)f=11== 0.417 HzT 2. 40(c) = 2 f = 2 0.417 = 2.62 rad saf444*P15.8Oscillatory MotionThe mass of the cube isjaefm = V = 2.7 10 3 kg m3 0.015 m3= 9.11 10 3 kgThe spring constant of the strip of steel isk=f=P15.9f=1= 22km14.3 NF== 52.0 N mx 0.027 5 m2k=x = A cos t52 kgk1=m 212T=orSolving for k,*P15.10=s 2 9.11 10 3 kg= 12.0 Hz1m= 2fk4 2 mA = 0.05 mT2=b4 2 7.00 kga2.60 sf2g=40.9 N m .a = A 2 cos tv = A sin tIf f = 3 600 rev min = 60 Hz , then = 120 s 1afv max = 0.05 120 m s = 18.8 m sP15.11(a)=k=m8.00 N m= 4.00 s 10.500 kgFrom this we find that(b)t=FG 1 IJ sin FG x IJ and whenH 4.00 K H 10.0 K1Using t =f2m s 2 = 7.11 km s 2aa fa = 160 sina 4.00t f cm sv = 40.0 cos 4.00t cm sv max = 40.0 cm s2amax = 160 cm s 2 .x = 6.00 cm, t = 0.161 s.a fa = 160 sin 4.00a0.161f =FG 1 IJ sin FG x IJH 4.00 K H 10.0 K1when x = 0 , t = 0 and whenx = 8.00 cm, t = 0.232 s.Therefore,t = 0.232 s .fx = 10.0 sin 4.00t cm .so position is given byv = 40.0 cos 4.00 0.161 = 32.0 cm sWe find(c)aamax = 0.05 12096.0 cm s 2 .Chapter 15P15.12445m = 1.00 kg , k = 25.0 N m, and A = 3.00 cm. At t = 0 , x = 3.00 cm(a)=k=m25.0= 5.00 rad s1.0022T=== 1.26 s 5.00so that,(b)bgv max = A = 3.00 10 2 m 5.00 rad s = 0.150 m sbamax = A 2 = 3.00 10 2 m 5.00 rad s(c)g2= 0.750 m s 2Because x = 3.00 cm and v = 0 at t = 0 , the required solution is x = A cos tafx = 3.00 cos 5.00t cmoraaffdx= 15.0 sin 5.00t cm sdtdva== 75.0 cos 5.00t cm s 2dtv=P15.13The 0.500 s must elapse between one turning point and the other. Thus the period is 1.00 s.=b2= 6. 28 sTfgaand v max = A = 6.28 s 0.100 m = 0.628 m s .P15.14(a)v max = AA=(b)Section 15.3P15.15(a)v max=vx = A sin t = FG v IJ sin tHKEnergy of the Simple Harmonic OscillatorEnergy is conserved for the block-spring system between the maximum-displacement andthe half-maximum points:aK + U f = aK + U f1b6.50 N mga0.100 mf2i32.5 mJ =k=m0+fb21m 0.300 m s2b=1m 0.300 m s2g+ 8.12 mJ26.50 N m= 3.46 rad s0.542 kg(b)=(c)amax = A 2 = 0.100 m 3.46 rad sbg2g2+1 2 11kA = mv 2 + kx 2222bge16.50 N m 5.00 10 2 m2m=a2 24.4 mJ9.0 10T == 1.20 m s 2j2=2f2m s22= 0.542 kg2 rad= 1.81 s3.46 rad s446P15.16Oscillatory Motion22== 25.1 rad sT0. 250m = 200 g , T = 0.250 s, E = 2.00 J ; =bg = 126 N m2a 2.00f= 0.178 m(a)(b)P15.17k = m 2 = 0.200 kg 25.1 rad sE=kA 2A=22E=k2126Choose the car with its shock-absorbing bumper as the system; by conservation of energy,11mv 2 = kx 2 :22k= 3.16 10 2 mmev=xjej2v max = A(c)a max = A 2 = 3.50 10 2 m 22.4 s 1(a)E=(b)P15.19(a)2kA 2 250 N m 3.50 10 mE==22(b)P15.18v = A2 x2 =ebj2ge250= 22.4 s 10.500j2= 28.0 mJkA2 x2m35.0e4.00 10 j e1.00 10 j = 1.02 m s50.0 101111mv = kA kx = a35.0 fLe 4.00 10 j e3.00 10 j O =MNPQ2222v=2 23222 22 22(d)P15.20v max = 0.784 m s= 17.5 m s 21 2 1kA = 35.0 N m 4.00 10 2 m22.(c)= 0.153 Jk=m=where5.00 10 6= 2.23 m s10 31 21kx = E mv 2 = 15.8 mJ22(a)k=(b)=(c)v max = A = 50.0 0. 200 = 1.41 m s at x = 0(d)amax = 2(e)E=(f)v = A 2 x 2 = 50.0(g)a = 2 x = 50.02 2F20.0 N== 100 N mx 0.200 mk= 50.0 rad sm1 2kA2f=soa fA = 50.0a0.200 f = 10.0 m s1= a100 fa0.200 f = 2.00 J22at x = A2FG 0.200 IJ =H 3 Ka80.2009f23.33 m s 2= 1.33 m s2= 1.13 Hz12.2 mJChapter 15P15.21(a)E=a f1 21kA , so if A = 2 A , E = k A 222=a f1k 2A22447= 4ETherefore E increases by factor of 4 .(b)kA , so if A is doubled, v max is doubled .m(c)a max =kA , so if A is doubled, a max also doubles .m(d)*P15.22v max =T = 2(a)y f = yi + v yi t +mis independent of A, so the period is unchanged .k1ayt 22111 m = 0 + 0 + 9.8 m s 2 t 22et=(b)j22 m s 2= 1.50 s9.8 mTake the initial point where she steps off the bridge and the final point at the bottom of hermotion.eK + Ug+ Usj = eK + Uig+ Usjf10 + mgy + 0 = 0 + 0 + kx 22165 kg 9.8 m s 2 36 m = k 25 m2k = 73. 4 N ma(c)The spring extension at equilibrium is x =f2F 65 kg 9.8 m s 2== 8.68 m , so this point isk73.4 N m11 + 8.68 m = 19.7 m below the bridge and the amplitude of her oscillation is36 19.7 = 16.3 m .k=m73.4 N m= 1.06 rad s65 kg(d)=(e)Take the phase as zero at maximum downward extension. We find what the phase was 25 mhigher when x = 8.68 m:In x = A cos t ,FGHt8.68 m = 16.3 m cos 1.06st = 2.01 sIJK16.3 m = 16.3 m cos 0t1.06 = 122 = 2.13 radsThen +2.01 s is the time over which the spring stretches.(f)total time = 1.50 s + 2.01 s = 3.50 s448P15.23Oscillatory MotionModel the oscillator as a block-spring system.v2 + 2x2 = 2 A2From energy considerations,v max = A and v =soFrom this we find x 2 =P15.243 2A4FG A IJH2KandA2x=2+ 2x2 = 2 A23A = 2.60 cm where A = 3.00 cm2The potential energy isa f1 2 1 2kx = kA cos 2 t .22Us =The rate of change of potential energy isa fa fdU s 1 21= kA 2 cos t sin t = kA 2 sin 2t .22dt(a)This rate of change is maximal and negative at2t =Then, t =, 2t = 2 + , or in general, 2t = 2n + for integer n.222af 4n + 14n + 1 =44 3.60 s 1afejFor n = 0 , this gives t = 0.218 s while n = 1 gives t = 1.09 s .All other values of n yield times outside the specified range.(b)Section 15.4P15.25dU sdt=max1 21kA = 3.24 N m 5.00 10 2 m22bgej e3.60 s j =2114.6 mWComparing Simple Harmonic Motion with Uniform Circular Motion(a)The motion is simple harmonic because the tire is rotating with constant velocity and youare looking at the motion of the bump projected in a plane perpendicular to the tire.(b)Since the car is moving with speed v = 3.00 m s , and its radius is 0.300 m, we have:=3.00 m s= 10.0 rad s .0.300 mTherefore, the period of the motion is:T=2=2b10.0 rad sg =0.628 s .Chapter 15P15.26The angle of the crank pin is = t .Its x-coordinate isPistonx = A cos = A cos tAwhere A is the distance from thecenter of the wheel to the crank pin.This is of the form x = A cos t + ,so the yoke and piston rod movewith simple harmonic motion.bSection 15.5P15.27(a)gFIG. P15.26The PendulumT = 2L=(b)x = AgT 24 2Lge9.80 m s ja12.0 sf2=Tmoon = 24 2L= 2g moon2= 35.7 m35.7 m1.67 m s 2= 29.1 sTT = 2LTgTTC = 2LCgCWe knowTT = TC = 2.00 sFor which, we seeLT LC=gT gCorP15.29The period in Tokyo isand the period in Cambridge isP15.28g C LC 0.994 2=== 1.001 5g T LT 0.992 7The swinging box is a physical pendulum with period T = 2I.mgdThe moment of inertia is given approximately byI=1mL2 (treating the box as a rod suspended from one end).3Then, with L 1.0 m and d T 213L,2mL2mgchL2= 2af2 1.0 m2L= 2= 1.6 s or T ~ 10 0 s .23g3 9.8 m sejx ( t)449450Oscillatory MotionP15.31=2:TT==P15.30g:LL=2=2= 1.42 s4. 43g9.80== 0.499 m224.43a fUsing the simple harmonic motion model:A = r = 1 m 15= 0. 262 m180g9.8 m s 2== 3.13 rad sL1m=(a)v max = A = 0. 262 m 3.13 s = 0.820 m s(b)a max = A 2 = 0.262 m 3.13 sba tan = r(c)g2= 2.57 m s 2a tan 2.57 m s 2== 2.57 rad s 2r1m=FIG. P15.31F = ma = 0.25 kg 2.57 m s 2 = 0.641 NMore precisely,(a)1mv 2and2 v max = 2 gL 1 cos = 0.817 m smgh =a(b)ah = L 1 cos ffI = mgL sin max =mgL sin 2mL=gsin i = 2.54 rad s 2La faf(c)P15.32Fmax = mg sin i = 0.250 9.80 sin 15.0 = 0.634 N(a)The string tension must support the weight of the bob, accelerate it upward, and also providethe restoring force, just as if the elevator were at rest in a gravity field 9.80 + 5.00 m s 2aT = 2L5.00 m= 2g14.8 m s 2T = 3.65 s(b)(c)g eff =5.00 mT = 2e9.80 m s2 5.00 m s 2j= 6.41 se9.80 m s j + e5.00 m s jT = 2 =2 25.00 m11.0 m s 22 2= 4.24 s= 11.0 m s 2fChapter 15P15.33Referring to the sketch we havexRFor small displacements,tan sin mgF=andx = kxRSince the restoring force is proportional to the displacement fromequilibrium, the motion is simple harmonic motion.F = mg sin tan =andComparing toF = m 2 x shows =P15.34T=(a)g.Rk=mtotal measured time50a fPeriod, T asfLength, L m41.000 0.750 0.50031.996 1.732 1.4222L4 2 LT = 2sog=gT2The calculated values for g are:(b)af10Period, T sg m s29.91j9.87From T 2 =Thus, g =0.50.759.76this agrees with the accepted value of g = 9.80 m s 2 within 0.5%.F 4 I L , the slope of TGH g JK22versus L graph =4 2= 4.01 s 2 m .g4 2= 9.85 m s 2 . This is the same as the value in (b).slopef = 0. 450 Hz , d = 0.350 m, and m = 2.20 kgT=1;fT = 2I =T2I;mgdmgd42=T2 =FG 1 IJH fK21.0L, mFIG. P15.34Thus, g ave = 9.85 m s 2(c)0.251.996 1.732 1.422eFIG. P15.33T2, s2The measured periods are:P15.354514 2 Imgdmgd42=a fa f =e0.450 s j2.20 9.80 0.350421 20.944 kg m 2FIG. P15.35452P15.36Oscillatory Motion(a)The parallel-axis theorem:I = I CM + Md 2 ==MFG 13 m IJH 12 Kaf11ML2 + Md 2 =M 1.00 m12122af+ M 1.00 m22ejM 13 m 213 mI= 2= 2= 2.09 sT = 212 Mg 1.00 mMgd12 9.80 m s 2(b)(a)fej1.00 m9.80 m s2difference == 2.01 s2.09 s 2.01 s= 4.08%2.01 sThe parallel axis theorem says directly I = I CM + md 2soT = 2eII= 2mgdCM+ md 2jmgdWhen d is very large T 2dgets large.gWhen d is very small T 2(b)FIG. P15.36For the simple pendulumT = 2P15.37aI CMgets large.mgdSo there must be a minimum, found byj bmgdgF 1IFG 1 IJ eImg + 2 bmgd g= 2 e I+ md j G J bmgd gH 2KH 2K e I+ md jmg2 md mgd=+=0+ md j bmgd g+ md j bmgd gIIeedTd=0=2 I CM + md 2dddde121 22 12CM1 22CM2 12CM3 23 2CM2 123 2This requires I CM md 2 + 2md 2 = 0orP15.38ICM = md 2 .We suppose the stick moves in a horizontal plane. Then,bfga112.00 kg 1.00 mmL2 =1212IT = 2I==4 2 IT2=e4 2 0.167 kg m 2a180 sf2j=2= 0.167 kg m 2203 N mCM+ md 2j1 22mdChapter 15P15.39eje(a)Id 2= ;dt 22= =ITe = I 2 = 5.00 10 7Section 15.6jFGH 0.2250 IJKFIG. P15.392= 3.16 10 4NmradDamped Oscillations11mv 2 + kx 2222dEd x= mv 2 + kxvdtdt2md x= kx bvdt 2dE= v kx bv + kvxdtdE= bv 2 < 0dtE=The total energy isTaking the time-derivative,Use Equation 15.31:aThus,P15.412I = 5.00 10 7 kg m 2(b)P15.40jT = 0.250 s, I = mr 2 = 20.0 10 3 kg 5.00 10 3 mfb i = 15.0g t = 1 000 = 5.50x1 000 Ae bt 2 m 5.50 b 1 000 g===e b15.0xiAx = Ae bt 2 m2mFG 5.50 IJ = 1.00 = bb1 000gH 15.0 K2mlnP15.42b= 1.00 10 3 s 12mbx = Ae bt 2 m cos t + Show thatis a solution of kx bwhere=bx = Ae bt 2 m cos t + FGHgg2dxd x=m 2dtdtFG IJH Kkbm2m(1)2.(2)IJ b gb gKd xb LFG b IJ cosbt + g Ae sinbt + gOP=AeH 2m K2m MdtNQLFG b IJ sinbt + g + Ae cosbt + gOP M AeH 2m KNQdxb= Ae bt 2 m cos t + Ae bt 2 m sin t + dt2m2 bt 2 m2 bt 2 mcontinued on next page(3)(4) bt 2 m bt 2 m2(5)453454Oscillatory MotionSubstitute (3), (4) into the left side of (1) and (5) into the right side of (1);2b g 2bm Ae cosbt + g + bAe sinbt + gbLFG b IJ cosbt + g Ae sinbt + gOP= M AeH 2m K2NQbcosbt + g+ Ae sinbt + g m Ae2Compare the coefficients of Aecosbt + g and Aesinbt + g :Fk b IbbF b Ibb= Gcosine-term: k +H 2m JK m = 4m mGH m 4m JK = k + 2m2m2 kAe bt 2 m cos t + + bt 2 m bt 2 m bt 2 m bt 2 m bt 2 m bt 2 m2 bt 2 m bt 2 m2sine-term:b = +22222af afbb + = b22bgSince the coefficients are equal, x = Ae bt 2 m cos t + is a solution of the equation.*P15.43The frequency if undamped would be 0 =(a)k=mWith damping2 = 0 FG b IJH 2m K2=2.05 10 4 N m= 44.0 s.10.6 kgFG 44 1 IJ FG 3 kg IJH s K H s 2 10.6 kg K= 1 933.96 0.02 = 44.0f=(b)b 44.0== 7.00 Hz2 2 s21sgIn x = A 0 e bt 2 m cos t + over one cycle, a time T =A0 e b 22 m22, the amplitude changes from A0 tofor a fractional decrease ofA 0 A 0 e b m= 1 e 3 a10.644.0 f = 1 e 0 .020 2 = 1 0.979 98 = 0.020 0 = 2.00% .A0(c)The energy is proportional to the square of the amplitude, so its fractional rate of decrease istwice as fast:E=We specify1 2 1 2 2 bt 2 m= E0 e bt m .kA = kA 0 e220.05E0 = E0 e 3 t 10.60.05 = e 3 t 10.6e + 3 t 10 .6 = 203t= ln 20 = 3.0010.6t = 10.6 sChapter 15Section 15.7P15.44(a)Forced OscillationsFor resonance, her frequency must matchf0 =(b)01=2 24.30 10 3 N m= 2.95 Hz .12.5 kgk1=m 2dxdv= A sin t , and a == A 2 cos t , the maximum accelerationdtdtis A 2 . When this becomes equal to the acceleration due to gravity, the normal forceexerted on her by the mattress will drop to zero at one point in the cycle:From x = A cos t , v =2A = gP15.45orb gF = 3.00 cos 2 t NA=g2=gkme9.80 m s jb12.5 kgg =A=2gm=k4.30 10 3 N mand2= 2 rad sT(a)=(b)k = 20.0 N mso2.85 cmT = 1.00 sIn this case,0 =k=m20.0= 3.16 rad s2.00The equation for the amplitude of a driven oscillator,FG F IJ eH mK0220j1a f34 2 3.162with b = 0, givesA=ThusP15.464552 1A = 0.050 9 m = 5.09 cm .F0 cos t kx = mbx = A cos t + d2xdt 20 ==km(1)g(2)bdx= A sin t + dtgd2x= A 2 cos t + dt 2b(3)g(4)b g e j bj cosbt + g = F cos tSubstitute (2) and (4) into (1):F0 cos t kA cos t + = m A 2 cos t + Solve for the amplitude:ekA mA20These will be equal, provided only that must be zero and kA mA 2 = F0Thus, A =F0mc hkm2g456P15.47Oscillatory MotionFrom the equation for the amplitude of a driven oscillator with no damping,F0 mA=e220j2e = 2 f = 20.0 s 1eF0 = mA F0 =P15.48220j20 =jFG 40.0 IJ e2.00 10 jb3 950 49.0g =H 9.80 K2e220j + b b m g2318 NA=2Fext me220j2=eFext m220j=Fext m2 2 0Fext m k Fext 6.30 N m1.70 N= =0.150 kgAm mA0.150 kg 0.440 mThus,22 =0 This yields = 8.23 rad s or = 4.03 rad sThen,f=2bgives either f = 1.31 Hzorfgaf = 0.641 HzThe beeper must resonate at the frequency of a simple pendulum of length 8.21 cm:f=*P15.50c hFext mA=With b = 0,P15.49k200== 49.0 s 240 .0m9.8012g1=L 29.80 m s 2= 1.74 Hz .0.082 1 mFor the resonance vibration with the occupants in the car, we have for the spring constant of thesuspensionf=12kme j d1 130 kg + 4b72.4 kg gi = 1.82 10F 4b72. 4 kg ge9.8 m s jx= == 1.56 10 mk = 4 2 f 2 m = 4 2 1.8 s 122Now as the occupants exitk1.82 105kg s225kg s 2Chapter 15457Additional ProblemsP15.51Let F represent the tension in the rod.pivot(a)At the pivot, F = Mg + Mg = 2 MgA fraction of the rods weight MgFG y IJ as well as theH LKPLweight of the ball pulls down on point P. Thus, thetension in the rod at point P isF = MgFG y IJ + Mg =H LKFGHMg 1 +yLIJKy.MFIG. P15.51(b)Relative to the pivot, I = I rod + I ball =14ML2 + ML2 = ML233Iwhere m = 2 M and d is the distance from themgdpivot to the center of mass of the rod and ball combination. Therefore,For the physical pendulum, T = 2d=For L = 2.00 m, T =P15.52(a)Total energy =M43c h + ML = 3L and T = 2L2M+Ma4f=2 2.00 m9.80 m s2b43ML2a 2 M f gc h3L443=2L.g2.68 s .fga1 2 1kA = 100 N m 0.200 m222= 2.00 JAt equilibrium, the total energy is:bgbgbg11m1 + m 2 v 2 = 16.0 kg v 2 = 8.00 kg v 2 .22Therefore,b8.00 kg gv2= 2.00 J , and v = 0.500 m s .This is the speed of m1 and m 2 at the equilibrium point. Beyond this point, the mass m 2moves with the constant speed of 0.500 m/s while mass m1 starts to slow down due to therestoring force of the spring.continued on next page458Oscillatory Motion(b)The energy of the m1 -spring system at equilibrium is:bgb11m1 v 2 = 9.00 kg 0.500 m s22This is also equal toTherefore,g2= 1.125 J .a f12k A , where A is the amplitude of the m1 -spring system.2a fa f1100 A 22= 1.125 or A = 0.150 m.m1= 1.885 skThe period of the m1 -spring system is T = 21T = 0.471 s after it passes the equilibrium point for the spring to become fully4stretched the first time. The distance separating m1 and m 2 at this time is:and it takesD=vP15.53F d xIGH dt JK2FG T IJ A = 0.500 m s a0.471 sf 0.150 m = 0.085 6 =H 4Ks= A 22BmaxPfmax = sn = s mg = mA 2A=8.56 cm .sg= 6.62 cm2nfBmgFIG. P15.53P15.54The maximum acceleration of the oscillating system is a max = A 2 = 4 2 Af 2 . The friction forceexerted between the two blocks must be capable of accelerating block B at this rate. Thus, if Block Bis about to slip,ef = fmax = sn = s mg = m 4 2 Af 2P15.55jors g.4 2 f 2A=Deuterium is the isotope of the element hydrogen with atoms having nuclei consisting of oneproton and one neutron. For brevity we refer to the molecule formed by two deuterium atoms as Dand to the diatomic molecule of hydrogen-1 as H.MD = 2MHD=HkMDkMH=MH=MD12fD =fH2= 0.919 10 14 HzChapter 15P15.5611mv 2 + I 2 ,22where is the rotation rate of the ball about itscenter of mass. Since the center of the ball movesalong a circle of radius 4R, its displacement fromequilibrium is s = 4R and its speed isdsd= 4Rv=. Also, since the ball rolls withoutdtdtslipping,459The kinetic energy of the ball is K =5Ra fFG IJH Kv=ds= Rdt=soRhFG IJH Kvd=4RdtsFIG. P15.56The kinetic energy is thenK==FGH1dm 4R2dtIJK2FG IJH K112mR 2 d10dt+FGH1 2mR 22 5IJ FG 4 d IJK H dt K22afWhen the ball has an angular displacement , its center is distance h = 4R 1 cos higher thanwhen at the equilibrium position. Thus, the potential energy is U g = mgh = 4mgR 1 cos . For smallafangles, 1 cos af2(see Appendix B). Hence, U g 2mgR 2 , and the total energy is2E = K +Ug =FG IJH K112mR 2 d10dt2+ 2mgR 2 .FG IJH K28 R d d F 5 g IJ .+ g = 0 , orThis reduces to= GH 28R K5 dtdtSince E = constant in time,2FG IJH K112 mR 2 d d 2dEd.+ 4mgR=0=5dtdt dt 2dt222With the angular acceleration equal to a negative constant times the angular position, this is in the5g.defining form of a simple harmonic motion equation with =28 RThe period of the simple harmonic motion is then T =2= 228 R.5g460P15.57Oscillatory Motion(a)LiLahaFIG. P15.57(a)(b)T = 2 1 dLdT=dtg L dtLgafWe need to find L t anddL. From the diagram in (a),dtFG IJH K1 dha h dL= ;.2 dt2 2 dtL = Li +But(1)dMdVdh== A . Therefore,dtdtdtIJK(2)FG 1 IJ FG dM IJ t = L LH 2 A K H dt K(3)FGH1 dM dLdh1 dM==;A dt dtdt2 A dtzLAlso,dL =LiiSubstituting Equation (2) and Equation (3) into Equation (1):FGH1g 2 a 2dT=dt(c)I FG dM IJJK H dt K1Li +12 a 2c htdMdt.Substitute Equation (3) into the equation for the period.2T=gLi +FG IJH K1dMt2dt2 aOr one can obtain T by integrating (b):FGHI FG dM IJzJK H dt K z L + dt c htL 2 OPLO F 1 I F dM I MT T =GH 2a JK GH dt JK M c h PMNM L + 2 1a FGH dM IJK t L PQPdtgNQL21 F dM I, so T =L +G Jt .g2 a H dt KgTidT =TiBut Ti = 2it1g 2 a 22i2i12 a 212 a 20dMdtdMdti2iChapter 15P15.58=(a)P15.59k2=mT2k = m =4 2 mm =(b)T2a fk T= m4 2FG T IJHTK2HyWe draw a free-body diagram of the pendulum.The force H exerted by the hinge causes no torqueabout the axis of rotation. = IHxhd 2= dt 2and = MgL sin + kxh cos = IkxLx2d dt 2kmmgFor small amplitude vibrations, use theapproximations: sin , cos 1, and x s = h .Therefore,FGHIJKMgL + kh 2d 2 = 2=Idt 2=bgbMgL + kh 2ML212v = A sin t + MgL + kh 2ML2v = A sin = v maxThis requires = 90 , sox = A cos t + 90And this is equivalent tox = A sin tNumerically we have=and v max = A20 m s = 10 s 1 AInak=m50 N m= 10 s 10.5 kgeafjf eA=2mjx = 2 m sin 10 s 1 tSo(b)= 2 fgIn x = A cos t + ,we have at t = 0(a)L sinFIG. P15.59f=*P15.602111mv 2 + kx 2 = kA 2 ,222impliesFGH1 21kx = 3 mv 22211 2 1 2 1 2kx + kx = kA3222x=continued on next pageIJK3A = 0.866 A = 1.73 m44 2x = A23hcos461462Oscillatory Motion(c)(d)gL=IngL=a2=9.8 m s 2e10 s j1 2f e= 0.098 0 mjx = 2 m sin 10 s 1 tthe particle is at x = 0 at t = 0 , at 10t = s , and so on.The particle is atx=1 mwhenwith solutionse10 s jt = 61= sin 10 s 1 t2ej1e10 s jt = + , and so on.6FIThe minimum time for the motion is t in 10 t = G J sH 6KFIt = G J s = 0.052 4 sH 60 K1P15.61(a)FIG. P15.60(d)At equilibrium, we haveF LI = 0 mg G 2 J + kx0 LH Kwhere x 0 is the equilibrium compression.After displacement by a small angle,FIG. P15.61F LIF LI = mg G 2 J + kxL = mg G 2 J + kbx0 L gL = kL2H KH KBut,1 = I = 3 mL2d 2.dt 2Sod 23k= .2mdtThe angular acceleration is opposite in direction and proportional to the displacement, so3kwe have simple harmonic motion with 2 =.m(b)f=1=2 23k1=m 2bg=3 100 N m5.00 kg1.23 HzChapter 15*P15.62463As it passes through equilibrium, the 4-kg object has speedv max = A =100 N mkA=2 m = 10.0 m s.m4 kgIn the completely inelastic collision momentum of the two-object system is conserved. So the new10-kg object starts its oscillation with speed given bybg b g bg4 kg 10 m s + 6 kg 0 = 10 kg v maxv max = 4.00 m s(a)The new amplitude is given by112mv max = kA 222b10 kg 4 m sg = b100 N mgA22A = 1.26 mThus the amplitude has decreased byT = 24 kgm= 2= 1.26 sk100 N mT = 210 2s = 1.99 s100The period has increased by1.99 m 1.26 m = 0.730 sThe old energy was112mv max = 4 kg 10 m s22The new mechanical energy is(c)The old period wasThe new period is(b)2.00 m 1.26 m = 0.735 m110 kg 4 m s2b gbbgbg2g2= 200 J= 80 JThe energy has decreased by 120 J .(d)P15.63The missing mechanical energy has turned into internal energy in the completely inelasticcollision.(a)T=(b)E=(c)At maximum angular displacement2= 2L= 3.00 sga fa f11mv 2 = 6.74 2.0622ah = L L cos = L 1 cos 2f= 14.3 Jmgh =1mv 22cos = 1 hLh=v2= 0.217 m2g = 25.5464P15.64Oscillatory MotionOne can write the following equations of motion:T kx = 0(describes the spring)mg T = ma = mafR T T = I2d xdt 2(for the hanging object)d 2 I d 2 x=dt 2 R dt 2(for the pulley)with I =FIG. P15.641MR 22Combining these equations gives the equation of motionFG m + 1 MIJ d x + kx = mg .H 2 K dt22afmgmg(wherearises because of the extension of the spring due tokkthe weight of the hanging object), with frequencyThe solution is x t = A sin t +f=2=k1=m + 1 M 2212(a)f = 3.56 Hz(b)For M = 0.250 kgf = 2.79 Hz(c)P15.65For M = 0For M = 0.750 kg100 N m.0.200 kg + 1 M2f = 2.10 HzSuppose a 100-kg biker compresses the suspension 2.00 cm.Then,k=F980 N== 4.90 10 4 N mx 2.00 10 2 mIf total mass of motorcycle and biker is 500 kg, the frequency of free vibration isf=121k=m 24.90 10 4 N m= 1.58 Hz500 kgIf he encounters washboard bumps at the same frequency, resonance will make the motorcyclebounce a lot. Assuming a speed of 20.0 m/s, we find these ridges are separated by20.0 m s1.58 s 1= 12.7 m ~ 10 1 m .In addition to this vibration mode of bouncing up and down as one unit, the motorcycle can alsovibrate at higher frequencies by rocking back and forth between front and rear wheels, by havingjust the front wheel bounce inside its fork, or by doing other things. Other spacing of bumps willexcite all of these other resonances.Chapter 15P15.66(a)For each segment of the springdK =Also,vx =xdm =andva f12dm v x .2mFIG. P15.66dx .Therefore, the total kinetic energy of the block-spring system isK=(b)=km effTherefore,P15.67(a) F = 2T sin jz FGHIJKx2v2 m0FGH2dx =FGHIJKm 21M+v .23IJK11m 2m eff v 2 =M+v223andT=11Mv 2 +222= 2M+ m3k.where = tan 1FG y IJH LKTherefore, for a small displacementysin tan =L(b)andFIG. P15.672TyF = L jThe total force exerted on the ball is opposite in direction and proportional to itsdisplacement from equilibrium, so the ball moves with simple harmonic motion. For aspring system, F = kxbecomes hereTherefore, the effective spring constant is2TandLF = =2Ty.Lk=m2T.mL465466P15.68Oscillatory Motion(a)Assuming a Hookes Law type spring,F = Mg = kxand empiricallyMg = 1.74x 0.113k = 1.74 N m 6% .soM , kg0.020 0Mg , N0.1960.040 00.2930.3920.050 00.3530.490.060 00.4130.5880.070 00.4710.6860.080 0(b)x, m0.170.4930.784We may write the equation as theoreticallyT2 =4 24 2M+ms3kkFIG. P15.68and empiricallyT 2 = 21.7 M + 0.058 9k=so4 2= 1.82 N m 3%21.7Time, s T , s7.03 0.703M , kg0.020 0T 2 , s20.4949.620.9620.040 00.92510.671.0670.050 01.13811.671.1670.060 01.36212.521.2520.070 01.56813.411.3410.080 01.798The k values 1.74 N m 6%andso(c)1.82 N m 3% differ by 4%they agree.Utilizing the axis-crossing point,ms = 3FG 0.058 9 IJ kg =H 21.7 K8 grams 12%in agreement with 7.4 grams.Chapter 15P15.69(a)K + U = 0Thus, K top + U top = K bot + U botMRwhere K top = U bot = 01 2I , but2h = R R cos = R 1 cos Therefore, mgh =afvRMR 2 mr 2and I =++ mR 222Substituting we find=222(b)MmMm2+r2R2r2R2+212dCM =afmR + M 0m+MMR 2 + 1 mr 2 + mR 22mgRWe require Ae bt 2 m =A2bt= ln 2or2mThe spring constant is irrelevant.or(b)22ImT gd CMT = 2T = 2(a)22mT = m + MP15.70e + bt 2 m = 20.100 kg st = 0.6932 0.375 kgb t = 5.20 sgWe can evaluate the energy at successive turning points, where1111 1 2cos t + = 1 and the energy is kx 2 = kA 2 e bt 2 m . We require kA 2 e bt 2 m =kA2222 2m ln 2 0.375 kg 0.693t === 2.60 s .ore + bt m = 2b0.100 kg sbga(c)mFIG. P15.692sovvf 1 FGH MR + mr + mR IJK R222L M mr + m OPvmgRa1 cos f = M +N 4 4R 2 Qa1 cos fv = 4 gRe + + 2jRg a1 cos fv=2amgR 1 cos =and467f1 2kA , the fractional rate of change of energy over time is22d 1dEdA1k 2 A dAdt 2 kAdt= 1= 2 1 2 dt = 2 dt2EA2 kA2 kAFrom E =eja ftwo times faster than the fractional rate of change in amplitude.FGHIJK468P15.71Oscillatory Motion(a)When the mass is displaced a distance x fromequilibrium, spring 1 is stretched a distance x1 andspring 2 is stretched a distance x 2 .By Newtons third law, we expectk1 x1 = k 2 x 2 .When this is combined with the requirement thatx = x1 + x 2 ,FIG. P15.71LM k OPxNk + k QL k k OP x = maF =MNk + k Q2x1 =we findThe force on either spring is given by11 21where a is the acceleration of the mass m.212This is in the formand(b)F = k eff x = maT = 2bso thatLetgIn this case each spring is distorted by the distance x which the mass is displaced. Therefore,the restoring force isgF = k1 + k 2 xP15.72bm k1 + k 2m= 2k effk1 k 2T = 2bk eff = k1 + k 2andmk1 + k 2g.represent the length below water at equilibrium and M the tubes mass: Fy = 0 Mg + r 2g = 0.Now with any excursion x from equilibrium Mg + r 2a xfg = Ma .Subtracting the equilibrium equation gives r 2 gx = Maa=F r g I x = xGH M JK22The opposite direction and direct proportionality of a to x imply SHM with angular frequency=T= r 2 gM2=FG 2 IJHrKMgChapter 15P15.73For max = 5.00 , the motion calculated by the Euler methodagrees quite precisely with the prediction of max cos t . Theperiod is T = 2.20 s .Time,t (s)0.0000.0040.0080.5440.5480.5521.0921.0961.1001.1041.6441.6481.6522.1922.1962.2002.204Angle, ()5.000 04.999 34.998 0Ang. speed(/s)0.000 00.163 10.326 2Ang. Accel. s2e j max cos t40.781 540.776 240.765 65.000 04.999 74.998 70.056 00.001 10.058 214.282 314.284 214.284 10.457 60.009 00.475 60.081 00.023 90.033 34.999 45.000 05.000 04.999 30.319 90.156 80.006 30.169 440.776 540.781 640.781 440.775 94.998 94.999 85.000 04.999 60.063 80.003 30.060 414.282 414.284 214.284 10.439 70.027 00.493 60.071 60.014 50.042 74.999 45.000 05.000 04.999 30.313 70.150 60.012 60.175 740.776 840.781 740.781 340.775 64.999 14.999 95.000 04.999 4For max = 100 , the simple harmonic motion approximation max cos t diverges greatly from the Euler calculation. Theperiod is T = 2.71 s , larger than the small-angle period by 23%.Time,Angle,t (s) ()0.000 100.000 00.00499.992 60.00899.977 61.096 84.744 91.100 85.218 21.104 85.684 01.348 99.996 01.352 100.000 81.356 99.998 32.19640.150 92.20041.045 52.20441.935 32.70499.998 52.708 100.000 82.71299.995 7Ang. speed(/s)0.000 01.843 23.686 5Ang. Accel. s2e j max cos t460.606 6460.817 3460.838 2100.000 099.993 599.973 9120.191 0118.327 2116.462 0465.948 8466.286 9466.588 699.995 499.999 899.991 13.053 31.210 00.633 2460.812 5460.805 7460.809 375.797 975.047 474.287 0224.867 7223.660 9222.431 8301.713 2307.260 7312.703 599.997 199.999 399.988 52.420 00.576 81.266 4460.809 0460.805 7460.812 912.642 211.507 510.371 2FIG. P15.73469470*P15.74Oscillatory Motion(a)The block moves with the board in what we take as the positive x direction, stretching thespring until the spring force kx is equal in magnitude to the maximum force of static mg.friction sn = s mg . This occurs at x = sk(b)Since v is small, the block is nearly at the rest at this break point. It starts almost immediatelyto move back to the left, the forces on it being kx and + k mg . While it is sliding the netforce exerted on it can be written as kx + k mg = kx +FGHIJKk k mg mg= k x k= kx relkkwhere x rel is the excursion of the block away from the point k mg.kConclusion: the block goes into simple harmonic motion centered about the equilibrium mgposition where the spring is stretched by k.k(d)The amplitude of its motion is its original displacement, A =bg s mg k mg. It first comes tokk2 k s mg k mg. Almost immediately at this point itA=kklatches onto the slowly-moving board to move with the board. The board exerts a force ofstatic friction on the block, and the cycle continues.rest at spring extension(c)The graph ofthe motionlooks like this:FIG. P15.74(c)(e)bg2 A 2 s k mg=.vkvThe time for which the block is springing back is one half a cycle of simple harmonic motion,The time during each cycle when the block is moving with the board isFGHIJK1mm=. We ignore the times at the end points of the motion when the speed of22kk2Athe block changes from v to 0 and from 0 to v. Since v is small compared to, thesetimes are negligible. Then the period isT=continued on next pagebg2 s k mgkv+m.kmkChapter 15(f)T=afbge2 0.4 0.25 0.3 kg 9.8 m s 2b0.024 m sgb12 N mgf=Thenbg2 s k mgj +0.3 kg= 3.06 s + 0.497 s = 3.56 s12 N m1= 0.281 Hz .T+mincreases as m increases, so the frequency decreases .k(g)T=(h)As k increases, T decreases and f increases .(i)As v increases, T decreases and f increases .(j)As s k increases, T increases and f decreases .(a)Newtons law of universal gravitation isF=Thus,F=Which is of Hookes law form with*P15.75471k=(b)kvbgGMmr=2FGHIJKGm 4 3r r2 3FG 4 GmIJ rH3 K4Gm3FG 4 IJ Gmr = maH 3KF 4Ia = G J Gr = rH 3KThe sack of mail moves without friction according to2Since acceleration is a negative constant times excursion from equilibrium, it executes SHMwith=4G3and periodT=The time for a one-way trip through the earth isg=sog4G=3R eb gand3G=34 GT=2We have also2GM e2Re=3G 4R e 23Re(a) 4.33 cm; (b) 5.00 cm s ;P15.6see the solutionP15.812.0 HzP15.1018.8 m s; 7.11 km s 22(c) 17.3 cm s ; (d) 3.14 s; 5.00 cmP15.44GR e3Re6.37 10 6 mT=== 2.53 10 3 s = 42.2 min .22g9.8 m sANSWERS TO EVEN PROBLEMSP15.2=(a) 15.8 cm; (b) 15.9 cm;(c) see the solution; (d) 51.1 m; (e) 50.7 m472Oscillatory Motion(a) 1.26 s; (b) 0.150 m s; 0.750 m s 2 ;15 cmsin 5t ;(c) x = 3 cmcos 5t ; v =s75 cma=cos 5ts2P15.42see the solutionP15.44(a) 2.95 Hz; (b) 2.85 cmP15.46see the solutionP15.48either 1.31 Hz or 0.641 HzP15.14F vI(a) ; (b) x = G J sin tHKP15.501.56 cmP15.16(a) 126 N m; (b) 0.178 mP15.52(a) 0.500 m s ; (b) 8.56 cmP15.18(a) 0.153 J; (b) 0.784 m s; (c) 17.5 m s 2P15.54A=P15.20(a) 100 N m; (b) 1.13 Hz;(c) 1.41 m s at x = 0 ;P15.56see the solution(f) 1.33 m s ; (g) 3.33 m s 2P15.58(a) k =P15.22(a) 1.50 s; (b) 73.4 N m;(c) 19.7 m below the bridge; (d) 1.06 rad s;(e) 2.01 s; (f) 3.50 sP15.60(a) x = 2 m sin 10t ; (b) at x 1.73 m;(c) 98.0 mm; (d) 52.4 msP15.24(a) 0.218 s and 1.09 s; (b) 14.6 mWP15.62P15.26The position of the piston is given byx = A cos t .(a) decreased by 0.735 m;(b) increased by 0.730 s;(c) decreased by 120 J; (d) see the solutionP15.64(a) 3.56 Hz ; (b) 2.79 Hz; (c) 2.10 HzP15.66(a)P15.68see the solution; (a) k = 1.74 N m 6% ;(b) 1.82 N m 3%; they agree;(c) 8 g 12%; it agreesP15.70(a) 5.20 s; (b) 2.60 s; (c) see the solutionP15.72see the solution; T =P15.74see the solution; (f) 0.281 Hz ;(g) decreases; (h) increases; (i) increases;(j) decreasesP15.12FGHFGHIJKIJKvsg4 2 f 22(d) 10.0 m s at x = A ; (e) 2.00 J;P15.28gC= 1.001 5gTP15.301.42 s; 0.499 mP15.32(a) 3.65 s; (b) 6.41 s; (c) 4.24 sP15.34(a) see the solution;(b), (c) 9.85 m s 2 ; agreeing with theaccepted value within 0.5%P15.36(a) 2.09 s; (b) 4.08%P15.38203 N mP15.40see the solutionFG IJH K4 2 mT; (b) m = m2TTaFGH2f a fIJKM+ m1m 23M+v ; (b) T = 223kFG 2 IJHrKMg16Wave MotionCHAPTER OUTLINE16.1Propagation of aDisturbanceSinusoidal WavesThe Speed of Waves onStringsReflection and TransmissionRate of Energy Transfer bySinusoidal Waves on StringsThe Linear Wave EquationANSWERS TO QUESTIONS16.6As the pulse moves down the string, the particles of the stringitself move side to side. Since the mediumhere, thestringmoves perpendicular to the direction of wavepropagation, the wave is transverse by definition.Q16.2To use a slinky to create a longitudinal wave, pull a few coilsback and release. For a transverse wave, jostle the end coilside to side.From v =Q16.416.416.5Q16.1Q16.316.216.3It depends on from what the wave reflects. If reflecting from aless dense string, the reflected part of the wave will be rightside up.T, we must increase the tension by a factor of 4.2 vAQ16.5Yes, among other things it depends on. v max = A = 2 fA =Q16.6Since the frequency is 3 cycles per second, the period isQ16.7Amplitude is increased by a factor ofQ16.8The section of rope moves up and down in SHM. Its speed is always changing. The wave continues onwith constant speed in one direction, setting further sections of the rope into up-and-down motion.Q16.9Each element of the rope must support the weight of the rope below it. The tension increases with. Here v is the speed of the wave.1second = 333 ms.32 . The wave speed does not change.height. (It increases linearly, if the rope does not stretch.) Then the wave speed v =Tincreaseswith height.Q16.10The difference is in the direction of motion of the elements of the medium. In longitudinal waves,the medium moves back and forth parallel to the direction of wave motion. In transverse waves, themedium moves perpendicular to the direction of wave motion.473474Wave MotionQ16.11Slower. Wave speed is inversely proportional to the square root of linear density.Q16.12As the wave passes from the massive string to the less massive string, the wave speed will increaseaccording to v =T. The frequency will remain unchanged. Since v = f , the wavelength mustincrease.Q16.13Higher tension makes wave speed higher. Greater linear density makes the wave move moreslowly.Q16.14The wave speed is independent of the maximum particle speed. The source determines themaximum particle speed, through its frequency and amplitude. The wave speed depends instead onproperties of the medium.Q16.15Longitudinal waves depend on the compressibility of the fluid for their propagation. Transversewaves require a restoring force in response to sheer strain. Fluids do not have the underlyingstructure to supply such a force. A fluid cannot support static sheer. A viscous fluid cantemporarily be put under sheer, but the higher its viscosity the more quickly it converts inputwork into internal energy. A local vibration imposed on it is strongly damped, and not a source ofwave propagation.Q16.16Let t = ts t p represent the difference in arrival times of the two waves at a station at distanced = v s ts = v p t pF1 1Ifrom the hypocenter. Then d = tG JHv v Ks1. Knowing the distance from the firstpstation places the hypocenter on a sphere around it. A measurement from a second station limits itto another sphere, which intersects with the first in a circle. Data from a third non-collinear stationwill generally limit the possibilities to a point.Q16.17The speed of a wave on a massless string would be infinite!SOLUTIONS TO PROBLEMSSection 16.1P16.1Propagation of a DisturbanceReplace x by x vt = x 4.5t6to gety=2x 4.5t + 3afChapter 16475P16.2FIG. P16.2P16.3a5.00 e a x + 5 t f is of the form f x + vt2fso it describes a wave moving to the left at v = 5.00 m s .P16.4(a)The longitudinal wave travels a shorter distance and is moving faster, so it will arrive atpoint B first.(b)The wave that travels through the Earth must traveleja distance of2 R sin 30.0 = 2 6.37 10 6 m sin 30.0 = 6.37 10 6 mat a speed of7 800 m/sTherefore, it takes6.37 10 6 m= 817 s7 800 m sThe wave that travels along the Earths surface must travelFG radIJ = 6.67 10H3 Ka distance ofs = R = Rat a speed of4 500 m/sTherefore, it takes6.67 10 6= 1 482 s4 500The time difference is665 s = 11.1 min6m476P16.5Wave Motionbg bwhere t is the travel time for the faster wave.afb g bga fb4.50 km sga17.3 sf = 23.6 sor t =a7.80 4.50f km sand the distance is d = b7.80 km sga 23.6 sf = 184 kmThen, 7.80 4.50 km s t = 4.50 km s 17.3 sSection 16.2P16.6.Sinusoidal WavesUsing data from the observations, we have = 1.20 mand f =8.0012.0 s8fFGH 12..00s IJK =0aTherefore, v = f = 1.20 m0.800 m sf=40.0 vibrations 4= Hz30.0 s3=P16.7v 42.5 cm s= 4= 31.9 cm = 0.319 mf3 Hzafav=425 cm= 42.5 cm s10.0 sfP16.8v = f = 4.00 Hz 60.0 cm = 240 cm s = 2.40 m sP16.9y = 0.020 0 m sin 2.11x 3.62t in SI unitsA = 2.00 cmk = 2.11 rad m=2= 2.98 mk = 3.62 rad sf== 0.576 Hz2bv = f =P16.10gaThe distance the waves have traveled is d = 7.80 km s t = 4.50 km s t + 17.3 sbg af 2 3.62== 1.72 m s2 k2.11g afy = 0.005 1 m sin 310 x 9.30t SI unitsv= 9.30== 0.030 0 m sk 310s = vt = 0.300 m in positive x - directionfChapter 16*P16.11af dbg bgiFrom y = 12.0 cm sin 1.57 rad m x 31.4 rad s tThe transverse velocity isy= A cos kx ttIts maximum magnitude is(a)A = 12 cm 31.4 rad s = 3.77 m say =(b)v yt=abafhaaffeA 2 = 0.12 m 31.4 s 1The maximum value isP16.12g A cos kx t = A 2 sin kx ttcfaf aj2= 118 m s 2fAt time t, the phase of y = 15.0 cm cos 0.157 x 50.3t at coordinate x isbg bg = 0.157 rad cm x 50.3 rad s t . Since 60.0 =B = A rad , or (since x A = 0 ),3rad , the requirement for point B is that3b0.157 rad cmgx b50.3 rad sgt = 0 b50.3 rad sgt rad .3BThis reduces to x B =P16.13 rad= 6.67 cm .3 0.157 rad cmbagfy = 0.250 sin 0.300 x 40.0t maCompare this with the general expression y = A sin kx t(a)A = 0.250 m(b) = 40.0 rad s(c)k = 0.300 rad m(d)=(e)v = f =(f)The wave moves to the right, in + x direction .22== 20.9 mk0.300 rad mFG IJ = FG 40.0 rad s IJ a20.9 mf =H 2 K H 2 K133 m sf477478P16.14Wave Motion(a)See figure at right.(b)T=2=y (cm)2= 0.125 s50.3100This agrees with the period found in the examplein the text.t (s)0.1100.2FIG. P16.141 = 0.785A = y max = 8.00 cm = 0.080 0 mTherefore,b gOr (where y 0 , t = 0 at t = 0 )(b)In general,Assumingb(a)y (mm)0.20.10.00.10.2t=00.20.4x (mm)FIG. P16.16(a)(b)2= 18.0 rad m 0.350 m11T= == 0.083 3 sf 12.0 sk=2= = 2 f = 2 12.0 s = 75.4 rad sfb gay = A sinb kx + t + g specializes toy = 0.200 m sinb18.0 x m + 75.4t s + gv = f = 12.0 s 0.350 m = 4.20 m s(c)at x = 0 , t = 0 we requireb g3.00 10 2 m = 0. 200 m sin + = 8.63 = 0.151 radsob g a0.200 mf sinb18.0 x m + 75.4t s 0.151 radgy x, t =gy = 0.080 0 sin 7.85 x + 6 t 0.785 mTherefore,P16.16=2a0.800 mf = 7.85 m = 2 f = 2 a3.00f = 6.00 rad sy = A sina kx + t fy = b0.080 0g sinb7.85 x + 6 t g my = 0.080 0 sinb7.85 x + 6 t + gyb x, 0g = 0 at x = 0.100 m0 = 0.080 0 sinb0.785 + gor(a)k=2then we require thatP16.15Chapter 16P16.17f FGH x + 4 tIJK8ay = 0.120 m sin(b)v=fa f FGH x + 4 tIJK8va0.200 s, 1.60 mf = 1.51 m sFIa = a 0.120 mfa 4 f sinG x + 4 tJH8Kaa0.200 s, 1.60 mf = 0ady:dta=(a)dv:dtx = 0.120 4 cos2 2=:82 = 4 =:Tk= = 16.0 mT = 0.500 sv=P16.18479(a) 16.0 m== 32.0 m sT 0.500 sb gbgyb0 , 0g = A sin = 0.020 0 my x , t = A sin kx + t + Let us write the wave function asdydt= A cos = 2.00 m s0, 0=Also,22== 80.0 sT0.025 0 sA 2 = xi2 +FG v IJ = b0.020 0 mg + FG 2.00 m s IJHKH 80.0 s Ki222A = 0.021 5 m(b)A sin 0.020 0= 2 = 2.51 = tan A cos 80 .0afYour calculators answer tan 1 2.51 = 1.19 rad has a negative sine and positive cosine,just the reverse of what is required. You must look beyond your calculator to find = 1.19 rad = 1.95 radbg(c)v y, max = A = 0.021 5 m 80.0 s = 5.41 m s(d) = v x T = 30.0 m s 0.025 0 s = 0.750 mbk=2=ga2= 8.38 m0.750 mb g bg bf = 80.0 sy x , t = 0.021 5 m sin 8.38 x rad m + 80.0 t rad s + 1.95 radg480P16.19Wave Motion(a)f=v=b1.00 m sg =2.00 mb0.500 Hzg = 2 f = 2 0.500 s = 3.14 rad s22= 3.14 rad m2.00 m(b)k=(c)y = A sin kx t + becomesbga0.100 mf sinb3.14x m 3.14t s + 0gy=(d)=For x = 0 the wave function requiresaf baf by = 0.100 m sin 3.14t s(e)(f)gy = 0.100 m sin 4.71 rad 3.14 t svy =gy= 0.100 m 3.14 s cos 3.14x m 3.14t stbg bgThe cosine varies between +1 and 1, sobv y 0.314 m sga0.100 mf sina1.00 rad 20.0tf(a)at x = 2.00 m , y =(b)P16.20y = 0.100 m sin 0.500 x 20.0t = A sin kx taf afso = 20.0 rad s and f =Section 16.3P16.212af= 3.18 HzThe Speed of Waves on StringsThe down and back distance is 4.00 m + 4.00 m = 8.00 m .afThe speed is thend total 4 8.00 mT== 40.0 m s =t0.800 sNow,=0.200 kg= 5.00 10 2 kg m4.00 mSoP16.22v=T = v 2 = 5.00 10 2 kg m 40.0 m sThe mass per unit length is: =jbeg2= 80.0 N0.060 0 kg= 1.20 10 2 kg m .5.00 mbgbThe required tension is: T = v 2 = 0.012 0 kg m 50.0 m sg2= 30.0 N .Chapter 16P16.23P16.24Tv=(a)1 350 kg m s 2=5.00 10 3 kg m= 520 m sa f = 2 f = 2 500 = 3 140 rad s , k =j bey = 2.00 10 4 m sin 16.0 x 3 140tv = 196 m s =(b)481 3 140== 16.0 rad mv196gT4.10 10 3 kg mT = 158 NTMgv=MgL L2= 2mtandP16.26T = Mg is the tension;Then,P16.25g==mLMgL L= is the wave speed.mt=eej1.60 m 4.00 10 3 kgLm== 1.64 m s 2Mt 2 3.00 kg 3.61 10 3 s 2jTv=T = v 2 = Av 2 = r 2 v 2ja feej b200 m sgT = 8 920 kg m3 7.50 10 4 m22T = 631 NP16.27Since is constant, =T22v2=T2P16.28T12v1andFv I=G JHv K212F 30.0 m s I a6.00 Nf =T =GH 20.0 m s JKThe period of the pendulum is T = 22113.5 N .LgLet F represent the tension in the string (to avoid confusion with the period) when the pendulum isvertical and stationary. The speed of waves in the string is then:v=MgMgLF= m =mLSince it might be difficult to measure L precisely, we eliminateso v =Mg T gTg=m 22M.mL=T g2482P16.29Wave MotionIf the tension in the wire is T, the tensile stress isStress =TAafT = A stress .soThe speed of transverse waves in the wire isv=T=aA StressmLf=StressmAL=StressmVolume=Stresswhere is the density. The maximum velocity occurs when the stress is a maximum:v max =P16.302.70 10 8 Pa= 185 m s .7 860 kg m 3mg = 2T sin From the free-body diagramT=The angle is found frommg2 sin cos =3L8L2=34FIG. P16.30 = 41.4(a)v=Tv=v=or(b)P16.31mg=2 sin 41. 4F 30.4GHIJkg KmsFI9.80 m sGGJ2e8.00 10 kg mj sin 41.4 JHK23mmm = 3.89 kgv = 60.0 = 30.4 m andThe total time is the sum of the two times.L=LvTLet A represent the cross-sectional area of one wire. The mass of one wire can be written both asm = V = AL and also as m = L .In each wiret=Then we have = A =Thus,t=LFor copper,For steel,The total time isd 24F d IGH 4T JKL a fb8 920ge1.00 10t = a 20.0 fMMMa4fa150fNL a fb7 860ge1.00 10t = a30.0fMMMa4fa150fN2120.137 + 0.192 = 0.329 sj OPPPQj OPPPQ123 2= 0.137 s123 2= 0.192 sChapter 16P16.32Refer to the diagrams. From the free-body diagram of point A: Fy = 0 T1 sin = Mg Fx = 0 T1 cos = TandCombining these equations to eliminate T1 gives the tension in theMg.string connecting points A and B as: T =tan v=T=Mgtan mL=DL/4L/4AdBMgLm tan dL/2MThe speed of transverse waves in this segment of string is then483MT1ATand the time for a pulse to travel from A to B ist=*P16.33L2v=f has units Hz = 1 s , so T =(a)MgmL tan .4MgFIG. P16.321has units of seconds, s . For the other T we have T = v 2 ,fkg m 2 kg m== N .m s2s2with unitsThe first T is period of time; the second is force of tension.(b)Section 16.4Reflection and TransmissionProblem 7 in Chapter 18 can be assigned with this section.Section 16.5P16.34f=P=P16.35Rate of Energy Transfer by Sinusoidal Waves on Stringsv=30.0= 60.0 Hz0.500FGH = 2 f = 120 rad sIJ aK11 0.180 2 A 2 v =12022 3.60f a0.100f a30.0f =221.07 kWSuppose that no energy is absorbed or carried down into the water. Then a fixed amount of power isspread thinner farther away from the source, spread over the circumference 2 r of an expandingcircle. The power-per-width across the wave frontP2 ris proportional to amplitude squared so amplitude is proportional toP.2 r484P16.36Wave MotionT = constant; v =T; P=1 2 A 2 v2(a)If L is doubled, v remains constant and P is constant .(b)If A is doubled and is halved, P 2 A 2 remains constant .(c)If and A are doubled, the product 2 A 2 A2remains constant, so2P remains constant .(d)If L and are halved, then 2 1is quadrupled, so P is quadrupled .2(Changing L doesnt affect P ).P16.37A = 5.00 10 2 mTv=Therefore,P= = 4.00 10 2 kg m1 2 A 2 v :22 =P = 300 WT = 100 N= 50.0 m sa f2 3002P=A 2 v 4.00 10 2 5.00 10 2ejej a50.0f2 = 346 rad s= 55.1 Hzf=2P16.38 = 30.0 g m = 30.0 10 3 kg m = 1.50 mf = 50.0 Hz: = 2 f = 314 s 12 A = 0.150 m:A = 7.50 10 2 m(a)FG 2 x tIJH Ky = e7.50 10 j sina 4.19 x 314t fy = A sin2314ja f e7.50 10 j FGH 4.19 IJK W11 2 A 2 v = 30.0 10 3 31422e(b)P16.39FIG. P16.38P=(a)v = f =(b)=22=m = 7.85 mk0.800(c)f=50.0= 7.96 Hz2(d)P=11 2 A 2 v = 12.0 10 3 50.0222 22 2 50.0m s = 62.5 m s= =k 0.8002 keja f a0.150f a62.5f W =2221.1 WP = 625 WChapter 16*P16.40FGHComparing y = 0.35 sin 10t 3x +k=(a)The rate of energy transport isjb11 2 A 2 v = 75 10 3 kg m 10 s22eg a0.35 mf 3.33 m s =2215.1 W .The energy per cycle isE = P T =P16.41IJ with y = A sinbkx t + g = A sinbt kx + g we haveK3 10 s= == 3.33 m s ., = 10 s , A = 0.35 m . Then v = f = 2 fm2 k 3 mP=(b)4485jb11 2 A 2 = 75 10 3 kg m 10 s22eg a0.35 mf222 m= 3.02 J .3Originally,1 2 A 2 v21TP0 = 2 A 22P0 =1P0 = 2 A 2 T2The doubled string will have doubled mass-per-length. Presuming that we hold tension constant, itcan carry power larger by 2 times.2 P0 =*P16.421 2 2 A T 22As for a strong wave, the rate of energy transfer is proportional to the square of the amplitude and tothe speed. We write P = FvA 2 where F is some constant. With no absorption of energy,22Fv bedrock A bedrock = Fv mudfill A mudfillv bedrockA= mudfill =v mudfillA bedrockThe amplitude increases by 5.00 times.25 v mudfill=5v mudfill486Wave MotionSection 16.6The Linear Wave Equationaf(a)A = 7.00 + 3.00 4.00 yields A = 40.0(b)P16.43In order for two vectors to be equal, they must have the same magnitude and the samedirection in three-dimensional space. All of their components must be equal. Thus,7.00 i + 0 j + 3.00k = A i + Bj + Ck requires A = 7.00 , B = 0 , and C = 3.00 .(c)In order for two functions to be identically equal, they must be equal for every value ofevery variable. They must have the same graphs. InafafA + B cos Cx + Dt + E = 0 + 7.00 mm cos 3.00 x + 4.00t + 2.00 ,the equality of average values requires that A = 0 . The equality of maximum valuesrequires B = 7.00 mm . The equality for the wavelength or periodicity as a function of xrequires C = 3.00 rad m . The equality of period requires D = 4.00 rad s , and theequality of zero-crossings requires E = 2.00 rad .*P16.442yThe linear wave equation isx=21 2 yv 2 t 2Ify = e b a x vt fthenyy= bve b a x vt f and= be b a x vt ftx2 yt 22yTherefore,P16.45= b 2 v 2 e b a x vt f andt1 2yThe linear wave equation isaTo show that y = ln b x vt22yx= b 2 e b a x vt fx 2, demonstrating that e b a x vt f is a solution22y=v 2 t 2f= v22yx 2is a solution, we find its first and second derivatives with respect to xand t and substitute into the equation.2yy1= bvt b x vtfa fay= ba x vt fx1=t 22ybxe ja f2v1 2y 1Then 2 2 = 2v tv x vt2=2a f1 bvafbx vtbf1ax vtf2=2b 2 x vt=2a=2yx 22v2ax vtf=21ax vtf2so the given wave function is a solution.Chapter 16P16.46(a)From y = x 2 + v 2 t 2 ,2yy= 2xxevaluatex 22yy= v 2 2ttDoes2yt 2equation.NoteSo(c)a1x + vt2at 2=2= 2v 21 2y?v 2 t 2=By substitution: 2 =(b)487f2+12 v 2 and this is true, so the wave function does satisfy the wavev2a1x vt2f 1 ax + vtf2f x + vt =f221 2111x + xvt + v 2 t 2 + x 2 xvt + v 2 t 2222222 2= x + v t as required.=af 1 ax vtf2and g x vt =2.y = sin x cos vt makes2yy= cos x cos vtxx 22yy= v sin x sin vttThen2yx2=t 2= sin x cos vt= v 2 sin x cos vt1 2 yv 2 t 21 2v sin x cos vt which is true as required.v2Note sin x + vt = sin x cos vt + cos x sin vtbecomes sin x cos vt =afafsin x vt = sin x cos vt cos x sin vt .a f a f1f a x + vtf = sina x + vt f2So sin x cos vt = f x + vt + g x vt withandafg x vt =a1sin x vt2f.Additional ProblemsP16.47Assume a typical distance between adjacent people ~ 1 m .Then the wave speed isv=x 1 m~~ 10 m st 0.1 sModel the stadium as a circle with a radius of order 100 m. Then, the time for one circuit around thestadium isT=e j22 r 2 10~= 63 s ~ 1 min .v10 m s488P16.48Wave MotionafCompare the given wave function y = 4.00 sin 2.00 x 3.00t cm to the general formafy = A sin kx t to find(a)(b)k=(c) = 2 f = 3.00 s 1 and f = 0.477 Hz(d)T=(e)P16.49amplitude A = 4.00 cm = 0.040 0 mThe minus sign indicates that the wave is traveling in the positive x -direction .(a)Let u = 10 t 3 x +2= 2.00 cm 1 and = cm = 0.031 4 m1= 2.09 sf4dudx= 10 3= 0 at a point of constant phasedtdtdx 10== 3.33 m s3dtThe velocity is in the positive x -direction .g abf FGH(b)y 0.100 , 0 = 0.350 m sin 0.300 +(c)k=(d)*P16.50(a)2IJK= 0.054 8 m = 5.48 cm4= 3 : = 0.667 m = 2 f = 10 : f = 5.00 Hzfa f FGH0.175 m = a0.350 mf sin b99.6 rad sgt sin b99.6 rad sgt = 0.5vy =y= 0.350 10 cos 10 t 3 x +t4aIJKa fafv y, max = 10 0.350 = 11.0 m sThe smallest two angles for which the sine function is 0.5 are 30 and 150, i.e., 0.523 6 radand 2.618 rad.99.6 rad s t1 = 0.523 6 rad , thus t1 = 5.26 msbgb99.6 rad sgt2= 2.618 rad , thus t 2 = 26.3 mst t 2 t1 = 26.3 ms 5.26 ms = 21.0 ms(b)P16.51Distance traveled by the wave =FG IJ t = FG 99.6 rad s IJ e21.0 10 sj =H k K H 1.25 rad m K3The equation v = f is a special case ofspeed = (cycle length)(repetition rate).ejbgThus, v = 19.0 10 3 m frame 24.0 frames s = 0.456 m s .1.68 m .Chapter 16P16.52Assuming the incline to be frictionless and taking the positive x-direction to be up the incline: Fx = T Mg sin = 0T = Mg sin or the tension in the string isv=The speed of transverse waves in the string is thenThe time interval for a pulse to travel the strings length is t =P16.53=g+ Usjtop0 + Mgx + 0 + 0 = 0 + 0 +x=b2 Mgkgeje+ E = K + U g + U s1 2kx2(a)T = kx = 2 Mg = 2 2.00 kg 9.80 m s 2 = 39.2 N(b)L = L0 + x = L0 +(c)v=2 Mgk39.2 NL = 0.500 m += 0.892 m100 N mv=T=TLm39.2 N 0.892 m5.0 10 3 kgv = 83.6 m sMgx =1 2kx2(a)T = kx = 2 Mg(b)L = L0 + x = L0 +(c)v=T=TL=m2 MgkFGH2 Mg2 MgL0 +mkIJKMg sin mL=Lm=L=vMgL sin Energy is conserved as the block moves down distance x:eK + UP16.54TjbottomMgL sin mmLMg sin 489490Wave MotionTv=From Equation 16.21, P ==80.0 N(a)(b)P16.55e5.00 103= 179 m sjkg 2.00 mFG IJH K1vv 2 A 2 and = 22FG IJH K12vvA 22P=2 2P=FH2=2 2 A 2 v 32IK b0.040 0 mg b179 m sga0.160 mf5.00 10 3 kg2.00 m232P = 1.77 10 4 W = 17.7 kWP16.56v=TNow v = f implies v =so thatkFG IJH K m=g k*P16.57v 2.gand in this case T = mg ; therefore, m =20.250 kg m=9.80 m s 2LM 18 s OPN 0.750 m Q11Let M = mass of block, m = mass of string. For the block,2= 14.7 kg . F = ma implies T =speed of a wave on the string is thenTv=t=(a)=v=MmmMm=M0.003 2 kg= 0.084 3 rad0.450 kgdmdx= A= AdLdxT=T=AT ax + baWith all SI units, v =(b)= rmrr 1=v = t =P16.58M 2 r=v x= 0 =v x=10 .0 =ef=Tj 10 x + 10 2 10 4b2 700ge0 + 10 je10 j224.024j 10 x + 10 2 cm 2324.0b2 700ge10eT3ms= 94.3 m sje j+ 10 2 10 4= 66.7 m s2mv b= m 2 r . TherChapter 16P16.59v=TwhereTherefore,dx, so thatBut v =dtT = xg , the weight of a length x, of rope.v = gxdxdt =gxt=andzL0P16.60dxgx=1x12gL= 20LgFG mxg IJ + Mg , so the wave speed is:H LKTTLF MgL IJ = dx .== xg + Gv=H m K dtmL F MgL IJ OP dx1 xg + b MgL m gt=Then t = z dt = z M xg + GgN H m KQOMgL I2 LFLF m+M MIt = MG Lg +t=2JK FGH MgL IJK PPJKg NHmmgGmHMQL F m 0IWhen M = 0 , as in the previous problem,t=2J= 2 LgGgm KHF 1 m 1 m + IF mIAs m 0 we expand m + M = M G 1 + J = M G 1 +H MKH 2 M 8 M JKF M + em M j em M j + M ILGJJto obtaint=2gGmHKLF1 m ImLt2GH 2 M JK = MggAt distance x from the bottom, the tension is T =t01212(b)1 2 x=L1 2L0(a)2212(a)x=01212(c)P16.6149121832The speed in the lower half of a rope of length L is the same function of distance (from theLbottom end) as the speed along the entire length of a rope of length.2LLwith L =Thus, the time required = 2g2FG IJH Kand the time required = 2FGHLL= 0.707 22ggIJK.It takes the pulse more that 70% of the total time to cover 50% of the distance.(b)By the same reasoning applied in part (a), the distance climbed in is given by d =LL, we find the distance climbed =.g41In half the total trip time, the pulse has climbed of the total length.4For =t=2g 2.4492Wave Motionv= 15.0== 5.00 m s in positive x -directionk 3.00v=15.0= 5.00 m s in negative x -direction3.00(c)v=15.0= 7.50 m s in negative x -direction2.00(d)P16.63(a)(b)P16.62v=12.012= 24.0 m s in positive x -directionTALLYoungs modulus for the wire may be written as Y =, where T is the tension maintained in thewire and L is the elongation produced by this tension. Also, the mass density of the wire may be.AThe speed of transverse waves in the wire is thenexpressed as =v=T=TA=YAc hLLL v 2=.LYIf the wire is aluminum and v = 100 m s, the strain isand the strain in the wire isjbe2.70 10 3 kg m3 100 m sL=L7.00 10 10 N m 2*P16.64(a)g2= 3.86 10 4 .Consider a short section of chain at the top of the loop. A freebody diagram is shown. Its length is s = R 2 and its mass isR2 . In the frame of reference of the center of the loop,Newtons second law is22mv 0R 2v 02T sin down =down = Fy = ma yRRa f2TRFIG. P16.64(a)2For a very short section, sin = and T = v 0 .T(b)The wave speed is v =(c)In the frame of reference of the center of the loop, each pulse moves with equal speedclockwise and counterclockwise.= v0 .vv0v0FIG. P16.64(c-1)continued on next pagev0vTChapter 16493In the frame of reference of the ground, once pulse moves backward at speed v 0 + v = 2 v 0and the other forward at v 0 v = 0 . The one pulse makes two revolutions while the loopmakes one revolution and the other pulse does not move around the loop. If it is generatedat the six-oclock position, it will stay at the six-oclock position.v0v0v0FIG. P16.64(c-2)P16.65(a)Assume the spring is originally stationary throughout, extended to have a length L muchgreater than its equilibrium length. We start moving one end forward with the speed v atwhich a wave propagates on the spring. In this way we create a single pulse of compressionthat moves down the length of the spring. For an increment of spring with length dx andmass dm, just as the pulse swallows it up, F = mabecomes kdx = adm orButkdmdx= a.dmk= so a = .dxdv vv2= when vi = 0. But L = vt , so a =.dt tLAlso, a =Equating the two expressions for a, we have(b)Using the expression from part (a) v =F T I F 2T IJv=G J =GH K H KF T I F 2T IJv = G J = GH K H 3 K(a)0012= v0= v00t left =L2vt right ==L2vL2 v0 2==L2 v0232 where v 0 120(b)1212P16.66kLt 02 2=t left + t right = 0.966 t 0=23v2or v =LkL2=mFG T IJH K120023= 0.354t 0 where t 0 t 02=k= 0.612 t 0Lv0kL.b100 N mga2.00 mf0.400 kg2= 31.6 m s .494Wave MotionafFG IJH K11 3 2 2 bx2A0 e 2 A 2 v = 2 A0 e 2 bx=222kkP x =P 0 =(c)P16.68(a)(b)P16.67P x= e 2 bxP 0v=af 3 2A02kafaf4 450 km= 468 km h = 130 m s9.50 hbgej a x f is a linear function, so it is of the formTo have a0f = we require b = . Then2130 m sv2d=== 1 730 mg9.80 m s 2*P16.69(a)0af aL f = x = mx + b0a f b x =Then(b)= mL + 0L 0Lm=soLLg 0 xL+ 0dxdx, the time required to move from x to x + dx is. The time required to movevdtfrom 0 to L isFrom v =t =z zz FGH bLdx L dx1==v 0 TT0zLaf x dx0I FG IJ dxFt =JK H L K GH TF x I1 Ft =GH L IJK GH b L g + JK 1T2Lt =e j3 T b g2Le je + + jt =3 T e je + j2L F + + It =G + JK3 TH1L0g12L 0 x+ 0L0L32LL000L32LL0LLL0LLL32 0320L0000000LLL 0IJKChapter 16495ANSWERS TO EVEN PROBLEMSP16.2see the solutionP16.4(a) the P wave; (b) 665 sP16.60.800 m sP16.82.40 m sP16.100.300 m in the positive x-directionP16.126.67 cmP16.14(a) see the solution; (b) 0.125 s; inagreement with the exampleP16.16(a) see the solution; (b) 18.0 m ; 83.3 ms ;75.4 rad s ; 4.20 m s ;(c) 0.2 m sin 18 x + 75.4t 0.151P16.40(a) 15.1 W ; (b) 3.02 JP16.42The amplitude increases by 5.00 timesP16.44see the solutionP16.46f a(a) see the solution;1122x + vt + x vt ;(b)2211(c) sin x + vt + sin x vt22aafafafP16.48(a) 0.021 5 m; (b) 1.95 rad; (c) 5.41 m s ;(d) y x , t =b gP16.50(a) 21.0 ms ; (b) 1.68 mP16.52t =P16.2030.0 N(a) 2Mg ; (b) L0 +P16.24(a) y = 0.2 mm sin 16 x 3 140t ;(b) 158 NP16.26631 NP16.28v=P16.30Fm I(a) v = G 30.4H s kg JKmLMg sin (a) see the solution; (b) 3.18 HzP16.22aTg2f bg(c)2 Mg;k2 Mg2 MgL0 +mkP16.56(a) v =FGHIJK14.7 kgP16.58Mmf(a) 0.040 0 m; (b) 0.031 4 m ;(c) 0.477 Hz; (d) 2.09 s;(e) positive x -directionfb0.021 5 mg sinb8.38x + 80.0 t + 1.95gP16.32f(a) y = 0.075 0 sin 4.19 x 314t ;(b) 625 WP16.54aP16.18g abP16.38eT7 10 x + 10 6jin SI units;(b) 94.3 m s; 66.7 m sm ; (b) 3.89 kgmL tan 4MgP16.341.07 kWP16.36(a), (b), (c) P is constant ;(d) P is quadrupledP16.60see the solutionP16.62(a) 5.00 i m s ; (b) 5.00 i m s ;(c) 7.50 i m s ; (d) 24.0 i m sP16.642(a) v 0 ; (b) v 0 ;(c) One travels 2 rev and the other doesnot move around the loop.496Wave MotionF 2T IJ = v 2 ;(a) v = GH KF 2T IJ = v 2 ; (b) 0.966tv = G3H 3 K12P16.660P16.6800120000130 m s ; 1.73 km17Sound WavesCHAPTER OUTLINE17.117.217.317.417.517.6Speed of Sound WavesPeriodic Sound WavesIntensity of Periodic SoundWavesThe Doppler EffectDigital Sound RecordingMotion Picture SoundANSWERS TO QUESTIONSQ17.1Sound waves are longitudinal because elements of themediumparcels of airmove parallel and antiparallel to thedirection of wave motion.Q17.2We assume that a perfect vacuum surrounds the clock. Thesound waves require a medium for them to travel to your ear.The hammer on the alarm will strike the bell, and the vibrationwill spread as sound waves through the body of the clock. If abone of your skull were in contact with the clock, you wouldhear the bell. However, in the absence of a surroundingmedium like air or water, no sound can be radiated away. Alarger-scale example of the same effect: Colossal storms ragingon the Sun are deathly still for us.What happens to the sound energy within the clock?Here is the answer: As the sound wave travels through thesteel and plastic, traversing joints and going around corners, itsenergy is converted into additional internal energy, raising thetemperature of the materials. After the sound has died away,the clock will glow very slightly brighter in the infrared portionof the electromagnetic spectrum.1meter from the sonic ranger, then the sensor would have to measure how long it2would take for a sound pulse to travel one meter. Since sound of any frequency moves at about343 m s, then the sonic ranger would have to be able to measure a time difference of under0.003 seconds. This small time measurement is possible with modern electronics. But it would bemore expensive to outfit sonic rangers with the more sensitive equipment than it is to print do not1use to measure distances less than meter in the users manual.2Q17.3If an object isQ17.4The speed of sound to two significant figures is 340 m s. Lets assume that you can measure time to1second by using a stopwatch. To get a speed to two significant figures, you need to measure a10time of at least 1.0 seconds. Since d = vt , the minimum distance is 340 meters.Q17.5The frequency increases by a factor of 2 because the wave speed, which is dependent only on themedium through which the wave travels, remains constant.497498Sound WavesQ17.6When listening, you are approximately the same distance from all of the members of the group. Ifdifferent frequencies traveled at different speeds, then you might hear the higher pitchedfrequencies before you heard the lower ones produced at the same time. Although it might beinteresting to think that each listener heard his or her own personal performance depending onwhere they were seated, a time lag like this could make a Beethoven sonata sound as if it werewritten by Charles Ives.Q17.7Since air is a viscous fluid, some of the energy of sound vibration is turned into internal energy. Atsuch great distances, the amplitude of the signal is so decreased by this effect you re unable to hearit.Q17.8We suppose that a point source has no structure, and radiates sound equally in all directions(isotropically). The sound wavefronts are expanding spheres, so the area over which the soundenergy spreads increases according to A = 4 r 2 . Thus, if the distance is tripled, the area increases bya factor of nine, and the new intensity will be one-ninth of the old intensity. This answer accordingto the inverse-square law applies if the medium is uniform and unbounded.For contrast, suppose that the sound is confined to move in a horizontal layer. (Thermalstratification in an ocean can have this effect on sonar pings.) Then the area over which the soundenergy is dispersed will only increase according to the circumference of an expanding circle:A = 2 rh , and so three times the distance will result in one third the intensity.In the case of an entirely enclosed speaking tube (such as a ships telephone), the areaperpendicular to the energy flow stays the same, and increasing the distance will not change theintensity appreciably.Q17.9He saw the first wave he encountered, light traveling at 3.00 10 8 m s . At the same moment,infrared as well as visible light began warming his skin, but some time was required to raise thetemperature of the outer skin layers before he noticed it. The meteor produced compressional wavesin the air and in the ground. The wave in the ground, which can be called either sound or a seismicwave, traveled much faster than the wave in air, since the ground is much stiffer againstcompression. Our witness received it next and noticed it as a little earthquake. He was no doubtunable to distinguish the P and S waves. The first air-compression wave he received was a shockwave with an amplitude on the order of meters. It transported him off his doorstep. Then he couldhear some additional direct sound, reflected sound, and perhaps the sound of the falling trees.Q17.10A microwave pulse is reflected from a moving object. The waves that are reflected back are Dopplershifted in frequency according to the speed of the target. The receiver in the radar gun detects thereflected wave and compares its frequency to that of the emitted pulse. Using the frequency shift,the speed can be calculated to high precision. Be forewarned: this technique works if you are eithertraveling toward or away from your local law enforcement agent!Q17.11As you move towards the canyon wall, the echo of your car horn would be shifted up in frequency;as you move away, the echo would be shifted down in frequency.Q17.12Normal conversation has an intensity level of about 60 dB.Q17.13A rock concert has an intensity level of about 120 dB.A cheering crowd has an intensity level of about 90 dB.Normal conversation has an intensity level of about 5060 dB.Turning a page in the textbook has an intensity level of about 1020 dB.Chapter 17499Q17.14One would expect the spectra of the light to be Doppler shifted up in frequency (blue shift) as thestar approaches us. As the star recedes in its orbit, the frequency spectrum would be shifted down(red shift). While the star is moving perpendicular to our line of sight, there will be no frequencyshift at all. Overall, the spectra would oscillate with a period equal to that of the orbiting stars.Q17.15For the sound from a source not to shift in frequency, the radial velocity of the source relative to theobserver must be zero; that is, the source must not be moving toward or away from the observer.The source can be moving in a plane perpendicular to the line between it and the observer. Otherpossibilities: The source and observer might both have zero velocity. They might have equalvelocities relative to the medium. The source might be moving around the observer on a sphere ofconstant radius. Even if the source speeds up on the sphere, slows down, or stops, the frequencyheard will be equal to the frequency emitted by the source.Q17.16Wind can change a Doppler shift but cannot cause one. Both v o and v s in our equations must beinterpreted as speeds of observer and source relative to the air. If source and observer are movingrelative to each other, the observer will hear one shifted frequency in still air and a different shiftedfrequency if wind is blowing. If the distance between source and observer is constant, there willnever be a Doppler shift.Q17.17If the object being tracked is moving away from the observer, then the sonic pulse would neverreach the object, as the object is moving away faster than the wave speed. If the object being trackedis moving towards the observer, then the object itself would reach the detector before reflectedpulse.Q17.18New-fallen snow is a wonderful acoustic absorber as it reflects very little of the sound that reaches it.It is full of tiny intricate air channels and does not spring back when it is distorted. It acts very muchlike acoustic tile in buildings. So where does the absorbed energy go? It turns into internalenergyalbeit a very small amount.Q17.19As a sound wave moves away from the source, its intensity decreases. With an echo, the sound mustmove from the source to the reflector and then back to the observer, covering a significant distance.Q17.20The observer would most likely hear the sonic boom of the plane itself and then beep, baap, boop.Since the plane is supersonic, the loudspeaker would pull ahead of the leading boop wavefrontbefore emitting the baap, and so forth.How are you? would be heard as ?uoy era woHQ17.21This system would be seen as a star moving in an elliptical path. Just like the light from a star in abinary star system, described in the answer to question 14, the spectrum of light from the star wouldundergo a series of Doppler shifts depending on the stars speed and direction of motion relative tothe observer. The repetition rate of the Doppler shift pattern is the period of the orbit. Informationabout the orbit size can be calculated from the size of the Doppler shifts.SOLUTIONS TO PROBLEMSSection 17.1Speed of Sound WavesbgafP17.1Since v light >> v sound : d 343 m s 16. 2 s = 5.56 kmP17.2v=B=2.80 10 10= 1.43 km s13.6 10 3500P17.3Sound Wavesa20.0 m 1.75 mf = 5.32 10Sound takes this time to reach the man:2343 m sso the warning should be shouted no later thanbefore the pot strikes.Since the whole time of fall is given by y =1 2gt :2s0.300 s + 5.32 10 2 s = 0.353 s18.25 m =19.80 m s 2 t 22ejt = 1.93 sthe warning needs to comeinto the fall, when the pot has fallen19.80 m s 2 1.58 s2to be above the ground byP17.41.93 s 0.353 s = 1.58 s20.0 m 12.2 m = 7.82 m(a)At 9 000 m, T =ejaf2= 12.2 mFG 9 000 IJ a1.00 Cf = 60.0 C so T = 30.0 C .H 150 KUsing the chain rule:afFGH IJKafdv dv dT dxdv dTvdv1, so dt = 247 s==v= v 0.607=dt dT dx dtdT dxv150247f z dvzvL 331.5 + 0.607a30.0f OPFv It = a 247 sf lnG J = a 247 sf ln MHv KN 331.5 + 0.607a30.0f Qtavfdt = 247 s0vifit = 27.2 s for sound to reach ground.(b)t=9 000h== 25.7 sv 331.5 + 0.607 30.0a fIt takes longer when the air cools off than if it were at a uniform temperature.*P17.5Let x 1 represent the cowboys distance from the nearer canyon wall and x 2 his distance from thefarther cliff. The sound for the first echo travels distance 2 x1 . For the second, 2 x 2 . For the third,2 x 2 2 x12 x1 + 2 x 2 2 x 22 x1 + 2 x 2 . For the fourth echo, 2 x1 + 2 x 2 + 2 x1 . Then= 1.92 s and= 1.47 s .340 m s340 m s2x21= 1.92 s + 1.47 s ; x 2 = 576 m.Thus x 1 = 340 m s 1.47 s = 250 m and2340 m s(a)(b)So x 1 + x 2 = 826 mb2 x 1 + 2 x 2 + 2 x 1 2 x1 + 2 x 2340 m sg=1.47 sChapter 17P17.6It is easiest to solve part (b) first:(b)The distance the sound travels to the plane is d s = h 2 +FG h IJH 2K2=h 5.2The sound travels this distance in 2.00 s, sobga f2a686 mf= 614 m .giving the altitude of the plane as h =ds =h 5= 343 m s 2.00 s = 686 m25(a)aThus, the speed of the plane is: v =Section 17.2P17.7*P17.8=*P17.9P17.1026 C= 346 m s273 CLet t represent the time for the echo to return. Then11vt = 346 m s 24 10 3 s = 4.16 m .22Let t represent the duration of the pulse:t =(c)307 m= 153 m s .2.00 s340 m sv== 5.67 mmf 60.0 10 3 s 1d=(b)h= 307 m .2Periodic Sound WavesThe sound speed is v = 331 m s 1 +(a)fThe distance the plane has traveled in 2.00 s is v 2.00 s =L = 10 =b10 10 1010==== 0.455 s .vff22 10 6 1 sg10 v 10 346 m s== 0.157 mmf22 10 6 1 sv 1 500 m s== 1.50 mmf10 6 s1 500 m s= 75.0 mIf f = 20 MHz , =2 10 7 sIf f = 1 MHz , =Pmax = v smaxsmaxej4.00 10 3 N m 2Pmax=== 1.55 10 10 mv1.20 kg m3 343 m s 2 10.0 10 3 s 1ejbga fej501502P17.11Sound WavesA = 2.00 m(a)2= 0.400 m = 40.0 cm15.7 858= 54.6 m sv= =k 15.7=a fbg a fej(b)(c)P17.12s = 2.00 cos 15.7 0.050 0 858 3.00 10 3 = 0.433 mv max = A = 2.00 m 858 s 1 = 1.72 mm sbgejF x 340 t IJ (SI units)P = a1.27 Paf sinGHm s K(a)The pressure amplitude is: Pmax = 1.27 Pa .(b)(c)k=(d)P17.13 = 2 f = 340 s, so f = 170 Hzv = f = 2.00 m 170 Hz = 340 m sk==2=2 vasmax =2faf2= 62.8 m 10.100 m= = 2 f =k== m , giving = 2.00 maTherefore,P17.142fb2 343 m sg = 2.16 1041sa0.100 mfP = a0.200 Paf sin 62.8 x m 2.16 102 v=b2 343 m sa0.100 mfg = 2.16 10ajb4fge0.200 PaPmax== 2. 25 10 8 m3v1.20 kg m 343 m s 2.16 10 4 s 1e=2j1maxP17.15ts .rad sa0.100 mf = 62.8 mTherefore, s = scosb kx t g = e 2.25 104FG 2 v IJ sH K2 a1.20fa343 f e5.50 10 j==Pmax = v smax = v2v 2 smax=Pmax8j emax20.840jm cos 62.8 x m 2.16 10 4 t s .65.81 m503Chapter 17P17.16(a)The sound pressure is extra tensile stress for one-half of each cycle. When it becomes0.500% 13.0 10 10 Pa = 6.50 10 8 Pa , the rod will break. Then, Pmax = v smaxafejsmax =(b)6.50 10 8 N m 2Pmax== 4.63 mm .v8.92 10 3 kg m3 5 010 m s 2 500 sajbeFrom s = smax cos kx tfafbgfgajbgb11122v smax = vvmax = 8.92 10 3 kg m 3 5 010 m s 14.5 m s222= 4.73 10 9 W m 2I=(c)*P17.17gs= smax sin kx ttv max = smax = 2 500 s 4.63 mm = 14.5 m sv=bgbafeg2Let P x represent absolute pressure as a function of x. The net forceto the right on the chunk of air is + P x A P x + x A . AtmosphericPxA .pressure subtracts out, leaving P x + x + P x A = P x + x APx Ax2sThe mass of the air is m = V = Ax and its acceleration is 2 . SotFIG. P17.17Newtons second law becomes2P sxA = Ax 2xts2sB= 2xxtaf a fa f afFGHafafIJKB 2s 2s= x 2 t 2Into this wave equation as a trial solution we substitute the wave function s x , t = smax cos kx twe finds= ksmax sin kx tx2s= k 2 smax cos kx t2xs= +smax sin kx tt2s= 2 smax cos kx tt 2BB 2s 2s=becomes k 2 smax cos kx t = 2 smax cos kx t x 2 t 2b gafafaThis is true providedB 4 2 2ffaB.f= 4 2 f 2 .The sound wave can propagate provided it has 2 f 2 = v 2 =speed v =ffaaaB; that is, provided it propagates with504Sound WavesSection 17.3*P17.18Intensity of Periodic Sound WavesThe sound power incident on the eardrum is = IA where I is the intensity of the sound andA = 5.0 10 5 m 2 is the area of the eardrum.(a)At the threshold of hearing, I = 1.0 10 12 W m 2 , andejej = 1.0 10 12 W m 2 5.0 10 5 m 2 = 5.00 10 17 W .(b)At the threshold of pain, I = 1.0 W m 2 , andeI=JKjej = 1.0 W m 2 5.0 10 5 m 2 = 5.00 10 5 W .P17.19FG I IJ = 10 logF 4.00 1066.0 dBGH 1.00 10HI KFII70.0 dB = 10 log GH 1.00 10 W m JKW m j10 bTherefore, I = e1.00 106 = 10 log120P17.20(a)1212(b)I=270 .0 102g=1.00 10 5 W m 2 .2Pmax, so2 vjbegePmax = 2 vI = 2 1.20 kg m3 343 m s 1.00 10 5 W m 2jPmax = 90.7 mPaP17.21I=(a)12 2 smax v2At f = 2 500 Hz , the frequency is increased by a factor of 2.50, so the intensity (at constanta f = 6.25 .Therefore, 6.25a0.600f = 3.75 W msmax ) increases by 2.50(b)P17.222.0.600 W m 2The original intensity is I 1 =(a)2122 2 smax v = 2 2 vf 2 smax2If the frequency is increased to f while a constant displacement amplitude is maintained,the new intensity isb g22I 2 = 2 2 v f smax socontinued on next pageb gFG IJH K22I 2 2 v f smaxf==22 2I1f2 vf smax2or I 2 =FG f IJH fK2I1 .Chapter 17(b)If the frequency is reduced to f =intensity is505fwhile the displacement amplitude is doubled, the new2I 2 = 2 2 vFG f IJ b2s gH 2K2max22= 2 2 vf 2 smax = I 1or the intensity is unchanged .*P17.23(a)For the low note the wavelength is =For the high note =v 343 m s== 2.34 m .f146.8 s343 m s= 0.390 m .880 s880 Hz= 5.99 nearly146.8 Hzequal to a small integer. This fact is associated with the consonance of the notes D and A.We observe that the ratio of the frequencies of these two notes is(b) = 10 dB logI=FGH 10I12W m2I = 75 dB gives I = 3.16 10JK5W m22Pmax2 vjbegPmax = 3.16 10 5 W m 2 2 1.20 kg m3 343 m s = 0.161 Pafor both low and high notes.(c)bg1122v smax = v 4 2 f 2 smax22Ismax =22 vf 2for the low note,I=smax =3.16 10 5 W m 212 1.20 kg m 343 m s 146.8 s236.24 10 5m = 4.25 10 7 m146.8for the high note,6.24 10 5smax =m = 7.09 10 8 m880=(d)*P17.24146.8880== 1.093 , the134.3 804.9wavelengths and displacement amplitudes are made 1.093 times larger, and the pressureamplitudes are unchanged.With both frequencies lower (numerically smaller) by the factorThe power necessarily supplied to the speaker is the power carried away by the sound wave:P=b1Av smax2eg22= 2 2 Avf 2 smaxj FGH 0.08 m IJK b343 m sgb600 1 sg e0.12 102= 2 2 1.20 kg m3 222jm2= 21.2 W506P17.25Sound Waves(a)I 1 = 1.00 10 12 W m 2 10 beorjI 1 = 1.00 10e4I 2 = 1.00 10W m12 1 10g = e1.00 10 12jW m 2 10 80.0 102W m 2 10 bj4.5 2 10g = e1.00 10 1252jW m 2 10 75.0 102orI 2 = 1.00 10W m = 3.16 10 W mWhen both sounds are present, the total intensity isI = I 1 + I 2 = 1.00 10 4 W m 2 + 3.16 10 5 W m 2 = 1.32 10 4 W m 2 .(b)The decibel level for the combined sounds is = 10 log*P17.26(a)(b)F 1.32 10GH 1.00 104W m2122W mI = 10 log 1.32 10 =ejJK881.2 dB .vand f is the same for all three waves. Since the speed is smallest in air, isf1 493 m ssmallest in air. It is larger by= 4.51 times in water and by331 m s5 950= 18.0 times in iron .331We have =From I =12v 2 smax ; smax =22I 02v 0, smax is smallest in iron, larger in water by7 860 5 9507 860 5 950 iron v iron== 5.60 times , and larger in air by= 331 times .1 000 1 4931.29 331 water v water(c)From I =2Pmax; Pmax = 2 Iv , Pmax is smallest in air, larger in water by2 v1 000 1 493= 59.1 times , and larger in iron by1.29 331(d)b7 860 5 950= 331 times .1.29 331g331 m s 2v v2=== 0.331 m in airf2 000 s1 493 m s5 950 m s== 1.49 m in water== 5.95 m in iron1 00 0 s1 000 s=smax =2I02 v 0=2 10 6 W m 2e1.29 kg m jb331 m sgb6 283 1 sg32smax =2 10 61= 1.84 10 10 m in water1 000 1 493 6 283smax == 1.09 10 8 m in air2 10 61= 3.29 10 11 m in iron7 860 5 950 6 283bbPmax = 2 Iv =gg2e106jejW m 2 1.29 kg m3 331 m s = 0.029 2 Pa in airb gb7 860gb5 950g =Pmax = 2 10 6 1 000 1 493 = 1.73 Pa in waterPmax = 2 10 69.67 Pa in ironChapter 17P17.27(a)120 dB = 10 dB logI = 1.00 W m 2 =r==4 ILMNM 10I12W m2OPQP4 r 26.00 We4 1.00 W m 2= 0.691 mjWe have assumed the speaker is an isotropic point source.(b)0 dB = 10 dB logFGH 1012IJKIW m2I = 1.00 10 12 W m 2r==4 I6.00 We4 1.00 10 -12 W m 2j= 691 kmWe have assumed a uniform medium that absorbs no energy.P17.28We begin with 2 = 10 logFG I IJ , and HI K201I, and I 1 =, giving 2I14 r224 r12Then, 2 1 = 10 logP17.29FG r IJHr K122= 20 log102Also, I 2 =FG I IJ , soHI KFI I = 10 log G J .HI KFr I=G J .Hr K= 10 log211212FG r IJHr K1.2Since intensity is inversely proportional to the square of the distance,a fa fa f2I4 =10.0P 21I 0. 4 and I 0. 4 = max == 0.121 W m 2 .1002 v2 1.20 343The difference in sound intensity level is = 10 logFG IHI4 km0.4 kmAt 0.400 km, 0. 4 = 10 logAt 4.00 km,IJ = 10a2.00f = 20.0 dB .KF 0.121 W m I = 110.8 dB .GH 10 W m JK2122af 4 = 0.4 + = 110.8 20.0 dB = 90.8 dB .Allowing for absorption of the wave over the distance traveled,bgaf = 4 7.00 dB km 3.60 km = 65.6 dB .4This is equivalent to the sound intensity level of heavy traffic.507508P17.30Sound WavesLet r1 and r2 be the distance from the speaker to the observer that hears 60.0 dB and 80.0 dB,respectively. Use the result of problem 28, 2 1 = 20 logThus, logFG r IJ , to obtain 80.0 60.0 = 20 logFG r IJ .Hr KHr K1122FG r IJ = 1 , so rHr K1= 10.0r2 . Also: r1 + r2 = 110 m , so1210.0r2 + r2 = 110 m giving r2 = 10.0 m , and r1 = 100 m .P17.31We presume the speakers broadcast equally in all directions.rAC = 3.00 2 + 4.00 2 m = 5.00 m1.00 10 3 WI=== 3.18 10 6 W m 224 r 24 5.00 m(a)a = 10 dB logfF 3.18 10 W m IGH 10 W m JK62122 = 10 dB 6.50 = 65.0 dBrBC = 4. 47 m(b)I=1.50 10 3 Waf4 4.47 m = 10 dB log2= 5.97 10 6 W m 2F 5.97 10 IGH 10 JK612 = 67.8 dBI = 3.18 W m 2 + 5.97 W m 2(c) = 10 dB logP17.32In I =F 9.15 10 I =GH 10 JK669.6 dB121, intensity I is proportional to 2 ,2r4 rso between locations 1 and 2:I 2 r12=.I 1 r222221b g , intensity is proportional to s , so II = ss .F s I F r I F 1I F r IThen, G J = G J or G J = G J , giving r = 2r = 2a50.0 mf = 100 m .H s K H r K H 2K H r KBut, r = a50.0 mf + d yields d = 86.6 m .In I =1v smax222max22212212212222121Chapter 17P17.33 = 10 logFG I IJH 10 KI = 10 b12 10g e10 12 j509W m2I a120 dB f = 1.00 W m 2 ; I a100 dB f = 1.00 10 2 W m 2 ; I a10 dB f = 1.00 10 11 W m 2(a) = 4 r 2 I so that r12 I 1 = r22 I 2r2(b)r2FI=r GHIFI=r GHI111212IJKIJK12af1.00= 30.0 m1.00 10 2af1.00= 9.49 10 5 m1.00 10 11= 3.00 m12= 3.00 maf e7.00 102ja2P17.35(a)E =t = 4 r 2 It = 4 100 m(b)P17.34 = 10 log(a)fThe sound intensity inside the church is given byF 7.00 10 I =GH 1.00 10 JK212W m 2 0.200 s = 1.76 kJ108 dBFG I IJHI KF I I101 dB = a10 dBf lnGH 10 W m JKI = 10 e10W m j = 10W m = 10 ln01210.11221.9022= 0.012 6 W m 2We suppose that sound comes perpendicularly out through the windows and doors. Then,the radiated power isejej = IA = 0.012 6 W m 2 22.0 m 2 = 0.277 W .Are you surprised by how small this is? The energy radiated in 20.0 minutes isbgaE =t = 0.277 J s 20.0 min(b)60fFGH 1.00.0 s IJK =min332 J .If the ground reflects all sound energy headed downward, the sound power, = 0.277 W ,covers the area of a hemisphere. One kilometer away, this area isbg2A = 2 r 2 = 2 1 000 m = 2 10 6 m 2 .The intensity at this distance isI=0. 277 W== 4.41 10 8 W m 2A 2 10 6 m 2and the sound intensity level isa41 10f FGH 14..00 10 = 10 dB ln8W m2122W mI=JK46.4 dB .510*P17.36Sound WavesAssume you are 1 m away from your lawnmower and receiving 100 dB sound from it. The intensityIof this sound is given by 100 dB = 10 dB log 12; I = 10 2 W m 2 . If the lawnmower10W m2radiates as a point source, its sound power is given by I =.4 r 2a f2 = 4 1 m 10 2 W m 2 = 0.126 WNow let your neighbor have an identical lawnmower 20 m away. You receive from it sound with0.126 Wintensity I == 2.5 10 5 W m 2 . The total sound intensity impinging on you is24 20 ma102fW m + 2.5 10 5 W m 2 = 1.002 5 10 2 W m 2 . So its level is210 dB log1.002 5 10 210 12= 100.01 dB .If the smallest noticeable difference is between 100 dB and 101 dB, this cannot be heard as achange from 100 dB.Section 17.4P17.37f= fThe Doppler Effectbv v gbv v gOS(a)(b)P17.38(a)a343 + 40.0f = 338 Hza343 + 20.0fa343 + 20.0f = 483 Hzf = 510a343 + 40.0fF 115 min I = 12.0 rad s = 2 f = 2 GH 60.0 s min JKv= A = b12.0 rad sge1.80 10 mj =f = 3203max(b)The heart wall is a moving observer.f= f(c)0.021 7 m sFG v + v IJ = b2 000 000 HzgFG 1 500 + 0.021 7 IJ =H v KH 1 500 KO2 000 028.9 HzNow the heart wall is a moving source.f = f FG v IJ = b2 000 029 HzgF 1 500 I =GH 1 500 0.021 7 JKHvv Ks2 000 057.8 HzChapter 17P17.39f=Approaching ambulance:511fb1 v vgSff =Departing ambulance:d1 b v vgiF vI F vI560G 1 J = 480G 1 + JH vK H vKSSSince f = 560 Hz and f = 480 HzSvS= 80.0v80.0 343m s = 26.4 m svS =1 0401 040a fP17.40(a)The maximum speed of the speaker is described by112mv max = kA 222v max =af20.0 N m0.500 m = 1.00 m s5.00 kgkA=mThe frequencies heard by the stationary observer range fromfmin = fFG v IJ to f Hv+v Kmaxmax=fFG v IJHvv Kmaxwhere v is the speed of sound.F 343 m s I =GH 343 m s + 1.00 m s JKF 343 m s I == 440 HzGH 343 m s 1.00 m s JKfmin = 440 Hzfmax(b) = 10 dB log439 Hz441 HzFG I IJ = 10 dB logF 4 r IGH I JKHI K200The maximum intensity level (of 60.0 dB) occurs at r = rmin = 1.00 m . The minimum intensitylevel occurs when the speaker is farthest from the listener (i.e., whenr = rmax = rmin + 2 A = 2.00 m).F I 10 dB logF IGH 4 I r JKGH 4 I r JKF 4 I r I = 10 dB logF r I .or= 10 dB log GGH r JKJKH 4 I rThis gives: 60.0 dB = 10 dB log a 4.00f = 6.02 dB , and = 54.0 dBThus, max min = 10 dB logmaxmin20 min20 minmin20 max20 max2max2minmin.512P17.41Sound Wavesf= fF 340 IGH 340 b9.80t g JK485a340f + a 485fd9.80t i = a512 fa340 fF 512 485 IJ 340 = 1.93 st =GH 485 K 9.80FG v IJHvv K485 = 512sfallffd1 =1 2gt f = 18.3 m :2t return =18.3= 0.053 8 s340The fork continues to fall while the sound returns.t total fall = t f + treturn = 1.93 s + 0.053 8 s = 1.985 sd total =bg1 2gt total fall = 19.3 m2afm10 C = 325 m ss C(a)v = 331 m s + 0.6(b)P17.42Approaching the bell, the athlete hears a frequency ofAfter passing the bell, she hears a lower frequency ofFG v + v IJH v KF v + b v g If = f GH v JKf= fOOThe ratio iswhich gives 6 v 6 v o = 5 v + 5 v o or*P17.43(a)f v vO 5==f v + vO 6vO =av 325 m s== 29.5 m s1111fSound moves upwind with speed 343 15 m s . Crests pass a stationary upwind point atfrequency 900 Hz.Then=v 328 m s== 0.364 mf900 s(b)By similar logic,=343 + 15 m sv== 0.398 mf900 s(c)The source is moving through the air at 15 m/s toward the observer. The observer isstationary relative to the air.f= fafFG v + v IJ = 900 HzFG 343 + 0 IJ =H 343 15 KH vv Ko941 Hzs(d)The source is moving through the air at 15 m/s away from the downwind firefighter. Herspeed relative to the air is 30 m/s toward the source.f= fFG v + v IJ = 900 HzF 343 + 30 I = 900 HzFG 373 IJ =GH 343 a15f JKH 358 KH vv Kos938 HzChapter 17*P17.44The half-angle of the cone of the shock wave is where = sin 1FG vHvsoundsourceIJ = sin FG 1 IJ = 41.8 .H 1.5 KK1As shown in the sketch, the angle between the direction of propagationof the shock wave and the direction of the planes velocity isThe half angle of the shock wave cone is given by sin =vS =P17.46 = sin 1P17.47(b)v lightsin v lightvS.2.25 10 8 m s= 2.82 10 8 m ssin 53.0afv1= sin 1= 46.4vS1.38sin =v1=; = 19.5vS 3.00tan =hh; x=xtan x=(a)=v shockFIG. P17.44 = 90 = 9041.8 = 48.2 .P17.4520 000 m= 5.66 10 4 m = 56.6 kmtan 19.5It takes the plane t =5.66 10 4 mx== 56.3 s to travel this distance.vS 3.00 335 m sbgxt=0hhObservera.v planeObserver hears the boomb.FIG. P17.47(a)513514Sound WavesSection 17.5Digital Sound RecordingSection 17.6Motion Picture Sound*P17.48For a 40-dB sound,40 dB = 10 dB logLMNM 10I = 10 8 W m 2 =2Pmax2 v12OPQPIW m2jbegPmax = 2 vI = 2 1.20 kg m 2 343 m s 10 8 W m 2 = 2.87 10 3 N m 22.87 10 3 N m 2(a)(b)For sounds of 40 dB or softer, too few digital words are available to represent the wave formwith good fidelity.(c)*P17.49code =In a sound wave P is negative half of the time but this coding scheme has no wordsavailable for negative pressure variations.If the source is to the left at angle from the direction you arefacing, the sound must travel an extra distance d sin to reach yourright ear as shown, where d is the distance between your ears. Thed sin . Thendelay time is t in v =t = sin 1*P17.5065 536 = 728.7 N m 2103 dB = 10 dB log(a)bg343 m s 210 10v t= sin 1d0.19 mLMMN 10I12W m2I = 2.00 10 2 W m 2 =6s= 22.3 left of center .earearFIG. P17.49OPPQ4 r 2=af4 1.6 m2 = 0.642 W(b)efficiency =sound output power 0.642 W== 0.004 28total input power150 WAdditional ProblemsP17.51Model your loud, sharp sound impulse as a single narrow peak in a graph of air pressure versustime. It is a noise with no pitch, no frequency, wavelength, or period. It radiates away from you in alldirections and some of it is incident on each one of the solid vertical risers of the bleachers. Supposethat, at the ambient temperature, sound moves at 340 m/s; and suppose that the horizontal width ofeach row of seats is 60 cm. Then there is a time delay of0.6 mb340 m sg = 0.002 scontinued on next pageChapter 17515between your sound impulse reaching each riser and the next. Whatever its material, each willreflect much of the sound that reaches it. The reflected wave sounds very different from the sharppop you made. If there are twenty rows of seats, you hear from the bleachers a tone with twentycrests, each separated from the next in time byaf = 0.004 s .b340 m sg2 0.6 mThis is the extra time for it to cross the width of one seat twice, once as an incident pulse and onceagain after its reflection. Thus, you hear a sound of definite pitch, with period about 0.004 s, frequency1~ 300 Hz0.003 5 swavelengthbg340 m sv== 1.2 m ~ 10 0 mf300 s=band durationgaf20 0.004 s ~ 10 1 s .v 343 m s== 0.232 mf 1 480 s 1(a)=(b)P17.52 = 81.0 dB = 10 dB logejLMNM 10I12W m2OPQPI = 10 12 W m 2 10 8.10 = 10 3.90 W m 2 = 1.26 10 4 W m 2 =smax =(c)P17.53 =2I= v 2e2 1.26 10 4 W m 2e1.20 kg m jb343 m sg4 e1 480 s j3v 343 m s== 0.246 mf 1 397 s 1Since cos 2 + sin 2 = 1 ,j1 2212v 2 smax2= 8.41 10 8 m = = 13.8 mmsin = 1 cos 2 (each sign applying half the time)afaP = Pmax sin kx t = v smax 1 cos 2 kx tf222P = v smax smax cos 2 kx t = v smax s 2ThereforeP17.54afThe trucks form a train analogous to a wave train of crests with speed v = 19.7 m s2and unshifted frequency f == 0.667 min 1 .3.00 min(a)(b)The cyclist as observer measures a lower Doppler-shifted frequency:19.7 + 4. 47v + vo= 0.667 min 1= 0.515 minf= fv19.7FG IJ eH KF v + v IJ = e0.667 minf = f GH v KojFGH a f IJKjFGH 19.7 + a.1.56f IJK =19 710.614 minThe cyclists speed has decreased very significantly, but there is only a modest increase inthe frequency of trucks passing him.516Sound WavesjafP17.55v=2dvt 1:d== 6.50 10 3 m s 1.85 s = 6.01 kmt2 2P17.56(a)The speed of a compression wave in a bar iseYv=(b)20.0 10 10 N m 27 860 kg m3= 5.04 10 3 m s .The signal to stop passes between layers of atoms as a sound wave, reaching the back end ofthe bar in timet=(c)=0.800 mL== 1.59 10 4 s .v 5.04 10 3 m sAs described by Newtons first law, the rearmost layer of steel has continued to moveforward with its original speed vi for this time, compressing the bar bybgejL = vi t = 12.0 m s 1.59 10 4 s = 1.90 10 3 m = 1.90 mm .(d)The strain in the rod is:(e)L 1.90 10 3 m== 2.38 10 3 .L0.800 mThe stress in the rod is: =YFG L IJ = e20.0 10HLK10jejN m 2 2.38 10 3 = 476 MPa .Since > 400 MPa , the rod will be permanently distorted.(f)We go through the same steps as in parts (a) through (e), but use algebraic expressionsrather than numbers:The speed of sound in the rod is v =Y.The back end of the rod continues to move forward at speed vi for a time of t =traveling distance L = vi t after the front end hits the wall.The strain in the rod is:L vi t== vi.LLYThe stress is then: = YFG L IJ = YvHLKL=L,vYiY= vi Y .For this to be less than the yield stress, y , it is necessary thatvi Y < y or vi <yY.With the given numbers, this speed is 10.1 m/s. The fact that the length of the rod dividesout means that the steel will start to bend right away at the front end of the rod. There it willyield enough so that eventually the remainder of the rod will experience only stress withinthe elastic range. You can see this effect when sledgehammer blows give a mushroom top toa rod used as a tent stake.Chapter 17P17.57(a)f= f517vbv v gdiverso 1 v diverf=vfFGH v diver = v 1 ffIJKwith v = 343 m s , f = 1 800 Hz and f = 2 150 Hzwe findFGHv diver = 343 1 (b)IJK1 800= 55.8 m s .2 150If the waves are reflected, and the skydiver is moving into them, we havef = f bv + v g f = f LM v OP bv + v gvNM bv v g QP vdiverdiverdiverso f = 1 800P17.58(a)f=a343 + 55.8f =a343 55.8ffvvu2 500 Hz .FG 1 1 IJH v u v + uK2bu v gfva v + u v + uf2uvff ===fv u1 eu v jv e1 eu v jj2a36.1fa 400f f ==340 1 a36.1f 340f =fvv u2(b)P17.59130 km h = 36.1 m sf f = fva f2222222285.9 Hzv vOwhere vOv vSand vS are measured relative to the medium in which the sound is propagated. In this case theocean current is opposite the direction of travel of the ships andWhen observer is moving in front of and in the same direction as the source, f = fbg= 64.0 km h b 10.0 km hg = 74.0 km h = 20.55 m svO = 45.0 km h 10.0 km h = 55.0 km h = 15.3 m s , andvSbTherefore, f = 1 200.0 Hz520 m15 3g 11520 m ss20..55 m ss =m1 204.2 Hz .518P17.60Sound WavesUse the Doppler formula, and remember that the bat is a moving source.If the velocity of the insect is v x ,40.4 = 40.0a340 + 5.00fb340 v g .a340 5.00fb340 + v gxxSolving,v x = 3.31 m s .Therefore, the bat is gaining on its prey at 1.69 m s .P17.61sin =av1=vS N Mh = v 12.8 safx = vS 10.0 sshock frontfhhv 1.28tan = = 1.28=xvS N Msin 1cos ==tan 1.28 = 38.6NM =P17.62(a)FIG. P17.62(a)=(c) =(d)(e)v 343 m s== 0.343 mf 1 000 s 1fIJ aKvv F v + v I a343 + 40.0f m s == GJ = 1 000 s = 0.383 mf f H v KF v v IJ = b1 000 Hzg a343 30.0f m s = 1.03 kHzf = fGa343 40.0f m sH vv KFGH343 40.0 m sv v v vS=== 0.303 mf fv1 000 s 1SOSshock frontFIG. P17.611= 1.60sin (b)vsx1Chapter 17P17.63t = LFG 1Hvair519IJKv v air1= L cuv cuv air v cubgeje331 m s 3.56 10 3 m sv air v cut =6.40 10 3 sL=v cu v air3 560 331 m sbgjL = 2.34 mP17.64PThe shock wavefront connects all observersfirst hearing the plane, including our observerO and the plane P, so here it is vertical. Theangle that the shock wavefront makes withthe direction of the planes line of travel isgiven bysin =340 m sv== 0.173vS 1 963 m sCso = 9.97 .OUsing the right triangle CPO, the angle isseen to beFIG. P17.64 = 90.0 = 90.09.97 = 80.0 .P17.65(a)F v I = sin F 331 I =GH 20.0 10 JKGH v JKF 1 533 IJ = 4.40 = sin GH 20.0 10 K = sin 11sound3obj(b)P17.6610.948311 1 2 = 10 log 120.0280.0 2 = 10 log 20.0 = +13.02 = 2 = 67.0 dBFG Y IJH KFTI=G J .H K12P17.67For the longitudinal wave v L =.12For the transverse wave vTIf we requirevLYm= 8.00 , we have T =andwhere =L64.0 vT=eje2massm=.volume r 2 Lj3102 r 2 Y 2.00 10 m 6.80 10 N m== 1.34 10 4 N .This gives T =64.064.0520P17.68Sound WavesThe total output sound energy is eE =t , where is the power radiated.eE eEeEeE===Thus, t =.2 IA4 d 2 I4 r IeBut, = 10 logjFG I IJ . Therefore, I = I e10 j and t =HI K00 10eE4 d 2 I 0 10 10.(a)If the source and the observer are moving away from each other, we have: S 0 = 180 ,and since cos180 = 1 , we get Equation 17.12 with negative values for both vO and vS .(b)P17.69If vO = 0 m s then f =vfv vS cos SAlso, when the train is 40.0 m from the intersection, and the car is 30.0 m from theintersection,cos S =so f =ga343 m s500 Hz343 m s 0.800 25.0 m sb45for f = 531 Hz .Note that as the train approaches, passes, and departs from the intersection, S varies from0 to 180 and the frequency heard by the observer varies from:affmax =fmin =P17.70343 m svf=500 Hz = 539 Hzv vS cos 0343 m s 25.0 m s343 m svf=500 Hz = 466 Hzv vS cos 180343 m s + 25.0 m safLet T represent the period of the source vibration, and E be the energy put into each wavefront.EThen av = . When the observer is at distance r in front of the source, he is receiving a sphericalTwavefront of radius vt, where t is the time since this energy was radiated, given by vt vS t = r . Then,t=a fThe area of the sphere is 4 vt2=4 v 2 r 2bv v gTbv v gS2r.v vS. The energy per unit area over the spherical wavefront2E avS=.A4 v 2 r 2The observer receives parcels of energy with the Doppler shifted frequencyvvf= f=, so the observer receives a wave with intensityv vST v vSis uniform with the valueFGHIJK bgI=FG E IJ f = FG T bv v gH A K GH 4 v ravS2 22IF v IJJK GH Tbv v g JK =Sav4 r2FG v v IJH v KS.Chapter 17P17.71(a)The time required for a sound pulse to travel distance L atLL. Using this expressionspeed v is given by t = =vY 521L2L1L3we findFIG. P17.71L1t1 =t2 ==Y1 1e7.00 101.50 m L1Y2 2=L110N me1.60 10e2j e2 700 kg m j3e1.50 m L1N m210j e11.3 103j= 1.96 10 4 L1 skg m3jjt 2 = 1.26 10 3 8.40 10 4 L1 sort3 =1.50 me11.0 1010N m3j e8 800 kg m j3t 3 = 4.24 10 4 sWe require t1 + t 2 = t 3 , or1.96 10 4 L1 + 1.26 10 3 8.40 10 4 L1 = 4.24 10 4 .This gives L1 = 1.30 m and L 2 = 1.50 1.30 = 0.201 m .The ratio of lengths is then(b)The ratio of lengthsL1= 6.45 .L2L1is adjusted in part (a) so that t1 + t 2 = t 3 . Sound travels the two pathsL2in equal time and the phase difference, = 0 .P17.72To find the separation of adjacent molecules, use a model where each molecule occupies a sphere ofradius r given by air =3or 1.20 kg m =4.82 10 26 kg43r3average mass per molecule43r3L 3e4.82 10 kg j OP,r=MMN 4 e1.20 kg m j PQ26313= 2.12 10 9 m .Intermolecular separation is 2r = 4. 25 10 9 m, so the highest possible frequency sound wave isfmax =343 m svv=== 8.03 10 10 Hz ~ 10 11 Hz . min 2r 4.25 10 9 m522Sound WavesANSWERS TO EVEN PROBLEMSP17.21.43 km sP17.36noP17.4(a) 27.2 s; (b) longer than 25.7 s, becausethe air is coolerP17.38(a) 2.17 cm s ; (b) 2 000 028.9 Hz ;(c) 2 000 057.8 HzP17.6(a) 153 m s ; (b) 614 mP17.40(a) 441 Hz; 439 Hz; (b) 54.0 dBP17.8(a) 4.16 m; (b) 0.455 s ; (c) 0.157 mmP17.42(a) 325 m s; (b) 29.5 m sP17.101.55 10 10 mP17.4448. 2P17.12(a) 1.27 Pa; (b) 170 Hz; (c) 2.00 m;(d) 340 m sP17.4646. 4P17.48(a) 7; (b) and (c) see the solutionP17.50(a) 0.642 W ; (b) 0.004 28 = 0.428%P17.52(a) 0.232 m; (b) 84.1 nm; (c) 13.8 mmP17.54(a) 0.515 min ; (b) 0.614 minP17.56(a) 5.04 km s ; (b) 159 s ; (c) 1.90 mm;(d) 0.002 38 ; (e) 476 MPa ;(f) see the solutionP17.58(a) see the solution; (b) 85.9 HzP17.60The gap between bat and insect is closingat 1.69 m s .P17.62(a) see the solution; (b) 0.343 m;(c) 0.303 m; (d) 0.383 m; (e) 1.03 kHzP17.6480.0P17.6667.0 dBP17.68t =e4jP17.14s = 22.5 nm cos 62.8 x 2.16 10 tP17.16(a) 4.63 mm; (b) 14.5 m s ;(c) 4.73 10 9 W m 2P17.18(a) 5.00 10 17 W ; (b) 5.00 10 5 WP17.20(a) 1.00 105F f I=G JHfK2W m ; (b) 90.7 mPa2I 1 ; (b) I 2 = I 1P17.22(a) I 2P17.2421.2 WP17.26(a) 4.51 times larger in water than in airand 18.0 times larger in iron;(b) 5.60 times larger in water than in ironand 331 times larger in air;(c) 59.1 times larger in water than in airand 331 times larger in iron;(d) 0.331 m; 1.49 m; 5.95 m; 10.9 nm;184 pm; 32.9 pm; 29.2 mPa; 1.73 Pa; 9.67 PaeE4 d I 0 10 2P17.28see the solutionP17.3010.0 m; 100 mP17.70see the solutionP17.3286.6 mP17.72~ 10 11 HzP17.34(a) 1.76 kJ ; (b) 108 dB1018Superposition and Standing WavesCHAPTER OUTLINE18.1Superposition andInterferenceStanding WavesStanding Waves in a StringFixed at Both EndsResonanceStanding Waves in AirColumnsStanding Waves in Rod andPlatesBeats: Interference in TimeNon-Sinusoidal WavePatternsANSWERS TO QUESTIONS18.418.518.618.718.8Q18.1No. Waves with other waveforms are also trains of disturbancethat add together when waves from different sources movethrough the same medium at the same time.Q18.2The energy has not disappeared, but is still carried by the wavepulses. Each particle of the string still has kinetic energy. This issimilar to the motion of a simple pendulum. The pendulumdoes not stop at its equilibrium position duringoscillationlikewise the particles of the string do not stop atthe equilibrium position of the string when these two wavessuperimpose.Q18.318.218.3No. A wave is not a solid object, but a chain of disturbance. Asdescribed by the principle of superposition, the waves movethrough each other.Q18.4They can, wherever the two waves are nearly enough in phase that their displacements will add tocreate a total displacement greater than the amplitude of either of the two original waves.When two one-dimensional sinusoidal waves of the same amplitude interfere, thiscondition is satisfied whenever the absolute value of the phase difference between the two waves isless than 120.Q18.5When the two tubes together are not an efficient transmitter of sound from source to receiver, theyare an efficient reflector. The incoming sound is reflected back to the source. The waves reflected bythe two tubes separately at the junction interfere constructively.Q18.6No. The total energy of the pair of waves remains the same. Energy missing from zones ofdestructive interference appears in zones of constructive interference.Q18.7Each of these standing wave patterns is made of two superimposed waves of identical frequenciestraveling, and hence transferring energy, in opposite directions. Since the energy transfer of thewaves are equal, then no net transfer of energy occurs.Q18.8Damping, and nonlinear effects in the vibration turn the energy of vibration into internal energy.Q18.9The air in the shower stall can vibrate in standing wave patterns to intensify those frequencies inyour voice which correspond to its free vibrations. The hard walls of the bathroom reflect soundvery well to make your voice more intense at all frequencies, giving the room a longer reverberationtime. The reverberant sound may help you to stay on key.523524Superposition and Standing WavesQ18.10The trombone slide and trumpet valves change the length of the air column inside the instrument,to change its resonant frequencies.Q18.11The vibration of the air must have zero amplitude at the closed end. For air in a pipe closed at oneend, the diagrams show how resonance vibrations have NA distances that are odd integersubmultiples of the NA distance in the fundamental vibration. If the pipe is open, resonancevibrations have NA distances that are all integer submultiples of the NA distance in thefundamental.FIG. Q18.11Q18.12What is needed is a tuning forkor other pure-tone generatorof the desired frequency. Strike thetuning fork and pluck the corresponding string on the piano at the same time. If they are preciselyin tune, you will hear a single pitch with no amplitude modulation. If the two pitches are a bit off,you will hear beats. As they vibrate, retune the piano string until the beat frequency goes to zero.Q18.13Air blowing fast by a rim of the pipe creates a shshshsh sound called edgetone noise, a mixture ofall frequencies, as the air turbulently switches between flowing on one side of the edge and theother. The air column inside the pipe finds one or more of its resonance frequencies in the noise. Theair column starts vibrating with large amplitude in a standing wave vibration mode. It radiatessound into the surrounding air (and also locks the flapping airstream at the edge to its ownfrequency, making the noise disappear after just a few cycles).Q18.14A typical standingwave vibration possibility for a bell is similar to that for the glass shown inFigure 18.17. Here six node-to-node distances fit around the circumference of the rim. Thecircumference is equal to three times the wavelength of the transverse wave of in-and-out bendingof the material. In other states the circumference is two, four, five, or higher integers times thewavelengths of the higherfrequency vibrations. (The circumference being equal to the wavelengthwould describe the bell moving from side to side without bending, which it can do withoutproducing any sound.) A tuned bell is cast and shaped so that some of these vibrations will havetheir frequencies constitute higher harmonics of a musical note, the strike tone. This tuning is lost ifa crack develops in the bell. The sides of the crack vibrate as antinodes. Energy of vibration may berapidly converted into internal energy at the end of the crack, so the bell may not ring for so long atime.Q18.15The bow string is pulled away from equilibrium and released, similar to the way that a guitar stringis pulled and released when it is plucked. Thus, standing waves will be excited in the bow string. Ifthe arrow leaves from the exact center of the string, then a series of odd harmonics will be excited.Even harmonies will not be excited because they have a node at the point where the string exhibitsits maximum displacement.Chapter 18525Q18.16Walking makes the persons hand vibrate a little. If the frequency of this motion is equal to thenatural frequency of coffee sloshing from side to side in the cup, then a largeamplitude vibration ofthe coffee will build up in resonance. To get off resonance and back to the normal case of a smallamplitude disturbance producing a smallamplitude result, the person can walk faster, walk slower,or get a larger or smaller cup. Alternatively, even at resonance he can reduce the amplitude byadding damping, as by stirring highfiber quickcooking oatmeal into the hot coffee.Q18.17Beats. The propellers are rotating at slightly different frequencies.Q18.18Instead of just radiating sound very softly into the surrounding air, the tuning fork makes thechalkboard vibrate. With its large area this stiff sounding board radiates sound into the air withhigher power. So it drains away the forks energy of vibration faster and the fork stops vibratingsooner. This process exemplifies conservation of energy, as the energy of vibration of the fork istransferred through the blackboard into energy of vibration of the air.Q18.19The difference between static and kinetic friction makes your finger alternately slip and stick as itslides over the glass. Your finger produces a noisy vibration, a mixture of different frequencies, likenew sneakers on a gymnasium floor. The glass finds one of its resonance frequencies in the noise.The thin stiff wall of the cup starts vibrating with large amplitude in a standing wave vibrationmode. A typical possibility is shown in Figure 18.17. It radiates sound into the surrounding air, andalso can lock your squeaking finger to its own frequency, making the noise disappear after just a fewcycles. Get a lot of different thinwalled glasses of fine crystal and try them out. Each will generallyproduce a different note. You can tune them by adding wine.Q18.20Helium is less dense than air. It carries sound at higher speed. Each cavity in your vocal apparatushas a standing-wave resonance frequency, and each of these frequencies is shifted to a higher value.Your vocal chords can vibrate at the same fundamental frequency, but your vocal tract amplifies byresonance a different set of higher frequencies. Then your voice has a different quacky quality.Warning: Inhaling any pressurized gas can cause a gas embolism which can lead to stroke ordeath, regardless of your age or health status. If you plan to try this demonstration in class, inhaleyour helium from a balloon, not directly from a pressurized tank.Q18.21Stick a bit of chewing gum to one tine of the second fork. If the beat frequency is then faster than 4beats per second, the second has a lower frequency than the standard fork. If the beats have sloweddown, the second fork has a higher frequency than the standard. Remove the gum, clean the fork,add or subtract 4 Hz according to what you found, and your answer will be the frequency of thesecond fork.SOLUTIONS TO PROBLEMSSection 18.1P18.1Superposition and Interferenceafafy = y1 + y 2 = 3.00 cos 4.00 x 1.60t + 4.00 sin 5.0 x 2.00t evaluated at the given x values.aaff(a)x = 1.00 , t = 1.00y = 3.00 cos 2.40 rad + 4.00 sin +3.00 rad = 1.65 cm(b)x = 1.00 , t = 0.500y = 3.00 cos +3.20 rad + 4.00 sin +4.00 rad = 6.02 cm(c)x = 0.500 , t = 0y = 3.00 cos +2.00 rad + 4.00 sin +2.50 rad = 1.15 cmafafafaf526Superposition and Standing WavesP18.2FIG. P18.2P18.3(a)y2(b)a f= f a x + vtf , so wave 2 travels in they1 = f x vt , so wave 1 travels in the +x directionTo cancel, y1 + y 2 = 0 :x direction5=2for the positive root,+5a 3 x 4t f + 2 a 3 x + 4t 6 fa3 x 4tf = a3x + 4t 6f3 x 4t = a3 x + 4t 6f22+228t = 6t = 0.750 s(at t = 0.750 s , the waves cancel everywhere)(c)for the negative root,x = 1.00 m6x = 6(at x = 1.00 m , the waves cancel always)P18.4Suppose the waves are sinusoidal.a4.00 cmf sinakx tf + a4.00 cmf sinakx t + 90.0f2a 4.00 cmf sina kx t + 45.0f cos 45.0So the amplitude is a8.00 cmf cos 45.0 = 5.66 cm .The sum isP18.5The resultant wave function has the formFG IJ sinFG kx t + IJH 2K H2KL 4 OP =FIA = 2 A cosG J = 2a5.00f cos MH 2KN 2 Qy = 2 A0 cos(a)(b)0f= 1 200== 600 Hz229.24 mChapter 18P18.6FG IJ = AH 2KFG IJH K1= cos 1= 60.0 =2232Thus, the phase difference is = 120 =3T1==This phase difference results if the time delay is3 3 f 3v3.00 m= 0.500 sTime delay =3 2.00 m s2 A 0 cos0sobP18.7g(a)If the end is fixed, there is inversion of the pulse upon reflection. Thus, when they meet,they cancel and the amplitude is zero .(b)P18.8If the end is free, there is no inversion on reflection. When they meet, the amplitude is2 A = 2 0.150 m = 0.300 m .(a)x = 9.00 + 4.00 3.00 = 13 3.00 = 0.606 mafv 343 m s== 1.14 mf300 Hzx 0.606== 0.530 of a wave ,1.14 = 2 0.530 = 3.33 rad=The wavelength isThus,aor(b)xfxvv343== 283 Hzf=2 x 2 0.606For destructive interference, we wantwhere x is a constant in this set up.P18.9527= 0.500 = fafWe suppose the mans ears are at the same level as the lower speaker. Sound from the upperspeaker is delayed by traveling the extra distance L2 + d 2 L .2n 1 with n = 1, 2 , 3 , He hears a minimum when this is2n1 2 vL2 + d 2 L =Then,fn1 2 vL2 + d 2 =+Lfabgbgbn 1 2g v=22L +d22L=ff22b gv2bn 1 2gv fd n 1 222+ L2 +fbg2 n 1 2 vLf2n = 1, 2 , 3 , This will give us the answer to (b). The path difference starts from nearly zero when the man is veryfar away and increases to d when L = 0. The number of minima he hears is the greatest integern1 2 vsolution to d fdf 1n = greatest integer + .v 2bcontinued on next pageg528Superposition and Standing Waves(a)afbg4.00 m 200 s 1df 1+ =+ = 2.92330 m s2v 2He hears two minima.(b)With n = 1,L=b gv2b1 2g v fd2 1 222f2a4.00 mf b330 m sg 4b200 sgb330 m sg 200 s22=2L = 9.28 mwith n = 2L=P18.10b gv2b3 2 g v fd2 3 222f2= 1.99 m .Suppose the mans ears are at the same level as the lower speaker. Sound from the upper speaker isdelayed by traveling the extra distance r = L2 + d 2 L .aHe hears a minimum when r = 2n 1Then,fFGH IJK with n = 1, 2, 3, 2IJ FG v IJKH f KF 1IF vIL + d = Gn J G J + LH 2KH f KF 1I F vI F 1IF vIL + d = G n J G J + 2G n J G J L + LH 2K H f K H 2KH f KF 1I F vI F 1IF vId G n J G J = 2G n J G J LH 2K H f K H 2KH f KFGHL2 + d 2 L = n 222122222222(1)Equation 1 gives the distances from the lower speaker at which the man will hear a minimum. Thepath difference r starts from nearly zero when the man is very far away and increases to d whenL = 0.(a)The number of minima he hears is the greatest integer value for which L 0 . This is the1 vsame as the greatest integer solution to d n , or2 fFGHIJ FG IJKH Knumber of minima heard = n max = greatest integer d(b)FG f IJ + 1H vK 2.From equation 1, the distances at which minima occur are given byLn =b g bv f g2bn 1 2gb v f gd2 n 1 222where n = 1, 2 , , n max .P18.11(a)Chapter 18f bbgaga f= b 25.0 rad cmga5.00 cmf b 40.0 rad sga 2.00 sf = 45.0 rad529 1 = 20.0 rad cm 5.00 cm 32.0 rad s 2.00 s = 36.0 rad1 = 9.00 radians = 516 = 156(b) = 20.0 x 32.0t 25.0 x 40.0t = 5.00 x + 8.00tAt t = 2.00 s , the requirement isa f af = 5.00 x + 8.00 2.00 = 2n + 1 for any integer n.For x < 3.20 , 5.00 x + 16.0 is positive, so we haveaf5.00 x + 16.0 = 2n + 1 , orx = 3.20 a2n + 1f5.00The smallest positive value of x occurs for n = 2 and isx = 3.20 P18.12a4 + 1f = 3.20 =5.000.058 4 cm .v 344 m s== 16.0 mf 21.5 HzFirst we calculate the wavelength:=Then we note that the path difference equals(a)9.00 m 1.00 m =12Therefore, the receiver will record a minimum in sound intensity.(b)We choose the origin at the midpoint between the speakers. If the receiver is located at point(x, y), then we must solve:Then,ax + 5.00fax + 5.00f2+ y2 2+ y2 =2=4ax 5.00fSquare both sides and simplify to get:20.0 x Upon squaring again, this reduces to:400 x 2 10.0 2 x +Substituting = 16.0 m , and reducing,orax 5.00fax 5.00f(When plotted this yields a curve called a hyperbola.)22121+ y2 + 2+ y2 =+ y24= 2 x 5.0016.09.00 x 2 16.0 y 2 = 144y2x2=116.0 9.002af2+ 2 y 2530Superposition and Standing WavesSection 18.2P18.13Standing Wavesaf af a fy = 1.50 m sin 0.400 x cos 200t = 2 A0 sin kx cos tTherefore, k =2== 0.400 rad mand = 2 f sof=2= 15.7 m0.400 rad m2The speed of waves in the medium is v = f =P18.14y = 0.030 0 m cos(a)200 rad s= 31.8 Hz2 rad=200 rad s= 500 m s2 f = =k 0.400 rad m2FG x IJ cosa40tfH 2Knodes occur where y = 0 :x= 2n + 122afa2n + 1f = , 3 , 5 , .F 0.400 IJ = 0.029 4 m= 0.030 0 m cosGH 2 Kso x =(b)P18.15y maxThe facing speakers produce a standing wave in the space between them, with the spacing betweennodes beingd NN =343 m sv=== 0. 214 m2 2 f 2 800 s 1ejIf the speakers vibrate in phase, the point halfway between them is an antinode of pressure at adistance from either speaker of1.25 m= 0.625 .2Then there isa node at0.625 0. 214= 0.518 m2a node at0.518 m 0.214 m = 0.303 ma node at0.303 m 0.214 m = 0.089 1 ma node at0.518 m + 0.214 m = 0.732 ma node at0.732 m + 0.214 m = 0.947 mand a node at0.947 m + 0.214 m = 1.16 m from either speaker.531Chapter 18P18.16y = 2 A 0 sin kx cos t2yx 22 y= 2 A0 k 2 sin kx cos tt 2= 2 A0 2 sin kx cos tSubstitution into the wave equation givesThis is satisfied, provided thatP18.172 A 0 k 2 sin kx cos t =v=kfafy = y + y = 3.00 sinb x g cosb0.600 t g + 3.00 sinb x g cosb0.600 t g cmy = a6.00 cmf sinb x g cosb0.600 t g(a)We can take cosb0.600 t g = 1 to get the maximum y.y= a6.00 cmf sina0.250 f =At x = 0.250 cm,FG 1 IJ e2 A Hv K20sin kx cos tjay1 = 3.00 sin x + 0.600t cm; y 2 = 3.00 sin x 0.600t cm12maxaf aAt x = 0.500 cm ,(c)ff aNow take cos 0.600 t = 1 to get y max :b4.24 cmfa fy max = 6.00 cm sin 0.500 = 6.00 cm(b)gaAt x = 1.50 cm,(d)y max = 6.00 cm sin 1.50 1 = 6.00 cmThe antinodes occur whenx=But k =2= , sonn = 1, 3 , 5 , 4bg = 2.00 cm= 0.500 cm as in (b)43= 1.50 cm as in (c)x2 =45= 2.50 cmx3 =4andP18.182(a)x1 =The resultant wave isy = 2 A sin kx +FGHIJ FGK Hcos t 22IJK= n2n sox=k2kwhich means that each node is shiftedto the left.2kThe nodes are located at(b)kx +The separation of nodes isx = n + 1LMa f OP LM n OPN k 2k Q N k 2k QThe nodes are still separated by half a wavelength.x = =k 2532Superposition and Standing WavesSection 18.3P18.19Standing Waves in a String Fixed at Both EndsL = 30.0 m ; = 9.00 10 3 kg m; T = 20.0 N ; f1 =whereFTIv=G JH Ksof1 =v2L12= 47.1 m s47.1= 0.786 Hz60.0f 2 = 2 f1 = 1.57 Hzf3 = 3 f1 = 2.36 Hzf 4 = 4 f1 = 3.14 Hzb gejT = 4 kg 9.8 m s 2 = 39.2 N=and the wave speed on the string isv=In its fundamental mode of vibration, we have = 2L = 2 5 m = 10 mThus,P18.21The tension in the string isIts linear density is*P18.20f=(a)T39.2 N=1.6 10 3 kg m= 156.5 m sa fv=156.5 m s= 15.7 Hz10 mLet n be the number of nodes in the standing wave resulting from the 25.0-kg mass. Thenn + 1 is the number of nodes for the standing wave resulting from the 16.0-kg mass. For2Lv, and the frequency is f = .standing waves, =nThus,f=n Tn2L and alsof=n + 1 Tn +12LThus,Tnn +1==nTn +1Therefore,4n + 4 = 5n , or n = 4Then,(b)m 8 10 3 kg== 1.6 10 3 kg mL5mf=42 2.00 mafb25.0 kg gg = 5b16.0 kggg 4b25.0 kg ge9.80 m s j =20.002 00 kg m350 HzThe largest mass will correspond to a standing wave of 1 loopan = 1f so1350 Hz =2 2.00 myieldingm = 400 kgafem 9.80 m s 2j0.002 00 kg mChapter 18*P18.22533The first string has linear density1.56 10 3 kg= 2.37 10 3 kg m.0.658 m1 =The second, 2 =6.75 10 3 kg= 7.11 10 3 kg m.0.950 mThe tension in both is T = 6.93 kg 9.8 m s 2 = 67.9 N . The speed of waves in the first string is67.9 NT== 169 m s12.37 10 3 kg mv1 =Tand in the second v 2 == 97.8 m s . The two strings vibrate at the same frequency, according to2n 1 v1 n 2 v 2=2 L12L 2n1 169 m s n 2 97.8 m s=2 0.658 m2 0.950 maf afn25= 2.50 = . Thus n1 = 2 and n 2 = 5 are the number of antinodes on each string in the lowest2n1resonance with a node at the junction.(b)The first string has 2 + 1 = 3 nodes and the second string 5more nodes, for a total of 8, or 6 other than the vibratorjunctionand pulley.(a)*P18.23The frequency isbg=2a0.658 mf2 169 m sFIG. P18.22(b)257 Hz .afFor the E-string on a guitar vibrating as a whole, v = f = 330 Hz 2 64.0 cm . When it is stopped at64.0 cmthe first fret we have 12 2 330 Hz 2 L F = v = 330 Hz 2 64.0 cm . So L F = 12. Similarly for the264.0 cm. The spacing between the firstsecond fret, 2 2 12 330 Hz 2 L F# = v = 330 Hz 2 64.0 cm . L F# =2 2 12and second frets isafafafafFG 1H264.0 cm1 12122 12IJ = 64.0 cmFG 1 1 IJ = 3.39 cm.KH 1.059 5 1.059 5 K2This is a more precise version of the answer to the example in the text.Now the eighteenth fret is distant from the bridge by L18 =much string vibrate: L19 =64.0 cm. And the nineteenth lets this2 18 1264.0 cm. The distance between them is2 19 12FG 1H264.0 cm18 121219 12IJ = 64.0 cm 1 FG 1 1 IJ =KK2 H21.51 121. 27 cm .534*P18.24Superposition and Standing WavesFor the whole string vibrating, d NN = 0.64 m =; = 1.28 m . The21speed of a pulse on the string is v = f = 330 1.28 m = 422 m s .s2(a)When the string is stopped at the fret, d NN = 0.64 m = ;32 = 0.853 mv 422 m sf= == 495 Hz . 0.853 m(b)The light touch at a point one third of the way along thestring damps out vibration in the two lowest vibrationstates of the string as a whole. The whole string vibrates inits third resonance possibility: 3d NN = 0.64 m = 3 = 0.427 mf=P18.25f1 =FG IJH KvT, where v =2Lv=;2FIG. P18.24(b)422 m s= 990 Hz .0.427 m121.2(a)If L is doubled, then f1 L1 will be reduced by a factor(b)If is doubled, then f1 1 2 will be reduced by a factor(c)P18.26FIG. P18.24(a)If T is doubled, then f1 T will increase by a factor of12.2.L = 60.0 cm = 0.600 m; T = 50.0 N ; = 0.100 g cm = 0.010 0 kg mfn =wherenv2LFG T IJ = 70.7 m sH KF 70.7 IJ = 58.9n = 20 000 Hz= nGH 1.20 K12v=fnLargest n = 339 f = 19.976 kHz .P18.27d NN = 0.700 m = 1.40 mf = v = 308 m s =Te1.20 10 j a0.700f(a)T = 163 N(b)f3 = 660 Hz3FIG. P18.27535Chapter 18P18.28af fv ; = 2L = fvFf IF fIF 392 IJ = 0.038 2 m G J L = L G 1 J = a0.350 mfG 1 H 440 KHf KH fK G = 2 0.350 m =LG L A = LGAGGGAAAGGAThus, L A = LG 0.038 2 m = 0.350 m 0.038 2 m = 0.312 m ,or the finger should be placed 31.2 cm from the bridge .LA =v1=2 fA 2 fAT; dL A =dT4 f A T;dL A 1 dT=2 TLAdL0.600 cmdT=2 A =2= 3.84%35.0 3.82 cmTLAaP18.29fIn the fundamental mode, the string above the rod has onlytwo nodes, at A and B, with an anti-node halfway between Aand B. Thus,AL2L= AB =or =.cos cos 2Since the fundamental frequency is f, the wave speed in thissegment of string isv = f =BL2Lf.cos MAlso, v =TT==m ABTLm cos Twhere T is the tension in this part of the string. Thus,2Lf=cos F4L2 f 2TLTLor=2m cos m cos cos and the mass of string above the rod is:T cos m=4Lf 2Mg[Equation 1]FIG. P18.29Now, consider the tension in the string. The light rod would rotate about point P if the string exertedany vertical force on it. Therefore, recalling Newtons third law, the rod must exert only a horizontalforce on the string. Consider a free-body diagram of the string segment in contact with the end ofthe rod.Mg Fy = T sin Mg = 0 T = sin Then, from Equation 1, the mass of string above the rod ism=FG Mg IJ cos =H sin K 4Lf2Mg4Lf 2 tan .536*P18.30Superposition and Standing WavesLet m = V represent the mass of the copper cylinder. The original tension in the wire isVg on the cylinder, to reduce the tension toT1 = mg = Vg . The water exerts a buoyant force water2T2 = Vg waterFG IJH KFG V IJ g = FG IJ Vg .H 2K H 2 KThe speed of a wave on the string changes fromf1 =v1=waterT1T1 1 toT2to f 2 =. The frequency changes fromT2 1 where we assume = 2L is constant.*P18.31f2T2==f1T1 water 28.92 1.00 2=8.92f 2 = 300 HzThen8.42= 291 Hz8.92af dbg i dbgiComparingy = 0.002 m sin rad m x cos 100 rad s twithy = 2 A sin kx cos t2k== m 1 , = 2.00 m , and = 2 f = 100 s 1 : f = 50.0 Hzwe find(a)= 1.00 m2L3.00 m== 3 loopsd NN 1.00 md NN =Then the distance between adjacent nodes isand on the string aree(b)b, b = 6.00 m2100 m sv= 16.7 Hzfb = a =b6.00 mIn the simplest standing wave vibration, d NN = 3.00 m =and(c)jv = f = 50 s 1 2 m = 100 m sFor the speed we haveIn v 0 =T0increases to, if the tension increases to Tc = 9T0 and the string does not stretch, the speedvc =Then9T0=3T0bg= 3 v 0 = 3 100 m s = 300 m s300 m sv= 6.00 mc = c =d NN = c = 3.00 m12fa50 sand one loop fits onto the string.Chapter 18Section 18.4P18.32537ResonanceThe natural frequency isf=9.80 m s 2= 0.352 Hz .2.00 mg1=L 211=T 2The big brother must push at this same frequency of 0.352 Hz to produce resonance.P18.33v=9.15 m= 3.66 m s2.50 s(a)The wave speed is(b)From the figure, there are antinodes at both ends of the pond, so the distance betweenadjacent antinodes= 9.15 m ,2isd AA =and the wavelength is = 18.3 mThe frequency is thenf=v=3.66 m s= 0.200 Hz18.3 mWe have assumed the wave speed is the same for all wavelengths.P18.34v = gd =The wave speed ise9.80 m s ja36.1 mf = 18.8 m s2The bay has one end open and one closed. Its simplest resonance is with a node of horizontalvelocity, which is also an antinode of vertical displacement, at the head of the bay and an antinodeof velocity, which is a node of displacement, at the mouth. The vibration of the water in the bay islike that in one half of the pond shown in Figure P18.33.Then,d NA = 210 10 3 m =and = 840 10 3 mTherefore, the period isT=41 840 10 3 m= == 4. 47 10 4 s = 12 h 24 min18.8 m sf vThis agrees precisely with the period of the lunar excitation , so we identify the extra-high tides asamplified by resonance.P18.35The distance between adjacent nodes is one-quarter of the circumference.d NN = d AA =so = 10.0 cm and f = 20.0 cm== 5.00 cm24v 900 m s== 9 000 Hz = 9.00 kHz . 0.100 mThe singer must match this frequency quite precisely for some interval of time to feed enoughenergy into the glass to crack it.538Superposition and Standing WavesSection 18.5P18.36Standing Waves in Air Columnsd AA = 0.320 m ; = 0.640 mv(a)(b)P18.37f= = 0.085 0 m; d AA = 42.5 mm(a)For the fundamental mode in a closed pipe, = 4L , asin the diagram.= 531 HzBut v = f , therefore L =So, L =(b)e4 240 s 1j= 0.357 mA=AFIG. P18.37jThe wavelength isN /2343 m sv== 0.715 m2 f 2 240 s 1eN /4Lv4fFor an open pipe, = 2L , as in the diagram.So, L =P18.38343 m sAv 343 m s== 1.31 mf261.6 sso the length of the open pipe vibrating in its simplest (A-N-A) mode isd A to A =A closed pipe has1 = 0.656 m2(N-A) for its simplest resonance,(N-A-N-A) for the second,andHere, the pipe length is*P18.39(N-A-N-A-N-A) for the third.5d N to A =5 5= 1.31 m = 1.64 m44afAssuming an air temperature of T = 37 C = 310 K , the speed of sound inside the pipe isbv = 331 m sg310 K= 353 m s .273 KIn the fundamental resonant mode, the wavelength of sound waves in a pipe closed at one end is = 4L . Thus, for the whooping craneaf = 4 5.0 ft = 2.0 10 1 ft and f =v=b353 m sg FG 3.281 ft IJ =2.0 10 ft H 1 m K157.9 Hz .Chapter 18P18.40539The air in the auditory canal, about 3 cm long, can vibrate with a node at the closed end andantinode at the open end,withd N to A = 3 cm =so = 0.12 mandf=4v 343 m s= 3 kHz0.12 mA small-amplitude external excitation at this frequency can, over time, feed energyinto a larger-amplitude resonance vibration of the air in the canal, making it audible.P18.41For a closed box, the resonant frequencies will have nodes at both sides, so the permitted1wavelengths will be L = n , n = 1, 2 , 3 , .2bi.e., L =gn nvnv=.and f =2L22fTherefore, with L = 0.860 m and L = 2.10 m, the resonant frequencies areafn = n 206 Hzaffor L = 0.860 m for each n from 1 to 9and fn = n 84.5 HzP18.42ffor L = 2.10 m for each n from 2 to 23.=The wavelength of sound isThe distance between water levels at resonance is d =t=andP18.43v2f Rt = r 2 d = r 2v2f r 2v.2 RfFor both open and closed pipes, resonant frequencies are equally spaced as numbers. The set ofresonant frequencies then must be 650 Hz, 550 Hz, 450 Hz, 350 Hz, 250 Hz, 150 Hz, 50 Hz. These areodd-integer multipliers of the fundamental frequency of 50.0 Hz . Then the pipe length isd NA =P18.44vf v 340 m s=== 1.70 m .4 4f4 50 sb gL= d AA = orn2Since =vfWith v = 343 m s andL=n2L=nFG v IJH2fKfor n = 1, 2 , 3 , for n = 1, 2 , 3 , f = 680 Hz,L=nF 343 m s I = na0.252 mfGH 2a680 Hzf JKfor n = 1, 2 , 3 , afPossible lengths for resonance are: L = 0.252 m, 0.504 m, 0.757 m, , n 0. 252 m .540P18.45Superposition and Standing WavesFor resonance in a narrow tube open at one end,f =n(a)bf = 384 Hzgvn = 1, 3 , 5 , .4LwarmairAssuming n = 1 and n = 3 ,384 =v3vand 384 =.4 0.2284 0.683aaff22.8 cm68.3 cmIn either case, v = 350 m s .(b)For the next resonance n = 5 , and L =begj5 v 5 350 m s== 1.14 m .4f4 384 s 1FIG. P18.45P18.46The length corresponding to the fundamental satisfies f =vv34: L1 === 0.167 m .4L4 f 4 512a fSince L > 20.0 cm, the next two modes will be observed, corresponding to f =or L 2 =P18.473v5v= 0.502 m and L3 == 0.837 m .4f4fWe suppose these are the lowest resonances of the enclosed air columns.= 0.670 m2For one,=v 343 m s== 1.34 mf256 s 1length = d AA =For the other,=v 343 m s== 0.780 mf440 s 1length = 0.390 mSo,(b)original length = 1.06 m = 2d AA = 2.12 m343 m s= 162 Hz2.12 m(a)P18.483v5vand f =.4L 24L3f=(a)For the fundamental mode of an open tube,L=(b)v = 331 m s 1 +343 m s v=== 0.195 m .2 2 f 2 880 s 1eja5.00f = 328 m s273We ignore the thermal expansion of the metal.f=The flute is flat by a semitone.v=328 m sv== 841 Hz2L 2 0.195 mafChapter 18Section 18.6541Standing Waves in Rod and Plates5 100v== 1.59 kHz2L2 1.60(a)f=(b)P18.49Since it is held in the center, there must be a node in the center as well as antinodes at theends. The even harmonics have an antinode at the center so only the odd harmonics area fa fpresent.(c)P18.50f=3 560v== 1.11 kHz2L2 1.60When the rod is clamped at one-quarter of its length, the vibration pattern reads ANANA and therod length is L = 2d AA = .Therefore, L =Section 18.7P18.51a fa fv 5 100 m s== 1.16 mf4 400 HzBeats: Interference in Timef v Tf new = 110540= 104.4 Hz600f = 5.64 beats sP18.52(a)The string could be tuned to either 521 Hz or 525 Hz from this evidence.(b)Tightening the string raises the wave speed and frequency. If the frequency were originally521 Hz, the beats would slow down.Instead, the frequency must have started at 525 Hz to become 526 Hz .(c)From f =v=T 2L=1 T2L FG IJH Kf2T2f=and T2 = 2f1T1f12T1 =FG 523 Hz IJH 526 Hz K2T1 = 0.989T1 .The fractional change that should be made in the tension is thenfractional change =T1 T2= 1 0.989 = 0.011 4 = 1.14% lower.T1The tension should be reduced by 1.14% .542P18.53Superposition and Standing WavesFor an echo f = fbv + v g the beat frequency is fbv v gssb= f f .Solving for fb .gives fb = fb2 v gsbv v g when approaching wall.2a1.33f= a 256fa343 1.33f = 1.99 Hz beat frequencys(a)fb(b)When he is moving away from the wall, v s changes sign. Solving for v s givesvs =*P18.542- foot pipes produces actual frequencies of 131 Hz and 196 Hz and a3combination tone at 196 131 Hz = 65.4 Hz , so this pair supplies the so-called missing fundamental.The 4 and 2-foot pipes produce a combination tone 262 131 Hz = 131 Hz , so this does not work.2The 2 and 2 - foot pipes produce a combination tone at 262 196 Hz = 65.4 Hz , so this works.32Also, 4, 2 , and 2 - foot pipes all playing together produce the 65.4-Hz combination tone.3Using the 4 and 2aSection 18.8P18.55a fa fa fa f5 343fb v== 3.38 m s .2 f fb2 256 5faaffNon-Sinusoidal Wave PatternsWe list the frequencies of the harmonics of each note in Hz:NoteAC#E1440.00554.37659.262880.001 108.71 318.5Harmonic31 320.01 663.11 977.841 760.02 217.52 637.0The second harmonic of E is close the the third harmonic of A, and the fourthharmonic of C# is close to the fifth harmonic of A.P18.56We evaluates = 100 sin + 157 sin 2 + 62.9 sin 3 + 105 sin 4+51.9 sin 5 + 29.5 sin 6 + 25.3 sin 7where s represents particle displacement in nanometersand represents the phase of the wave in radians. As advances by 2 , time advances by (1/523) s. Here is theresult:FIG. P18.5652 200.02 771.93 296.3Chapter 18543Additional ProblemsP18.57f = 87.0 Hzspeed of sound in air: v a = 340 m s(a)jaefv = f b = 87.0 s 1 0.400 mb =v = 34.8 m s(b) a = 4Lva = a fUVWL=340 m sva== 0.977 m4 f 4 87.0 s 1ejFIG. P18.57*P18.58(a)Use the Doppler formulaf= fbv v g .bv v g0sWith f1 = frequency of the speaker in front of student andf 2 = frequency of the speaker behind the student.m +f b343b343sm1s.500m sg = 458 Hz gb343 m s 1.50 m sg = 454 Hzf = a 456 Hzfb343 m s + 0gaf1 = 456 Hz2Therefore, fb = f1 f 2 = 3.99 Hz .(b)The waves broadcast by both speakers have =v 343 m s== 0.752 m . The standing wavef456 s= 0.376 m . The student walks from one maximum to the next in20.376 m1= 0.251 s , so the frequency at which she hears maxima is f = = 3.99 Hz .time t =T1.50 m sbetween them has d AA =P18.59Moving away from station, frequency is depressed:343f = 180 2.00 = 178 Hz :178 = 180343 vSolving for v givesTherefore,v=a2.00fa343fa f178v = 3.85 m s away from stationMoving toward the station, the frequency is enhanced:343f = 180 + 2.00 = 182 Hz :182 = 180343 v2.00 343Solving for v gives4=182Therefore,v = 3.77 m s toward the stationa fa f544P18.60Superposition and Standing Wavesv=a48.0fa2.00f = 141 m s4.80 10 3d NN = 1.00 m ; = 2.00 m ; f =a =P18.61v= 70.7 Hzv a 343 m s== 4.85 mf70.7 HzCall L the depth of the well and v the speed of sound.1fb343f 4 = a2n 1f 4vf = a2n4 51.5 s m sgeja2n + 1fb343 m sgv= a 2n + 1f=L = 2an + 1f 144f4e60.0 s ja2n 1fb343 m sg = a2n + 1fb343 m sg4e51.5 s j4e60.0 s jaL = 2n 1Then for some integer n1112and for the next resonance2Thus,1and we require an integer solution toThe equation gives n =112n + 1 2n 1=60.051.5111.5= 6.56 , so the best fitting integer is n = 7 .17ThenL=andL=af bg = 21.6 m4e51.5 s j2a7 f + 1 b343 m sg= 21.4 m4e60.0 s j2 7 1 343 m s11suggest the best value for the depth of the well is 21.5 m .P18.62The second standing wave mode of the air in the pipe reads ANAN, with d NA =so = 2.33 mandf=v=343 m s= 147 Hz2.33 mFor the string, and v are different but f is the same.0.400 m= d NN =22so = 0.400 mafafTv = f = 0. 400 m 147 Hz = 58.8 m s =ejbT = v 2 = 9.00 10 3 kg m 58.8 m sg2= 31.1 N 1.75 m=34Chapter 18P18.63(a)Since the first node is at the weld, the wavelength in the thin wire is 2L or 80.0 cm. Thefrequency and tension are the same in both sections, sof=(b)12LT=12fthin wire.so L =(a)12 0.400a4.60= 59.9 Hz .2.00 10 3fAs the thick wire is twice the diameter, the linear density is 4 times that of the thin wire. = 8.00 g mP18.64545TL =LM 1 OPN a2fa59.9f Q4.60= 20.0 cm half the length of the8.00 10 3For the block: Fx = T Mg sin 30.0 = 0so T = Mg sin 30.0 =(b)1Mg .2The length of the section of string parallel to the incline ish= 2 h . The total length of the string is then 3h .sin 30.0FIG. P18.64m3h(c)The mass per unit length of the string is=(d)The speed of waves in the string isv=(e)In the fundamental mode, the segment of length h vibrates as one loop. The distancebetween adjacent nodes is then d NN =f=3 Mgh2mv=1 3 Mgh=2h2m3 Mg8mh= h .The period of the standing wave of 3 nodes (or two loops) isT=(h)FG Mg IJ FG 3 h IJ =H 2 KH m KWhen the vertical segment of string vibrates with 2 loops (i.e., 3 nodes), then h = 2the wavelength is(f)== h , so the wavelength is = 2h .2The frequency is(g)Tejfb = 1.02 f f = 2.00 10 2 f =e2.00 10 j23 Mg8mh1 2m= =h=3 Mghf v2mh3 MgFG IJ andH 2K546Superposition and Standing Waves(a)f=n2LsoP18.65Tf LL 1===f L 2L 2The frequency should be halved to get the same number of antinodes for twice thelength.(b)nT=nTFG IJ = LM n OPH K Nn + 1 QL n OP TT = MNn + 1 QT F nf L I=GJT H n fL KTn=Tnso222The tension must be(c)FG IJH KT3=T22P18.662f n L T =f nL Tso2T 9=T 16to get twice as many antinodes.0.010 0 kg= 5.00 10 3 kg m :For the wire, =2.00 mTv=e200 kg m s j2=5.00 10 3 kg mv = 200 m sIf it vibrates in its simplest state, d NN = 2.00 m =:2f=vb200 m sg = 50.0 Hz=4.00 m(a)The tuning fork can have frequencies 45.0 Hz or 55.0 Hz .(b)If f = 45.0 Hz , v = f = 45.0 s 4.00 m = 180 m s .bbgThen, T = v 2 = 180 m sg e5.00 102or if f = 55.0 Hz , T = v 2 = f 2P18.67j = b55.0 sg a 4.00 mf e5.00 103kg m = 162 N2223jkg m = 242 N .We look for a solution of the formafafbg= A sina 2.00 x 10.0t f cos + A cosa 2.00 x 10.0t f sin 5.00 sin 2.00 x 10.0t + 10.0 cos 2.00 x 10.0 t = A sin 2.00 x 10.0 t + This will be true if both5.00 = A cos and 10.0 = A sin ,requiringa5.00f + a10.0f22= A2A = 11.2 and = 63.4aThe resultant wave 11.2 sin 2.00 x 10.0t + 63.4fis sinusoidal.Chapter 182and = 2 f =2 vFG 2 x IJ cosFG 2 vt IJH K H Kb gy x , t = 2 A sin kx cos t = 2 A sinWith k =For the fundamental vibration, 1 = 2Lsoy1 x , t = 2 A sin(c)For the second harmonic 2 = L andy 2 x , t = 2 A sin(d)P18.69(a)(b)P18.68In general, n =(a)Let represent the angle each slanted ropemakes with the vertical.:b gFG x IJ cosFG vt IJHLK H L Kb gFG 2 x IJ cosFG 2 vt IJH LK H L Kb g2LandnFG n x IJ cosFG n vt IJH LK H L Kyn x , t = 2 A sinIn the diagram, observe that:sin =1.00 m 2=1.50 m 3or = 41.8 .Considering the mass, Fy = 0 : 2T cos = mgb12.0 kg ge9.80 m s j =or T =FIG. P18.6922 cos 41.878.9 NTFor the standing wave pattern shown (3 loops),32=2 2.00 m= 1.33 m3Thus, the required frequency isd AA =d=or*P18.70v=f=av==78.9 N= 281 m s0.001 00 kg mThe speed of transverse waves in the string is(b)f281 m s= 211 Hz1.33 m= 7.05 10 3 m is the distance between antinodes.2Then = 14.1 10 3 mand f =v=3.70 10 3 m s14.1 103m547= 2.62 10 5 Hz .The crystal can be tuned to vibrate at 2 18 Hz , so that binary counters canderive from it a signal at precisely 1 Hz.FIG. P18.70548Superposition and Standing WavesANSWERS TO EVEN PROBLEMSP18.2see the solutionP18.380.656 m; 1.64 mP18.45.66 cmP18.403 kHz; see the solutionP18.60.500 sP18.42t =P18.8(a) 3.33 rad; (b) 283 HzP18.10(a) The number is the greatestf1+ ;integer dv2P18.44L = 0.252 m, 0.504 m, 0.757 m, ,n 0.252 m for n = 1, 2 , 3 , P18.460.502 m; 0.837 mP18.48(a) 0.195 m; (b) 841 mP18.501.16 mP18.52(a) 521 Hz or 525 Hz; (b) 526 Hz;(c) reduce by 1.14%FG IJH Kd bn 1 2g b v f g=2bn 1 2 gb v f g2(b) Ln22wheren = 1, 2 , , n maxP18.12;2(b) along the hyperbola 9 x 2 16 y 2 = 144(a) x =af r 2v2 Rfaf224-foot and 2 -foot ; 2 and 2 - foot; and33all three togetherP18.14(a) 2n + 1 m for n = 0 , 1, 2 , 3 , ;(b) 0.029 4 mP18.54P18.16see the solutionP18.56see the solutionP18.18see the solutionP18.58(a) and (b) 3.99 beats sP18.2015.7 HzP18.604.85 mP18.22(a) 257 Hz; (b) 6P18.6231.1 NP18.24(a) 495 Hz; (b) 990 HzP18.64(a)P18.2619.976 kHzP18.283.84%3 Mgh1mMg ; (b) 3h; (c); (d);2m23h3 Mg2mh(e); (g) h;; (f)8mh3 Mge(h) 2.00 10 2j3 Mg8mhP18.30291 HzP18.320.352 HzP18.66(a) 45.0 Hz or 55.0 Hz; (b) 162 N or 242 NP18.34see the solutionP18.68see the solutionP18.36(a) 531 Hz; (b) 42.5 mmP18.70262 kHz19TemperatureCHAPTER OUTLINE19.1Temperature and the ZerothLaw of ThermodynamicsThermometers and theCelsius Temperature ScaleThe Constant-Volume GasThermometer and theAbsolute Temperature ScaleThermal Expansion ofSolids and LiquidsMacroscopic Description ofan Ideal GasANSWERS TO QUESTIONS19.319.419.5Q19.1Two objects in thermal equilibrium need not be in contact.Consider the two objects that are in thermal equilibrium inFigure 19.1(c). The act of separating them by a small distancedoes not affect how the molecules are moving inside eitherobject, so they will still be in thermal equilibrium.Q19.2The coppers temperature drops and the water temperaturerises until both temperatures are the same. Then the metal andthe water are in thermal equilibrium.Q19.319.2The astronaut is referring to the temperature of the lunarsurface, specifically a 400F difference. A thermometer wouldregister the temperature of the thermometer liquid. Since thereis no atmosphere in the moon, the thermometer will not read arealistic temperature unless it is placed into the lunar soil.Q19.4Rubber contracts when it is warmed.Q19.5Thermal expansion of the glass bulb occurs first, since the wall of the bulb is in direct contact withthe hot water. Then the mercury heats up, and it expands.Q19.6If the amalgam had a larger coefficient of expansion than your tooth, it would expand more than thecavity in your tooth when you take a sip of your ever-beloved coffee, resulting in a broken orcracked tooth! As you ice down your now excruciatingly painful broken tooth, the amalgam wouldcontract more than the cavity in your tooth and fall out, leaving the nerve roots exposed. Isnt it nicethat your dentist knows thermodynamics?Q19.7The measurements made with the heated steel tape will be too shortbut only by a factor of5 10 5 of the measured length.Q19.8(a)One mole of H 2 has a mass of 2.016 0 g.(b)One mole of He has a mass of 4.002 6 g.(c)One mole of CO has a mass of 28.010 g.Q19.9The ideal gas law, PV = nRT predicts zero volume at absolute zero. This is incorrect because theideal gas law cannot work all the way down to or below the temperature at which gas turns toliquid, or in the case of CO 2 , a solid.549550TemperatureQ19.10Call the process isobaric cooling or isobaric contraction. The rubber wall is easy to stretch. The airinside is nearly at atmospheric pressure originally and stays at atmospheric pressure as the wallmoves in, just maintaining equality of pressure outside and inside. The air is nearly an ideal gas tostart with, but PV = nRT soon fails. Volume will drop by a larger factor than temperature as thewater vapor liquefies and then freezes, as the carbon dioxide turns to snow, as the argon turns toslush, and as the oxygen liquefies. From the outside, you see contraction to a small fraction of theoriginal volume.Q19.11Cylinder A must be at lower pressure. If the gas is thin, it will be at one-third the absolute pressureof B.Q19.12At high temperature and pressure, the steam inside exerts large forces on the pot and cover. Stronglatches hold them together, but they would explode apart if you tried to open the hot cooker.Q19.13(a)The water level in the cave rises by a smaller distance than the water outside, as the trappedair is compressed. Air can escape from the cave if the rock is not completely airtight, and alsoby dissolving in the water.(b)The ideal cave stays completely full of water at low tide. The water in the cave is supportedby atmospheric pressure on the free water surface outside.(a)(b)FIG. Q19.13Q19.14Absolute zero is a natural choice for the zero of a temperature scale. If an alien race had bodies thatwere mostly liquid wateror if they just liked its taste or its cleaning propertiesit is conceivablethat they might place one hundred degrees between its freezing and boiling points. It is veryunlikely, on the other hand, that these would be our familiar normal ice and steam points, becauseatmospheric pressure would surely be different where the aliens come from.Q19.15As the temperature increases, the brass expands. This would effectively increase the distance, d,from the pivot point to the center of mass of the pendulum, and also increase the moment of inertiaof the pendulum. Since the moment of inertia is proportional to d 2 , and the period of a physicalIpendulum is T = 2, the period would increase, and the clock would run slow.mgdQ19.16As the water rises in temperature, it expands. The excess volume would spill out of the coolingsystem. Modern cooling systems have an overflow reservoir to take up excess volume when thecoolant heats up and expands.Q19.17The coefficient of expansion of metal is larger than that of glass. When hot water is run over the jar,both the glass and the lid expand, but at different rates. Since all dimensions expand, there will be acertain temperature at which the inner diameter of the lid has expanded more than the top of thejar, and the lid will be easier to remove.Chapter 19Q19.18The sphere expands when heated, so that it no longer fits through thering. With the sphere still hot, you can separate the sphere and ring byheating the ring. This more surprising result occurs because the thermalexpansion of the ring is not like the inflation of a blood-pressure cuff.Rather, it is like a photographic enlargement; every linear dimension,including the hole diameter, increases by the same factor. The reasonfor this is that the atoms everywhere, including those around the innercircumference, push away from each other. The only way that theatoms can accommodate the greater distances is for thecircumferenceand corresponding diameterto grow. This propertywas once used to fit metal rims to wooden wagon and horse-buggywheels. If the ring is heated and the sphere left at room temperature,the sphere would pass through the ring with more space to spare.551FIG. Q19.18SOLUTIONS TO PROBLEMSSection 19.1Temperature and the Zeroth Law of ThermodynamicsNo problems in this sectionSection 19.2Thermometers and the Celsius Temperature ScaleSection 19.3The Constant-Volume Gas Thermometer and the Absolute Temperature ScaleP19.1Since we have a linear graph, the pressure is related to the temperature as P = A + BT , where A andB are constants. To find A and B, we use the dataaf1.635 atm = A + a78.0 CfB0.900 atm = A + 80.0 C B(1)(2)Solving (1) and (2) simultaneously,we findA = 1.272 atmandB = 4.652 10 3 atm CTherefore,P = 1.272 atm + 4.652 10 3 atm C T(a)P = 0 = 1.272 atm + 4.652 10 3 atm C TAt absolute zerowhich givesejejT = 274 C .(b)At the freezing point of water P = 1.272 atm + 0 = 1.27 atm .(c)And at the boiling point P = 1.272 atm + 4.652 10 3 atm C 100 C = 1.74 atm .ejaf552P19.2TemperatureP1V = nRT1and P2 V = nRT2imply thatP2 T2=P1 T1afaf(a)P2 =0.980 atm 273 K + 45.0 KP1T2== 1.06 atm273 + 20.0 KT1(b)T3 =T1 P3P1afa293 K fa0.500 atmf = 149 K ==0.980 atm124 CFIG. P19.2af99TC + 32.0 F = 195.81 + 32.0 = 320 F55TF =T = TC + 273.15 = 195.81 + 273.15 = 77.3 K(a)To convert from Fahrenheit to Celsius, we useTC =and the Kelvin temperature is found asP19.4(a)(b)P19.3T = TC + 273 = 310 K(b)In a fashion identical to that used in (a), we find TC = 20.6 CT = 253 KandFG 212 F 32.0 F IJ =H 100 C 0.00 C KP19.6(a)T = 450 C = 450 C(b)P19.5bT = 450 C = 450 K810 Faf100 C = aa60.0 Sf + bSubtracting, 100 C = aa75.0 SfRequire0.00 C = a 15.0 S + baa = 1.33 C S .fThen 0.00 C = 1.33 15.0 S C+ bb = 20.0 C .bgSo the conversion is TC = 1.33 C S TS + 20.0 C .P19.7(a)T = 1 064 + 273 = 1 337 K melting pointT = 2 660 + 273 = 2 933 K boiling point(b)g af55TF 32.0 = 98.6 32.0 = 37.0 C99T = 1 596 C = 1 596 K . The differences are the same.Chapter 19Section 19.4P19.8553Thermal Expansion of Solids and Liquids = 1.10 10 5 C 1 for steeleajfL = 518 m 1.10 10 5 C 1 35.0 C 20.0 C = 0.313 mP19.9The wire is 35.0 m long when TC = 20.0 C .bL = Li T Tiga fa f for Cu.L = a35.0 mfe1.70 10 aCf jc35.0 C a 20.0 C fh =1 = 20.0 C = 1.70 10 5 C15afeja+3.27 cmfP19.10L = Li T = 25.0 m 12.0 10 6 C 40.0 C = 1.20 cmP19.11For the dimensions to increase, L = Li Tafa1.00 10 2 cm = 1.30 10 4 C 1 2.20 cm T 20.0 CfT = 55.0 C*P19.12jaefafL = Li T = 22 10 6 C 2.40 cm 30 C = 1.58 10 3 cmafafafafL = Li T = 9.00 10 6 C 1 30.0 cm 65.0 C = 0.176 mmL = Li T = 9.00 10 6 C 1 1.50 cm 65.0 C = 8.78 10 4 cm(c)*P19.14(a)(b)P19.13V = 3Vi T = 3 9.00 10 6 C 1eFjGH 30.0a fa1.50f4IaJK2fcm3 65.0 C = 0.093 0 cm3The horizontal section expands according to L = Li T .jaefafx = 17 10 6 C 1 28.0 cm 46.5 C 18.0 C = 1.36 10 2 cmThe vertical section expands similarly byjaeFIG. P19.14fafy = 17 10 6 C 1 134 cm 28.5 C = 6.49 10 2 cm .The vector displacement of the pipe elbow has magnituder = x 2 + y 2 =a0.136 mmf + a0.649 mmf22= 0.663 mmand is directed to the right below the horizontal at angle = tan 1FG y IJ = tan FG 0.649 mm IJ = 78.2H 0.136 mm KH x K1r = 0.663 mm to the right at 78.2 below the horizontal554P19.15TemperaturebgbL Al 1 + Al T = LBrass 1 + Brass TL Al LBrassT =LBrass Brass L Al Al(a)T =ga10.01 10.00fa10.00fe19.0 10 j a10.01fe24.0 10 j66T = 199 C so T = 179 C. This is attainable.T =(b)a10.02 10.00fa10.00fe19.0 10 j a10.02fe24.0 10 j66T = 396 C so T = 376 C which is below 0 K so it cannot be reached.P19.16g a50.0 CfjbeA = 2 17.0 10 6 C 1 0.080 0 mA = 2Ai T :(a)2A = 1.09 10 5 m 2 = 0.109 cm 2(b)The length of each side of the hole has increased. Thus, this represents an increase in thearea of the hole.bgeeP19.17V = 3 Vi T = 5.81 10 4 3 11.0 10 6P19.18(a)afjjb50.0 galga20.0f =0.548 gala5.050 cm = 5.000 cm 1 + 24.0 10 6 C 1 T 20.0 CL = Li 1 + T :fT = 437 C(b)L Al = LBrass for some T , orWe must getbgbLi , Al 1 + Al T = Li , Brass 1 + Brass Tegjej5.000 cm 1 + 24.0 10 6 C 1 T = 5.050 cm 1 + 19.0 10 6 C 1 TSolving for T , T = 2 080 C ,T = 3 000 CsoThis will not work because aluminum melts at 660 C .P19.19bagf(a)V f = Vi 1 + T = 100 1 + 1.50 10 4 15.0 = 99.8 mL(b)Vacetone = Vi TbbVflask = Vi TggacetonePyrexb= 3Vi TgPyrexfor same Vi , T ,Vacetone acetone1.50 10 41===6Vflask flask6.40 10 23 3.20 10ejThe volume change of flask isabout 6% of the change in the acetones volume .Chapter 19P19.20(a),(b)555The material would expand by L = Li T ,L= T , but instead feels stressLiF YL== YT = 7.00 10 9 N m 2 12.0 10 6 CALiea f a30.0 Cfj1= 2.52 10 6 N m 2 . This will not break concrete.P19.21(a)bgV = Vt t T VAl Al T = t 3 Al Vi Tejeja= 9.00 10 4 0.720 10 4 C 1 2 000 cm3 60.0 CV = 99.4 cm3(b)foverflows.The whole new volume of turpentine isjaef2 000 cm3 + 9.00 10 4 C 1 2 000 cm3 60.0 C = 2 108 cm3so the fraction lost is99.4 cm3= 4.71 10 232 108 cmand this fraction of the cylinders depth will be empty upon cooling:af4.71 10 2 20.0 cm = 0.943 cm .*P19.22The volume of the sphere isVPb =af4 3 4 r = 2 cm333= 33.5 cm3 .The amount of mercury overflowing isejoverflow = VHg + VPb Vglass = Hg VHg + PbVPb glassVglass Twhere Vglass = VHg + VPb is the initial volume. Thenej ejej e11LO= Ma182 27f10118 cm + a87 27f1033.5 cm P 40 C = 0.812 cmCCNQjoverflow = Hg glass VHg + Pb glass VPb T = Hg 3 glass VHg + 3 Pb 3 glass VPb T6P19.23In633F YL=require L = Li TALiF= YTAF500 N=T =42AY2.00 10 m 20.0 10 10 N m 2 11.0 10 6 CeT = 1.14 Cjejej3556*P19.24TemperatureModel the wire as contracting according to L = Li T and then stretching according toL YFstress = = Y= Li T = YT .ALi Lie145 C = 396 NCj(a)F = YAT = 20 10 10 N m 2 4 10 6 m 2 11 10 6(b)T =3 10 8 N m 2stress== 136 CY20 10 10 N m 2 11 10 6 CejTo increase the stress the temperature must decrease to 35 C 136 C = 101 C .(c)*P19.25The original length divides out, so the answers would not change.The area of the chip decreases according toA = A1 T = A f AibagA f = Ai 1 + T = Ai 1 + 2TfThe star images are scattered uniformly, so the number N of stars that fit is proportional to the area.afjaefThen N f = N i 1 + 2T = 5 342 1 + 2 4.68 10 6 C 1 100 C 20 C = 5 336 star images .Section 19.5Macroscopic Description of an Ideal GasafeP19.27jej9.00 atm 1.013 10 5 Pa atm 8.00 10 3 m 3PV== 2.99 mol8.314 N mol K 293 KRT(a)n=(b)P19.26N = nN A(a)Initially, PVi = ni RTiia= a 2.99 molfe6.02 10fafja1.00 atmfV = n Ra10.0 + 273.15f KP b0. 280V g = n Ra 40.0 + 273.15f K23molecules mol = 1.80 10 24 moleculesiFinally, Pf V f = n f RT ffiii0.280 Pfgiving313.15 K=1.00 atm 283.15 KPf = 3.95 atmorPf = 4.00 10 5 Pa abs. .Dividing these equations,(b)a fP a1.02fb0.280V g = n Ra85.0 + 273.15f KAfter being drivendiiPd = 1.121Pf = 4.49 10 5 PaP19.28a fa fa fa f3 150 0.1003 PV== 884 balloons34 r P 4 0.150 3 1.20If we have no special means for squeezing the last 100 L of helium out of the tank, the tank will befull of helium at 1.20 atm when the last balloon is inflated. The number of balloons is then reduced0.100 m3 3to to 884 = 877 .34 0.15 mPV = NP V =ea4 3 r NP :3jfN=Chapter 19P19.29The equation of state of an ideal gas is PV = nRT so we need to solve for the number of moles to find N.ejafagafaf1.01 10 5 N m 2 10.0 m 20.0 m 30.0 mPV== 2. 49 10 5 moln=RT8.314 J mol K 293 Kbe5N = nN A = 2.49 10 mol 6.022 10*P19.30(a)PVi = ni RTi =i23fjmolecules mol = 1.50 10 29 moleculesmiRTiMej3356MPVi 4.00 10 kg 1.013 10 N 4 6.37 10 m mole Ki=mi =RTi8.314 Nm 50 Kmolem2 3= 1.06 10 21 kg(b)Pf V fPVii=n f RT fni RTiF 1.06 10 kg + 8.00 10 kg I TGHJK 50 K1.06 10 kgF 1 IJ = 56.9 KT = 100 K GH 1.76 KnRT F 9.00 g I F 8.314 J I F=GP=JK GH mol K JK GH 2.00 773 K m IJK =V 10H 18.0 g mol212 1 =20f21fP19.313aP19.3331.61 MPa = 15.9 atmfa f(a)T2 = T1P2= 300 K 3 = 900 KP1(b)P19.32557T2 = T1P2 V2= 300 2 2 = 1 200 KP1V1 Fy = 0 :a fa fbg out gV in gV 200 kg g = 0boutge in 400 m3j = 200 kgThe density of the air outside is 1.25 kg m3 .nP=From PV = nRT ,V RTThe density is inversely proportional to the temperature, and the densityof the hot air isjFGH 283 K IJKTe1.25 kg m jFGH 1 283 K IJK e400 m j = 200 kgTe in = 1.25 kg m3Thenin33in283 K1= 0.400Tin283 K0.600 =Tin = 472 KTinFIG. P19.33558*P19.34P19.35TemperatureConsider the air in the tank during one discharge process. We suppose that the process is slowenough that the temperature remains constant. Then as the pressure drops from 2.40 atm to1.20 atm, the volume of the air doubles. During the first discharge, the air volume changes from 1 Lto 2 L. Just 1 L of water is expelled and 3 L remains. In the second discharge, the air volume changesfrom 2 L to 4 L and 2 L of water is sprayed out. In the third discharge, only the last 1 L of watercomes out. Were it not for male pattern dumbness, each person could more efficiently use his deviceby starting with the tank half full of water.(a)PV = nRTejejbga fm = nM = a 41.6 molfb 28.9 g molg =1.013 10 5 Pa 1.00 m3PVn=== 41.6 molRT8.314 J mol K 293 K(b)1.20 kg , in agreement with the tabulated density of31.20 kg m at 20.0C.*P19.36ej2The void volume is 0.765Vtotal = 0.765 r 2 = 0.765 1.27 10 2 m 0.2 m = 7.75 10 5 m3 . Now forthe gas remaining PV = nRTn=P19.37(a)bn=PV = nRTejga5523PV 12.5 1.013 10 N m 7.75 10 m== 3.96 10 2 molRT8.314 Nm mole K 273 + 25 KfPVRTafe335PVM 1.013 10 Pa 0.100 m 28.9 10 kg mol=m = nM =RT8.314 J mol K 300 Kbfgajm = 1.17 10 3 kgej(b)(c)F = PA = 1.013 10 5 N m 2 0.100 m(d)P19.38Fg = mg = 1.17 10 3 kg 9.80 m s 2 = 11.5 mNThe molecules must be moving very fast to hit the walls hard.jaeAt depth,P = P0 + ghAt the surface,P0 V f = nRT f :Thereforeand2= 1.01 kNPVi = nRTiP0 V fbP + ghgVF T I FG P + gh IJGH T JK H P KF 293 K IJ FG 1.013 10= 1.00 cm GH 278 K K GHV f = ViVfff0i=TfTi0i03V f = 3.67 cm35ejejaf IJJKPa + 1 025 kg m 3 9.80 m s 2 25.0 m1.013 10 5 PaChapter 19P19.39mfPV = nRT :mi=nfm f = misoPf V f RTi Pf=RT f PVi Pii=niFP IGH P JKfim = mi m f = miP19.40559F P P I = 12.0 kgFG 41.0 atm 26.0 atm IJ =GH P JKH 41.0 atm Kif4.39 kgiMy bedroom is 4 m long, 4 m wide, and 2.4 m high, enclosing air at 100 kPa and 20 C = 293 K . Thinkof the air as 80.0% N 2 and 20.0% O 2 .Avogadros number of molecules has massa0.800fb28.0 g molg + a0.200fb32.0 g molg = 0.028 8 kg molF mIPV = nRT = G J RTH MKPVM e1.00 10 N m je38.4 m jb0.028 8 kg molg== 45.4 kgm=RTb8.314 J mol K ga293 K fThen5gives*P19.4123~ 10 2 kgThe CO 2 is far from liquefaction, so after it comes out of solution it behaves as an ideal gas. Its molarmass is M = 12.0 g mol + 2 16.0 g mol = 44.0 g mol . The quantity of gas in the cylinder ism sample6.50 g== 0.148 moln=M44.0 g molbThenPV = nRTgivesV=gbganRT 0.148 mol 8.314 J mol K 273 K + 20 K=P1.013 10 5 N m 2ejejejf FG 1 N m IJ F 10 L I =H 1 J K GH 1 m JK333.55 LP19.4210 9 Pa 1.00 m 3 6.02 10 23 molecules molPVN A== 2.41 10 11 moleculesN=RT8.314 J K mol 300 KP19.43P0 V = n1 RT1 =bFG m IJ RTH MKF m IJ RTP V = n RT = GHMKP VM F 1m m =GH T T1 IJKR021211222012gaf560P19.44Temperature(a)Initially the air in the bell satisfies P0 Vbell = nRTiorafP0 2.50 m A = nRTi(1)When the bell is lowered, the air in the bell satisfiesafPbell 2.50 m x A = nRT f(2)where x is the height the water rises in the bell. Also, the pressure in the bell, once it islowered, is equal to the sea water pressure at the depth of the water level in the bell.aaffPbell = P0 + g 82.3 m x P0 + g 82.3 m(3)The approximation is good, as x < 2.50 m. Substituting (3) into (2) and substituting nR from(1) into (2),afafP0 + g 82.3 m 2.50 m x A = P0 VbellTfTi.Using P0 = 1 atm = 1.013 10 5 Pa and = 1.025 10 3 kg m3LOfMM T FGH1 + ga82.3 mf IJK PPPN TQLM 277.15 K F e1.025 10 kg m je9.80 m s ja82.3 mf IG1 +JJ= a 2.50 mfM1 293.15 K G1.013 10 N mKHMNax = 2.50 m 1 1f00331225OPPPQx = 2.24 m(b)If the water in the bell is to be expelled, the air pressure in the bell must be raised to thewater pressure at the bottom of the bell. That is,afPbell = P0 + g 82.3 m5ejajef= 1.013 10 Pa + 1.025 10 3 kg m 3 9.80 m s 2 82.3 m5Pbell = 9.28 10 Pa = 9.16 atmAdditional ProblemsP19.45The excess expansion of the brass isba f afa Lf = 2.66 10 m L = 19.0 11.0 10 64(a)The rod contracts more than tape toa length reading 0.950 0 m 0.000 266 m = 0.949 7 m(b)ga Cf a0.950 mfa35.0 CfLrod Ltape = brass steel Li T0.950 0 m + 0.000 266 m = 0.950 3 m1Chapter 19P19.46At 0C, 10.0 gallons of gasoline has mass,=frommVjbem = V = 730 kg m3 10.0 galgFGH 0.003 80 m IJK = 27.7 kg1.00 gal3The gasoline will expand in volume bybgafV = Vi T = 9.60 10 4 C 1 10.0 gal 20.0 C 0.0 C = 0.192 galAt 20.0C,10.192 gal = 27.7 kg10.0 gal = 27.7 kgF 10.0 gal I = 27.2 kgGH 10.192 gal JKThe extra mass contained in 10.0 gallons at 0.0C is27.7 kg 27.2 kg = 0.523 kg .P19.47Neglecting the expansion of the glass,h =h =VTAb0.250 cm 2g e1.82 10 e 2.00 10 cmj343324jaf C 1 30.0 C = 3.55 cmFIG. P19.47P19.48(a)The volume of the liquid increases as V = Vi T . The volume of the flask increases asVg = 3Vi T . Therefore, the overflow in the capillary is Vc = Vi T 3 ; and in thebcapillary Vc = Ah .Therefore, h =(b)Vi 3 T .Abgb gFor a mercury thermometer Hg = 1.82 10 4 C 1and for glass,3 = 3 3.20 10 6 C 1Thus 3 or << .g561562P19.49TemperatureThe frequency played by the cold-walled flute is fi =vv=. i 2 LiWhen the instrument warms upff =fivvv===. f 2L f 2Li 1 + T 1 + TafThe final frequency is lower. The change in frequency isFGHIJK11 + TTvT1 + T2 Lif = f i f f = f i 1 FGHIJKa fb343 m sge24.0 10 Cja15.0 Cf =f 2a0.655 mff =v2 Li60.094 3 HzThis change in frequency is imperceptibly small.P19.50(a)P0 V P V =TTV = V + AhP = P0 +kkhAFG P + kh IJ aV + Ahf = P V FG T IJH AKHTKe1.013 10 N m + 2.00 10 N m hje5.00 10 m + e0.010 0 m jhj= e1.013 10 N m je5.00 100523353FIG. P19.50232m3KjFGH 523 K IJK2932 000 h 2 + 2 013 h 397 = 0(b)20C05h=250Ch2 013 2 689= 0.169 m4 000eja2.00 10 3 N m 0.169kh5= 1.013 10 Pa +P = P +A0.010 0 m 2P = 1.35 10 5 PafChapter 19P19.51(a)=mmand d = 2 dVVVFor very small changes in V and , this can be expressed as = m V= T .V VThe negative sign means that any increase in temperature causes the density to decreaseand vice versa.(b)*P19.52For water we have =1.000 0 g cm 3 0.999 7 g cm 3== 5 10 5 C 1 .3T1.000 0 g cm 10.0 C 4.0 CjaefThe astronauts exhale this much CO 2 :n=m sampleM=FGHIaJKgFGHIJK1.09 kg1 000 g1 mol= 520 mol .3 astronauts 7 daysastronaut day 1 kg44.0 gfbThen 520 mol of methane is generated. It is far from liquefaction and behaves as an ideal gas.P=P19.53(a)bfganRT 520 mol 8.314 J mol K 273 K 45 K= 6.57 10 6 Pa=V150 10 3 m3We assume that air at atmospheric pressure is above thepiston.mg+ P0AIn equilibriumPgas =Therefore,nRT mg=+ P0hAAh=ornRTmg + P0 Awhere we have used V = hA as the volume of the gas.(b)From the data given,h=e20.0 kg 9.80 m s= 0.661 mbga0.200 mol 8.314 J K mol 400 K2j e5+ 1.013 10 N m2jef0.008 00 m 2jFIG. P19.53563564P19.54TemperatureThe angle of bending , between tangents to the two ends of the strip, isequal to the angle the strip subtends at its center of curvature. (The anglesare equal because their sides are perpendicular, right side to the right sideand left side to left side.)(a)Li + L1 = r1The definition of radian measure givesFIG. P19.54Li + L 2 = r2andbL 2 L1 = r2 r1By subtraction,g 2 Li T 1 Li T = r=b2g 1 Li Trbg(b)In the expression from part (a), is directly proportional to T and also to 2 1 .Therefore is zero when either of these quantities becomes zero.(c)The material that expands more when heated contracts more when cooled, so the bimetallicstrip bends the other way. It is fun to demonstrate this with liquid nitrogen.(d)=bg2 2 1 Li T2 r=eejfa f0.500 mm= 1.45 10 2 = 1. 45 10 2 radP19.55ja2 19 10 6 0.9 10 6 C 1 200 mm 1 CFG 180 IJ =H rad K0.830From the diagram we see that the change in area isA = w + w + w .Since and w are each small quantities, the product w willbe very small. Therefore, we assume w 0.Sincew = wTwe then haveA = wT + wTand = T ,FIG. P19.55and since A = w , A = 2AT .The approximation assumes w 0, or T 0 . Another way of stating this is T << 1 .Chapter 19P19.56LTi = 2 ig(a)Li =soTi2 g4 2565a1.000 sf e9.80 m s j = 0.248 2 m=224 2bfgaL = Li T = 19.0 10 6 C 1 0.284 2 m 10.0 C = 4.72 10 5 m0.248 3 mLi + L= 2= 1.000 095 0 sg9.80 m s 2T f = 2T = 9.50 10 5 seIn one week, the time lost is time lost = 1 week 9.50 10 5 s lost per second(b)btime lost = 7.00 d weekgFGH 86 400ds IJK FGH 9.50 101.005time lost = 57.5 s lostP19.57zI = r 2 dma f b gafI aT f= a1 + T fI bT gI aT f I bT g 2TI bT gr T = r Ti 1 + Tand since2for T << 1 we findiithusi(a)I= 2 17.0 10 6 C 1 100 C = 0.340%IWith = 24.0 10 6 C 1T = 100 Cwe find for Al:P19.58T = 100 Cand(b)With = 17.0 10 6 C 1 andwe find for Cu:(a)I= 2 24.0 10 6 C 1 100 C = 0.480%I(b)P = P0 + gdgP0 Vi=PfjaeB = gV B=jaefP V = P0 VigP0 VibP + gdg0Since d is in the denominator, B must decrease as the depth increases.(The volume of the balloon becomes smaller with increasing pressure.)(c)bafafggP0 Vi P0 + gdP01 Bd===gP0Vi P02 B0P0 + gdP0 + gd = 2 P0d=1.013 10 5 N m 2P0== 10.3 mg 1.00 10 3 kg m3 9.80 m s 2ejejjs lostsIJK566*P19.59TemperatureThe effective coefficient is defined by Ltotal = effective Ltotal T where Ltotal = LCu + LPb andLtotal = LCu + LPb = xLtotal + 1 x Ltotal . Then by substitutiona fbg Cu LCu T + Pb LPb T = eff LCu + LPb Ta fgx = Cu x + Pb 1 x = effbCux=*P19.60 Pbeff20 10617 106 Pb1 C 29 10 6 1 C1 C 29 1061 C9= 0.75012=(a)No torque acts on the disk so its angular momentum is constant. Its moment of inertiadecreases as it contracts so its angular speed must increase .(b)I i i = I f f =11112MRi2 i = MR 2 f = M Ri + RiT f = MRi2 1 Tf22222 f = i 1 TP19.61=25.0 rad se1 e17 106j1 C 830 Cj2=2f25.0 rad s= 25.7 rad s0.972After expansion, the length of one of the spans isafafL f = Li 1 + T = 125 m 1 + 12 10 6 C 1 20.0 C = 125.03 m .L f , y, and the original 125 m length of this span form a right triangle with y as the altitude. Using thePythagorean theorem gives:a125.03 mf2af= y 2 + 125 m2yielding y = 2.74 m .P19.62afAfter expansion, the length of one of the spans is L f = L 1 + T . L f , y, and the original length L ofthis span form a right triangle with y as the altitude. Using the Pythagorean theorem givesL2f = L2 + y 2 ,SinceP19.63(a)ory = L2f L2 = L2Let m represent the sample mass. The number of moles is n =Then, =mmRT or PM = RT .MVmPM.=VRTea f 1 = L 2T + T2y L 2T .T << 1 ,So PV = nRT becomes PV =(b)a1 + T fjbg1.013 10 5 N m 2 0.032 0 kg molPM=== 1.33 kg m 3RT8.314 J mol K 293 Kbgafmmand the density is = .VMChapter 19FG nR IJ TH PK(b)V=dV nR V==dTP TThus,(a)From PV = nRT , the volume is:Therefore, when pressure is held constant,P19.64At T = 0 C = 273 K , this predicts=FG 1 IJ dV = FG 1 IJ V , or =H V K dT H V K T 1T1= 3.66 10 3 K 1273 K He = 3.665 10 3 K 1 and air = 3.67 10 3 K 1Experimental values are:They agree within 0.06% and 0.2%, respectively.P19.65For each gas alone, P1 =N 1 kTN kTN kTand P2 = 2and P3 = 3 , etc.VVVFor all gasesbgP1V1 + P2 V2 + P3 V3 = N 1 + N 2 + N 3 kT andbN1g+ N 2 + N 3 kT = PVAlso, V1 = V2 = V3 = = V , therefore P = P1 + P2 + P3 .P19.66(a)567Using the Periodic Table, we find the molecular masses of the air components to beb ga fb gM N 2 = 28.01 u , M O 2 = 32.00 u , M Ar = 39.95 ubgand M CO 2 = 44.01 u .Thus, the number of moles of each gas in the sample is75b g 28.01.52 g = 2.696 molg mol23.15 gnbO g == 0.723 4 mol32.00 g mol1.28 gna Ar f == 0.032 0 mol39.95 g moln N2 =2b gn CO 2 =0.05 g= 0.001 1 mol44.01 g molThe total number of moles is n 0 = ni = 3.453 mol . Then, the partial pressure of N 2 isb gP N2 =2.696 mol1.013 10 5 Pa = 79.1 kPa .3.453 molejSimilarly,b gP O 2 = 21.2 kPacontinued on next pagea fP Ar = 940 PabgP CO 2 = 33.3 Pa568Temperature(b)Solving the ideal gas law equation for V and using T = 273.15 + 15.00 = 288.15 K , we findV=Then, =(c)afb100 10 3 kgm== 1.22 kg m3 .V 8.166 10 2 m3The 100 g sample must have an appropriate molar mass to yield n 0 moles of gas: that isa fM air =*P19.67fga3.453 mol 8.314 J mol K 288.15 Kn 0 RT= 8.166 10 2 m3 .=P1.013 10 5 Pa100 g= 29.0 g mol .3.453 molConsider a spherical steel shell of inner radius r and much smaller thickness t, containing helium atpressure P. When it contains so much helium that it is on the point of bursting into twohemispheres, we have P r 2 = 5 10 8 N m 2 2 rt . The mass of the steel ise sV = s 4 r 2 t = s 4 r 2Pr10 9 Paj. For the helium in the tank, PV = nRT becomesm4P r 3 = nRT = He RT = 1 atmVballoon .3M HeThe buoyant force on the balloon is the weight of the air it displaces, which is described bym41 atmVballoon = air RT = P r 3 . The net upward force on the balloon with the steel tank hangingM air3from it is+ m air g m He g m s g =M air P 4 r 3 g M He P 4 r 3 g s P 4 r 3 g3 RT3 RT10 9 PaThe balloon will or will not lift the tank depending on whether this quantity is positive or negative,M air M He 9 s . At 20C this quantity iswhich depends on the sign of3 RT10 Pab=ga28.9 4.00f 10 kg mol 7 860 kg m3b8.314 J mol K g293 K10 N m3932= 3.41 10 6 s 2 m 2 7.86 10 6 s 2 m 2where we have used the density of iron. The net force on the balloon is downward so the heliumballoon is not able to lift its tank.Chapter 19With piston alone:T = constant, so PV = P0 V0orP19.68569P Ahi = P0 Ah0b g b gFh IP=P G JHh KWith A = constant,00iP = P0 +But,mp gAFIG. P19.68where m p is the mass of the piston.P0 +Thus,hi =which reduces tomp g= P0Ah01+mp gP0 AFG h IJHh K0i50.0 cm=1+e20.0 kg 9.80 m s 2a= 49.81 cmj1.013 10 5 Pa 0.400 mf2With the man of mass M on the piston, a very similar calculation (replacing m p by m p + M ) gives:h =h0em1+p +MP0 Ajg50.0 cm=1+e95.0 kg 9.80 m s 2aj1.013 10 5 Pa 0.400 m= 49.10 cmf2Thus, when the man steps on the piston, it moves downward byh = hi h = 49.81 cm 49.10 cm = 0.706 cm = 7.06 mm .(b)P = const, so(a)(b)zTidL= dT :LaAhi Ah =TTihi49.81= 293 K= 297 KT = Ti49.10horFG IJH KgivingP19.69V V=T TiTifL f = 1.00 m edT =zLiLiFGHIJKFG IJ = T H KLfdL lnLLia2.00 10 5 C 1 100 CfL f = Li eT= 1.002 002 mafL f = 1.00 m 1 + 2.00 10 5 C 1 100 C = 1.002 000 m :afL f = 1.00 m ea2.00 10 2 C 1 100 Caf(or 24C)L f L fLf= 2.00 10 6 = 2.00 10 4%= 7.389 mfL f = 1.00 m 1 + 0.020 0 C 1 100 C = 3.000 m :L f L fLf= 59. 4%570P19.70TemperatureAt 20.0C, the unstretched lengths of the steel and copper wires area f afa f a20.0 Cf = 1.999 56 mL a 20.0 C f = a 2.000 mf 1 + 17.0 10 aCf a 20.0 C f = 1.999 32 mLs 20.0 C = 2.000 m 1 + 11.0 10 6 C16c1Under a tension F, the length of the steel and copper wires areLMNLs = Ls 1 +FYAOPQLMNLc = Lc 1 +sFYAOPQwhere Ls + L c = 4.000 m .cSince the tension, F, must be the same in each wire, solve for F:F=b L + L g bLscLsYs As+s + LcLcYc A cg.When the wires are stretched, their areas becomee= e1.000 10jmjAs = 1.000 10 3 mAc322eja f1 + e17.0 10 ja 20.0 f1 + 11.0 10 6 20.0262= 3.140 10 6 m 2= 3.139 10 6 m 2Recall Ys = 20.0 10 10 Pa and Yc = 11.0 10 10 Pa . Substituting into the equation for F, we obtainF=b4.000 m 1.999 56 m + 1.999 32 m1.999 56 me20.0 1010jejPa 3.140 10 6 m 2 + 1.999 32 mge11.0 1010jeF = 125 NTo find the x-coordinate of the junction,bLgMM 20.0 10N eLs = 1.999 56 m 1 +125 N10jeN m 2 3.140 10 6Thus the x-coordinate is 2.000 + 1.999 958 = 4.20 10 5 m .OP= 1.999 958 mm jPQ2jPa 3.139 10 6 m 2Chapter 19ej e7.86 10(a) = r 2 = 5.00 10 4 m(b)P19.71f1 =21vTand v =so f1 =2L2L3jkg m3 = 6.17 10 3 kg mTb g = e6.17 10 ja2 0.800 200f2Therefore, T = 2Lf1(c)32= 632 NFirst find the unstressed length of the string at 0C:FGHL = L natural 1 +eA = 5.00 10 4Therefore,IJKmjTLso L natural =+ T AYAY12= 7.854 10 7 m 2 and Y = 20.0 10 10 PaT632== 4.02 10 3 , and710AY7.854 1020.0 10ejejL natural =a0.800 mfe1 + 4.02 10 j3= 0.796 8 m .afThe unstressed length at 30.0C is L30 C = L natural 1 + 30.0 C 0.0 C ,bg eja fLM1 + T OP , where T is the tension in the string at 30.0C,N AY QL L 1OP = e7.854 10 je20.0 10 jLM 0.800 1OP = 580 N .T = AY MNLQN 0.797 06 Qor L30 C = 0.796 8 m 1 + 11.0 10 6 30.0 = 0.797 06 m .Since L = L30 C71030 CTo find the frequency at 30.0C, realize thataf1T=so f1 = 200 Hzf1T*P19.72f580 N= 192 Hz .632 NSome gas will pass through the porous plug from the reaction chamber 1 to the reservoir 2 as thereaction chamber is heated, but the net quantity of gas stays constant according ton i1 + n i 2 = n f 1 + n f 2 .Assuming the gas is ideal, we apply n =PVto each term:RTb gb ga f a f a f a fF 5 IJ = P FG 1 + 4 IJ1 atmGP = 1.12 atmH 300 K K H 673 K 300 K KPf V0Pf 4V0Pi 4V0PV0i+=+300 K R 300 K R673 K R 300 K Rff571572P19.73TemperatureLet 2 represent the angle the curved rail subtends. We haveaLi + L = 2R = Li 1 + Tandsin =Thus,=Li2R=afLi2Rf afLi1 + T = 1 + T sin 2RFIG. P19.73afbgand we must solve the transcendental equation = 1 + T sin = 1.000 005 5 sin Homing in on the non-zero solution gives, to four digits, = 0.018 16 rad = 1.040 5Now,h = R R cos =aLi 1 cos 2 sin fThis yields h = 4.54 m , a remarkably large value compared to L = 5.50 cm .*P19.74(a)Let xL represent the distance of the stationary line below thetop edge of the plate. The normal force on the lower part of theplate is mg 1 x cos and the force of kinetic friction on it is k mg 1 x cos up the roof. Again, k mgx cos acts down theroof on the upper part of the plate. The near-equilibrium of theplate requires Fx = 0a fa fa fmotionf ktf kbxLtemperature rising k mgx cos + k mg 1 x cos mg sin = 0FIG. P19.74(a)2 k mgx cos = mg sin k mg cos 2 k x = k tan x=1 tan 2 2 kand the stationary line is indeed below the top edge by xL =(b)With the temperature falling, the plate contracts faster than theroof. The upper part slides down and feels an upward frictionalforce k mg 1 x cos . The lower part slides up and feelsdownward frictional force k mgx cos . The equation Fx = 0is then the same as in part (a) and the stationary line is aboveLtan the bottom edge by xL =.1k2a fFGH(c)IJKStart thinking about the plate at dawn, as the temperaturestarts to rise. As in part (a), a line at distance xL below the topedge of the plate stays stationary relative to the roof as long asthe temperature rises. The point P on the plate at distance xLabove the bottom edge is destined to become the fixed pointwhen the temperature starts falling. As the temperature rises,this point moves down the roof because of the expansion of thecentral part of the plate. Its displacement for the day iscontinued on next pageIJKFGHtan L1.2kmotionf ktf kbxLtemperature fallingFIG. P19.74(b)xLxLPFIG. P19.74(c)Chapter 19fgaL L F tan I OPbT T g gML 2 G 1 NM 2 H JK QPF L tan IJ bT T g .= b gGH Kb= b573L = 2 1 L xL xL T2121hkkhccAt dawn the next day the point P is farther down the roof by the distance L . It representsthe displacement of every other point on the plate.b(d)12gFGH L tan IJK bT T g = FGH 24 10khc6IJK11 1.20 m tan 18.5 15 10 632 C = 0.275 mmCC0. 42If 2 < 1 , the diagram in part (a) applies to temperature falling and the diagram in part (b)applies to temperature rising. The weight of the plate still pulls it step by step down theroof. The same expression describes how far it moves each day.(e)ANSWERS TO EVEN PROBLEMSP19.2(a) 1.06 atm ; (b) 124 CP19.32(a) 900 K; (b) 1 200 KP19.4(a) 37.0 C = 310 K ; (b) 20.6 C = 253 KP19.34see the solutionP19.6TC = 1.33 C S TS + 20.0 CP19.363.96 10 2 molP19.80.313 mP19.383.67 cm3P19.101.20 cmP19.40between 10 1 kg and 10 2 kgP19.1215.8 mP19.422.41 10 11 moleculesP19.140.663 mm to the right at 78.2 below thehorizontalP19.44(a) 2.24 m; (b) 9.28 10 5 PaP19.460.523 kgP19.48(a) see the solution; (b) << P19.50(a) 0.169 m; (b) 1.35 10 5 PaP19.526.57 MPaP19.54(a) =; (b) see the solution;r(c) it bends the other way; (d) 0.830bg2P19.16(a) 0.109 cm ; (b) increaseP19.18(a) 437C ; (b) 3 000C ; noP19.20(a) 2.52 10 6 N m 2 ; (b) noP19.220.812 cm3P19.24(a) 396 N; (b) 101 C ; (c) no changeP19.26(a) 2.99 mol ; (b) 1.80 10 24 moleculesP19.56(a) increase by 95.0 s ; (b) loses 57.5 sP19.28884 balloonsP19.58P19.30(a) 1.06 10 21 kg ; (b) 56.9 K(a) B = gP0 Vi P0 + gd(c) 10.3 mb2g 1 Li Tbg1up; (b) decrease;574TemperatureP19.60(a) yes; see the solution; (b) 25.7 rad sP19.62y L 2TP19.64(a) see the solution;(b) 3.66 10 3 K 1 , within 0.06% and 0.2%of the experimental valuesP19.66af12(a) 79.1 kPa for N 2 ; 21.2 kPa for O 2 ;940 Pa for Ar; 33.3 Pa for CO 2 ;(b) 81.7 L; 1.22 kg m3 ; (c) 29.0 g molP19.68(a) 7.06 mm; (b) 297 KP19.70125 N ; 42.0 mP19.721.12 atmP19.74(a), (b), (c) see the solution; (d) 0.275 mm;(e) see the solution20Heat and the First Law ofThermodynamicsCHAPTER OUTLINE20.120.2ANSWERS TO QUESTIONSHeat and Internal EnergySpecific Heat andCalorimetryLatent HeatWork and Heat inThermodynamic ProcessesThe First Law ofThermodynamicsSome Applications ofthe First Law ofThermodynamicsEnergy TransferMechanisms20.520.620.7Temperature is a measure of molecular motion. Heat is energyin the process of being transferred between objects by randommolecular collisions. Internal energy is an objects energy ofrandom molecular motion and molecular interaction.Q20.2The T is twice as great in the ethyl alcohol.Q20.3The final equilibrium temperature will show no significantincrease over the initial temperature of the water.Q20.420.320.4Q20.1Some water may boil away. You would have to very preciselymeasure how much, and very quickly measure thetemperature of the steam; it is not necessarily 100C .Q20.5The fingers are wetted to create a layer of steam between the fingers and the molten lead. The steamacts as an insulator and can prevent or delay serious burns. The molten lead demonstration isdangerous, and we do not recommend it.Q20.6Heat is energy being transferred, not energy contained in an object. Further, a large-mass object, oran object made of a material with high specific heat, can contain more internal energy than a highertemperature object.Q20.7There are three properties to consider here: thermal conductivity, specific heat, and mass. With dryaluminum, the thermal conductivity of aluminum is much greater than that of (dry) skin. Thismeans that the internal energy in the aluminum can more readily be transferred to the atmospherethan to your fingers. In essence, your skin acts as a thermal insulator to some degree (pun intended).If the aluminum is wet, it can wet the outer layer of your skin to make it into a good conductor ofheat; then more internal energy from the aluminum can get into you. Further, the water itself, withadditional mass and with a relatively large specific heat compared to aluminum, can be a significantsource of extra energy to burn you. In practical terms, when you let go of a hot, dry piece ofaluminum foil, the heat transfer immediately ends. When you let go of a hot and wet piece ofaluminum foil, the hot water sticks to your skin, continuing the heat transfer, and resulting in moreenergy transfer to you!Q20.8Write 1 000 kg 4 186 J kg C 1 C = V 1.3 kg m3 1 000 J kg C 1 C to find V = 3. 2 10 3 m 3 .bga f ejb575ga f576Heat and the First Law of ThermodynamicsQ20.9The large amount of energy stored in concrete during the day as the sun falls on it is released atnight, resulting in an higher average evening temperature than the countryside. The cool air in thesurrounding countryside exerts a buoyant force on the warmer air in the city, pushing it upwardand moving into the city in the process. Thus, evening breezes tend to blow from country to city.Q20.10If the system is isolated, no energy enters or leaves the system by heat, work, or other transferprocesses. Within the system energy can change from one form to another, but since energy isconserved these transformations cannot affect the total amount of energy. The total energy isconstant.Q20.11(a) and (b) both increase by minuscule amounts.Q20.12The steam locomotive engine is a perfect example of turning internal energy into mechanicalenergy. Liquid water is heated past the point of vaporization. Through a controlled mechanicalprocess, the expanding water vapor is allowed to push a piston. The translational kinetic energy ofthe piston is usually turned into rotational kinetic energy of the drive wheel.Q20.13Yes. If you know the different specific heats of zinc and copper, you can determine the fraction ofeach by heating a known mass of pennies to a specific initial temperature, say 100C , and dumpingthem into a known quantity of water, at say 20C . The final temperature T will reveal the metalcontent:a f aaffmpennies xcCu + 1 x c Zn 100 C T = mH 2O c H 2O T 20 C .Since all quantities are known, except x, the fraction of the penny that is copper will be found byputting in the experimental numbers m pennies , m H 2O , T final , c Zn , and c Cu .afQ20.14The materials used to make the support structure of the roof have a higher thermal conductivitythan the insulated spaces in between. The heat from the barn conducts through the rafters and meltsthe snow.Q20.15The tile is a better thermal conductor than carpet. Thus, energy is conducted away from your feetmore rapidly by the tile than by the carpeted floor.Q20.16The question refers to baking in a conventional oven, not to microwaving. The metal has muchhigher thermal conductivity than the potato. The metal quickly conducts energy from the hot oveninto the center of potato.Q20.17Copper has a higher thermal conductivity than the wood. Heat from the flame is conducted throughthe copper away from the paper, so that the paper need not reach its kindling temperature. Thewood does not conduct the heat away from the paper as readily as the copper, so the energy in thepaper can increase enough to make it ignite.Q20.18In winter the interior of the house is warmer than the air outside. On a summer day we want theinterior to stay cooler than the exterior. Heavy draperies over the windows can slow down energytransfer by conduction, by convection, and by radiation, to make it easier to maintain the desireddifference in temperature.Q20.19You must allow time for the flow of energy into the center of the piece of meat. To avoid burning theoutside, the meat should be relatively far from the flame. If the outer layer does char, the carbon willslow subsequent energy flow to the interior.Chapter 20577Q20.20At night, the Styrofoam beads would decrease the overall thermal conductivity of the windows, andthus decrease the amount of heat conducted from inside to outside. The air pockets in the Styrofoamare an efficient insulator. During the winter day, the influx of sunlight coming through the windowwarms the living space.An interesting asidethe majority of the energy that goes into warming a home fromsunlight through a window is not the infrared light given off by the sun. Glass is a relatively goodinsulator of infrared. If not, the window on your cooking oven might as well be just an open hole!Glass is opaque to a large portion of the ultraviolet range. The glass molecules absorb ultravioletlight from the sun and re-emit the energy in the infrared region. It is this re-emitted infraredradiation that contributes to warming your home, along with visible light.Q20.21In winter the produce is protected from freezing. The heat capacity of the earth is so high that soilfreezes only to a depth of a few decimeters in temperate regions. Throughout the year thetemperature will stay nearly constant all day and night. Factors to be considered are the insulatingproperties of soil, the absence of a path for energy to be radiated away from or to the vegetables, andthe hindrance to the formation of convection currents in the small, enclosed space.Q20.22The high mass and specific heat of the barrel of water and its high heat of fusion mean that a largeamount of energy would have to leak out of the cellar before the water and the produce froze solid.Evaporation of the water keeps the relative humidity high to protect foodstuffs from drying out.Q20.23The sunlight hitting the peaks warms the air immediately around them. This air, which is slightlywarmer and less dense than the surrounding air, rises, as it is buoyed up by cooler air from thevalley below. The air from the valley flows up toward the sunny peaks, creating the morning breeze.Q20.24Sunlight hits the earth and warms the air immediately above it. This warm, less-dense air rises,creating an up-draft. Many raptors, like eagles, hawks and falcons use updrafts to aid in hunting.These birds can often be seen flying without flapping their wingsjust sitting in an updraft withwings extended.Q20.25The bit of water immediately over the flame warms up and expands. It is buoyed up and risesthrough the rest of the water. Colder, more dense water flows in to take its place. Convectioncurrents are set up. This effectively warms the bulk of the water all at once, much more rapidly thanit would be by heat being conducted through the water from the flame.Q20.26The porcelain of the teacup is a thermal insulator. That is, it is a thermal conductor of relatively lowconductivity. When you wrap your hands around a cup of hot tea, you make A large and L small inT Tcfor the rate of energy transfer by heat from tea into you. When you holdthe equation P = kA hLthe cup by the handle, you make the rate of energy transfer much smaller by reducing A andincreasing L. The air around the cup handle will also reduce the temperature where you aretouching it. A paper cup can be fitted into a tubular jacket of corrugated cardboard, with thechannels running vertically, for remarkably effective insulation, according to the same principles.Q20.27As described in the answer to question 20.25, convection currents in the water serve to bring more ofthe heat into the water from the paper cup than the specific heats and thermal conductivities ofpaper and water would suggest. Since the boiling point of water is far lower than the kindlingtemperature of the cup, the extra energy goes into boiling the water.Q20.28Keep them dry. The air pockets in the pad conduct energy by heat, but only slowly. Wet pads wouldabsorb some energy in warming up themselves, but the pot would still be hot and the water wouldquickly conduct and convect a lot of energy right into you.578Heat and the First Law of ThermodynamicsQ20.29The person should add the cream immediately when the coffee is poured. Then the smallertemperature difference between coffee and environment will reduce the rate of energy loss duringthe several minutes.Q20.30The cup without the spoon will be warmer. Heat is conducted from the coffee up through the metal.The energy then radiates and convects into the atmosphere.Q20.31Convection. The bridge deck loses energy rapidly to the air both above it and below it.Q20.32The marshmallow has very small mass compared to the saliva in the teachers mouth and thesurrounding tissues. Mostly air and sugar, the marshmallow also has a low specific heat compared toliving matter. Then the marshmallow can zoom up through a large temperature change whilecausing only a small temperature drop of the teachers mouth. The marshmallow is a foam withclosed cells and it carries very little liquid nitrogen into the mouth. The liquid nitrogen still on themarshmallow comes in contact with the much hotter saliva and immediately boils into cold gaseousnitrogen. This nitrogen gas has very low thermal conductivity. It creates an insulating thermalbarrier between the marshmallow and the teachers mouth (the Leydenfrost effect). A similar effectcan be seen when water droplets are put on a hot skillet. Each one dances around as it slowlyshrinks, because it is levitated on a thin film of steam. The most extreme demonstration of this effectis pouring liquid nitrogen into ones mouth and blowing out a plume of nitrogen gas. We stronglyrecommended that you read of Jearl Walkers adventures with this demonstration rather than tryingit.Q20.33(a)Warm a pot of coffee on a hot stove.(b)Place an ice cube at 0C in warm waterthe ice will absorb energy while melting, but notincrease in temperature.(c)Let a high-pressure gas at room temperature slowly expand by pushing on a piston. Workcomes out of the gas in a constant-temperature expansion as the same quantity of heat flowsin from the surroundings.(d)Warm your hands by rubbing them together. Heat your tepid coffee in a microwave oven.Energy input by work, by electromagnetic radiation, or by other means, can all alikeproduce a temperature increase.(e)Davys experiment is an example of this process.(f)This is not necessarily true. Consider some supercooled liquid water, unstable but withtemperature below 0C . Drop in a snowflake or a grain of dust to trigger its freezing intoice, and the loss of internal energy measured by its latent heat of fusion can actually push itstemperature up.Q20.34Heat is conducted from the warm oil to the pipe that carries it. That heat is then conducted to thecooling fins and up through the solid material of the fins. The energy then radiates off in alldirections and is efficiently carried away by convection into the air. The ground below is left frozen.Chapter 20579SOLUTIONS TO PROBLEMSSection 20.1P20.1Heat and Internal EnergyTaking m = 1.00 kg , we havebjagefU g = mgh = 1.00 kg 9.80 m s 2 50.0 m = 490 J .ButT f = Ti + T =P20.2bgba10.0 + 0.117f CgU g = Q = mcT = 1.00 kg 4 186 J kg C T = 490 J so T = 0.117 CThe container is thermally insulated, so no energy flows by heat:Q=0Eint = Q + Winput = 0 + Winput = 2mghandThe work on the falling weights is equal to the work done on thewater in the container by the rotating blades. This work results inan increase in internal energy of the water:2mgh = Eint = m water cTT =ejaf2 1.50 kg 9.80 m s 2 3.00 m2mgh88.2 J==837 J Cm water c0.200 kg 4 186 J kg Cbg= 0.105 CSection 20.2P20.3FIG. P20.2Specific Heat and CalorimetryQ = mc silver Tbag1.23 kJ = 0.525 kg c silver 10.0 Cfc silver = 0.234 kJ kg CFromQ = mcTwe findP20.4T =1 200 JQ== 62.0 Cmc 0.050 0 kg 387 J kg CbgThus, the final temperature is 87.0C .*P20.5We imagine the stone energy reservoir has a large area in contact with air and is always at nearly thesame temperature as the air. Its overnight loss of energy is described byQ mcT=tt6 000 J s 14 h 3 600 s h3.02 10 8 J kg CPtm==== 1.78 10 4 kgc T850 J 20 C850 J kg C 18 C 38 CP=bbga fbgagfaf580*P20.6Heat and the First Law of ThermodynamicsThe laser energy output:ejPt = 1.60 10 13 J s 2.50 10 9 s = 4.00 10 4 J .The teakettle input:bgQ = mcT = 0.800 kg 4 186 J kg C 80 C = 2.68 10 5 J .This is larger by 6.70 times.P20.7Qcold = Q hotamcT ff20.0 kg b 4 186 J kg C gdTwatera= mcTironfi bgbgd 25.0 C = 1.50 kg 448 J kg C T f 600 CiT f = 29.6 CP20.8Let us find the energy transferred in one minute.Q = m cup c cup + m water c water Tbgbg bgbQ = 0. 200 kg 900 J kg C + 0.800 kg 4 186 J kg Cg a1.50 Cf = 5 290 JIf this much energy is removed from the system each minute, the rate of removal isP=P20.9(a)Qt=5 290 J= 88.2 J s = 88.2 W .60.0 sQcold = Q hotbmwcwgdidid+ m c c c T f Tc = mCu cCu T f TCu munk c unk T f Tunkiwhere w is for water, c the calorimeter, Cu the copper sample, and unk the unknown.g a20.0 10.0f C= b50.0 g gb0.092 4 cal g C ga 20.0 80.0f C b70.0 g gc a 20.0 100 f C2.44 10 cal = e5.60 10 g C jcbgb250 g 1.00 cal g C + 100 g 0.215 cal g Cunk33unkor c unk = 0.435 cal g C .(b)The material of the sample is beryllium .Chapter 20P20.10b f gbmghg = mcTa0.600fe3.00 10(a)3jaje581f = b3.00 g gb0.092 4 cal g CgaT fkg 9.80 m s 2 50.0 m4.186 J calT = 0.760 C ; T = 25.8 CNo . Both the change in potential energy and the heat absorbed are proportional to the(b)mass; hence, the mass cancels in the energy relation.*P20.11We do not know whether the aluminum will rise or drop in temperature. The energy the water canJ6 C = 6 279 J . The energy the copper can putabsorb in rising to 26C is mcT = 0.25 kg 4 186kg CJ74 C = 2 864 J . Since 6 279 J > 2 864 J , the finalout in dropping to 26C is mcT = 0.1 kg 387kg Ctemperature is less than 26C . We can write Q h = Q c asQ water + Q Al + QCu = 00.25 kg 4 186JJT f 20 C + 0.4 kg 900T f 26 Ckg Ckg Cd+0.1 kg 387idiJT f 100 C = 0kg Cdi1 046.5T f 20 930 C + 360T f 9 360 C + 38.7T f 3 870 C = 01 445.2T f = 34 160 CT f = 23.6 CP20.12Qcold = Q hotdid id ibm c + m c gT bm c + m c gT = m c T + m cbm c + m c + m c gT = bm c + m c gT + m c Tbm c + m c gT + m c TT =m Al c Al T f Tc + m c c w T f Tc = m h c w T f ThAl Alc wAl Alc wfP20.13Al AlfAl Alc wfAl Alh wc wcch w Thh w fc wch w hh w hm Al c Al + m c c w + m h c wejThe rate of collection of energy is P = 550 W m 2 6.00 m 2 = 3 300 W . The amount of energyrequired to raise the temperature of 1 000 kg of water by 40.0C is:bgafQ = mcT = 1 000 kg 4 186 J kg C 40.0 C = 1.67 10 8 JThus,Pt = 1.67 10 8 Jort =1.67 10 8 J= 50.7 ks = 14.1 h .3 300 W582*P20.14Heat and the First Law of ThermodynamicsVessel one contains oxygen according to PV = nRT :nc =ej353PV 1.75 1.013 10 Pa 16.8 10 m== 1.194 mol .8.314 Nm mol K 300 KRTVessel two contains this much oxygen:nh =(a)ej2. 25 1.013 10 5 22.4 10 3a f8.314 450mol = 1.365 mol .The gas comes to an equilibrium temperature according toamcT fn McdTcoldcfa= mcTifhotdi 300 K + n h Mc T f 450 K = 0The molar mass M and specific heat divide out:1.194T f 358.2 K + 1.365T f 614.1 K = 0Tf =(b)The pressure of the whole sample in its final state isP=Section 20.3P20.15972.3 K= 380 K2.5592.559 mol 8.314 J 380 KnRT== 2.06 10 5 Pa = 2.04 atm .Vmol K 22.4 + 16.8 10 3 m3afLatent HeatThe heat needed is the sum of the following terms:fbg a+b heat to reach melting point g + b heat to vaporizeg + a heat to reach 110 C fQ needed = heat to reach melting point + heat to meltThus, we havega f ej+b 4 186 J kg C ga100 C f + e 2.26 10 J kg j + b 2 010 J kg C ga10.0 C fbQ needed = 0.040 0 kg 2 090 J kg C 10.0 C + 3.33 10 5 J kg6Q needed = 1.22 10 5 JP20.16Qcold = Q hotbmwcwgdid+ m c c c T f Ti = m s L v + c w T f 100bgifgaJ kg + b 4 186 J kg C ga50.0 C 100 C fb0.250 kg 4 186 J kg C + 0.050 0 kg 387 J kg C 50.0 C 20.0 C= m s 2.26 10 6ms =3.20 10 4 J= 0.012 9 kg = 12.9 g steam2.47 10 6 J kgChapter 20P20.17583The bullet will not melt all the ice, so its final temperature is 0C.ThenFG 1 mvH22+ mc TIJKbullet= mwL fwhere m w is the melt water massmw =mw =P20.18(a)jbe0.500 3.00 10 3 kg 240 m sg2bga+ 3.00 10 3 kg 128 J kg C 30.0 C3.33 105fJ kg86.4 J + 11.5 J= 0.294 g333 000 J kgejejQ1 = heat to melt all the ice = 50.0 10 3 kg 3.33 10 5 J kg = 1.67 10 4 Jb= e50.0 10Q 2 = heat to raise temp of ice to 100 C3jbgagfkg 4 186 J kg C 100 C = 2.09 10 4 JThus, the total heat to melt ice and raise temp to 100C = 3.76 10 4 JQ3 =heat available= 10.0 10 3 kg 2.26 10 6 J kg = 2.26 10 4 Jas steam condensesejejThus, we see that Q3 > Q1 , but Q3 < Q1 + Q 2 .Therefore, all the ice melts but T f < 100 C . Let us now find T fQcold = Q hote50.0 10 kg je3.33 10 J kg j + e50.0 10 kg jb4 186 J kg CgdT 0 Ci= e10.0 10 kg je 2.26 10 J kg j e10.0 10 kg jb 4 186 J kg C gdT 100 C i3353f36fFrom which, T f = 40.4 C .(b)Q1 = heat to melt all ice = 1.67 10 4 J [See part (a)]heat given up= 10 3 kg 2.26 10 6 J kg = 2.26 10 3 JQ2 =as steam condenseseQ3jeheat given up as condensed== e10steam cools to 0 Cj3jbgafkg 4 186 J kg C 100 C = 419 JNote that Q 2 + Q3 < Q1 . Therefore, the final temperature will be 0C with some iceremaining. Let us find the mass of ice which must melt to condense the steam and cool thecondensate to 0C.mL f = Q 2 + Q3 = 2.68 10 3 JThus, m =2.68 10 3 J= 8.04 10 3 kg = 8.04 g .53.33 10 J kgTherefore, there is 42.0 g of ice left over .584P20.19Heat and the First Law of Thermodynamicse j1.00 kg b0.092 0 cal g C ga 293 77.3f C = mb 48.0 cal g gQ = mCu cCu T = m N 2 L vapN2m = 0.414 kg*P20.20The original gravitational energy of the hailstone-Earth system changes entirely into additionalinternal energy in the hailstone, to produce its phase change. No temperature change occurs, eitherin the hailstone, in the air, or in sidewalk. Thenmgy = mLy=P20.21FGHIJKL 3.33 10 5 J kg 1 kg m 2 s 2== 3.40 10 4 m1Jg9.8 m s 2(a)Since the heat required to melt 250 g of ice at 0C exceeds the heat required to cool 600 g ofwater from 18C to 0C, the final temperature of the system (water + ice) must be 0C .(b)Let m represent the mass of ice that melts before the system reaches equilibrium at 0C.Qcold = Q hotbgJ kg j = b0.600 kg gb 4 186 J kg C ga0 C 18.0 C fmL f = m w c w 0 C Tiem 3.33 105m = 136 g, so the ice remaining = 250 g 136 g = 114 gP20.22The original kinetic energy all becomes thermal energy:FG IJ eH Kjb1115.00 10 3 kg 500 m smv 2 + mv 2 = 2222g2= 1.25 kJ .Raising the temperature to the melting point requiresbfgaQ = mcT = 10.0 10 3 kg 128 J kg C 327 C 20.0 C = 393 J .Since 1 250 J > 393 J , the lead starts to melt. Melting it all requiresejejQ = mL = 10.0 10 3 kg 2.45 10 4 J kg = 245 J .Since 1 250 J > 393 + 245 J , it all melts. If we assume liquid lead has the same specific heat as solidlead, the final temperature is given bybgd1.25 10 3 J = 393 J + 245 J + 10.0 10 3 kg 128 J kg C T f 327 CT f = 805 CiChapter 20Section 20.4Work and Heat in Thermodynamic ProcesseszfP20.23PWif = PdVifThe work done on the gas is the negative of the area under thecurve P = V 2 between Vi and V f .zf1Wif = V 2 dV = V f3 Vi33ieeP = V 2ijOjV f = 2Vi = 2 1.00 m 3 = 2.00 m 3j LNMe2.00 m j + e1.00 m j OQP =15.00 atm m6 1.013 10 5 Pa atmWif = 3eP20.24585jez3 33 3V31.00 m32.00 mFIG. P20.231.18 MJW = PdV(a)feja e 4.00 10 Pa ja3.00 2.00f m e 2.00 10 Paja 4.00 3.00f mW = 6.00 10 6 Pa 2.00 1.00 m3 +6363+Wi f = 12.0 MJW f i = +12.0 MJ(b)FIG. P20.24FG nR IJ dTH PKaifaP20.25W = P V = PP20.26W = PdV = P dV = PV = nRT = nR T2 T1zzffiP20.27fa f Ti = nRT = 0.200 8.314 280 = 466 JfbiDuring the heating process P =gFG P IJ V .HV Kiizf(a)W = PdV = iW =Vi2 3ViFG P IJ VHV K 2ii(b)z FGH3ViViIJKPiVdVVi=Pi9Vi2 Vi2 = 4PVii2ViejPV = nRTLMF P I V OPV = nRTMNGH V JK PQF P IJ VT =GH nRV Kii2iiTemperature must be proportional to the square of volume, rising to nine times its originalvalue.586Heat and the First Law of ThermodynamicsSection 20.5The First Law of ThermodynamicsP20.28(a)W = PV = 0.800 atm 7.00 L 1.013 10 5 Pa atm 10 3 m 3 L = +567 J(b)Eint = Q + W = 400 J + 567 J = 167 JP20.29afafejejEint = Q + WQ = Eint W = 500 J 220 J = 720 JThe negative sign indicates that positive energy is transferred from the system by heat.P20.30(a)Q = W = Area of triangleQ=(b)jaf14.00 m 3 6.00 kPa = 12.0 kJ2eQ = W = 12.0 kJFIG. P20.30P20.31BCCAABP20.32WBCQW0Eint+bQ = E since W = 0gbE < 0 and W > 0 , so Q < 0g++bW < 0 , E > 0 since E < 0 for B C A ; so Q > 0g= P bV V g = 3.00 atmb0.400 0.090 0 g mintBCintintBC3BP(atm)= 94.2 kJEint = Q + Wint3.0afEint, C Eint, B = 100 94.2 kJEint, C Eint, B = 5.79 kJ1.0Since T is constant,BCA0.090 0.20D0.40Eint, D Eint, C = 0bagFIG. P20.32fWDA = PD VA VD = 1.00 atm 0.200 1.20 m3= +101 kJafEint, A Eint, D = 150 kJ + +101 kJ = 48.7 kJdi di dNow, Eint, B Eint, A = Eint, C Eint, B + Eint, D Eint, C + Eint, A Eint, DEint, B Eint, A = 5.79 kJ + 0 48.7 kJ = 42.9 kJi1.2V(m 3)587Chapter 20*P20.331 2r . The arrow in Figure P20.332looks like a semicircle when the scale makes 1.2 L fill the samespace as 100 kPa. Its area isThe area of a true semicircle isafafje50030011 2.4 L 200 kPa = 2.4 10 3 m3 2 10 5 N m 2 .22eP(kPa)jThe work on the gas is0zA1.2B3.6BW = PdV = area under the arch shown in the graphAFG 1 2.4a200f J + 3 10 N mH2= b754 J + 1 440 Jg = 2 190 J5=24.8 10 3 m3FIG. P20.33IJKEint = Q + W = 5 790 J 2 190 J = 3.60 kJSection 20.6P20.34(a)Some Applications of the First Law of ThermodynamicsFG V IJ = P V lnFG V IJHV KHV KF W I = b0.025 0g expLM 3 000 OP =so V = V expG +MN 0.025 0e1.013 10 j PQH P V JKfW = nRT lnfiifPf V ffi5fej=g1.013 10 5 Pa 0.025 0 m 3(b)P20.35Tf =(a)(b)b1.00 mol 8.314 J K mol(a)305 KafV1 V2=T1 T2V3.00300 K = 900 KT2 = 2 T1 =1.00V1aP20.360.007 65 m 3Eint = Q PV = 12.5 kJ 2.50 kPa 3.00 1.00 m 3 = 7.50 kJnR=ffW = PV = P 3VTefLjMM eN= 1.013 10 5 N m 2 3 24.0 10 6 C 1OF .00IjGH 2.70 110 kg m JK a18.0 CfPPkgQW = 48.6 mJbgbgaf(b)Q = cmT = 900 J kg C 1.00 kg 18.0 C = 16.2 kJ(c)Eint = Q + W = 16. 2 kJ 48.6 mJ = 16.2 kJ336.0V (L)588P20.37Heat and the First Law of ThermodynamicsOPa f LMMN ejej PQF 18.0 g I =W = a1.00 molfb8.314 J K molga373 K f + e1.013 10 N m jGH 10 g m JKQ = mL = 0.018 0 kg e 2.26 10 J kg j = 40.7 kJbgW = PV = P Vs Vw = P nRT18.0 g+PP1.00 g cm 3 10 6 cm 3 m352633.10 kJ6vEint = Q + W = 37.6 kJP20.38(a)The work done during each step of the cycle equals thenegative of the area under that segment of the PV curve.W = WDA + W AB + WBC + WCDbgbgW = Pi Vi 3Vi + 0 3 Pi 3Vi Vi + 0 = 4PVii(b)(c)P20.39The initial and final values of T for the system are equal.Therefore, Eint = 0 and Q = W = 4PVi .iW = 4PVi = 4nRTi = 4 1.00 8.314 273 = 9.08 kJi(a)PVi = Pf V f = nRT = 2.00 mol 8.314 J K mol 300 K = 4.99 10 3 Jia fafa fbganRT 4.99 10 J=0.400 atmPiVf =(c)f3Vi =(b)FIG. P20.38nRT 4.99 10 3 J 1== Vi = 0.041 0 m31.20 atmPf3FG V IJ = e4.99 10 j lnFG 1 IJ =H 3KHV KzW = PdV = nRT lnf3+5.48 kJiEint = 0 = Q + WQ = 5.48 kJP20.40Eint, ABC = Eint, AC(a)(conservation of energy)Eint, ABC = Q ABC + WABC(First Law)Q ABC = 800 J + 500 J = 1 300 J(b)WCD = PC VCD , VAB = VCD , and PA = 5 PC11Then, WCD = PA VAB = W AB = 100 J55(+ means that work is done on the system)(c)WCDA = WCD so that QCA = Eint, CA WCDA = 800 J 100 J = 900 J( means that energy must be removed from the system by heat)(d)Eint, CD = Eint, CDA Eint, DA = 800 J 500 J = 1 300 Jand QCD = Eint, CD WCD = 1 300 J 100 J = 1 400 JFIG. P20.40Chapter 20Section 20.7P20.41Energy Transfer MechanismsTLP L 10.0 W 0.040 0 mk=== 2.22 10 2 W m CAT1.20 m 2 15.0 CP = kAbgaP20.42fkAT b0.800 W m C ge3.00 m ja 25.0 C fP=== 1.00 10P20.43In the steady state condition,PAu = PAgso thatk Au A AuIn this caseA Au = A Ag26.00 10 3 mLFG T IJH x KAu4= k Ag A AgW = 10.0 kWFG T IJH x KAgx Au = x Agaf= aT 30.0fTAu = 80.0 TTAgandFIG. P20.43where T is the temperature of the junction.Therefore,k Au 80.0 T = k Ag T 30.0aP=ATkLii*P20.45fe6.00 m ja50.0 Cf2=iej2 4.00 10 3 m0.800 W m C + 5.00 10 3 m0.023 4 W m C= 1.34 kWWe suppose that the area of the transistor is so small that energy flow by heat from the transistordirectly to the air is negligible compared to energy conduction through the mica.P = kAbT T ghcLTh = Tc +P20.46aT = 51.2 CAndP20.44fej1.50 W 0.085 2 10 3 mPL= 35.0 C += 67.9 CkA0.075 3 W m C 8.25 6.25 10 6 m 2bgafFrom Table 20.4,(a)R = 0.890 ft 2 F h Btu(b)The insulating glass in the table must have sheets of glass less thanestimate the R-value of a 0.250-inch air space asThen for the double glazingLMNRb = 0.890 +(c)bg1inch thick. So we80.250times that of the thicker air space.3.50FG 0.250 IJ 1.01 + 0.890OP ft F h =H 3.50 KQ Btu21.85ft 2 F h.BtuSince A and T2 T1 are constants, heat flow is reduced by a factor of1.85= 2.08 .0.890589590P20.47Heat and the First Law of ThermodynamicsjLNM eej OQPa0.965fb5 800 K g2P = AeT 4 = 5.669 6 10 8 W m 2 K 4 4 6.96 10 8 m4P = 3.77 10 26 WP20.48Suppose the pizza is 70 cm in diameter and = 2.0 cm thick, sizzling at 100C. It cannot lose heat byconduction or convection. It radiates according to P = AeT 4 . Here, A is its surface area,afA = 2 r 2 + 2 r = 2 0.35 m2afaf+ 2 0.35 m 0.02 m = 0.81 m 2 .Suppose it is dark in the infrared, with emissivity about 0.8. Theneja fajeP = 5.67 10 8 W m 2 K 4 0.81 m 2 0.80 373 Kf4= 710 W ~ 10 3 W .If the density of the pizza is half that of water, its mass isjaef a0.02 mf = 4 kg .m = V = r 2 = 500 kg m3 0.35 m2Suppose its specific heat is c = 0.6 cal g C . The drop in temperature of the pizza is described by:dQ = mc T f TiP=dT fdtP20.49dT fdQ= mc0dtdt=710 J sP== 0.07 C s ~ 10 1 K smc4 kg 0.6 4 186 J kg Cb gbgP = AeT 42.00 W = 5.67 10 8 W m 2 K 4 0.250 10 6 m 2 0.950 T 4ejajeeT = 1.49 10 14 K 4P20.50ij14f= 3.49 10 3 KWe suppose the earth below is an insulator. The square meter must radiate in the infrared as muchenergy as it absorbs, P = AeT 4 . Assuming that e = 1.00 for blackbody blacktop:eja fje1 000 W = 5.67 10 8 W m 2 K 4 1.00 m 2 1.00 T 4eT = 1.76 10 10 KP20.51j4 14= 364 K (You can cook an egg on it.)The sphere of radius R absorbs sunlight over the area of its day hemisphere, projected as a flat circleperpendicular to the light: R 2 . It radiates in all directions, over area 4 R 2 . Then, in steady state,Pin = Poutejeje 1 340 W m 2 R 2 = e 4 R 2 T 4The emissivity e, the radius R, and all cancel.L 1 340 W mTherefore, T = MMN 4e5.67 10 W m282OPK jPQ414= 277 K = 4 C .Chapter 20591Additional ProblemsP20.5277.3 K = 195.8C is the boiling point of nitrogen. It gains no heat to warm as a liquid, but gains heatto vaporize:bgejQ = mL v = 0.100 kg 2.01 10 5 J kg = 2.01 10 4 J .The water first loses heat by cooling. Before it starts to freeze, it can losebgbgafQ = mcT = 0.200 kg 4 186 J kg C 5.00 C = 4.19 10 3 J .ejThe remaining 2.01 10 4 4.19 10 3 J = 1.59 10 4 J that is removed from the water can freeze amass x of water:Q = mL fe1.59 10 4 J = x 3.33 10 5 J kgjx = 0.047 7 kg = 47.7 g of water can be frozenP20.53The increase in internal energy required to melt 1.00 kg of snow isbgejEint = 1.00 kg 3.33 10 5 J kg = 3.33 10 5 JThe force of friction isbgejf = n = mg = 0. 200 75.0 kg 9.80 m s 2 = 147 NAccording to the problem statement, the loss of mechanical energy of the skier is assumed to beequal to the increase in internal energy of the snow. This increase in internal energy isafEint = fr = 147 N r = 3.33 10 5 Jr = 2.27 10 3 m .andP20.54(a)The energy thus far gained by the copper equals the energy loss by the silver. Your downparka is an excellent insulator.Qcold = Q hotord i = m c dT T ib9.00 g gb387 J kg Cga16.0 Cf = b14.0 g gb234 J kg CgdTdT 30.0 Ci = 17.0 CmCu c Cu T f Tifso(b)Ag AgCufi Agf 30.0 CiFG dT IJH dt K= mCu c CuAgT f , Ag = 13.0 C .Differentiating the energy gain-and-loss equation gives: m Ag c AgFG dT IJH dt KFG dT IJH dt KAgFG IJ = 9.00 gb387 J kg Cg b+0.500 C sgH K 14.0 gb234 J kg Cg0.532 C s b negative sign decreasing temperatureg=Ag=AgmCu c Cu dTm Ag c Ag dtCuAgFG dT IJH dt KCu592P20.55Heat and the First Law of Thermodynamics(a)Before conduction has time to become important, the energy lost by the rod equals theenergy gained by the helium. Therefore,mL v He = mc T Alorsob g c hbVL g = cVc T hcVc T hV =bL ge2.70 g cm je62.5 cm jb0.210 cal g Cga295.8 CfV =e0.125 g cm je2.09 10 J kg jb1.00 cal 4.186 Jgb1.00 kg 1 000 g gv HeAlAlHev He3He334VHe = 1.68 10 4 cm3 = 16.8 liters(b)The rate at which energy is supplied to the rod in order to maintain constant temperaturesis given bydT295.8 KP = kA= 31.0 J s cm K 2.50 cm 2= 917 Wdx25.0 cmThis power supplied to the helium will produce a boil-off rate ofFG IJ bH KjFGHgeafejejIJK917 W 10 3 g kgP== 351 cm 3 s = 0.351 L sL v0.125 g cm3 2.09 10 4 J kge*P20.56jAt the equilibrium temperature Teq the diameters of the sphere and ring are equal:ej5.01 cm + 5.01 cme 2.40 10ej1 C jeT T j = 5.00 cm + 5.00 cme1.70 10d s + d s Al Teq Ti = d r + d r Cu Teq 15 C5eqi5je1 C Teq 15 C0.01 C + 1.202 4 10 4 Teq 1.202 4 10 4 Ti = 8.5 10 5 Teq 1.275 10 3 C1.127 5 10 2 C + 3.524 10 5 Teq = 1.202 4 10 4 Ti319.95 C + Teq = 3.412 0TiAt the equilibrium temperature, the energy lost is equal to the energy gained:ejem s c Al Teq Ti = m r cCu Teq 15 Cejje10.9 g 0.215 cal g C Teq Ti = 25 g 0.092 4 cal g C Teq 15 C2.343 5Teq 2.343 5Ti = 34.65 C 2.31Teq4.653 5Teq = 34.65 C + 2.343 5TiSolving by substitution,bg4.653 5 3.412 0Ti 319.95 C = 34.65 C + 2.343 5Ti15.877 7Ti 1 488.89 C = 34.65 C + 2.343 5Ti1 523.54 C= 113 C13.534(b)Ti =(a)Teq = 319.95 + 3.412 0 112.57 = 64.1 CafjjChapter 20P20.57b gQ = mcT = V cT so that when a constant temperature difference T is maintained,the rate of adding energy to the liquid is P =P.RTand the specific heat of the liquid is c =P20.58(a)Work done by the gas is the negative of the area under thePV curveW = Pi(b)FG IJH KdQdV=cT = RcTdtdtFG V V IJ =H2 Kii+PVii.2zIn this case the area under the curve is W = PdV . Since theprocess is isothermal,FG V IJ = nRTH 4KF dV IJ bPV g = PV lnFG V 4 IJ = PV ln 4W = z GHVKHV KPV = PVi = 4PiiiiVi 4andi ii iiViiFIG. P20.58i i= +1.39 PVii(c)P20.59The area under the curve is 0 and W = 0 .Call the initial pressure P1 . In the constant volume process 1 2 the work is zero.P1V1 = nRT1P2 V2 = nRT2soFGHIJ a fKP2 V2 T21=; T2 = 300 K1 = 75.0 K4P1V1 T1Now in 2 3z3bgW = PdV = P2 V3 V2 = P3V3 + P2V22afbgaW = nRT3 + nRT2 = 1.00 mol 8.314 J mol K 300 K 75.0 KW = 1.87 kJf593594*P20.60Heat and the First Law of ThermodynamicsThe initial moment of inertia of the disk isjaf11114MR 2 = VR 2 = R 2 tR 2 = 8 920 kg m3 28 m 1.2 m = 1.033 10 10 kg m 22222eThe rotation speeds up as the disk cools off, according toI i i = I f f2111MRi2 i = MR 2 f = MRi2 1 T ff22211 f = i= 25 rad s21 T1 17 10 6 1 C 830 Ccc(a)hhej2= 25.720 7 rad sThe kinetic energy increases by11111I f 2 I i i2 = I i i f I i i2 = I i i f if222221102= 1.033 10 kg m 25 rad s 0.720 7 rad s = 9.31 10 10 J2dibbgfga(b)Eint = mcT = 2.64 10 7 kg 387 J kg C 20 C 850 C = 8.47 10 12 J(c)As 8.47 10 12 J leaves the fund of internal energy, 9.31 10 10 J changes into extra kineticenergy, and the rest, 8.38 10 12 J is radiated.*P20.61The loss of mechanical energy isGM E m 11= 670 kg 1.4 10 4 m smvi2 +RE22ej2+6.67 10 11 Nm 2 5.98 10 24 kg 670 kgkg 2 6.37 10 6 m= 6.57 10 10 J + 4.20 10 10 J = 1.08 10 11 JOne half becomes extra internal energy in the aluminum: Eint = 5.38 10 10 J. To raise itstemperature to the melting point requires energymcT = 670 kg 900afhJ660 15 C = 4.07 10 8 J .kg CcTo melt it, mL = 670 kg 3.97 10 5 J kg = 2.66 10 8 J . To raise it to the boiling point,bgbgmcT = 670 1 170 2 450 600 J = 1.40 10 9 J . To boil it, mL = 670 kg 1.14 10 7 J kg = 7.64 10 9 J .Thenbgdi5.38 10 10 J = 9.71 10 9 J + 670 1 170 T f 2 450 C J CT f = 5.87 10 4 CChapter 20afb595g(a)Fv = 50.0 N 40.0 m s = 2 000 W(b)P20.62Energy received by each object is 1 000 W 10 s = 10 4 J = 2 389 cal . The specific heat of ironga fbis 0.107 cal g C , so the heat capacity of each object is 5.00 10 3 0.107 = 535.0 cal C.T =P20.632 389 cal= 4.47 C535.0 cal CThe power incident on the solar collector isj aefPi = IA = 600 W m 2 0.300 m2= 170 W .For a 40.0% reflector, the collected power is Pc = 67.9 W. The totalenergy required to increase the temperature of the water to theboiling point and to evaporate it is Q = cmT + mLV :bgafQ = 0.500 kg 4 186 J kg C 80.0 C + 2.26 10 6 J kg = 1.30 10 6 J .The time interval required is t =P20.64QPc=1.30 10 6 J= 5.31 h .67.9 WFIG. P20.63From Q = mLV the rate of boiling is described byP=Q LV m=ttPm=t LVModel the water vapor as an ideal gasFG m IJ RTH MKP V m F RT I=G Jtt H M KP F RT IP Av =G JL HMKP0 V0 = nRT =00v=Vbgaf1 000 W 8.314 J mol K 373 KP RT=MLV P0 A0.018 0 kg mol 2.26 10 6 J kg 1.013 10 5 N m 2 2.00 10 4 m 2v = 3.76 m sbgejejej596P20.65Heat and the First Law of ThermodynamicsEnergy goes in at a constant rate P . For the period from50.0 min to 60.0 min, Q = mcTaaf bfgbT( C)fgaP 10.0 min = 10 kg + mi 4 186 J kg C 2.00 C 0 C(1)P 10.0 min = 83.7 kJ + 8.37 kJ kg mibgafeSubstitute P =emi 3.33 10 5 J kg50.0 minemi 3.33 10 5 J kgj3.001.00For the period from 0 to 50.0 min, Q = mi L fP 50.0 min = mi 3.33 10 5 J kg2.000.00j20.0j into Equation (1) to findb40.060.0 t (min)FIG. P20.65g= 83.7 kJ + 8.37 kJ kg mi5.0083.7 kJ= 1.44 kgmi =66.6 8.37 kJ kgaP20.66(a)fThe block starts with K i =bgb11mvi2 = 1.60 kg 2.50 m s22g2= 5.00 JAll this becomes extra internal energy in ice, melting some according to Q = m ice L f . Thus,the mass of ice that melts ism ice =For the block:Q K i5.00 J=== 1.50 10 5 kg = 15.0 mg .LfL f 3.33 10 5 J kgQ = 0 (no energy flows by heat since there is no temperature difference)W = 5.00 JEint = 0 (no temperature change)andK = 5.00 JFor the ice,Q=0W = +5.00 JEint = +5.00 Jand(b)K = 0Again, K i = 5.00 J and m ice = 15.0 mgFor the block of ice: Q = 0; Eint = +5.00 J ; K = 5.00 Jso W = 0 .For the copper, nothing happens: Q = Eint = K = W = 0 .continued on next pageChapter 20Again, K i = 5.00 J. Both blocks must rise equally in temperature.(c)T =Q = mcT :Q 5.00 J== 4.04 10 3 Cmc 2 1.60 kg 387 J kg CbgbgAt any instant, the two blocks are at the same temperature, so for both Q = 0.For the moving block:K = 5.00 JandEint = +2.50 JsoW = 2.50 JFor the stationary block:K = 0andEint = +2.50 JsoW = +2.50 JFor each object in each situation, the general continuity equation for energy, in the formK + Eint = W + Q , correctly describes the relationship between energy transfers andchanges in the objects energy content.P20.67A = Aend walls + A ends of attic + A side walls + Aroofaf LMN 1 4.00 m a4.00 mf tan 37.0OPQ2F 4.00 m IJ+2a10.0 m 5.00 mf + 2a10.0 mfGH cos37.0 KA = 2 8.00 m 5.00 m + 2 2 A = 304 m 2P=ejajef4.80 10 4 kW m C 304 m 2 25.0 CkAT== 17.4 kW = 4.15 kcal sL0.210 mbgbgThus, the energy lost per day by heat is 4.15 kcal s 86 400 s = 3.59 10 5 kcal day .The gas needed to replace this loss isP20.683.59 10 5 kcal day9 300 kcal m3= 38.6 m3 day .FG IJH KTLAdx= kAdtxzzLe8.00t4.00L0xdx = kT dt2 8.00x2= kTt4.00je3.33 10 5 J kg 917 kg m3t = 3.66 10 4 s = 10.2 hF b0.080 0 mg b0.040 0 mgjGG2H22IJJ = b2.00 W m Cga10.0 CftK597598P20.69Heat and the First Law of ThermodynamicsW = W AB + WBC + WCD + WDAzzzPzBCDAABCDP2W = PdV PdV PdV PdVW = nRT1zzBzCzDFG V IJ P bVHV KB21CFG V IJ P bVHV Kg VB nRT2 ln21CP1A VDgVB P1VP=and 2 = 2V1 P2VC P1FP IFP IFP IDAV1Now P1VA = P2 VB and P2 VC = P1VD , so only the logarithmicterms do not cancel out.Also,CAdVdV P2 dV nRT2 P1 dVVVABCDW = nRT1 lnBV2FIG. P20.69FP IFP I W = nRT1 lnG P1 J nRT2 lnG P2 J = +nRT1 lnG P2 J nRT2 lnG P2 J = nRbT2 T1 g lnG P2 JH KH KH KH KH K11121Moreover P1V2 = nRT2 and P1V1 = nRT1W =P20.70bg FGH P IJKP P1 V2 V1 ln21For a cylindrical shell of radius r, height L, and thickness dr, the equation for thermal conduction,dQdT= kAdtdxbecomesUnder equilibrium conditions,dT = FGH1dQdt 2 kLI FG dr IJJK H r KgdQis constant; therefore,dtTb Ta = andFGHI FG IJJK H K1dQblndt 2 kLab gb gdQ 2 kL Ta Tb=dtln b aBut Ta > Tb , soP20.71bdQdT= k 2 rLdtdrFrom problem 70, the rate of energy flow through the wall isb gb gdQ 2 kL Ta Tb=dtln b aejbga5dQ 2 4.00 10 cal s cm C 3 500 cm 60.0 C=dtln 256 cm 250 cmbgfdQ= 2. 23 10 3 cal s = 9.32 kWdtThis is the rate of energy loss from the plane by heat, and consequently isthe rate at which energy must be supplied in order to maintain a constanttemperature.FIG. P20.71VChapter 20P20.72599Qcold = Q hotbdTQ Al = Q water + Qcaloorm Al c Alf TiiAlgbgd= m w c w + m c c c T f Tiiwb0.200 kg gc a+39.3 Cf = 0.400 kgb4 186 J kg Cg + 0.040 0 kgb630 J kg Cg a3.70 CfAlc Al =6.29 10 3 J= 800 J kg C7.86 kg Ceajfb(a)P = AeT 4 = 5.67 10 8 W m 2 K 4 5.1 10 14 m 2 0.965 5 800 K(b)*P20.73g4Tavg = 0.1 4 800 K + 0.9 5 890 K = 5.78 10 3 KbgbgThis is cooler than 5 800 K bye5 800 5 781= 0.327% .5 800gj ej bW 0.9e5.1 10 j0.965b5 890g = 3.17 10 WP = 5.67 10 8 W m 2 K 4 0.1 5.1 10 14 m 2 0.965 4 800 K(c)+5.67 10 8= 3.16 10 22 W14This is larger than 3.158 10 22 W by44221.29 10 20 W= 0.408%.3.16 10 22 WANSWERS TO EVEN PROBLEMSP20.20.105CP20.22liquid lead at 805CP20.487.0CP20.24(a) 12.0 MJ ; (b) +12.0 MJP20.6The energy input to the water is 6.70 timeslarger than the laser output of 40.0 kJ.P20.26nR T2 T1P20.888.2 WP20.28(a) 567 J ; (b) 167 JP20.10(a) 25.8C ; (b) noP20.30(a) 12.0 kJ; (b) 12.0 kJP20.3242.9 kJP20.34(a) 7.65 L; (b) 305 Kbm=Al c Alg+ m c c w Tc + m h c w ThbgP20.12TfP20.14(a) 380 K ; (b) 206 kPaP20.36(a) 48.6 mJ ; (b) 16.2 kJ; (c) 16.2 kJP20.1612.9 gP20.38(a) 4PVi ; (b) +4PVi ; (c) 9.08 kJiiP20.18(a) all the ice melts; 40.4C ;(b) 8.04 g melts; 0CP20.40(a) 1 300 J ; (b) 100 J ; (c) 900 J ; (d) 1 400 JP20.2034.0 kmP20.4210.0 kWm Al c Al + m c c w + m h c w600Heat and the First Law of ThermodynamicsP20.62P20.441.34 kWP20.46(a) 0.890 ft 2 F h Btu ; (b) 1.85(c) 2.08P20.48see the solution (a)P20.6810.2 hP20.70see the solutionP20.72800 J kg C(a) 64.1C ; (b) 113CP20.58(a) 15.0 mg ; block: Q = 0; W = 5.00 J ;Eint = 0 ; K = 5.00 J ;ice: Q = 0; W = 5.00 J ; Eint = 5.00 J ; K = 0(b) 15.0 mg ; block: Q = 0; W = 0 ;Eint = 5.00 J ; K = 5.00 J ;metal: Q = 0; W = 0 ; Eint = 0 ; K = 0(c) 0.004 04C ; moving block: Q = 0;W = 2.50 J ; Eint = 2.50 J ; K = 5.00 J ;stationary block: Q = 0; W = 2.50 J ;Eint = 2.50 J ; K = 0(a) 13.0C ; (b) 0.532 C sP20.56P20.6647.7 gP20.543.76 m s364 KP20.52P20.64(a) ~ 10 3 W ; (b) ~ 10 1 K sP20.50ft 2 F h;Btu(a) 2 000 W ; (b) 4.47CP20.60101PVi ; (b) 1.39 PVi ; (c) 0ii2(a) 9.31 10 J ; (b) 8.47 10(c) 8.38 10 12 J12J;21The Kinetic Theory of GasesANSWERS TO QUESTIONSCHAPTER OUTLINE21.1Molecular Model of an IdealGasMolar Specific Heat of anIdeal GasAdiabatic Processes for anIdeal GasThe Equipartition of EnergyThe Boltzmann DistributionLawDistribution of MolecularSpeedsMean Free Path21.321.421.521.621.7Q21.1The molecules of all different kinds collide with the walls of thecontainer, so molecules of all different kinds exert partialpressures that contribute to the total pressure. The moleculescan be so small that they collide with one another relativelyrarely and each kind exerts partial pressure as if the other kindsof molecules were absent. If the molecules collide with oneanother often, the collisions exactly conserve momentum andso do not affect the net force on the walls.Q21.2The helium must have the higher rms speed. According toEquation 21.4, the gas with the smaller mass per atom musthave the higher average speed-squared and thus the higherrms speed.Q21.321.2Yes. As soon as the gases are mixed, they come to thermalequilibrium. Equation 21.4 predicts that the lighter heliumatoms will on average have a greater speed than the heaviernitrogen molecules. Collisions between the different kinds ofmolecules gives each kind the same average kinetic energy oftranslation.Q21.4If the average velocity were non-zero, then the bulk sample of gas would be moving in the directionof the average velocity. In a closed tank, this motion would result in a pressure difference within thetank that could not be sustained.Q21.5The alcohol evaporates, absorbing energy from the skin to lower the skin temperature.Q21.6Partially evacuating the container is equivalent to letting the remaining gas expand. This means thatthe gas does work, making its internal energy and hence its temperature decrease. The liquid in thecontainer will eventually reach thermal equilibrium with the low pressure gas. This effect of anexpanding gas decreasing in temperature is a key process in your refrigerator or air conditioner.Q21.7Since the volume is fixed, the density of the cooled gas cannot change, so the mean free path doesnot change. The collision frequency decreases since each molecule of the gas has a lower averagespeed.Q21.8The mean free path decreases as the density of the gas increases.Q21.9The volume of the balloon will decrease. The pressure inside the balloon is nearly equal to theconstant exterior atmospheric pressure. Then from PV = nRT , volume must decrease in proportionto the absolute temperature. Call the process isobaric contraction.601602The Kinetic Theory of GasesQ21.10The dry air is more dense. Since the air and the water vapor are at the same temperature, they havethe same kinetic energy per molecule. For a controlled experiment, the humid and dry air are at thesame pressure, so the number of molecules per unit volume must be the same for both. The watermolecule has a smaller molecular mass (18.0 u) than any of the gases that make up the air, so thehumid air must have the smaller mass per unit volume.Q21.11Suppose the balloon rises into air uniform in temperature. The air cannot be uniform in pressurebecause the lower layers support the weight of all the air above them. The rubber in a typical balloonis easy to stretch and stretches or contracts until interior and exterior pressures are nearly equal. Soas the balloon rises it expands. This is an isothermal expansion, with P decreasing as V increases bythe same factor in PV = nRT . If the rubber wall is very strong it will eventually contain the helium athigher pressure than the air outside but at the same density, so that the balloon will stop rising.More likely, the rubber will stretch and break, releasing the helium to keep rising and boil out ofthe Earths atmosphere.Q21.12A diatomic gas has more degrees of freedomthose of vibration and rotationthan a monatomicgas. The energy content per mole is proportional to the number of degrees of freedom.Q21.13(a)Average molecular kinetic energy increases by a factor of 3.(b)The rms speed increases by a factor of(c)Average momentum change increases by(d)Rate of collisions increases by a factor of(e)Pressure increases by a factor of 3.3.3.3 since the mean free path remains unchanged.Q21.14They can, as this possibility is not contradicted by any of our descriptions of the motion of gases. Ifthe vessel contains more than a few molecules, it is highly improbable that all will have the samespeed. Collisions will make their speeds scatter according to the Boltzmann distribution law.Q21.15Collisions between molecules are mediated by electrical interactions among their electrons. On anatomic level, collisions of billiard balls work the same way. Collisions between gas molecules areperfectly elastic. Collisions between macroscopic spheres can be very nearly elastic. So the hardsphere model is very good. On the other hand, an atom is not solid, but has small-mass electronsmoving through empty space as they orbit the nucleus.Q21.16As a parcel of air is pushed upward, it moves into a region of lower pressure, so it expands and doeswork on its surroundings. Its fund of internal energy drops, and so does its temperature. Asmentioned in the question, the low thermal conductivity of air means that very little heat will beconducted into the now-cool parcel from the denser but warmer air below it.Q21.17A more massive diatomic or polyatomic molecule will generally have a lower frequency of vibration.At room temperature, vibration has a higher probability of being excited than in a less massivemolecule. The absorption of energy into vibration shows up in higher specific heats.SOLUTIONS TO PROBLEMSSection 21.1P21.1Molecular Model of an Ideal GasF = NmP=a8.00 sin 45.0 8.00 sin 45.0v= 500 5.00 10 3 kgt30.0 seF= 1.57 N m 2 = 1.57 PaAjfms= 0.943 NChapter 21e5.00 10 j 2e4.68 102623P21.2P21.3F=jbkg 300 m s1.00 s14.0 NFand P = == 17.6 kPa .A 8.00 10 4 m 2603g = 14.0 NWe first find the pressure exerted by the gas on the wall of the container.NkT 3 N A k BT 3 RT 3 8.314 N m mol K 293 K==== 9.13 10 5 PaP=33VVV8.00 10 mThus, the force on one of the walls of the cubical container isbefgajejF = PA = 9.13 10 5 Pa 4.00 10 2 m 2 = 3.65 10 4 N .P21.4FGHIJK2 N mv 2, so that3V2Use Equation 21.2, P =K av =mv 2 3 PV=where N = nN A = 2 N A22NK av =3 8.00 atm 1.013 10 5 Pa atm 5.00 10 3 m 33 PV=2 2N A2 2 mol 6.02 10 23 molecules molbagafejefejjK av = 5.05 10 21 J moleculeP21.5P=2NKE3Vd iEquation 21.2ejej353 PV3 1.20 10 4.00 10== 2.00 10 24 moleculesN=2 KE23.60 10 22d in=P21.6ej242.00 10 moleculesN== 3.32 molN A 6.02 10 23 molecules molOne mole of helium contains Avogadros number of molecules and has a mass of 4.00 g. Let us call mthe mass of one atom, and we haveN A m = 4.00 g mol4.00 g molm== 6.64 10 24 g moleculeor6.02 10 23 molecules molm = 6.64 10 27 kgP21.7aff54PV 1.013 10 Pa 3 0.150 m=N=k BT1.38 10 23 J K 293 K(a)PV = Nk BT :(b)K=(c)For helium, the atomic mass iseja3= 3.54 10 23 atomsja f33k BT = 1.38 10 23 293 J = 6.07 10 21 J22em=4.00 g molmolecules mol27kg molecule6.02 10m = 6.64 1013mv 2 = k BT :2223 v rms =3 k BT= 1.35 km sm= 6.64 10 24 g molecule604P21.8The Kinetic Theory of Gasesv=vOv He3 k BTmM He4.001===32.08.00MOvO =1 350 m s8.00= 477 m sjaf(a)K=33k BT = 1.38 10 23 J K 423 K = 8.76 10 21 J22(b)P21.9K=12mv rms = 8.76 10 21 J2esov rms =For helium,m=1.75 10 20 Jm(1)4.00 g mol6.02 10 23 molecules mol= 6.64 10 24 g moleculem = 6.64 10 27 kg molecule39.9 g molm=Similarly for argon,6.02 10 23 molecules mol= 6.63 10 23 g moleculem = 6.63 10 26 kg moleculeSubstituting in (1) above,we find for helium,and for argon,P21.10(a)v rms = 1.62 km sv rms = 514 m sPV = nRT =Nmv 23The total translational kinetic energy is33PV = 3.00 1.013 10 5 5.00 10 3 = 2.28 kJ22jej3 k T 3 RT 3a8.314fa300fmv==== 6. 21 10222N2e6.02 10 jF 1 N m I FG 1 J IJ = 1 J m1 Pa = a1 PafGH 1 Pa JK H 1 N m KEtrans =(b)2eB23AP21.11(a)(b)Nmv 2= Etrans :2221J3For a monatomic ideal gas, Eint =3nRT2For any ideal gas, the energy of molecular translation is the same,Etrans =Thus, the energy per volume is33nRT = PV .22Etrans3=P .2VChapter 21Section 21.2P21.12Eint =EintP21.13Molar Specific Heat of an Ideal Gas3nRT233= nRT = 3.00 mol 8.314 J mol K 2.00 K = 74.8 J22afbfgaWe us the tabulated values for C P and C Vbfga(a)(b)Eint = nCV T = 1.00 mol 20.4 J mol K 120 K = 2.45 kJ(c)P21.14Q = nC P T = 1.00 mol 28.8 J mol K 420 300 K = 3.46 kJW = Q + Eint = 3.46 kJ + 2.45 kJ = 1.01 kJbfgaThe piston moves to keep pressure constant. Since V =nRT, thenPnRTfor a constant pressure process.P2QQQ==Q = nC P T = n CV + R T so T =7nRn CV + R n 5 R 2 + RV =b gb g bnR F 2Q I 2Q 2 QVV =G J= =P H 7nR K 7 P 7 nRTe4.40 10 Jja5.00 Lf2V == 2.52 L7 a1.00 molfb8.314 J mol K ga300 K fandg3Thus,P21.15V f = Vi + V = 5.00 L + 2.52 L = 7.52 Ln = 1.00 mol, Ti = 300 K(b)Since V = constant, W = 0(a)Eint = Q + W = 209 J + 0 = 209 J(c)Eint = nCV T = nsoFG 3 RIJ TH2 KT =afbf2 209 J2 Eint== 16.8 K3nR3 1.00 mol 8.314 J mol KaT = Ti + T = 300 K + 16.8 K = 317 Kg605606P21.16The Kinetic Theory of Gases(a)Consider heating it at constant pressure. Oxygen and nitrogen are diatomic, so C P =Q = nC P T =eFG IJH KN m je100 m ja1.00 K f =77 PVTnRT =22 T527 1.013 10Q=2300 K(b)(a)Uggy=e1.18 10 5 Jj9.80 m s 2 2.00 mWe assume that the bulb does not expand. Then this is a constant-volume heating process.PVThe quantity of the gas is n = i . The energy input is Q = P t = nCV T soRTiPt PtRTi=.nCVPVCViFGHThe final temperature is T f = Ti + T = Ti 1 +The final pressure is Pf = PiP21.18(a)(b)FGHFGHTf= Pi 1 +TiIJKPtR.PVCViIJKPtR.PVCViI = 1.18 atmJs mol K 1.013 10 N 4 a0.05 mf 12.5 J KF 1.00 mol I = 719 J kg K = 0.719 kJ kg K55C = R = b8.314 J mol K gG22H 0.028 9 kg JKF PV IJm = Mn = M GH RT KF 200 10 Pae0.350 m j Im = b0.028 9 kg molgGGH b8.314 J mol K ga300 K f JJK = 0.811 kgPf = 1 atm 1 +3.60 J 4 s 8.314 J m 2 3 mol K35V3(c)118 kJ= 6.03 10 3 kgT =(b)3U g = mgym=*P21.173We consider a constant volume process where no work is done.bfgaQ = mCV T = 0.811 kg 0.719 kJ kg K 700 K 300 K = 233 kJ(d)7R2We now consider a constant pressure process where the internal energy of the gas isincreased and work is done.FG 7R IJ T = mFG 7C IJ Tb gH2KH 5KL7OQ = 0.811 kg M b0.719 kJ kg K gPa 400 K f = 327 kJN5QQ = mC P T = m CV + R T = mVChapter 21P21.19Consider 800 cm3 of (flavored) water at 90.0 C mixing with 200 cm3 of diatomic ideal gas at 20.0C:Qcold = Q hotdidTa fi = bV gm air c P , air T f Ti , air = m w c w ToraT fw m air c P , air= Ti , airfmwcwwa90.0 C 20.0 Cfb V gccair P , airw wwwhere we have anticipated that the final temperature of the mixture will be close to 90.0C.7The molar specific heat of air isC P, air = R27 R71.00 molSo the specific heat per gram is= 8.314 J mol K= 1.01 J g Cc P, air =2 M228.9 gIFG IJ bgFGHJKH K1.20 10 g cm je 200 cm j b1.01 J g C ga70.0 C faT f = ee1.00 g cm je800 cm j b4.186 J kg CgaT f 5.05 10 C33wor3333wThe change of temperature for the water is between 10 3 C and 10 2 C .P21.20bQ = nC P Tgisobaricb+ nCV TgisovolumetricIn the isobaric process, V doubles so T must double, to 2Ti .In the isovolumetric process, P triples so T changes from 2Ti to 6Ti .Q =nP21.21607FG 7 RIJ b2T T g + nFG 5 RIJ b6T 2T g = 13.5nRT =H2 KH2 Kiiiii13.5 PVIn the isovolumetric process A B , W = 0 and Q = nCV T = 500 J500 J = nFG 3R IJ bTH2KBTB = 300 K +g TA or TB = TA +a2 500 J3nRfa f= 340 K3a1.00 molfb8.314 J mol K g2 500 JIn the isobaric process B C ,Q = nC P T =Thus,bg5nRTC TB = 500 J .2af2 500 J1 000 J= 340 K = 316 K5nR5 1.00 mol 8.314 J mol K(a)TC = TB (b)The work done on the gas during the isobaric process isabfbg agfbgaWBC = PB V = nR TC TB = 1.00 mol 8.314 J mol K 316 K 340 JorWBC = +200 JThe work done on the gas in the isovolumetric process is zero, so in totalWon gas = +200 J .f608*P21.22The Kinetic Theory of Gases(a)At any point in the heating process, Pi = kVi and P = kV =nRTiPf =Vi22Vi = 2 Pi and T f =Pf V fnRzf(b)The work input is W = PdV = i=2 Pi 2Vi= 4Ti .nRznRTi2ViViVi2nRT V 2VdV = 2 i2VibP21.23(a)a2Vi=VinRTi2Vi2e4Vi23 Vi2 = nRTi .2jg515R 4Ti Ti = + nRTi . The heat input22The change in internal energy, is Eint = nCV T = nis Q = Eint W =PinRTiV . At the end,V=ViVi2f18nRTi = 9 1 mol RTi .2The heat required to produce a temperature change isQ = n1C 1 T + n 2 C 2 TThe number of molecules is N 1 + N 2 , so the number of moles of the mixture is n1 + n 2 andQ = n1 + n 2 CT ,bgn C + n 2C 2C= 1 1.n1 + n 2so(b)mQ = n i C i T =i =1F n I CTGH JKmi =1im n i Cii =1mC= nii =1Section 21.3P21.24(a)(b)(c)Adiabatic Processes for an Ideal GasPVii=TfPf V fTi=Pf V fPViisoFG P IJ FG V IJ = a20.0fa0.118fH P KH V KffTfiTiSince the process is adiabatic,Since = 1.40 =C P R + CV=,CVCVbEint = nCV T = 0.016 0 molandiVii=FPI=G JHP KVff1=FG 1.00 IJH 20.0 K57= 0.118= 2.35Q=0CV =5R and T = 2.35Ti Ti = 1.35Ti2gFGH 5 IJK b8.314 J mol K g 1.35a300 K f =2W = Q + Eint = 0 + 135 J = +135 J .135 JChapter 21P21.25(a)PVi = Pf V fiPf(b)FV I=PG JHV KiTi =FG 12.0 IJH 30.0 K= 5.00 atmfe1.40= 1.39 atmjej353PVi 5.00 1.013 10 Pa 12.0 10 mi== 365 KnR2.00 mol 8.314 J mol KTf =(c)iPf V f=nRbegje1.39 1.013 10 Pa 30.0 10 3 m 35b2.00 mol 8.314 J mol Kgj=253 KThe process is adiabatic: Q = 0C P R + CV5=, CV = R2CVCV5= nCV T = 2.00 mol8.314 J mol K2 = 1.40 =EintFG bHgIJK a253 K 365 K f =4.66 kJW = Eint Q = 4.66 kJ 0 = 4.66 kJP21.26Vi = F 2.50 10GH 22mIJK20.500 m = 2.45 10 4 m 3The quantity of air we find from PVi = nRTiin=eje1.013 10 5 Pa 2.45 10 4 m 3PVii=RTi8.314 J mol K 300 Kbgafjn = 9.97 10 3 molAdiabatic compression: Pf = 101.3 kPa + 800 kPa = 901.3 kPa(a)PVi = Pf V fiV f = ViFPIGH P JK1i= 2. 45 10 4 m 3fFG 101.3 IJH 901.3 K57V f = 5.15 10 5 m 3(b)Pf V f = nRT fTfFPI=T=TG JPVP HP KF 101.3 IJ b g == 300 K GH 901.3 KiPf V fi ii1PfiifF P Ib=T G JHP Ki1 1gif5 7 1Tf(c)560 KThe work put into the gas in compressing it is Eint = nCV TeW = 9.97 10 3 molW = 53.9 Jcontinued on next pagej 5 b8.314 J mol K ga560 300f K2609610The Kinetic Theory of GasesNow imagine this energy being shared with the inner wall as the gas is held at constantvolume. The pump wall has outer diameter 25.0 mm + 2.00 mm + 2.00 mm = 29.0 mm , andvolumeLM e14.5 10 mj e12.5 10 mj OP4.00 10NQand mass V = e7.86 10 kg m je6.79 10 m j = 53.3 g23233632m = 6.79 10 6 m 33The overall warming process is described by53.9 J = nC V T + mcTej 5 b8.314 J mol K gdT 300 K i2+e53.3 10 kg jb 448 J kg K gdT 300 K i53.9 J = b0.207 J K + 23.9 J K gdT 300 K i53.9 J = 9.97 10 3 molff3ffffT ff 300 K = 2.24 KP21.27TfTiFV I=G JHV K 1i=fFG 1 IJH 2K0. 400If Ti = 300 K , then T f = 227 K .*P21.28(a)InPVii=Pf V fwe have PfFV I=PG JHV KifPfThenPVi Pf V fi=TiTfiF 0.720 m I=PGH 0.240 m JK3iT f = Ti1.403Pf V fPVii= 4.66 Pia f 1 = 1.553= Ti 4.66The factor of increase in temperature is the same as the factor of increase in internal energy,Eint, f= 1.55 .according to Eint = nCV T . ThenEint, i(b)FV I==G JInTPVHV KF 0.720 m I2=GH V JKTfPf V fiVfii ifVi3FV I=G JHV K0. 40f30.720 m= 2 1 0 . 4 = 2 2.5 = 5.66VfVf =0.720 m 3= 0.127 m35.66if 1we have611Chapter 21P21.29(a)(b)PSee the diagram at the right.PBVB = PC VCB3 Pi3 PVi = PVCiiAdiabatice j e j= 2.19a 4.00 L f = 8.77 LVC = 3 1 Vi = 3 5 7 Vi = 2.19ViVC(c)PiPBVB = nRTB = 3 PVi = 3nRTiiafCAVCVi = 4 LTB = 3Ti = 3 300 K = 900 K(d)In AB, Q AB = nCV V = nFIG. P21.29After one whole cycle, TA = Ti = 300 K .(e)V(L)FG 5 RIJ b3T T g = a5.00fnRTH2 KiiiQBC = 0 as this process is adiabaticg a fbPC VC = nRTC = Pi 2.19Vi = 2.19 nRTisoTC = 2.19TiQCA = nC P T = nFG 7 RIJ bT 2.19T g = a4.17fnRTH2 KiiFor the whole cycle,iafafQ ABCA = Q AB + QBC + QCA = 5.00 4.17 nRTi = 0.829 nRTib E gint ABCA= 0 = Q ABCA + W ABCAafafW ABCA = Q ABCA = 0.829 nRTi = 0.829 PViiafejejW ABCA = 0.829 1.013 10 5 Pa 4.00 10 3 m 3 = 336 JP21.30(a)See the diagram at the right.(b)PBVB = PC VCPB3PiAdiabatic3 PVi = PVCiiVC = 3 1 Vi = 3 5 7 Vi = 2.19Vi(c)PBVB = nRTB = 3 PVi = 3nRTiiTB = 3Ti(d)After one whole cycle, TA = TiPiACVCViFIG. P21.30continued on next pageafV L612The Kinetic Theory of Gases(e)In AB, Q AB = nCV T = nFG 5 RIJ b3T T g = a5.00fnRTH2 KiiiQBC = 0 as this process is abiabaticb gF7 I= nC T = nG RJ bT 2.19T g = 4.17nRTH2 KPC VC = nRTC = Pi 2.19Vi = 2.19nRTi so TC = 2.19TiQCAPiiiFor the whole cycle,afQ ABCA = Q AB + QBC + QCA = 5.00 4.17 nRTi = 0.830nRTib E gint ABCA= 0 = Q ABCA + W ABCAW ABCA = Q ABCA = 0.830nRTi = 0.830 PViiP21.31(a)The work done on the gas iszVbWab = PdV .VaFor the isothermal process,Wab = nRTaz FGHVbVaIJK1dVVFG V IJ = nRT lnFG V IJ .HV KHV K= 5.00 molb8.314 J mol K ga 293 K f lna10.0 fWab = nRTa lnThus, WabbaabFIG. P21.31Wab = 28.0 kJ .(b)For the adiabatic process, we must first find the final temperature, Tb . Since air consistsprimarily of diatomic molecules, we shall use air = 1.40 and C V , air =af5 R 5 8.314== 20.8 J mol K .22Then, for the adiabatic preocessTbFV I=T G JHV K 1aaa f= 293 K 10.0b0. 400= 736 K .Thus, the work done on the gas during the adiabatic process isbg = b0 + nC T g = nC bT T g= 5.00 molb 20.8 J mol K ga736 293f K = 46.0 kJWab Q + EintorWabcontinued on next pageabVabVba.613Chapter 21(c)For the isothermal process, we havePb Vb = PaVa .Thus, Pb = PaFG V IJ = 1.00 atma10.0f =HV Ka10.0 atm .bFor the adiabatic process, we have Pb Vb = Pa Va .Thus, Pb = PaP21.32FG V IJHV Kaa f= 1.00 atm 10.0b1.40= 25.1 atm .We suppose the air plus burnt gasoline behaves like a diatomicideal gas. We find its final absolute pressure:e21.0 atm 50.0 cm3FG 1 IJH 8KPf = 21.0 atmj7575e= Pf 400 cm3j75= 1.14 atmNow Q = 0dand W = Eint = nCV T f TiW =i555nRT f nRTi = Pf V f PVii222diFIG. P21.32F 1.013 10 N m I 105W = 1.14 atme 400 cm j 21.0 atme50.0 cm j G2H 1 atm JK e33W = 150 JThe output work is W = +150 JThe time for this stroke isP=FGH1 1 min4 2 500W150 J== 25.0 kWt6.00 10 3 sI FG 60 s IJ = 6.00 10JK H 1 min K3s526m3 cm 3j614The Kinetic Theory of GasesSection 21.4P21.33The Equipartition of EnergyThe heat capacity at constant volume is nC V . An ideal gas of diatomic molecules has three degrees offreedom for translation in the x, y, and z directions. If we take the y axis along the axis of a molecule,then outside forces cannot excite rotation about this axis, since they have no lever arms. Collisionswill set the molecule spinning only about the x and z axes.(a)If the molecules do not vibrate, they have five degrees of freedom. Random collisions put1equal amounts of energy k BT into all five kinds of motion. The average energy of one25molecule is k BT . The internal energy of the two-mole sample is2NFG 5 k TIJ = nN FG 5 k TIJ = nFG 5 RIJ T = nC T .H2 KH2 K H2 KBThe molar heat capacity is C V =ABV5R and the samples heat capacity is2nCV = nFG 5 RIJ = 2 molFG 5 b8.314 J mol K gIJH2 KH2KnCV = 41.6 J KFor the heat capacity at constant pressure we haveg FGH 5 R + RIJK = 7 nR = 2 molFGH 7 b8.314 J mol K gIJK222bnC P = n C V + R = nnC P = 58.2 J K(b)In vibration with the center of mass fixed, both atoms are always moving in oppositedirections with equal speeds. Vibration adds two more degrees of freedom for two moreterms in the molecular energy, for kinetic and for elastic potential energy. We haveFG 7 RIJ =H2 KF9 I= nG RJ =H2 KnCV = nandP21.3458.2 J KnC P74.8 J KFG k T IJ = f FG nRT IJH2K H 2K1 F dE I 1= GJ = fRn H dT K 2(1)Eint = NfB(2)CVint(3)C P = CV + R =(4) =CP f + 2=CVfbg1f +2 R2Chapter 21P21.35Rotational Kinetic Energy =1 2I2615ClI = 2mr 2 , m = 35.0 1.67 10 27 kg , r = 10 10 mI = 1.17 10 45 kg m 2 K rot =Cl = 2.00 10 12 s 1FIG. P21.351 2I = 2.33 10 21 J2Section 21.5The Boltzmann Distribution LawSection 21.6Distribution of Molecular SpeedsP21.36(a)The ratio of the number at higher energy to the number at lower energy is e E kBT whereE is the energy difference. Here,aE = 10.2 eV10fFGH 1.601eV J IJK = 1.63 101918Jand at 0C,jaefk BT = 1.38 10 23 J K 273 K = 3.77 10 21 J .Since this is much less than the excitation energy, nearly all the atoms will be in the groundstate and the number excited ise1.63 10j FGH 3.77 10 JJ IJK = e2.70 10 je182.70 10 25 exp2521433.This number is much less than one, so almost all of the time no atom is excited .(b)At 10 000C,ejk BT = 1.38 10 23 J K 10 273 K = 1.42 10 19 J .The number excited is1.63 10e2.70 10 j expFGH 1.42 10 JJ IJK = e2.70 10 je2518192511.5= 2.70 10 20 .616The Kinetic Theory of Gases(a)v av =(b)P21.37ev j ni v iN2av= ni vi2P21.38(a)(b)P21.39= 54.9 m 2 s 2Nev j2so v rms =(c)af af af af af a f11 2 + 2 3 + 3 5 + 4 7 + 3 9 + 2 12 = 6.80 m s15=av= 54.9 = 7.41 m sv mp = 7.00 m sVrms, 35Vrms, 373 RTM 35=3 RTM 37=F 37.0 g mol IGH 35.0 g mol JK35The lighter atom,12= 1.03Cl , moves faster.dN v= 0 to finddvIn the Maxwell Boltzmann speed distribution function takeF m I4 N GH 2 k T JKFGH32exp Bmv 22 k BTI F 2v 2mv I = 0JK GH 2k T JK3Band solve for v to find the most probable speed.Reject as solutionsv = 0 and v = Retain only2Thenv mp =mv 2=0k BT2 k BTmP21.40The most probable speed is v mp =P21.41(a)From v av =kgf=eT=At 0C,je 6.64 10 27 kg 1.12 10 4 m se8 1.38 10je 6.64 10 27 kg 2.37 10 3 m s(b)P21.426.64 1027132 m s .8 k BTmwe find the temperature as T =ejae2 1.38 10 23 J K 4.20 K2 k BT=mj23J mol Kj2= 1.06 10 3 K132mv rms0 = k BT022e8 1.38 10 23 J mol KAt the higher temperature,b1m 2 v rms02g2j=a3k BT2fT = 4T0 = 4 273 K = 1 092 K = 819 C .j2= 2.37 10 4 KChapter 21*P21.43(a)(b)From the Boltzmann distribution law, the number density of molecules with gravitationalenergy mgy is n 0 e mgy k BT . These are the molecules with height y, so this is the number pervolume at height y as a function of y.b g=en yn0 mgy k BT=e= e MgyeN A k BT= e MgyjeRTje 28 .9 10 3 kg mol 9.8 m s 2 11 10 3 mj b8.314 J molK ga 293 K f= e 1.279 = 0.278*P21.44(a)We calculateze mgyk BTzdy =e mgyk BT=FG mgdy IJ F k T IH k T K GH mg JKk TT=a0 1f = kmgmgBBy=00k BT mgyemgk BTBB0Using Table B.6 in the appendixzye mgyk BTdy =0zzThen y =ye mgy0k BTdy=e mgyk BTdyF k TIbmg k T g GH mg JK1!2=2B.Bbk T mg gBk BT mg2=k BT.mg0(b)Section 21.7P21.45(a)y=bk BT8.314 J 283 K s 2RT=== 8.31 10 3 mM N A g Mg mol K 28.9 10 3 kg 9.8 mgMean Free PathFG N IJ RT and N = PVN so thatRTHN Ke1.00 10 ja133fa1.00fe6.02 10 j =N=a8.314fa300fAPV =A10(b)=f=233.21 10 12 molecules11.00 m3V==2 122 12nV d 2N d 23.21 10 12 molecules 3.00 10 10 me= 779 km(c)617v= 6.42 10 4 s 1jej a 2f212618P21.46The Kinetic Theory of GasesThe average molecular speed isv=8 k BT8 k B N AT=m N Amv=8 RTMb e 2.016 10g8 8.314 J mol K 3.00 Kv=3kg moljv = 178 m s(a)The mean free path is1=22 d n V=1ej2 0. 200 10 9 m21 m3= 5.63 10 18 mThe mean free time isv(b)=5.63 10 18 m= 3.17 10 16 s = 1.00 10 9 yr .178 m sNow nV is 10 6 times larger, to makesmaller by 10 6 times:= 5.63 10 12 m .Thus,P21.47v= 3.17 10 10 s = 1.00 10 3 yr .From Equation 21.30, =For an ideal gas, nV =k BTTherefore, =P21.48=2 d 2 nVd = 3.60 10 10 m2 d 2 nVNP=V k BT2 d 2 P11, as required.nV =nV =Pk BT1.013 10 5e1.38 10 ja293f23= 2.51 10 25 m3 = 6.93 10 8 m, or about 193 molecular diameters .Chapter 21P21.49k BTUsing P = nV k BT , Equation 21.30 becomes =(a)(b)e1.38 10=e23jaJ K 293 Kje52 1.013 10 Pa 3.10 10Equation (1) shows that P1P2 =(c)For3110= P2jm2.(1)2 Pd 2f2= 9.36 10 8 mTaking P1a1.00 atmfe9.36 10from (a) and with18j=m1.00 m2= 1.00 m, we find9.36 10 8 atm .= 3.10 10 10 m , we havea1.00 atmfe9.36 10P =33.10 10108j=mm302 atm .Additional ProblemsP21.50(a)n=PV (1.013 10 5 Pa)( 4.20 m 3.00 m 2.50 m)== 1.31 10 3 mol( 8.314 J mol K )( 293 K )RTejeN = nN A = 1.31 10 3 mol 6.02 10 23 molecules moljN = 7.89 10 26 moleculesjbeg(b)m = nM = 1.31 10 3 mol 0.028 9 kg mol = 37.9 kg(c)133m 0 v 2 = k BT = 1.38 10 23 J k 293 K = 6.07 10 21 J molecule222(d)For one molecule,m0 =f0.028 9 kg molM== 4.80 10 26 kg moleculeN A 6.02 10 23 molecules molv rms =(e),(f)jaeej=2 6.07 10 21 J molecule4.80 1026kg moleculeFG 5 RIJ T = 5 PVH2 K 25= e1.013 10 Paje31.5 m j =2503 m sEint = nCV T = nEint537.98 MJ619620P21.51The Kinetic Theory of Gases(a)Pf = 100 kPaVf =nRT f=PfT f = 400 Kbga2.00 mol 8.314 J mol K 400 K3100 10 Paa ff = 0.066 5 m3= 66.5 Lafbga fW = PV = nRT = a 2.00 molfb8.314 J mol K ga100 K f =Eint = 3.50 nRT = 3.50 2.00 mol 8.314 J mol K 100 K = 5.82 kJ1.66 kJQ = Eint W = 5.82 kJ + 1.66 kJ = 7.48 kJ(b)T f = 400 KPf = PiV f = Vi =bfganRTi 2.00 mol 8.314 J mol K 300 K== 0.049 9 m 3 = 49.9 LPi100 10 3 PaF T I = 100 kPaFG 400 K IJ =GH T JKH 300 K KfzW = PdV = 0 since V = constant133 kPaiEint = 5.82 kJ as in part (a)(c)Q = Eint W = 5.82 kJ 0 = 5.82 kJPf = 120 kPaT f = 300 KV f = ViF P I = 49.9 LFG 100 kPa IJ =GH P JKH 120 kPa Kia fEint = 3.50 nRT = 0 since T = constant41.6 LfF I = nRT lnFG P IJGH JKHP KF 100 kPa IJ = +909 JW = a 2.00 molfb8.314 J mol K ga300 K f lnGH 120 kPa KzzVfW = PdV = nRTiViVfdV= nRTi lnVViiifQ = Eint W = 0 910 J = 909 J(d)Pf = 120 kPa =C P CV + R 3.50 R + R 4.50 9====3.50 R3.50 7CVCVFPIF 100 kPa IJ ==: soV = V G J = 49.9 L GH 120 kPa KHP KF P V IJ = 300 K FG 120 kPa IJ FG 43.3 L IJ = 312 KT =T GH 100 kPa K H 49.9 L KH PV KE = a3.50fnRT = 3.50a 2.00 molfb8.314 J mol K ga12.4 K f = 722 JQ = 0 badiabatic process g1Pf V ffPViiifffi iintW = Q + Eint = 0 + 722 J = +722 Jiif7943.3 LChapter 21P21.52(a)The average speed v av is just the weighted average of all the speeds.af a f a f a f a f a f a f=a 2 + 3 + 5 + 4 + 3 + 2 + 1f2 v + 3 2 v + 5 3 v + 4 4v + 3 5 v + 2 6 v + 1 7 vv av =(b)3.65 vFirst find the average of the square of the speeds,2v av=a f + 3a2vf + 5a3vf + 4a4vf + 3a5vf + 2a6 vf + 1a7 vf2v2222222+3+5+4+3+2+12= 15.95 v 2 .2The root-mean square speed is then v rms = v av = 3.99 v .(c)The most probable speed is the one that most of the particles have;i.e., five particles have speed 3.00 v .(d)PV =12Nmv av3afFGH220 m 15.95 vmv 2= 106Therefore, P =3VV(e)112mv av = m 15.95 v 2 = 7.98mv 2 .22ejzf(a)PV = k . So, W = PdV = ki(b).The average kinetic energy for each particle isK=P21.53IJKzfidV Pf V f PVii= 1VdEint = dQ + dW and dQ = 0 for an adiabatic process.diTherefore, W = + Eint = nCV T f Ti .To show consistency between these 2 equations, consider that =Therefore,C1= V. 1 RUsing this, the result found in part (a) becomesdW = Pf V f PViiAlso, for an ideal gasi CR .VPV= nT so that W = nCV T f Ti .RdiCPand C P CV = R .CV621622*P21.54The Kinetic Theory of GasesdW = nCV T f Ti(a)i32 500 J = 1 mol 8.314 J mol K T f 500 K2diT f = 300 KPVi = Pf V fi(b)F nRT IJPGH P Kiib gPiPf = Pi*P21.55b g 1 1=TfI b5 3gb3 2FTGH T JKfPfiffiTiF nRT I=P GH P JKTi Pi1 = T f Pf1fFT I b gP =PG JHT KgF 300 IJ = 1.00 atm= 3.60 atmGH 500 Kffi 1i5 2Let the subscripts 1 and 2 refer to the hot and cold compartments, respectively. The pressure ishigher in the hot compartment, therefore the hot compartment expands and the cold compartmentcontracts. The work done by the adiabatically expanding gas is equal and opposite to the work doneby the adiabatically compressed gas.nRnRT1i T1 f = T2i T2 f 1 1did T1 f + T2 f = T1i + T2 i = 800 KConsider the adiabatic changes of the gases.P1i V1i = P1 f V1f and P2i V2i = P2 f V2 fP1i V1iP2 iV2 i=P1 f V1fP2 f V2 fF IGH JKV1 fP1i=P2 iV2 f, since V1i = V2i and P1 f = P2 fFGHnRT1 f P1 fnRT1i V1i=nRT2 i V2 inRT2 f P2 fFGHFT=GHTT1 fT1i=T2iT2 fT1 fT2 f1i2iIJKIJKIJK, using the ideal gas law, since V1i = V2i and P1 f = P2 f1=FG 550 K IJH 250 K K1 1.4= 1.756 (2)Solving equations (1) and (2) simultaneously givesT1 f = 510 K, T2 f = 290 K .i(1)Chapter 21*P21.56623The work done by the gas on the bullet becomes its kinetic energy:b11mv 2 = 1.1 10 3 kg 120 m s22The work on the gas is2= 7.92 J .1Pf V f PVi = 7.92 J.i 1dFV I .GH V JKL FV I O1So 7.92 J =P MV G J V P .PQ0. 40 M H V KNgiAlso Pf V f = PViiiPf = PifiififAnd V f = 12 cm3 + 50 cm 0.03 cm 2 = 13.5 cm3 .Then Pi =P21.57a fLM13.5 cm c hN7.92 J 0.40 10 6 cm 3 m 3312 1.4013 .5= 12 cm OQP35.74 10 6 Pa = 56.6 atm .The pressure of the gas in the lungs of the diver must be the same as the absolute pressure of thewater at this depth of 50.0 meters. This is:jejaF 1.00 atm IJ = 5.98 atmPaGH 1.013 10 Pa KefP = P0 + gh = 1.00 atm + 1.03 10 3 kg m 3 9.80 m s 2 50.0 morP = 1.00 atm + 5.05 10 55If the partial pressure due to the oxygen in the gas mixture is to be 1.00 atmosphere (or the fraction11of the total pressure) oxygen molecules should make up onlyof the total number of5.985.98molecules. This will be true if 1.00 mole of oxygen is used for every 4.98 mole of helium. The ratio byweight is thena4.98 mol Hefb4.003 g mol Hegg =b1.00 mol O gb2 15.999 g mol O gg2P21.58(a)0.623 .2Maxwells speed distribution function isNvF m I= 4 N GH 2 k T JK3 2v 2 e mv22 k BTBWithN = 1.00 10 4 ,0.032 kgM== 5.32 10 26 kgm=N A 6.02 10 23T = 500 Kandk B = 1.38 10 23 J molecule Kejthis becomes N v = 1.71 10 4 v 2 eej 3 .85 10 6 v 2To the right is a plot of this function for the range0 v 1 500 m s.FIG. P21.58(a)continued on next page624The Kinetic Theory of Gases(b)The most probable speed occurs where N v is a maximum.From the graph, v mp 510 m s(c)v av =8 k BT=meja f = e5.32 10 j8 1.38 10 23 50026575 m sAlso,v rms =(d)3 k BT=meja f =3 1.38 10 23 5005.32 1026624 m s300 m s v 600 m sThe fraction of particles in the rangez600N v dvis300whereN = 10 4Nand the integral of N v is read from the graph as the area under the curve.This is approximately 4 400 and the fraction is 0.44 or 44% .P21.59(a)Since pressure increases as volume decreases (and vice versa),LM OPN Q1 dVdV< 0 and > 0.V dPdP(b)For an ideal gas, V =FGHIJK1 d nRTnRTand 1 = .PV dP PIf the compression is isothermal, T is constant and1 = (c)IJKFor an adiabatic compression, PV = C (where C is a constant) and2 =(d)FGH11nRT 2 = .VPP1 = =FG IJH K1 d CV dP P1=FG IJH K1 1 C1 1P1 = 1 +1 =.PV P b1 g+1 P11== 0.500 atm 12.00 atmPafCP5and for a monatomic ideal gas, = , so that3CV2 =1=P53a1= 0.300 atm 12.00 atmfChapter 21P21.60(a)BThe speed of sound is v =where B = VdP.dVAccording to Problem 59, in an adiabatic process, this is B =a fa f12= P .nRT M PMm s nM===where m s is the sample mass. Then, the speed of soundVVV RTRTAlso, =Bin the ideal gas is v =b= PgaFG RT IJ =H PM K1.40 8.314 J mol K 293 Kv=RT.Mf=(b)344 m s0.028 9 kg molThis nearly agrees with the 343 m/s listed in Table 17.1.(c)We use k B = kB N AT k BTRTRand M = mN A : v ===.MmN AmNAThe most probable molecular speed is2k BT,m8k BT, and the rms speed ismthe average speed is3k BT.mAll are somewhat larger than the speed of sound.P21.61n=(a)(b)(c)1.20 kgm== 41.5 molM 0.028 9 kg molVi =PfPiafbVfViso V fFP IF 400 IJ= V G J = e0.514 m jGH 200 KHPKfi2Pf V f2= 2.06 m 3i333=VfViVi123(e)3e400 10 Paje2.06 m j = 2.38 10 KnRa41.5 molfb8.314 J mol K gF P I 2V = 2 F P I V VW = z PdV = C z V dV = GjG Je3 HV KH V JK 32 F 200 10 Pa I LW = GMe2.06 m j a0.514 mf OQP = 4.80 10 J3 H 0.514 m J NKTf =Vf(d)fga41.5 mol 8.314 J mol K 298 KnRTi== 0.514 m 3Pi200 10 3 Pa=625aEint = nCV T = 41.5 moli12i3 3 23 2 Vfi12iVi32f3 2i3 2fLMN 5 b8.314 J mol K gOPQe2.38 10253j 298 KEint = 1.80 10 6 JQ = Eint W = 1.80 10 6 J + 4.80 10 5 J = 2.28 10 6 J = 2.28 MJ626P21.62The Kinetic Theory of Gasesb111mvi2 mv 2 = 0.142 kgf222The ball loses energyg a47.2f a42.5f22m 2 s 2 = 29.9 Jg a19.4 mf = 0.083 4 mPV 1.013 10 Pae0.083 4 m j=n=RTb8.314 J mol K ga293 K f = 3.47 molbV = 0.037 0 mThe air volume is235and its quantity is3The air absorbs energy according toQ = nC P TT =SoP21.63afN v v = 4 NNote thatThus,Q=nC P 3. 47 molF mv IGH 2k T JKF 2k T IJv =GH m KF m IN a vf = 4 N GH 2 k T JKF v I eeN avf=GJN ev j H v KF m IGH 2 k T JK32BForB12vv2ee v22v mpj21 v 2 v mpjmpmpv mp50a f = FG 1 IJN e v j H 50 KNv vv3 2Bvv=0. 296 CBmpvc hb8.314 J mol K g =2v 2 exp2And29.9 J722eb g1 1 502= 1.09 10 3mpThe other values are computed similarly, with thefollowing results:vv mp15011012121050afNv ve jN v v mp1.09 10 32.69 10 20.5291.000.1991.01 10 411.25 10 1 082To find the last value, note:a50f e2 1 2 50010 log 2 500 e= 2 500 e 2 499aln10 fb2 499 ln10 g = 10 log 2 50010 2 499 ln10 = 10 log 2 500 2 499 ln 10 = 10 1 081.904Chapter 21P21.64(a)627The effect of high angular speed is like the effect of a very high gravitational field on anatmosphere. The result is:The larger-mass molecules settle to the outside while the region at smaller r has a higherconcentration of low-mass molecules.(b)Consider a single kind of molecules, all of mass m. To cause the centripetal acceleration ofthe molecules between r and r + dr , the pressure must increase outward according to Fr = mar . Thus,afbge jPA P + dP A = nmA dr r 2where n is the number of molecules per unit volume and A is the area of any cylindricalsurface. This reduces to dP = nm 2 rdr .But also P = nk BT , so dP = k BTdn . Therefore, the equation becomeszznrdn m 2dn m 2==rdr givingrdr ornk BTnk BT 0n0afln nnn0F IGH JKm 2 r 2=k BT 2FG n IJ = m rH n K 2k T2ln0P21.65Then,=0and solving for n: n = n 0 e mrB22First find v av as v av =2v av2r2 2 2 k BT.z1m.v 2 N v dv . Let a =N02 k BT4 N 1 2 a 3 2N0zv 4 e av2dv= 4 a 3 2 1 2 3 k BT=am38a22The root-mean square speed is then v rms = v av =3 k BT.mTo find the average speed, we havev av*P21.66e4Na 3 2 1 21=vN v dv =N0Nzjz2v 3 e av dv =04 a 3 2 1 2=2a 28 k BT.mdPfor the function implied by PV = nRT = constant , and also for the differentdVfunction implied by PV = constant . We can use implicit differentiation:We want to evaluateFrom PV = constantFrom PV = constantTherefore,The theorem is proved.PFG dP IJH dV KFG dP IJH dV KdVdP+V=0dVdVPV 1 + V FG dP IJH dV KdP=0dV=adiabatFG dP IJH dV Kisotherm=isotherm=adiabatPVPV628P21.67The Kinetic Theory of Gasesejej1.013 10 5 Pa 5.00 10 3 m 3PV== 0.203 molRT8.314 J mol K 300 K(a)n=(b)TB = TAbfgaFG P IJ = 300 K FG 3.00 IJ =H 1.00 KHP KB900 KATC = TB = 900 KVC = VA(c)C15.0 LAaFIG. P21.67fbfgaga33nRTA = 0.203 mol 8.314 J mol K 300 K = 760 J2233= Eint, C = nRTB = 0.203 mol 8.314 J mol K 900 K = 2.28 kJ22Eint, A =Eint, BFG T IJ = 5.00 LFG 900 IJ =H 300 KHT Ka(d)ABCP (atm)1.003.001.00fbV(L)5.005.0015.00fEint (kJ)0.7602.282.28T(K)300900900(e)For the process AB, lock the piston in place and put the cylinder into an oven at 900 K. ForBC, keep the sample in the oven while gradually letting the gas expand to lift a load on thepiston as far as it can. For CA, carry the cylinder back into the room at 300 K and let the gascool without touching the piston.(f)For AB:afEint = Eint, B Eint, A = 2.28 0.760 kJ = 1.52 kJW= 0Q = Eint W = 1.52 kJFor BC:FG V IJHV KEint = 0 , W = nRTB lnafbCBgaf a fW = 0. 203 mol 8.314 J mol K 900 K ln 3.00 = 1.67 kJQ = Eint W = 1.67 kJFor CA:afW = PV = nRT = a0.203 molfb8.314 J mol K ga 600 K f =Eint = Eint, A Eint, C = 0.760 2.28 kJ = 1.52 kJ1.01 kJQ = Eint W = 1.52 kJ 1.01 kJ = 2.53 kJ(g)We add the amounts of energy for each process to find them for the whole cycle.Q ABCA = +1.52 kJ + 1.67 kJ 2.53 kJ = 0.656 kJW ABCA = 0 1.67 kJ + 1.01 kJ = 0.656 kJb E gint ABCA= +1.52 kJ + 0 1.52 kJ = 0P21.68(a)00moleculesb10 000 g gFGH 1.18.0mol IJK FGH 6.02 10 mol IJK =g1.0023Chapter 216293.34 10 26 molecules(b)After one day, 10 1 of the original molecules would remain. After two days, the fractionwould be 10 2 , and so on. After 26 days, only 3 of the original molecules would likelyremain, and after 27 days , likely none.(c)The soup is this fraction of the hydrosphere:F 10.0 kg I .GH 1.32 10 kg JK21Therefore, todays soup likely contains this fraction of the original molecules. The number oforiginal molecules likely in the pot again today is:F 10.0 kg I 3.34 10GH 1.32 10 kg JK e21P21.6926jmolecules = 2.53 10 6 molecules .1GmMGM. Since the free-fall acceleration at the surface is g = 2 , this canmv 2 =2RERE1GmM2also be written as: mv == mgRE .2RE(a)For escape,(b)For O 2 , the mass of one molecule ism=0.032 0 kg mol6.02 10 23 molecules mol= 5.32 10 26 kg molecule .FG 3 k T IJ , the temperature isH 2 Ke5.32 10 kgje9.80 m s je6.37 10 mj =mgR=T=15 k15e1.38 10J mol K jBThen, if mgRE = 102662E23BP21.70(a)1.60 10 4 K .For sodium atoms (with a molar mass M = 32.0 g mol )13mv 2 = k BT22FG IJH K1 M3v 2 = k BT2 NA2v rms =(b)t=dv rms=3 RT=M0.010 m= 20 ms0.510 m sbge3 8.314 J mol K 2.40 10 4 K23.0 103kgj=0.510 m s630The Kinetic Theory of GasesANSWERS TO EVEN PROBLEMSP21.217.6 kPaP21.42819CP21.45.05 10 21 J moleculeP21.44(a) see the solution; (b) 8.31 kmP21.66.64 10 27 kgP21.46(a) 5.63 10 18 m; 1.00 10 9 yr ;(b) 5.63 10 12 m; 1.00 10 3 yrP21.8477 m sP21.48193 molecular diametersP21.10(a) 2.28 kJ; (b) 6.21 10 21 JP21.50(a) 7.89 10 26 molecules; (b) 37.9 kg ;P21.1274.8 JP21.147.52 LP21.16(a) 118 kJ ; (b) 6.03 10 3 kgP21.18(a) 719 J kg K ; (b) 0.811 kg ; (c) 233 kJ;(d) 327 kJ(c) 6.07 10 21 J molecule ; (d) 503 m s;(e) 7.98 MJ ; (f) 7.98 MJ13.5 PVP21.22(a) 4Ti ; (b) 9 1 mol RTiP21.24(a) 5.15 10 5 m 3 ; (b) 560 K ; (c) 2.24 KP21.28(a) see the solution; (b) 2.19Vi ; (c) 3Ti ;(d) Ti ; (e) 0.830 PViiP21.3225.0 kWP21.34see the solutionP21.36(a) No atom, almost all the time;(b) 2.70 10 20P21.38(a) 1.03 ; (b)P21.40132 m s(a) 300 K ; (b) 1.00 atmP21.565.74 10 6 PaP21.58(a) see the solution; (b) 5.1 10 2 m s;(c) v av = 575 m s ; v rms = 624 m s ; (d) 44%P21.60(a) see the solution; (b) 344 m s nearlyagreeing with the tabulated value;(c) see the solution; somewhat smallerthan eachP21.620.296Csee the solutionP21.66see the solutionP21.68(a) 3.34 10 26 molecules ; (b) during the27th day; (c) 2.53 10 6 moleculesP21.70(a) 0.510 m s ; (b) 20 ms(a) 1.55 ; (b) 0.127 m3P21.30P21.54(a) 0.118 ; (b) 2.35 ; (c) 0; 135 J ; 135 JP21.26(a) 3.65 v; (b) 3.99 v; (c) 3.00 v;mv 2; (e) 7.98mv 2(d) 106VP21.64P21.20P21.52a35fClFGHIJK22Heat Engines, Entropy, and theSecond Law of ThermodynamicsCHAPTER OUTLINE22.122.222.322.422.522.622.722.8Heat Engines and theSecond Law ofThermodynamicsHeat Pumps andRefrigeratorsReversible and IrreversibleProcessesThe Carnot EngineGasoline and Diesel EnginesEntropyEntropy Changes inIrreversible ProcessesEntropy on a MicroscopicScaleANSWERS TO QUESTIONSQ22.1First, the efficiency of the automobile engine cannot exceed theCarnot efficiency: it is limited by the temperature of burningfuel and the temperature of the environment into which theexhaust is dumped. Second, the engine block cannot beallowed to go over a certain temperature. Third, any practicalengine has friction, incomplete burning of fuel, and limits setby timing and energy transfer by heat.Q22.2It is easier to control the temperature of a hot reservoir. If itcools down, then heat can be added through some externalmeans, like an exothermic reaction. If it gets too hot, then heatcan be allowed to escape into the atmosphere. To maintainthe temperature of a cold reservoir, one must remove heat ifthe reservoir gets too hot. Doing this requires either an evencolder reservoir, which you also must maintain, or anendothermic process.Q22.3A higher steam temperature means that more energy can be extracted from the steam. For aconstant temperature heat sink at Tc , and steam at Th , the efficiency of the power plant goes asTh TcT= 1 c and is maximized for a high Th .ThThQ22.4No. Any heat engine takes in energy by heat and must also put out energy by heat. The energy thatis dumped as exhaust into the low-temperature sink will always be thermal pollution in the outsideenvironment. So-called steady growth in human energy use cannot continue.Q22.5No. The first law of thermodynamics is a statement about energy conservation, while the second is astatement about stable thermal equilibrium. They are by no means mutually exclusive. For theparticular case of a cycling heat engine, the first law implies Q h = Weng + Q c , and the second lawimplies Q c > 0.Q22.6Take an automobile as an example. According to the first law or the idea of energy conservation, itmust take in all the energy it puts out. Its energy source is chemical energy in gasoline. During thecombustion process, some of that energy goes into moving the pistons and eventually into themechanical motion of the car. Clearly much of the energy goes into heat, which, through the coolingsystem, is dissipated into the atmosphere. Moreover, there are numerous places where friction, bothmechanical and fluid, turns mechanical energy into heat. In even the most efficient internalcombustion engine cars, less than 30% of the energy from the fuel actually goes into moving the car.The rest ends up as useless heat in the atmosphere.631632Heat Engines, Entropy, and the Second Law of ThermodynamicsQ22.7Suppose the ambient temperature is 20C. A gas can be heated to the temperature of the bottom ofthe pond, and allowed to cool as it blows through a turbine. The Carnot efficiency of such an engineT 80== 22%.is about e c =Th 373Q22.8No, because the work done to run the heat pump represents energy transferred into the house byheat.Q22.9A slice of hot pizza cools off. Road friction brings a skidding car to a stop. A cup falls to the floor andshatters. Your cat dies. Any process is irreversible if it looks funny or frightening when shown in avideotape running backwards. The free flight of a projectile is nearly reversible.Q22.10Below the frost line, the winter temperature is much higher than the air or surface temperature. Theearth is a huge reservoir of internal energy, but digging a lot of deep trenches is much moreexpensive than setting a heat-exchanger out on a concrete pad. A heat pump can have a muchhigher coefficient of performance when it is transferring energy by heat between reservoirs at closeto the same temperature.Q22.11(a)When the two sides of the semiconductor are at different temperatures, an electric potential(voltage) is generated across the material, which can drive electric current through anexternal circuit. The two cups at 50C contain the same amount of internal energy as the pairof hot and cold cups. But no energy flows by heat through the converter bridging betweenthem and no voltage is generated across the semiconductors.(b)A heat engine must put out exhaust energy by heat. The cold cup provides a sink to absorboutput or wasted energy by heat, which has nowhere to go between two cups of equallywarm water.Q22.12Energy flows by heat from a hot bowl of chili into the cooler surrounding air. Heat lost by the hotstuff is equal to heat gained by the cold stuff, but the entropy decrease of the hot stuff is less than theentropy increase of the cold stuff. As you inflate a soft car tire at a service station, air from a tank athigh pressure expands to fill a larger volume. That air increases in entropy and the surroundingatmosphere undergoes no significant entropy change. The brakes of your car get warm as you cometo a stop. The shoes and drums increase in entropy and nothing loses energy by heat, so nothingdecreases in entropy.Q22.13(a)For an expanding ideal gas at constant temperature, S =(b)For a reversible adiabatic expansion Q = 0 , and S = 0 . An ideal gas undergoing anirreversible adiabatic expansion can have any positive value for S up to the value given inpart (a).FG IJH KVQ= nR ln 2 .V1TQ22.14The rest of the Universe must have an entropy change of +8.0 J/K, or more.Q22.15Even at essentially constant temperature, energy must flow by heat out of the solidifying sugar intothe surroundings, to raise the entropy of the environment. The water molecules become lessordered as they leave the liquid in the container to mix into the whole atmosphere andhydrosphere. Thus the entropy of the surroundings increases, and the second law describes thesituation correctly.Chapter 22633Q22.16To increase its entropy, raise its temperature. To decrease its entropy, lower its temperature.Remove energy from it by heat is not such a good answer, for if you hammer on it or rub it with ablunt file and at the same time remove energy from it by heat into a constant temperature bath, itsentropy can stay constant.Q22.17An analogy used by Carnot is instructive: A waterfall continuously converts mechanical energy intointernal energy. It continuously creates entropy as the organized motion of the falling water turnsinto disorganized molecular motion. We humans put turbines into the waterfall, diverting some ofthe energy stream to our use. Water flows spontaneously from high to low elevation and energyspontaneously flows by heat from high to low temperature. Into the great flow of solar radiationfrom Sun to Earth, living things put themselves. They live on energy flow, more than just on energy.A basking snake diverts energy from a high-temperature source (the Sun) through itself temporarily,before the energy inevitably is radiated from the body of the snake to a low-temperature sink (outerspace). A tree builds organized cellulose molecules and we build libraries and babies who look liketheir grandmothers, all out of a thin diverted stream in the universal flow of energy crashing downto disorder. We do not violate the second law, for we build local reductions in the entropy of onething within the inexorable increase in the total entropy of the Universe. Your roommates exerciseputs energy into the room by heat.Q22.18(a)Entropy increases as the yeast dies and as energy is transferred from the hot oven into theoriginally cooler dough and then from the hot bread into the surrounding air.(b)Entropy increases some more as you metabolize the starches, converting chemical energyinto internal energy.Q22.19Either statement can be considered an instructive analogy. We choose to take the first view. Allprocesses require energy, either as energy content or as energy input. The kinetic energy which itpossessed at its formation continues to make the Earth go around. Energy released by nuclearreactions in the core of the Sun drives weather on the Earth and essentially all processes in thebiosphere. The energy intensity of sunlight controls how lush a forest or jungle can be and howwarm a planet is. Continuous energy input is not required for the motion of the planet. Continuousenergy input is required for life because energy tends to be continuously degraded, as heat flowsinto lower-temperature sinks. The continuously increasing entropy of the Universe is the index toenergy-transfers completed.Q22.20The statement is not true. Although the probability is not exactly zero that this will happen, theprobability of the concentration of air in one corner of the room is very nearly zero. If some billionsof molecules are heading toward that corner just now, other billions are heading away from thecorner in their random motion. Spontaneous compression of the air would violate the second law ofthermodynamics. It would be a spontaneous departure from thermal and mechanical equilibrium.Q22.21Shaking opens up spaces between jellybeans. The smaller ones more often can fall down into spacesbelow them. The accumulation of larger candies on top and smaller ones on the bottom implies asmall increase in order, a small decrease in one contribution to the total entropy, but the second lawis not violated. The total entropy increases as the system warms up, its increase in internal energycoming from the work put into shaking the box and also from a bit of gravitational energy loss as thebeans settle compactly together.634Heat Engines, Entropy, and the Second Law of ThermodynamicsSOLUTIONS TO PROBLEMSSection 22.1Heat Engines and the Second Law of ThermodynamicsWengP22.2e=Q c = Q h Weng = 360 J 25.0 J = 335 JQh=25.0 J= 0.069 4 or 6.94%360 J(a)(b)P22.1Weng = Q h Q c = 200 Je=WengQc=1QhQh(1)= 0.300(2)From (2), Q c = 0.700 Q h(3)Solving (3) and (1) simultaneously,we have(a)(b)P22.3Q h = 667 J andQ c = 467 J .(a)We have e =WengQh=Qh QcQh=1QcQh= 0.250with Q c = 8 000 J, we have Q h = 10.7 kJ(b)Weng = Q h Q c = 2 667 Jand from P =*P22.4Wengt, we have t =WengP=2 667 J= 0.533 s .5 000 J sWe have Q hx = 4Q hy , Weng x = 2Weng y and Q cx = 7Q cy . As well as Q hx = Weng x + Q cx andQ hy = Weng y + Q cy . Substituting, 4Q hy = 2Weng y + 7Q cy4Q hy = 2Weng y + 7Q hy 7Weng y5Weng y = 3Q hy(b)ey =(a)ex =Weng yQ hyWeng xQ hx==3= 60.0%52Weng y4Q hy=af20.600 = 0.300 = 30.0%4Chapter 22*P22.5(a)The input energy each hour ise7.89 103jbg 601min = 1.18 10 J hhF 1 L IJ = 29.4 L hJ hjGH 4.03 10 J K9J revolution 2 500 rev mineimplying fuel input 1.18 10 97Q h = Weng + Q c . For a continuous-transfer process we may divide by time to have(b)Q h Weng Q c=+tttWeng Q h Q cUseful power output ==tttF 7.89 10 J 4.58 10 J I 2 500 rev 1 min = 1.38 10GH revolution revolution JK 1 min 60 sF 1 hp IJ = 185 hpP = 1.38 10 WGH 746 W KP1.38 10 J s F 1 rev IP = ==b2 500 rev 60 sg GH 2 rad JK = 527 N mQ4.58 10 J F 2 500 rev I=GJ = 1.91 10 Wtrevolution H 60 s K3=5eng3c5ejejQ c = mL f = 15 10 3 kg 1.18 10 4 J kg = 177 JThe heat to melt 15.0 g of Hg isThe energy absorbed to freeze 1.00 g of aluminum isejejQ h = mL f = 10 3 kg 3.97 10 5 J / kg = 397 JWeng = Q h Q c = 220 Jand the work output ise=Section 22.2WengQh=220 J= 0.554 , or 55.4%397 JTh Tc 933 K 243.1 K== 0.749 = 74.9%933 KThThe theoretical (Carnot) efficiency isP22.7Weng(d)P22.655eng(c)3Heat Pumps and RefrigeratorsbgCOP refrigerator =QcW(a)If Q c = 120 J and COP = 5.00 , then W = 24.0 J(b)Heat expelled = Heat removed + Work done.Q h = Q c + W = 120 J + 24 J = 144 J635636P22.8Heat Engines, Entropy, and the Second Law of ThermodynamicsQcQ. Therefore, W = c .W3.00The heat removed each minute isCOP = 3.00 =bbgbgbf bfgagageQC= 0.030 0 kg 4 186 J kg C 22.0 C + 0.030 0 kg 3.33 10 5 J kgt+ 0.030 0 kg 2 090 J kg C 20.0 C = 1.40 10 4 J minor,Qc= 233 J s.tThus, the work done per sec = P =233 J s= 77.8 W .3.00(a)FG 10.0 Btu IJ FG 1055 J IJ FG 1 h IJ FG 1 W IJ =H h W K H 1 Btu K H 3 600 s K H 1 J s K(b)Coefficient of performance for a refrigerator:(c)P22.9jWith EER 5, 52.93Btu 10 000 Btu h=:Ph WaCOPfP=refrigerator10 000 Btu h5 BtuhW= 2 000 W = 2.00 kWafbgCost = e3.00 10 kWhjb0.100 $ kWhg = $300P t = 2.00 kW 1 500 h = 3.00 10 3 kWhEnergy purchased is3With EER 10, 10Btu 10 000 Btu h=:Ph WP=10 000 Btu h10 BtuhW= 1 000 W = 1.00 kWafb gCost = e1.50 10 kWhjb0.100 $ kWhg = $150P t = 1.00 kW 1 500 h = 1.50 10 3 kWhEnergy purchased is3Thus, the cost for air conditioning isSection 22.3half as much with EER 10Reversible and Irreversible ProcessesNo problems in this sectionSection 22.4P22.10The Carnot EngineWengTandWhen e = e c , 1 c =ThQh(a)QhWengtQht=1TcThFH IK t e1.50 10 Wjb3 600 sg==Wengt5T11 Tch293773Q h = 8.69 10 8 J = 869 MJ(b)Qc = Qh F W I t = 8.69 10 e1.50 10 jb3 600g = 3.30 10GH t JKeng858J = 330 MJChapter 22P22.11Tc = 703 KTh = 2 143 KT 1 440== 67.2%Th 2 143(a)ec =(b)Q h = 1.40 10 5 J , Weng = 0.420 Q hP=Wengt=5.88 10 4 J= 58.8 kW1sT 120 K== 0.253Th 473 KThe Carnot efficiency of the engine isec =At 20.0% of this maximum efficiency,e = 0.200 0.253 = 0.050 6From the definition of efficiencyWeng = Q h eandP22.12Qh =aIsothermal expansion atWenge=10.0 kJ= 197 kJ0.050 6Th = 523 KIsothermal compression atP22.13fTc = 323 KGas absorbs 1 200 J during expansion.(a)(b)FG T IJ = 1 200 JFG 323 IJ = 741 JH 523 KHT K= Q Q = b1 200 741g J = 459 JcQc = QhWenghhcTcThec = 1 0.300 = 1 From which,*P22.15We useasP22.14Th = 819 K = 546 C573 KThThe efficiency isec = 1 ThenTc=ThQht(a)QctQht=aaffQ c Th273 + 100 K= 15. 4 W= 19.6 Wt Tc273 + 20 KQ h = Weng + Q cThe useful power output is(b)QcTc=1ThQhQh =F Q I t = mLGH t JKhVWengtm==QhtQct= 19.6 W 15.4 W = 4.20 WgFGHIJKQ h t3 600 s= 19.6 J s= 3.12 10 2 kgt LV2.26 10 6 J kgb637638Heat Engines, Entropy, and the Second Law of ThermodynamicsaaffThe Carnot summer efficiency ise c ,s = 1 273 + 20 KTc=1= 0.530Th273 + 350 KAnd in winter,P22.16ec ,w = 1 283= 0.546623FG 0.546 IJ = 0.330 or 33.0%H 0.530 KF P V I = F PV I .In an adiabatic process, P V = PV . Also, GH T JK GH T JKF P Ib gDividing the second equation by the first yields T = T G JHPKThen the actual winter efficiency is0.320P22.17(a)fffi ifi ififSince = 1 .i 1 25= = 0.400 and we havefor Argon,53bT f = 1 073 K(b)ifgFGH 1300 10 Pa IJK.50 10 Pa30. 4006= 564 K .Eint = nCV T = Q Weng = 0 Weng , so Weng = nCV T ,and the power output isP==Wengt=nCV Tortb80.0 kg ge1.00 mol0 .039 9 kgjc hb8.314 J mol K gb564 1 073gK3260.0 s5P = 2.12 10 W = 212 kWTc564 K=1= 0.475 or 47.5%1 073 KTh(c)P22.18eC = 1 (a)emax = 1 (b)P=WengtTc278=1= 5.12 10 2 = 5.12%Th293= 75.0 10 6 J sFrom e =(c)jbegWeng = 75.0 10 6 J s 3 600 s h = 2.70 10 11 J hTherefore,WengQhwe findQh =Wenge=2.70 10 11 J h5.12 10 2= 5.27 10 12 J h = 5.27 TJ hAs fossil-fuel prices rise, this way to use solar energy will become a good buy.Chapter 22*P22.19(a)e=Weng1 + Weng2e1 Q1 h + e 2 Q 2 hQ h1=Q h1Now Q 2 h = Q1 c = Q1 h Weng 1 = Q h1 e1Q1 h .So e =(b)be 1 Q 1 h + e 2 Q 1 h e1 Q 1 hQ1 he = e 1 + e 2 e1 e 2 = 1 g=e1 + e 2 e 1 e 2 .FGHTiTT+1 c 1 iThTiThIJ FG 1 T IJ = 2 TKH T K TciihTcT T TT1+ i + c c = 1 cTiTh Ti ThThThe combination of reversible engines is itself a reversible engine so it has the Carnotefficiency.(c)With Weng2 = Weng1 , e =1FGHTcT= 2 1 iThThWeng1 + Weng2Q1 hIJK2TTc=1 iThTh2Ti = Th + Tc0Ti =(d)b1Th + Tc2e1 = e 2 = 1 gTiT=1 cThTiTi2 = Tc Thb gTi = ThTcP22.2012The work output is Weng =We are told e =0.200 =andWengQhbb5.00 m s1mtQh2gbb5.00 m s1Substitute Q h = m t20.200Then,26.50 m s300 K 1= mteC = 1 2ThQh300 K= 0.2001ThFGGH1212gg2.bgm b5.00 m sgt.2m t 6.50 m s300 K= 0.338Th300 KTh == 453 K0.6621g12m train 5.00 m s .222IJJK=2Weng1Q1 h= 2 e1639640P22.21Heat Engines, Entropy, and the Second Law of ThermodynamicsFor the Carnot engine, e c = 1 Tc300 K=1= 0.600 .750 KThWengAlso,ec =soQh =andQ c = Q h Weng = 250 J 150 J = 100 J .Qh.Wengec=150 J= 250 J .0.600FIG. P22.21(a)Qh =WengeS=150 J= 214 J0.700Q c = Q h Weng = 214 J 150 J = 64.3 J(b)Q h, net = 214 J 250 J = 35.7 JQ c, net = 64.3 J 100 J = 35.7 JThe net flow of energy by heat from the cold to the hotreservoir without work input, is impossible.WengQ c = Q h Weng =Weng =and(d)For engine S:so(c)FIG. P22.21(b)Q h = Q c + Weng = 233 J + 100 J = 333 J .Qc1eS1=eS Weng .100 J= 233 J .10 .700 1Q h, net = 333 J 250 J = 83.3 JWnet = 233 J 150 J = 83.3 JQ c,net = 0The output of 83.3 J of energy from the heat engineby work in a cyclic process without any exhaust byheat is impossible.FIG. P22.21(d)(e)Both engines operate in cycles, soSS = SCarnot = 0 .For the reservoirs,S h = QhThThus, Stotal = SS + SCarnot + S h + S c = 0 + 0 A decrease in total entropy is impossible.and S c = +QcTc.83.3 J0+= 0.111 J K .750 K 300 KChapter 22P22.22(a)641First, consider the adiabatic process D A :FG V IJ = 1 400 kPaFG 10.0 L IJH 15.0 L KHV KFG nRT IJ V = FG nRT IJ VHV K HV KF V IJ = 720 K FG 10.0 IJ =T =T GH 15.0 KHV KPD VD = PA VA soPD = PADAlsoD53AADAA 1orDA= 712 kPa .D23A549 K .DNow, consider the isothermal process C D : TC = TD = 549 K .FG V IJ = LMP FG V IJ OPFG V IJ = P VH V K MN H V K PQH V K V V1 400 kPaa10.0 L f== 445 kPa24.0 La15.0 L fPC = PDDCADAADCCA 1D53PC23Next, consider the adiabatic process B C : PBVB = PC VC .But, PC =PA VAVC VD 1FG V IJ V = F P V I V which reduces to V = V VVH V K GH V V JKF V IJ = 1 400 kPaFG 10.0 L IJ = 875 kPa .Finally, P = P GH 16.0 L KHV KAHence, PABBABACA 1DBCA C15.0 LDABStateABCD(b)FG V IJ .HV K10.0 La 24.0 L f== 16.0 Lfrom above. Also considering the isothermal process, PB = PAP(kPa)1 400875445712For the isothermal process A B :so Q = W = nRT lnV(L)10.016.024.015.0T(K)720720549549Eint = nCV T = 0FG V IJ = 2.34 molb8.314 J mol K ga720 K f lnFG 16.0 IJ =H 10.0 KHV KB+6.58 kJ .AFor the adiabatic process B C :bLM 3 b8.314 J mol K gOPa549 720f K =N2Q= 0 + a 4.98 kJf = 4.98 kJ .gEint = nCV TC TB = 2.34 moland W = Q + Eintcontinued on next pageQ= 04.98 kJAB.642Heat Engines, Entropy, and the Second Law of ThermodynamicsEint = nCV T = 0For the isothermal process C D :and Q = W = nRT lnFG V IJ = 2.34 molb8.314 J mol K ga549 K f lnFG 15.0 IJ =H 24.0 KHV KD5.02 kJ .CFinally, for the adiabatic process D A :bgEint = nCV TA TD = 2.34 molQ= 0LM 3 b8.314 J mol K gOPa720 549f K =N2Q+4.98 kJand W = Q + Eint = 0 + 4.98 kJ = +4.98 kJ .ProcessABBCCDD AABCDAQ(kJ)+6.5805.020+1.56W(kJ)6.584.98+5.02+4.981.56Eint (kJ)04.980+4.980The work done by the engine is the negative of the work input. The output work Weng isgiven by the work column in the table with all signs reversed.(c)e=WengQhec = 1 P22.23aCOPfrefrigP22.24aCOPfheat pumpP22.25(a)==W ABCD 1.56 kJ== 0.237 or 23.7%6.58 kJQ ABTc549=1= 0.237 or 23.7%720ThTc270== 9.00T 30.0=Qc + WW=Th 295== 11.8T25For a complete cycle, Eint = 0 andW = Qh Q c = Q cLM bQ g OPMN Q 1PQ .hcWe have already shown that for a Carnot cycle (and only for a Carnot cycle)Therefore,(b)W = QcLM T T OPN T QhcCOP =Qc=Th.Tc.cWe have the definition of the coefficient of performance for a refrigerator,Using the result from part (a), this becomesQhTc.Th TcCOP =QcW.Chapter 22P22.26643COP = 0.100COPCarnot cycleQhorWQhWFQ IFI1= 0.100GGH W JKH Carnot efficiency JKF T IJ = 0.100FG 293 K IJ = 1.17= 0.100GH 293 K 268 K KHT T Kh= 0.100Carnot cyclehhcFIG. P22.26Thus, 1.17 joules of energy enter the room by heat for each joule of work done.P22.27P22.28QcTc4.00== 0.013 8 =T 289W W = 72.2 J per 1 J energy removed by heat.aCOPfCarnot refrig=A Carnot refrigerator runs on minimum power.Q t Q tQQFor it: h = c so h = c .ThTcThTcSolving part (b) first:FG IJ bH KIJ eKgFGHjFGHIJK(b)(a)P22.29Q h Q c Th298 K1h== 8.00 MJ h= 8.73 10 6 J h= 2. 43 kW273 K3 600 stt Tc8.00 10 6 J hW Qh Qc== 2.43 kW = 204 Wttt3 600 s he=W= 0.350QhW = 0.350Q hQh = W + QcQ c = 0.650Q hQ c 0.650Q h== 1.86COP refrigerator =W 0.350Q hb*P22.30gTo have the same efficiencies as engines, 1 between reservoirs with the same ratiobecomesThpThp Tcp=TcpThpTcpThp==1Tcrthe pump and refrigerator must operateThrTcr, which we define as r. Now COPp = 1.50COPrThrThp3 Tcr23r23 rThror=,r= .,=2 Thr Tcr3Thp rThp 2 Thr rThr 1 r 1 r(a)COPr =2r= 31r 1(b)COPp =11== 3.0021r 1 3(c)e=1r =123= 2.002= 33.3%3644Heat Engines, Entropy, and the Second Law of ThermodynamicsSection 22.5P22.31(a)Gasoline and Diesel EnginesPVi = Pf V fiFV IF 50.0 cm I= P G J = e3.00 10 PajGH 300 cm JKHV KPf(b)i36izVi= 244 kPa3fW = PdV1.40P = PiViFG V IJHVKiIntegrating,F 1 IJ PV LM1 FG V IJW =GH 1 K MN H V K 1ii ifOPPQ = a2.50fe3.00 10 Paje5.00 1065= 192 JP22.32Compression ratio = 6.00 , = 1.40FG V IJHV KF 1 IJe=1GH 6.00 K 1(a)2Efficiency of an Otto-engine e = 1 1(b)P22.330 . 400= 51. 2% .If actual efficiency e = 15.0% losses in system are e e = 36.2% .eOtto = 1 1bV V g1 1=121a6.20fb7 51g=11a6.20f0. 400eOtto = 0.518We have assumed the fuel-air mixture to behave like a diatomic gas.Now e =WengQh=Weng tQh t746 W 1 hpQ h Weng t== 102 hpte0.518Qh= 146 kWtQ h = Weng + Q cQctQct=Q h WengttF 746 W I =GH 1 hp JK= 146 10 3 W 102 hp70.8 kWL F 50.0 cm I OPm jM1 GMN H 300 cm JK PQ3330. 400Chapter 22P22.34(a), (b) The quantity of gas isn=ejej100 10 3 Pa 500 10 6 m 3PA VA= 0.020 5 mol=RTA8.314 J mol K 293 KEint, A =bfga555nRTA = PA VA = 100 10 3 Pa 500 10 6 m 3 = 125 J222ejeFG V IJ = e100 10 Paja8.00fHV KIn process AB, PB = PAe3Aj1.40= 1.84 10 6 PaBjej1.84 10 6 Pa 500 10 6 m3 8.00PBVB= 673 K=TB =nR0.020 5 mol 8.314 J mol KEint, B =sobbgbggbfga55nRTB = 0.020 5 mol 8.314 J mol K 673 K = 287 J22Eint, AB = 287 J 125 J = 162 J = Q Wout = 0 WoutW AB = 162 JProcess BC takes us to:bgbgbg0.020 5 mol 8.314 J mol K 1 023 KnRTC== 2.79 10 6 PaVC62.5 10 6 m 355Eint, C = nRTC = 0.020 5 mol 8.314 J mol K 1 023 K = 436 J22Eint, BC = 436 J 287 J = 149 J = Q Wout = Q 0PC =bgbgbgQBC = 149 JIn process CD:FG V IJ = e2.79 10 PajFG 1 IJ = 1.52 10 PaH 8.00 KHV Ke1.52 10 Paje500 10 m j = 445 KP V==nRb0.020 5 molgb8.314 J mol K g55= nRT = b0.020 5 molgb8.314 J mol K ga 445 K f =22PD = PC6C5D5TD1.4063D DEint, DD190 JEint, CD = 190 J 436 J = 246 J = Q Wout = 0 WoutWCD = 246 JandEint, DA = Eint, A Eint, D = 125 J 190 J = 65.0 J = Q Wout = Q 0QDA = 65.0 Jcontinued on next page645646Heat Engines, Entropy, and the Second Law of ThermodynamicsFor the entire cycle, Eint, net = 162 J + 149 246 65.0 = 0 . The net work isWeng = 162 J + 0 + 246 J + 0 = 84.3 JQ net = 0 + 149 J + 0 65.0 J = 84.3 JThe tables look like:StateABCDAT(K)2936731 023445293P(kPa)1001 8402 790152100V(cm3 )50062.562.5500500ProcessABBCCDDAABCDAQ(J)0149065.084.3output W(J)1620246084.3Eint (J)16214924665.00Eint (J)125287436190125(c)The input energy is Q h = 149 J , the waste is Q c = 65.0 J , and Weng = 84.3 J .(d)The efficiency is: e =(e)Let f represent the angular speed of the crankshaft. ThenWengQh=84.3 J= 0.565 .149 Jobtain work in the amount of 84.3 J/cycle:1 000 J s =f=Section 22.6P22.35FG f IJ b84.3 J cyclegH 2K2 000 J s= 23.7 rev s = 1.42 10 3 rev min84.3 J cycleEntropyFor a freezing process,S =bgefis the frequency at which we2j5Q 0.500 kg 3.33 10 J kg== 610 J K .273 KTChapter 22P22.36647At a constant temperature of 4.20 K,20.5 kJ kgLvQ==T4.20 K4.20 KS = 4.88 kJ kg KS =FG IJH KF 353 IJ = 46.6 cal K =S = 250 g b1.00 cal g C g lnGH 293 Kz zfP22.37S =i*P22.38(a)TfTfdQmcdT== mc lnTTTiTi195 J KThe process is isobaric because it takes place under constant atmospheric pressure. Asdescribed by Newtons third law, the stewing syrup must exert the same force on the air asthe air exerts on it. The heating process is not adiabatic (energy goes in by heat), isothermal(T goes up), isovolumetic (it likely expands a bit), cyclic (it is different at the end), orisentropic (entropy increases). It could be made as nearly reversible as you wish, by notusing a kitchen stove but a heater kept always just incrementally higher in temperaturethan the syrup. The process would then also be eternal, and impractical for food production.(b)The final temperature is220 F = 212 F + 8 F = 100 C + 8 FFG 100 0 C IJ = 104 C .H 212 32 F KFor the mixture,bgaQ = m1 c 1 T + m 2 c 2 T = 900 g 1 cal g C + 930 g 0.299 cal g C 104.4 C 23 C= 9.59 10 4 cal = 4.02 10 5 J(c)Consider the reversible heating process described in part (a):z zbffgbg TTTF 4.186 J IJ FG 1 C IJ lnFG 273 + 104 IJ= 900a1f + 930a0.299f bcal C gGH 1 cal K H 1 K K H 273 + 23 K= b 4 930 J K g0.243 = 1.20 10 J KS =idQ=Tm1 c 1 + m 2 c 2 dT= m1 c 1 + m 2 c 2 lni3fif648*P22.39Heat Engines, Entropy, and the Second Law of ThermodynamicsWe take data from the description of Figure 20.2 in section 20.3, and we assume a constant specificheat for each phase. As the ice is warmed from 12C to 0C, its entropy increases byzfS =izz273 K273 Kmc ice dTdQ273 K== mc ice T 1 dT = mc ice ln T 261 KTT261 K261 KbfgabgFGH FGH 273 IJK IJK261S = 0.027 0 kg 2 090 J kg C ln 273 K ln 261 K = 0.027 0 kg 2 090 J kg C lnS = 2.54 J KAs the ice melts its entropy change isej5Q mL f 0.027 0 kg 3.33 10 J kgS = === 32.9 J KTT273 KAs liquid water warms from 273 K to 373 K,zfS =iFG T IJ = 0.027 0 kgb4 186 J kg Cg lnFG 373 IJ = 35.3 J KH 273 KHT Kmc liquid dTf= mc liquid lnTiAs the water boils and the steam warms,S =S =FG IJH KTfmL v+ mc steam lnTTie0.027 0 kg 2.26 10 6 J kg373 Kj + 0.027 0 kgb2 010 J kg Cg lnFG 388 IJ = 164 J K + 2.14 J KH 373 KThe total entropy change isa2.54 + 32.9 + 35.3 + 164 + 2.14f J K =236 J K .We could equally well have taken the values for specific heats and latent heats from Tables 20.1 and20.2. For steam at constant pressure, the molar specific heat in Table 21.2 implies a specific heat of1 mol= 1 970 J kg K , nearly agreeing with 2 010 J kg K .35.4 J mol K0.018 kgbSection 22.7gFGHIJKEntropy Changes in Irreversible ProcessesFGH1 000 1 000Q 2 Q1=T2 T1290 5 700IJKP22.40S =J K = 3.27 J KP22.41The car ends up in the same thermodynamic state as it started, so it undergoes zero changes inentropy. The original kinetic energy of the car is transferred by heat to the surrounding air, addingto the internal energy of the air. Its change in entropy isS =12mv 2T=a f750 20.02932J K = 1.02 kJ K .Chapter 22P22.42649c iron = 448 J kg C ; c water = 4 186 J kg Cbgdi bgbgdQcold = Q hot :4.00 kg 4 186 J kg C T f 10.0 C = 1.00 kg 448 J kg C T f 900 Cwhich yieldsT f = 33.2 C = 306.2 KS =z306.2 K283 Kizc water m water dT 306.2 K c iron m iron dT+TT1 173 KFG 306.2 IJ + c m lnFG 306.2 IJH 283 KH 1 173 KS = b 4 186 J kg K gb 4.00 kg gb0.078 8g + b 448 J kg K gb1.00 kg ga 1.34fS = c water m water lnironironS = 718 J KP22.43Sitting here writing, I convert chemical energy, in ordered molecules in food, into internal energythat leaves my body by heat into the room-temperature surroundings. My rate of energy output isequal to my metabolic rate,2 500 kcal d =FGHIJK2 500 10 3 cal 4.186 J= 120 W .86 400 s1 calMy body is in steady state, changing little in entropy, as the environment increases in entropy at therateS Q T Q t 120 W==== 0. 4 W K ~ 1 W K .293 KttTWhen using powerful appliances or an automobile, my personal contribution to entropy productionis much greater than the above estimate, based only on metabolism.P22.44bV=(b)Eint = nCV T =(c)W =0(d)Sargon =bsogafgejF 40.0 gm I L 3 b8.314 J mol K gOa200 Cf =GH 39.9 g mol JK MN 2PQiS bath =2.50 kJQ = Eint = 2.50 kJFG IJH KF 40.0 g I L 3 b8.314 J mol K gO lnFG 273 IJ ==GPQ H 473 KH 39.9 g mol JK MN 2zf(e)gb40.0 g 8.314 J mol K 473 KnRTi= 39.4 10 3 m3 = 39.4 L=3Pi39.9 g mol 100 10 Pa(a)TfdQ= nCV lnTTi6.87 J K2.50 kJ= +9.16 J K273 KThe total change in entropy isStotal = Sargon + S bath = 6.87 J K + 9.16 J K = +2.29 J KStotal > 0 for this irreversible process.650P22.45Heat Engines, Entropy, and the Second Law of ThermodynamicsS = nR lnF V I = R ln 2 =GH V JKf5.76 J KiThere is no change in temperature .FIG. P22.45P22.46F V I = b0.044 0ga2fR ln 2GH V JKS = 0.088 0a8.314f ln 2 = 0.507 J KS = nR lnfiFIG. P22.46P22.47For any infinitesimal step in a process on an ideal gas,dEint = dQ + dW :dQ = dEint dW = nCV dT + PdV = nCV dT +anddQdTdV= nCV+ nRTTVIf the whole process is reversible,S =zfiAlso, from the ideal gas law,aS = 1.00 molTfTi=dQr=Tz FGHfinCVIJKnRTdVVFG IJH KPf V fPVii2 00fb0 040 0fLMN 3 b8.314 J mol K gOPQ lnFGH aa1..00fb0..025 0gg IJK + a1.00 molfb8.314 J mol K g lnFGH 0..040 0 IJK20 025 0= 18.4 J KP22.48F T I + nR lnF V IGH T JKGH V JKL5O F 2P 2V IJ + a1.00 molfb8.314 J mol K g lnFG 2V IJ= a1.00 molfM b8.314 J mol K gP lnGHVKN2Q H PV KS = nC V lnFG IJH KTfVfdTdV+ nR= nCV ln+ nR lnTVTiViffiiS = 34.6 J KChapter 22Section 22.8Entropy on a Microscopic ScaleA 12 can only be obtained one way 6 + 6A 7 can be obtained six ways: 6 + 1 , 5 + 2 , 4 + 3 , 3 + 4 , 2 + 5 , 1 + 6(a)The table is shown below. On the basis of the table, the most probable result of a toss is2 heads and 2 tails .(b)The most ordered state is the least likely state. Thus, on the basis of the table this iseither all heads or all tails .(c)P22.50(a)(b)P22.49The most disordered is the most likely state. Thus, this is 2 heads and 2 tails .ResultAll heads3H, 1T2H, 2T1H, 3TAll tailsPossible CombinationsHHHHTHHH, HTHH, HHTH, HHHTTTHH, THTH, THHT, HTTH, HTHT, HHTTHTTT, THTT, TTHT, TTTHTTTTTotal14641(a)ResultAll red2R, 1G1R, 2GAll greenPossible CombinationsRRRRRG, RGR, GRRRGG, GRG, GGRGGGTotal1331(b)P22.51651ResultAll red4R, 1G3R, 2GPossible CombinationsRRRRRRRRRG, RRRGR, RRGRR, RGRRR, GRRRRRRRGG, RRGRG, RGRRG, GRRRG, RRGGR,RGRGR, GRRGR, RGGRR, GRGRR, GGRRRGGGRR, GGRGR, GRGGR, RGGGR, GGRRG,GRGRG, RGGRG, GRRGG, RGRGG, RRGGGRGGGG, GRGGG, GGRGG, GGGRG, GGGGRGGGGGTotal152R, 3G1R, 4GAll green101051Additional ProblemsP22.52The conversion of gravitational potential energy into kinetic energy as the water falls is reversible.But the subsequent conversion into internal energy is not. We imagine arriving at the same finalstate by adding energy by heat, in amount mgy, to the water from a stove at a temperatureinfinitesimally above 20.0C. Then,S =zejejaf332dQ Q mgy 5 000 m 1 000 kg m 9.80 m s 50.0 m= === 8.36 10 6 J K .TTT293 K652P22.53Heat Engines, Entropy, and the Second Law of ThermodynamicsH ETso if all the electric energy is converted into internal energy, the steady-statetcondition of the house is described by H ET = Q .Pelectric =(a)Therefore,For a heat pump,(b)Pelectric =aCOPfQ= 5 000 Wt=CarnotTh 295 K== 10.92T27 KafActual COP = 0.6 10.92 = 6.55 =QhW=Q h tW tTherefore, to bring 5 000 W of energy into the house only requires input powerPheat pump =P22.54W Q h t 5 000 W=== 763 WtCOP6.56feQ c = mcT + mL + mcT =bgajbgaQ c = 0.500 kg 4 186 J kg C 10 C + 0.500 kg 3.33 10 5 J kg + 0.500 kg 2 090 J kg C 20 CQ c = 2.08 10 5 JQcWW=P22.55bg= COPc refrigerator =bQ c Th TcTcTcTh Tcg = e2.08 10 Jj 20.0 C a20.0 Cf =a273 20.0f K5S hot =1 000 J600 KScold =+750 J350 K(a)SU = S hot + Scold = 0.476 J K(b)ec = 1 T1= 0.417T2bgWeng = e c Q h = 0.417 1 000 J = 417 J(c)Wnet = 417 J 250 J = 167 JbgT1 SU = 350 K 0.476 J K = 167 J32.9 kJfChapter 22*P22.56(a)653The energy put into the engine by the hot reservoir is dQ h = mcdTh . The energy put into theLM FMN GHa fcold reservoir by the engine is dQ c = mcdTc = 1 e dQ h = 1 1 IJ OPmcdT . ThenK PQhdTc dTh=TcThzTcThdTdT=T T TTfTczTfh ln TlnTfTcT= ln T T fhTfTc= lnTfThT f2 = Tc Thb gT f = ThTc(b)d12idThen Q h = Weng + Q c .d i d i= mceT T T T T + T j= mceT 2 T T + T j = mce T T jFV IFor an isothermal process,Q = nRT lnG JHV KTherefore,Q = nRb3T g ln 2F 1IQ = nRbT g lnG JandH 2K3= nRbT 3T gFor the constant volume processes, Q = E23= nRb3T T gandQ = E2Weng = mc Th T f mc T f Tch(a)h chP22.57h chccch2c211i3i2int, 24int, 4iiiiThe net energy by heat transferred is thenQ = Q1 + Q 2 + Q 3 + Q 4or(b)iThe hot reservoir loses energy Q h = mc Th T f . The cold reservoir gains Q c = mc T f Tc .FIG. P22.57Q = 2nRTi ln 2 .A positive value for heat represents energy transferred into the system.Therefore,aQ h = Q1 + Q 4 = 3nRTi 1 + ln 2fSince the change in temperature for the complete cycle is zero,Eint = 0 and Weng = QTherefore, the efficiency isec =WengQh=2 ln 2Q== 0.273Q h 3 1 + ln 2af654P22.58Heat Engines, Entropy, and the Second Law of Thermodynamics(a)Wengt8= 1.50 10 WaelectricalL OP, Q = mL = MfMN 0.150 PQt ,Wengtand L = 33.0 kJ g = 33.0 10 6 J kgLM W t OP tN 0.150 Q Le1.50 10 Wjb86 400 s daygm== 2 620 metric tons day0.150e33.0 10 J kg je10 kg metric tonjCost = b$8.00 metric tongb 2 618 metric tons day gb365 days yr gm=eng86(b)3Cost = $7.65 million year(c)First find the rate at which heat energy is discharged into the water. If the plant is 15.0%efficient in producing electrical energy then the rate of heat production isQctThen,Qct==FWGH tengI FG 1 1IJ = e1.50 10 WjFG 1 1IJ = 8.50 10JK H e KH 0.150 K88W.mcTandtQc8.50 10 8 J sm= t == 4.06 10 4 kg s .t cT4 186 J kg C 5.00 CbP22.59WengT=ec = 1 c =ThQhQ h = Weng + Q c :WengtQht:QhtQctQctQ c = mcT :Qct==fgaP ThP=Th Tc1 Tc ThbQhtgWengt=P ThP TcP =Th TcTh Tc=FG m IJ cT = P TH t K T TchcP Tcm=Th Tc cTtbgejaf1.00 10 9 W 300 Km== 5.97 10 4 kg st 200 K 4 186 J kg C 6.00 CbgafChapter 22P22.60WengT=ec = 1 c =ThQhWengtQhtQhP==P ThTh Tce1 jF Q I P = P TQ=Gt H t JK T TtTcThchchcQ c = mcT , where c is the specific heat of water.QcTherefore,tm=tandP22.61(a)=FG m IJ cT = P TH t K T TchcP TcTh Tc cTbgaf af598.6 F = a98.6 32.0f C = a37.0 + 273.15f K = 310.15 K9dQdTF 310.15 IJ = 54.86 cal KS=z= b 453.6 g gb1.00 cal g K g z= 453.6 lnGH 274.82 KTTQa310.15 274.82f = 51.67 cal KS== a 453.6 fa1.00f35.0 F =535.0 32.0 C = 1.67 + 273.15 K = 274.82 K9310.15ice water274.82bodyTbody310.15Ssystem = 54.86 51.67 = 3.19 cal K(b)a453.6fa1fbTFg eja fb 274.82 = 70.0 10 3 1 310.15 TFgThus,b70.0 + 0.453 6g 10 T = a70.0fa310.15f + b0.453 6ga274.82f 103F3and TF = 309.92 K = 36.77 C = 98.19 FFG 309.92 IJ = 54.52 cal KH 274.82 KF 310.15 IJ = 51.93 cal K= e70.0 10 j lnGH 309.92 KSice water = 453.6 lnS body3Ssys = 54.52 51.93 = 2.59 cal K which is less than the estimate in part (a).655656P22.62Heat Engines, Entropy, and the Second Law of Thermodynamics(a)For the isothermal process AB, the work on the gas isW AB = PA VA lnFG V IJHV KBAe50 0j FGH 10..0 IJKjeW AB = 5 1.013 10 5 Pa 10.0 10 3 m3 lnW AB = 8.15 10 3 Jwhere we have used 1.00 atm = 1.013 10 5 Pa1.00 L = 1.00 10 3 m 3andjaeFIG. P22.62fWBC = PB V = 1.013 10 5 Pa 10.0 50.0 10 3 m3 = +4.05 10 3 JWCA = 0 and Weng = W AB WBC = 4.11 10 3 J = 4.11 kJ(b)Since AB is an isothermal process, Eint, AB = 0andQ AB = W AB = 8.15 10 3 JFor an ideal monatomic gas,CV =3R5Rand C P =22TB = TA =ejej1.013 10 5 50.0 10 3PBVB5.05 10 3==nRRRejej1.013 10 5 10.0 10 3PC VC1.01 10 3==TC =nRRRAlso,QCA = nC V T = 1.00FG 3 RIJ FG 5.05 10H 2 KH3so the total energy absorbed by heat is Q AB + QCA = 8.15 kJ + 6.08 kJ = 14.2 kJ .(c)QBC = nC P T =QBC =(d)e=a51.013 10 52WengQhe=f55nRT = PB VBC22j a10.0 50.0f 10WengQ AB + QCA=3= 1.01 10 4 J = 10.1 kJ4.11 10 3 J= 0.289 or 28.9%1.42 10 4 JIJK 1.01 10 3= 6.08 kJR657Chapter 22*P22.63Like a refrigerator, an air conditioner has as its purpose the removal of energy by heat from the coldreservoir.Tc280 K== 14.020 KTh TcIts ideal COP isCOPCarnot =(a)0.400 14.0 = 5.60 =a fIts actual COP is5.60QcQh Qc=Q c tQ h t Q c tQhQQ 5.60 c = ctttaf5.60 10.0 kW = 6.60WengQQcand c = 8.48 kWttQh Qc= 10.0 kW 8.48 kW = 1.52 kWtt(b)Q h = Weng + Q c :(c)The air conditioner operates in a cycle, so the entropy of the working fluid does not change.The hot reservoir increases in entropy by=tQhTh=e10.0 103jbJ s 3 600 s300 Kg = 1.20 105J KThe cold room decreases in entropy byS = QcTc=e8.48 103jbJ s 3 600 s280 Kg = 1.09 10The net entropy change is positive, as it must be:+1.20 10 5 J K 1.09 10 5 J K = 1.09 10 4 J KTc280 K== 11.225 KTh TcThe new ideal COP isCOPCarnot =We suppose the actual COP is(d)0.400 11.2 = 4.48a fAs a fraction of the original 5.60, this isdrop by 20.0% .zVfP22.64(a)W=Vi(b)PdV = nRTz2 ViVia f4. 48= 0.800 , so the fractional change is to5.60FG IJH K2VidV= 1.00 RT ln= RT ln 2VViThe second law refers to cycles.5J K658Heat Engines, Entropy, and the Second Law of ThermodynamicsAt point A, PVi = nRTiiandn = 1.00 molAt point B, 3PVi = nRTBiP22.65soTB = 3TiandTC = 6TisoTD = 2Tib3 P gb2V g = nRTAt point D, P b 2V g = nRTAt point C,iiiCiDThe heat for each step in the cycle is found using C V =CP =5R:23Rand2bg= nC b6T 3T g = 7.50nRT= nC b 2T 6T g = 6nRT= nC bT 2T g = 2.50nRTQ AB = nCV 3Ti Ti = 3nRTiQBCQCDQDAPiiViiPiiQentering = Q h = Q AB + QBC = 10.5nRTiTherefore,(b)Qleaving = Q c = QCD + QDA = 8.50nRTi(c)Actual efficiency,e=(d)Carnot efficiency,ec = 1 z zfS =i(a)Qh QcQh= 0.190TcT= 1 i = 0.8336TiThzFG IJH KffTfnC P dTdQT== nC P T 1 dT = nC P ln T T f = nC P ln T f ln Ti = nC P lniTTTiiiS = nC P ln*P22.67ii(a)*P22.66FIG. P22.65iF PVGH nRfdiIJKnR= nC P ln 3PViThe ideal gas at constant temperature keeps constant internal energy. As it puts out energyby work in expanding it must take in an equal amount of energy by heat. Thus its entropyincreases. Let Pi , Vi , Ti represent the state of the gas before the isothermal expansion. LetPC , VC , Ti represent the state after this process, so that PVi = PC VC . Let Pi , 3Vi , T f representithe state after the adiabatic compression.b gThenPC VC = Pi 3ViSubstitutingPC =givesPViVC 1 = Pi 3 ViiThenVC 1 = 3 Vi 1 andcontinued on next pagePViiVCejVC= 3Vib 1gChapter 22The work output in the isothermal expansion iszzCCW = PdV = nRTi V 1 dV = nRTi lniiFG V IJ = nRT lne3HV KCii659 IH 1 JK ln 3b g j = nRTi FG 1This is also the input heat, so the entropy change isS =FGHIJKQ= nRln 3 1TSinceC P = C V = CV + Rwe haveb 1gCandCP =P22.68= R , CV =R 1R 1S = nC P ln 3Then the result is(b)VThe pair of processes considered here carry the gas from the initial state in Problem 66 to thefinal state there. Entropy is a function of state. Entropy change does not depend on path.Therefore the entropy change in Problem 66 equals Sisothermal + S adiabatic in this problem.Since Sadiabatic = 0, the answers to Problems 66 and 67 (a) must be the same.Simply evaluate the maximum (Carnot) efficiency.eC =T 4.00 K== 0.014 4277 KThThe proposal does not merit serious consideration.P22.69The heat transfer over the paths CD and BA is zerosince they are adiabatic.bgOver path BC: QBC = nC P TC TB > 0bgOver path DA: QDA = nCV TA TD < 0PAdiabaticProcessesBCTherefore, Q c = QDA and Q h = Q BCDThe efficiency is thenbTQbT1 LT T Oe=1 MP NT T Qe=1Qcg T gCADhDACB TA C VC=1BPVi3ViFIG. P22.69V660P22.70Heat Engines, Entropy, and the Second Law of Thermodynamics(a)Use the equation of state for an ideal gasnRTP1.00 8.314 600VA == 1.97 10 3 m 3525.0 1.013 10V=VCa fa fej1.00a8.314fa 400f==1.013 10 532.8 10 3 m3FIG. P22.70Since AB is isothermal,PA VA = PBVBand since BC is adiabatic,PBVB = PC VCCombining these expressions, VBLF P I V OP= MG JNMH P K V QPCCAb g1 1ALF 1.00 I e32.8 10 m j= MGMMH 25.0 JK 1.97 10 mNOPbPPQLF 25.0 I e1.97 10 m j= MGMMH 1.00 JK 32.8 10 mNOPbPPQ3 1.403331 0. 400gVB = 11.9 10 3 m3Similarly,VDLF P I V OP= MG JNMH P K V QPb gA 1 1ACCorPA VA = PBVBandPB = PA331 0 . 400gVD = 5.44 10 3 m3Since AB is isothermal,3 1.403Also, CD is an isothermal and PDFG V IJ = 25.0 atmF 1.97 10GH 11.9 10HV KF V IJ = 1.00 atmF 32.8 10=P GGH 5.44 10HV KABCCD3333IJKm IJ=m Km3= 4.14 atmm3336.03 atmSolving part (c) before part (b):ec = 1 Tc400 K=1= 0.333600 KTh(c)For this Carnot cycle,(b)Energy is added by heat to the gas during the process AB. For the isothermal process,Eint = 0 .and the first law givesQ AB = W AB = nRTh lnFG V IJHV KBAbga.9f FGH 1197 IJK = 8.97 kJ1.orQ h = Q AB = 1.00 mol 8.314 J mol K 600 K lnThen, frome=WengQhafthe net work done per cycle is Weng = e c Q h = 0.333 8.97 kJ = 2.99 kJ .Chapter 22(a)20.0C(b)P22.71S = mc ln661(c)(d)TfT1+ mc lnTfT2gLMNb= 1.00 kg 4.19 kJ kg K lnTfT1+ lnOP = b4.19 kJ K g lnFG 293 293 IJH 283 303 KT QTf2S = +4.88 J KYes . Entropy has increased.ANSWERS TO EVEN PROBLEMSP22.2(a) 667 J ; (b) 467 JP22.4(a) 30.0% ; (b) 60.0%P22.655.4%P22.877.8 WP22.10(a) 869 MJ ; (b) 330 MJP22.12197 kJP22.14546CP22.1633.0%P22.18(a) 5.12%; (b) 5.27 TJ h;(c) see the solutionP22.34(a), (b) see the solution;(c) Q h = 149 J ; Q c = 65.0 J ; Weng = 84.3 J ;(d) 56.5%; (e) 1.42 10 3 rev minP22.364.88 kJ kg KP22.38(a) isobaric; (b) 402 kJ; (c) 1.20 kJ KP22.403.27 J KP22.42718 J KP22.44(a) 39.4 L ; (b) 2.50 kJ; (c) 2.50 kJ;(d) 6.87 J K ; (e) +9.16 J KP22.460.507 J KP22.4834.6 J KP22.50(a) 2 heads and 2 tails ;(b) All heads or all tails;(c) 2 heads and 2 tailsP22.20453 KP22.22(a), (b) see the solution;(c) 23.7%; see the solutionP22.2411.8P22.528.36 MJ KP22.261.17 JP22.5432.9 kJP22.28(a) 204 W ; (b) 2.43 kWP22.56see the solutionP22.30(a) 2.00 ; (b) 3.00 ; (c) 33.3%P22.58(a) 2.62 10 3 tons d ; (b) $7.65 million yr ;P22.32(a) 51.2% ; (b) 36.2%(c) 4.06 10 4 kg s662P22.60Heat Engines, Entropy, and the Second Law of ThermodynamicsP TcTh Tc cTbgP22.62(a) 4.11 kJ ; (b) 14.2 kJ; (c) 10.1 kJ; (d) 28.9%P22.64see the solutionP22.66nC P ln 3P22.68no; see the solutionP22.70(a)ABCDP, atm25.04.141.006.03(b) 2.99 kJ ; (c) 33.3%V, L1.9711.932.85.4423Electric FieldsCHAPTER OUTLINE23.1Properties of ElectricChargesCharging Objects byInductionCoulombs LawThe Electric FieldElectric Field of aContinuous ChargeDistributionElectric Field LinesMotion of Charged Particlesin a Uniform Electric FieldANSWERS TO QUESTIONS23.323.423.523.623.7Q23.1A neutral atom is one that has no net charge. This means that ithas the same number of electrons orbiting the nucleus as it hasprotons in the nucleus. A negatively charged atom has one ormore excess electrons.Q23.2When the comb is nearby, molecules in the paper are polarized,similar to the molecules in the wall in Figure 23.5a, and thepaper is attracted. During contact, charge from the comb istransferred to the paper by conduction. Then the paper has thesame charge as the comb, and is repelled.Q23.323.2The clothes dryer rubs dissimilar materials together as ittumbles the clothes. Electrons are transferred from one kind ofmolecule to another. The charges on pieces of cloth, or onnearby objects charged by induction, can produce strongelectric fields that promote the ionization process in thesurrounding air that is necessary for a spark to occur. Then youhear or see the sparks.Q23.4To avoid making a spark. Rubber-soled shoes acquire a charge by friction with the floor and coulddischarge with a spark, possibly causing an explosion of any flammable material in the oxygenenriched atmosphere.Q23.5Electrons are less massive and more mobile than protons. Also, they are more easily detached fromatoms than protons.Q23.6The electric field due to the charged rod induces charges on near and far sides of the sphere. Theattractive Coulomb force of the rod on the dissimilar charge on the close side of the sphere is largerthan the repulsive Coulomb force of the rod on the like charge on the far side of the sphere. Theresult is a net attraction of the sphere to the rod. When the sphere touches the rod, charge isconducted between the rod and the sphere, leaving both the rod and the sphere like-charged. Thisresults in a repulsive Coulomb force.Q23.7All of the constituents of air are nonpolar except for water. The polar water molecules in the air quitereadily steal charge from a charged object, as any physics teacher trying to perform electrostaticsdemonstrations in the summer well knows. As a resultit is difficult to accumulate large amounts ofexcess charge on an object in a humid climate. During a North American winter, the cold, dry airallows accumulation of significant excess charge, giving the potential (pun intended) for a shocking(pun also intended) introduction to static electricity sparks.12Electric FieldsQ23.8Similarities: A force of gravity is proportional to the product of the intrinsic properties (masses) oftwo particles, and inversely proportional to the square of the separation distance. An electrical forceexhibits the same proportionalities, with charge as the intrinsic property.Differences: The electrical force can either attract or repel, while the gravitational force asdescribed by Newtons law can only attract. The electrical force between elementary particles isvastly stronger than the gravitational force.Q23.9No. The balloon induces polarization of the molecules in the wall, so that a layer of positive chargeexists near the balloon. This is just like the situation in Figure 23.5a, except that the signs of thecharges are reversed. The attraction between these charges and the negative charges on the balloonis stronger than the repulsion between the negative charges on the balloon and the negative chargesin the polarized molecules (because they are farther from the balloon), so that there is a net attractiveforce toward the wall. Ionization processes in the air surrounding the balloon provide ions to whichexcess electrons in the balloon can transfer, reducing the charge on the balloon and eventuallycausing the attractive force to be insufficient to support the weight of the balloon.Q23.10The electric field due to the charged rod induces a charge in the aluminum foil. If the rod is broughttowards the aluminum from above, the top of the aluminum will have a negative charge induced onit, while the parts draping over the pencil can have a positive charge induced on them. Thesepositive induced charges on the two parts give rise to a repulsive Coulomb force. If the pencil is agood insulator, the net charge on the aluminum can be zero.Q23.11So the electric field created by the test charge does not distort the electric field you are trying tomeasure, by moving the charges that create it.Q23.12With a very high budget, you could send first a proton and then an electron into an evacuatedregion in which the field exists. If the field is gravitational, both particles will experience a force inthe same direction, while they will experience forces in opposite directions if the field is electric.On a more practical scale, stick identical pith balls on each end of a toothpick. Charge one pithball + and the other , creating a large-scale dipole. Carefully suspend this dipole about its center ofmass so that it can rotate freely. When suspended in the field in question, the dipole will rotate toalign itself with an electric field, while it will not for a gravitational field. If the test device does notrotate, be sure to insert it into the field in more than one orientation in case it was aligned with theelectric field when you inserted it on the first trial.Q23.13The student standing on the insulating platform is held at the same electrical potential as thegenerator sphere. Charge will only flow when there is a difference in potential. The student whounwisely touches the charged sphere is near zero electrical potential when compared to the chargedsphere. When the student comes in contact with the sphere, charge will flow from the sphere to himor her until they are at the same electrical potential.Q23.14An electric field once established by a positive or negative charge extends in all directions from thecharge. Thus, it can exist in empty space if that is what surrounds the charge. There is no material atpoint A in Figure 23.23(a), so there is no charge, nor is there a force. There would be a force if acharge were present at point A, however. A field does exist at point A.Q23.15If a charge distribution is small compared to the distance of a field point from it, the chargedistribution can be modeled as a single particle with charge equal to the net charge of thedistribution. Further, if a charge distribution is spherically symmetric, it will create a field at exteriorpoints just as if all of its charge were a point charge at its center.Chapter 233Q23.16The direction of the electric field is the direction in which a positive test charge would feel a forcewhen placed in the field. A charge will not experience two electrical forces at the same time, but thevector sum of the two. If electric field lines crossed, then a test charge placed at the point at whichthey cross would feel a force in two directions. Furthermore, the path that the test charge wouldfollow if released at the point where the field lines cross would be indeterminate.Q23.17Both figures are drawn correctly. E1 and E 2 are the electric fields separately created by the pointcharges q1 and q 2 in Figure 23.14 or q and q in Figure 23.15, respectively. The net electric field is thevector sum of E1 and E 2 , shown as E. Figure 23.21 shows only one electric field line at each pointaway from the charge. At the point location of an object modeled as a point charge, the direction ofthe field is undefined, and so is its magnitude.Q23.18The electric forces on the particles have the same magnitude, but are in opposite directions. Theelectron will have a much larger acceleration (by a factor of about 2 000) than the proton, due to itsmuch smaller mass.Q23.19The electric field around a point charge approaches infinity as r approaches zero.Q23.20Vertically downward.Q23.21Four times as many electric field lines start at the surface of the larger charge as end at the smallercharge. The extra lines extend away from the pair of charges. They may never end, or they mayterminate on more distant negative charges. Figure 23.24 shows the situation for charges +2q and q.Q23.22At a point exactly midway between the two changes.Q23.23Linear charge density, , is charge per unit length. It is used when trying to determine the electricfield created by a charged rod.Surface charge density, , is charge per unit area. It is used when determining the electric fieldabove a charged sheet or disk.Volume charge density, , is charge per unit volume. It is used when determining the electricfield due to a uniformly charged sphere made of insulating material.Q23.24Yes, the path would still be parabolic. The electrical force on the electron is in the downwarddirection. This is similar to throwing a ball from the roof of a building horizontally or at some anglewith the vertical. In both cases, the acceleration due to gravity is downward, giving a parabolictrajectory.Q23.25No. Life would be no different if electrons were + charged and protons were charged. Oppositecharges would still attract, and like charges would repel. The naming of + and charge is merely aconvention.Q23.26If the antenna were not grounded, electric charges in the atmosphere during a storm could place theantenna at a high positive or negative potential. The antenna would then place the television setinside the house at the high voltage, to make it a shock hazard. The wire to the ground keeps theantenna, the television set, and even the air around the antenna at close to zero potential.Q23.27People are all attracted to the Earth. If the force were electrostatic, people would all carry chargewith the same sign and would repel each other. This repulsion is not observed. When we changedthe charge on a person, as in the chapter-opener photograph, the persons weight would changegreatly in magnitude or direction. We could levitate an airplane simply by draining away its electriccharge. The failure of such experiments gives evidence that the attraction to the Earth is not due toelectrical forces.4Electric FieldsQ23.28In special orientations the force between two dipoles can be zero or a force of repulsion. In generaleach dipole will exert a torque on the other, tending to align its axis with the field created by the firstdipole. After this alignment, each dipole exerts a force of attraction on the other.SOLUTIONS TO PROBLEMSSection 23.1*P23.1(a)Properties of Electric ChargesThe mass of an average neutral hydrogen atom is 1.007 9u. Losing one electron reduces itsmass by a negligible amount, toej1.007 9 1.660 10 27 kg 9.11 10 31 kg = 1.67 10 27 kg .Its charge, due to loss of one electron, isej0 1 1.60 10 19 C = +1.60 10 19 C .(b)By similar logic, charge = +1.60 10 19 Cejmass = 22.99 1.66 10 27 kg 9.11 10 31 kg = 3.82 10 26 kg(c)charge of Cl = 1.60 10 19 Cejmass = 35.453 1.66 10 27 kg + 9.11 10 31 kg = 5.89 10 26 kg(d)ejcharge of Ca ++ = 2 1.60 10 19 C = +3.20 10 19 Cej ejmass = 40.078 1.66 10 27 kg 2 9.11 10 31 kg = 6.65 10 26 kg(e)eemass = 14.007 1.66 10 27(f)jkg j + 3e9.11 10charge of N 3 = 3 1.60 10 19 C = 4.80 10 19 Ce31jjkg = 2.33 10 26 kgcharge of N 4 + = 4 1.60 10 19 C = +6.40 10 19 Cej ejmass = 14.007 1.66 10 27 kg 4 9.11 10 31 kg = 2.32 10 26 kg(g)We think of a nitrogen nucleus as a seven-times ionized nitrogen atom.ejcharge = 7 1.60 10 19 C = 1.12 10 18 Cemass = 14.007 1.66 10(h)27j ejkg 7 9.11 10 31 kg = 2.32 10 26 kgcharge = 1.60 10 19 Cbgmass = 2 1.007 9 + 15.999 1.66 10 27 kg + 9.11 10 31 kg = 2.99 10 26 kgChapter 23F 10.0 grams I FG 6.02 10GH 107.87 grams mol JK H(a)N=(b)P23.2# electrons added =atomsmol23IJ FG 47 electrons IJ =K H atom K2.62 10 24Q1.00 10 3 C== 6.25 10 15e 1.60 10 19 C electron2.38 electrons for every 10 9 already present .orSection 23.2Charging Objects by InductionSection 23.3Coulombs LawP23.3If each person has a mass of 70 kg and is (almost) composed of water, then each person containsNF 70 000 grams I FG 6.02 10GH 18 grams mol JK Hmoleculesmol23IJ FG 10 protons IJ 2.3 10K H molecule K28protons .With an excess of 1% electrons over protons, each person has a chargeejeje3.7 10 j= e9 10 jq = 0.01 1.6 10 19 C 2.3 10 28 = 3.7 10 7 C .SoF = keq1 q 2r27 290.6 2N = 4 10 25 N ~ 10 26 N .This force is almost enough to lift a weight equal to that of the Earth:ejMg = 6 10 24 kg 9.8 m s 2 = 6 10 25 N ~ 10 26 N .*P23.4The force on one proton is F =e8.99 10P23.5(a)(b)9F 1.6 10N m C jGH 2 10Fe =Fg =2k e q1 q 2r2r2r21915e8.99 10=Gm1 m 2k e q1 q 2Cm9e6.67 10=IJKaway from the other proton. Its magnitude is2= 57.5 N .jeN m 2 C 2 1.60 10 19 Ce3.80 101110jm2jej2= 1.59 10 9 NN m 2 C 2 1.67 10 27 kge3.80 1010jm2j2= 1.29 10 45 NThe electric force is larger by 1.24 10 36 times .(c)If k eq=mq1 q 2r2=Gm1 m 2r2brepulsiongwith q1 = q 2 = q and m1 = m 2 = m , then6.67 10 11 N m 2 kg 2G== 8.61 10 11 C kg .ke8.99 10 9 N m 2 C 256Electric FieldsP23.6We find the equal-magnitude charges on both spheres:q1 q 2F = ke= ker2q2r2a1.00 10 4 N= 1.05 10 3 C .8.99 10 9 N m 2 C 2fF= 1.00 mkeq=rsoThe number of electron transferred is thenN xfer =1.05 10 3 C= 6.59 10 15 electrons .1.60 10 19 C e The whole number of electrons in each sphere isF 10.0 g I e6.02 10GH 107.87 g mol JKN tot =23jejatoms mol 47 e atom = 2.62 10 24 e .The fraction transferred is thenf=P23.7FGHIJKN xfer6.59 10 15== 2.51 10 9 = 2.51 charges in every billion.24N tot2.62 10F1 = k eF2 = k ee8.99 10=q1 q 2r2q1 q 2r2=e8.99 10jejeN m 2 C 2 7.00 10 6 C 2.00 10 6 C99N m2j = 0.503 Na0.500 mfC je7.00 10 C je 4.00 10 C j= 1.01 Na0.500 mf26262Fx = 0.503 cos 60.0+1.01 cos 60.0 = 0.755 NFy = 0.503 sin 60.01.01 sin 60.0 = 0.436 Naf afF = 0.755 N i 0.436 N j = 0.872 N at an angle of 330FIG. P23.7P23.8F = keP23.9(a)q1 q 2r2e8.99 10=9jeN m 2 C 2 1.60 10 19 Cej2 6.37 10 6 mj e6.02 10 j223 22= 514 kNThe force is one of attraction . The distance r in Coulombs law is the distance betweencenters. The magnitude of the force isF=(b)k e q1 q 2r2e= 8.99 10 9 N m 2 C 212.0 10 C je18.0 10 C j=jea0.300 mf9922.16 10 5 N .The net charge of 6.00 10 9 C will be equally split between the two spheres, or3.00 10 9 C on each. The force is one of repulsion , and its magnitude isF=k e q1 q 2r2e= 8.99 10 9 N m 2 C 23.00 10 C je3.00 10 C j=jea0.300 mf9928.99 10 7 N .Chapter 23P23.10x from the left end of the rod. This beadLet the third bead have charge Q and be located distancewill experience a net force given byF=b gk e 3q Qx2i+7b g e i j .ad x fke q Q2The net force will be zero if31=2xdxaf2, or d x =x3.This gives an equilibrium position of the third bead of x = 0.634d .The equilibrium is stable if the third bead has positive charge .kee2e(a)F=(b)P23.11We have F =r2= 8.99 10 N m2210Cjm2j2= 8.22 10 8 Nmv 2from whichrv=P23.12e1.60 10C je0.529 10199ej=8.22 10 8 N 0.529 10 10 mFr=m9.11 1031The top charge exerts a force on the negative chargekgk e qQd 2 + 2x2ch2.19 10 6 m s .which is directed upward and to theFG d IJ to the x-axis. The two positive charges together exert forceH 2x KF 2 k qQ I FG a xfi IJ2 k qQdGGJJ G= ma or for x << , a x.2md 8e + x j K GH e + x j JJKHleft, at an angle of tan 1ed24(a)d24212e3The acceleration is equal to a negative constant times the excursion from equilibrium, as in16 k e qQa = 2 x , so we have Simple Harmonic Motion with 2 =.md 3T=(b)22=2md 3, where m is the mass of the object with charge Q .k e qQv max = A = 4ak e qQmd 38Electric FieldsSection 23.4The Electric FieldFor equilibrium,Fe = FgorqE = mg j .Thus,P23.13E=e jemgj.qjej9.11 10 31 kg 9.80 m s 2mgj=j = 5.58 10 11 N C j19q1.60 10Ce(a)(b)P23.14E=1.67 10 27 kg 9.80 m s 2mgE=j=j=q1.60 10 19 Ceejbejje Fy = 0 : QEj + mge jj = 0m =P23.15ejje1.02 10j7jNC jg24.0 10 6 C 610 N CQE== 1.49 gramsg9.80 m s 2The point is designated in the sketch. The magnitudes of the electric fields,E1 , (due to the 2.50 10 6 C charge) and E 2 (due to the 6.00 10 6 Ccharge) areE1 =E2 =ke qr2keqr2e8.99 10==e8.99 109jeN m 2 C 2 2.50 10 6 Cd9j2jead + 1.00 mfN m 2 C 2 6.00 10 6 Cj2(1)FIG. P23.15(2)Equate the right sides of (1) and (2)to getad + 1.00 mford + 1.00 m = 1.55dwhich yieldsd = 1.82 mord = 0.392 m .2= 2.40d 2The negative value for d is unsatisfactory because that locates a point between the charges whereboth fields are in the same direction.Thus,d = 1.82 m to the left of the 2.50 C charge .9Chapter 23P23.16If we treat the concentrations as point charges,E+ = keE = keqeCj b1a40.0 mgf e jj = 3.60 10000a40.0 Cf e jj = 3.60 10C jb1 000 mg= 8.99 10 9 N m 2 C 2r2qe= 8.99 10 9 N m 2r2222e jafe jaf5N C j downward5N C j downwardE = E + + E = 7.20 10 5 N C downward*P23.17The first charge creates at the origin fieldk eQ+Q x = 0to the right.a2Suppose the total field at the origin is to the right. Then q mustbe negative:k eQa2i+keqe ij = 2 k Q iaa3 a fe22qxFIG. P23.17q = 9Q .In the alternative, the total field at the origin is to the left:k eQaP23.182(a)i+ke q9a2e ij = 2 k Q e ijae2q = +27Q .e8.99 10 je7.00 10 j = 2.52 10ra0.500fk q e8.99 10 je 4.00 10 j=== 1.44 10ra0.500fE1 =69ke q=22NC5NC69E25e22Ex = E2 E1 cos 60 = 1. 44 10 5 2.52 10 5 cos 60.0 = 18.0 10 3 N C5FIG. P23.183Ey = E1 sin 60.0 = 2.52 10 sin 60.0 = 218 10 N CE = 18.0 i 218 j 10 3 N C = 18.0 i 218 j kN C(b)P23.19(a)ejej8.99 10 je3.00 10 jE =e jj = e a0.100fe jj = e2.70 10r8.99 10 je6.00 10 jk qE =e ij = e a0.300fe ij = e5.99 10rE = E + E = e5.99 10 N C ji e 2.70 10 N C j j121e99k e q1222221e3jeje3.00i 13.5 jj NF = qE = 5.00 10 9 C 599 i 2 700 j N CeNC j2NCi992jF = 3.00 10 6 i 13.5 10 6 j N =e36.0i 436 jj mNj322(b)jeF = qE = 2.00 10 6 C 18.0 i 218 j 10 3 N C = 36.0 i 436 j 10 3 N =jFIG. P23.1910Electric FieldsP23.20E=(a)keqr2e8.99 10 je2.00 10 j = 14 400 N Ca1.12fandE = 2b14 400 g sin 26.6 = 1.29 1069=2Ex = 0yFIG. P23.20ejej(b)F = qE = 3.00 10 6 1. 29 10 4 j = 3.86 10 2 j N(a)E=k e q1r1 +r12E = 3.06ke qa2ke q2r22r2 +i + 5.06F = qE = 5.91(b)P23.22ke q 2a2k e q3keqa2r32r3 =b g i + k b3 qg ei cos 45.0+ j sin 45.0j + k b4qg jke 2qea2j = 5.91keqa2e2a2a2at 58.8at 58.8The electric field at any point x isE=a fax af cx a afh ex a jke q2ke q2=k e q 4ax22 2.b g4a k e qWhen x is much, much greater than a, we find E P23.23NCE = 1.29 10 4 j N C .soP23.214(a)One of the charges creates at P a field E =the x-axis as shown.x3k eQ nR2 + x2.at an angle toWhen all the charges produce field, for n > 1 , the componentsperpendicular to the x-axis add to zero.The total field is(b)P23.24E=b g cos =nk e Q n i2R +x2ek eQx i2R + x2j3 2.FIG. P23.23A circle of charge corresponds to letting n grow beyond all bounds, but the result does notdepend on n. Smearing the charge around the circle does not change its amount or itsdistance from the field point, so it does not change the field .keqr2r=keqa2e ij + a2kaq e ij + a3kaq e ij + = ka qi FGH1 + 21 + 31 + IJK =ffe2e2e222 2ke qi6a2Chapter 23Section 23.5P23.25Electric Field of a Continuous Charge Distributioneb ga f a f a f ajej8.99 10 9 22.0 10 6ke Qkek eQ===E=0.290 0.140 + 0.290d +dd +dd +dfafE = 1.59 10 6 N C , directed toward the rod.P23.26E=zk e dqx2E = ke 0zx0P23.27E=ex, where dq = 0 dxFG 1 IJH xKdxx=x0ke 0x0The direction is i or left for 0 > 0e8.99 10 je75.0 10 jx = 6.74 10 xex + 0.100 jex + 0.010 0j69j+ a2= ke 0 2k e xQ2FIG. P23.253 2=52 322322(a)At x = 0.050 0 m ,E = 2.41 10 7 i N C = 24.1i MN C(c)At x = 0.300 m ,E = 6.40 10 6 i N C = 6. 40 i MN C(d)P23.29E = 6.64 10 6 i N C = 6.64i MN C(b)P23.28At x = 0.010 0 m ,At x = 1.00 m ,E = 6.64 10 5 i N C = 0.664i MN CzE = dE =E=LM k x dxe ij OPz MMN x PPQ = k x i z xeexk e Qx+ a2e3x020 0j0 03FGHdx = k e 0 x 0 i x03 2For a maximum,dE= Qk edxLMMM exNax 2 + a 2 3 x 2 = 0 or x =212+ a2j3 2OPP=0x +a j PeQ3x 22 522.Substituting into the expression for E givesE=k eQa2e aj322 32=k eQ332a2=2 k eQ3 3a2=Q6 3 0 a 2.12x2x0I=JKke 0i2 x0e j1112Electric FieldsP23.30FGHxE = 2 k e 1 eE = 2 8.99 102x +R92IJKje7.90 103FjGG 1 HIJJ = 4.46 10 FG 1 H+ a0.350f Kxx22(a)At x = 0.050 0 m ,At x = 0.100 m ,E = 3.24 10 8 N C = 324 MN C(c)At x = 0.500 m ,E = 8.07 10 7 N C = 80.7 MN C(d)At x = 2.00 m ,E = 6.68 10 8 N C = 6.68 MN C(a)From Example 23.9: E = 2 k e 1 x2E = 3.83 10 8 N C = 383 MN C(b)P23.31IJ+ 0.123 Kx8FGH=IJKxx2 + R2Q= 1.84 10 3 C m 2R 2jaefE = 1.04 10 8 N C 0.900 = 9.36 10 7 N C = 93.6 MN Cbappx: E = 2 k e = 104 MN C about 11% high(b)FjGHeE = 1.04 10 8 N C 1 I = 1.04 10J ecm K30.0 cmjbg8N C 0.004 96 = 0.516 MN C30.0 + 3.00Q5.20 10 6appx: E = k e 2 = 8.99 10 9= 0.519 MN C about 0.6% high2r0.30eP23.32g22bj a fLMNML= 2 k M1 MNThe electric field at a distance x isThis is equivalent toExxEx = 2 k e 1 For large x,R2x2OPQP1e1+<< 1 andx2 + R2g1 + R2 x2OPPQR2R21+ 2x22xIF1JJ = 2 k e1 + R e2 x j 1j= 2 k G 1 GH 1 + R e2 x j K1 + R e2 x jk Qe1 x jF RI== k QG x +H 2 JK1 + R e2x j2ExsoQSubstitute =, R2But for x >> R ,1x2 + R2 2Ex1x2, soEx e22e2k eQx22e2e22for a disk at large distances222Chapter 23zzDue to symmetryEy = dEy = 0 , and Ex = dE sin = k ewhereEx =zdq sin dq = ds = rd ,so that,P23.33zke k sin d = e cos r 0raf=0r22k e rqLand r = .L2 8.99 10 9 N m 2 C 2 7.50 10 6 C 2 k qEx = e2 =.2L0.140 mFIG. P23.33=whereeThus,jefa13j7Ex = 2.16 10 N C .Solving,ejSince the rod has a negative charge, E = 2.16 10 7 i N C = 21.6 i MN C .P23.34(a)We define x = 0 at the point where we are to find the field. One ring, with thickness dx, hasQdxand produces, at the chosen point, a fieldchargehkexQdxdE =i.3 2hx2 + R2ejThe total field isz zE=dE =d+hek Qi ex + R jE=2hb 1 2gdall chargek eQxdxh x2 + Rj2 3 2d+h2 1 22e(b)i=zek eQi d + h 2x + R22h x=dk Qi= ehx=dLMMM dNej3 22 xdxOPP+R jead + hf + R j PQ1212 122212Think of the cylinder as a stack of disks, each with thickness dx, chargeper-area =dE =So,Qdx. One disk produces a field R2h2 k eQdx R2hFGG 1 H exIJi .+R j JKF2 k Qdx G1R h GH exx22 12IJJ iE=z dE = z+R j KOP = 2k Qi LMx2 k Qi L1E=M z dx 2 z ex + R j 2xdxP R h MR h MMNNQ2 k Qi LE=MNd + h d ead + hf + R j + ed + R j OPQR h2 k Qi LOE=R h MNh + ed + R j ead + hf + R j PQd+hall chargex =de2d+hd+hde2x2 122x=d2 1 222e2e2Qdx, and chargeh22 122e21222 122212e221 x +Rd+hd212j1 2 d+hdOPPPQ14Electric FieldsP23.35(a)The electric field at point P due to each element of length dx, isk dqdE = 2 e 2 and is directed along the line joining the element tox +ypoint P. By symmetry,Ex = dEx = 0and sincedq = dx ,E = Ey = dEy = dE cos wherecos =zzzze2Therefore,E = 2 k e y0dx2x +yj2 32y2x + y2.FIG. P23.352 k e sin 0.y= 0 = 90Ey =2k e .y(b)P23.36For a bar of infinite length,(a)The whole surface area of the cylinder is A = 2 r 2 + 2 rL = 2 r r + L .j beanda fgQ = A = 15.0 10 9 C m 2 2 0.025 0 m 0.025 0 m + 0.060 0 m = 2.00 10 10 C(b)For the curved lateral surface only, A = 2 rL .j befgaQ = A = 15.0 10 9 C m 2 2 0.025 0 m 0.060 0 m = 1.41 10 10 Cj beg b0.060 0 mg =2(c)P23.37Q = V = r 2 L = 500 10 9 C m 3 0.025 0 m(a)Every object has the same volume, V = 8 0.030 0 mejeaf35.89 10 11 C= 2.16 10 4 m3 .jFor each, Q = V = 400 10 9 C m 3 2.16 10 4 m3 = 8.64 10 11 C(b)We must count the 9.00 cm 2 squares painted with charge:(i)6 4 = 24 squaresej ejj ejj ejj ejQ = A = 15.0 10 9 C m 2 24.0 9.00 10 4 m 2 = 3.24 10 10 C(ii)34 squares exposedeQ = A = 15.0 10 9 C m 2 34.0 9.00 10 4 m 2 = 4.59 10 10 C(iii)34 squareseQ = A = 15.0 10 9 C m 2 34.0 9.00 10 4 m 2 = 4.59 10 10 C(iv)32 squareseQ = A = 15.0 10 9 C m 2 32.0 9.00 10 4 m 2 = 4.32 10 10 C(c)(i)total edge length:e= e80.0 10Q = = 80.0 10 12(ii)Q=continued on next pagebgC mj24 b0.030 0 mg =C mj44 b0.030 0 mg == 24 0.030 0 m125.76 10 11 C1.06 10 10 CChapter 23ej bgej bg(iii)(iv)Section 23.6Q = = 80.0 10 12 C m 64 0.030 0 m = 1.54 10 10 CQ = = 80.0 10 12 C m 40 0.030 0 m = 0.960 10 10 CElectric Field LinesP23.39P23.38FIG. P23.38P23.40(a)(b)P23.41(a)FIG. P23.39q1 61== 3q 2 18q1 is negative, q 2 is positiveThe electric field has the general appearance shown. It is zeroat the center , where (by symmetry) one can see that the threecharges individually produce fields that cancel out.In addition to the center of the triangle, the electric field lines in thesecond figure to the right indicate three other points near themiddle of each leg of the triangle where E = 0 , but they are moredifficult to find mathematically.(b)You may need to review vector addition in Chapter Three. Theelectric field at point P can be found by adding the electric fieldvectors due to each of the two lower point charges: E = E 1 + E 2 .The electric field from a point charge is E = k eqr2r.As shown in the solution figure at right,E1 = k eE2 = keqa2qa2to the right and upward at 60to the left and upward at 60E = E1 + E 2 = k e= 1.73 k eqa2jqa2ecos 60 i + sin 60 jj + e cos 60 i + sin 60 jj = kFIG. P23.41eqa2ej2 sin 60 j1516Electric FieldsSection 23.7Motion of Charged Particles in a Uniform Electric FieldqEmF = qE = maa=v f = vi + atP23.42vf =qEtme1.602 10 ja520fe48.0 10 j ==4.39 10 6 m se1.602 10 ja520fe48.0 10 j =2.39 10 3 m s19veelectron:9319.11 10in a direction opposite to the field19vp =proton:91.67 10 27in the same direction as the fielda fqE 1.602 10 19 640== 6.14 10 10 m s 2m1.67 10 27a=v f = vi + at(c)x f xi =(d)K=(a)196.00 10 5qE 1.602 10a=== 5.76 10 13 m s so a = 5.76 10 13 i m s 2m1.67 10 27(b)P23.44(a)(b)P23.43v f = vi + 2 a x f xie1vi + v f t2dixf =eejejeedejej2= 1.20 10 15 Jjjjbgv i = 2.84 10 6 i m sv f = vi + atej0 = 2.84 10 6 + 5.76 10 13 tt = 4.93 10 8 sThe required electric field will be in the direction of motion .Work done = Kso, Fd = which becomesE=1mvi2 (since the final velocity = 0 )2eEd = KandjieP23.45t = 1.95 10 5 s11.20 10 6 1.95 10 5 = 11.7 m211mv 2 = 1.67 10 27 kg 1.20 10 6 m s220 = vi2 + 2 5.76 10 13 0.070 0(c)j1.20 10 6 = 6.14 10 10 tK.edChapter 23P23.4617The acceleration is given byda fiv 2 = vi2 + 2 a x f x i orfv 2 = 0 + 2a h .fSolvinga=Now F = ma :ThereforeqE = (a)v2f2hFGH.mv 2f2h mg j + qE = mv 2 jfIJK2h.+ mg j .Gravity alone would give the bead downward impact velocityjaef2 9.80 m s 2 5.00 m = 9.90 m s .To change this to 21.0 m/s down, a downward electric field must exert a downwardelectric force.FGHIJKFGH(b)(a)t=(b)P23.4721.00 10 3 kg N s 2m vfg =q=E 2h1.00 10 4 N C kg may =*P23.48OPPQ2 9.80 m s 2 = 3.43 C0.050 0x== 1.11 10 7 s = 111 nsv x 4.50 10 5ejej199.60 10 3qE 1.602 10== 9.21 10 11 m s 227m1.67 10ey f yi = v yi t +(c)I LM b21.0 m sgJK M 2a5.00 mfNj1ayt 2 :2yf =19.21 10 11 1.11 10 72ejeej2= 5.68 10 3 m = 5.68 mmjejv yf = v yi + a y t = 9.21 10 11 1.11 10 7 = 1.02 10 5 m sv x = 4.50 10 5 m sjbge je j F = e 2 10 kg m s je jj = 1 10 m s j .a=and moves with acceleration:eje jm2 10kgIts x-component of velocity is constant at e1.00 10 m sj cos 37 = 7.99 10 m s . Thus it moves in aeThe particle feels a constant force: F = qE = 1 10 6 C 2 000 N C j = 2 10 3 N j321316524parabola opening downward. The maximum height it attains above the bottom plate is described bydi22v yf = v yi + 2 a y y f yi :e0 = 6.02 10 4 m sy f = 1.81 10 4 m .continued on next pagej e2 10213jdm s2 y f 0i18Electric FieldsSince this is less than 10 mm, the particle does not strike the top plate, but moves in a symmetricparabola and strikes the bottom plate after a time given byy f = yi + v yi t +1ayt 22ej0 = 0 + 6.02 10 4 m s t +11 10 13 m s 2 t 22ejsince t > 0 ,t = 1.20 10 8 s .The particles range isx f = xi + v x t = 0 + 7.99 10 4 m s 1.20 10 8 s = 9.61 10 4 m .ejejIn sum,The particle strikes the negative plate after moving in a parabola with a height of 0.181 mmand a width of 0.961 mm.P23.49vi = 9.55 10 3 m say =(a)e^ja fj19720eE 1.60 10== 6.90 10 10 m s 227m1.67 10evi2 sin 2= 1. 27 10 3 m so thatayR=e9.55 10 j3 26.90 10sin 210sin 2 = 0.961t=(b)FIG. P23.49= 1.27 10 3 = 36.9RR=vix vi cos 90.0 = 53.1If = 36.9 , t = 167 ns .If = 53.1 , t = 221 ns .Additional Problems*P23.50The two given charges exert equal-size forces of attraction on eachother. If a third charge, positive or negative, were placed betweenthem they could not be in equilibrium. If the third charge were at apoint x > 15 cm , it would exert a stronger force on the 45 C thanon the 12 C , and could not produce equilibrium for both. Thusthe third charge must be at x = d < 0 . Its equilibrium requiresbk e q 12 Cd2g = k qb45 Cga15 cm + dfe215 cm + d = 1.94dFG 15 cm + d IJH d K2=dqx=012 C15 cmx+45 CFIG. P23.5045= 3.7512d = 16.0 cm .The third charge is at x = 16.0 cm . The equilibrium of the 12 C requiresbg = k b45 Cg12 Ca16.0 cmfa15 cmfk e q 12 C2e2q = 51.3 C .All six individual forces are now equal in magnitude, so we have equilibrium as required, and this isthe only solution.Chapter 23jbegThe proton moves with accelerationwhile the e has accelerationP23.5119C 640 N CqE 1.60 10== 6.13 10 10 m s 2ap =m1.673 10 27 kgae =(a)e1.60 10jbC 640 N C9.110 10FGHIJKg = 1.12 1014m s 2 = 1 836 a p .1a p t 2 ), knowing:2The distance from the positive plate to where the meeting occurs equals the distance the1sodium ion travels (i.e., d Na = a Na t 2 ). This is found from:2eEeE111122t2 +t2 .4.00 cm = a Na t + aCl t :4.00 cm =222 22.99 u2 35. 45 u111This may be written as4.00 cm = a Na t 2 + 0.649 a Na t 2 = 1.65 a Na t 222214.00 cm2= 2.43 cm .sod Na = a Na t =21.65FGH(a)kg111a p t 2 + a e t 2 = 1 837 a p t 2 .22214.00 cm2= 21.8 m .d = apt =21 837Thus,P23.5231We want to find the distance traveled by the proton (i.e., d =4.00 cm =(b)19IJKbFGHIJKgFGHIJKThe field, E1 , due to the 4.00 10 9 C charge is in the xdirection.8.99 10 9 N m 2 C 2 4.00 10 9 CkeqE1 = 2 r =i2r2.50 meajefjFIG. P23.52(a)= 5.75 i N CLikewise, E 2 and E3 , due to the 5.00 10 9 C charge and the 3.00 10 9 C charge areE2 =E3ke qr2e8.99 10r=e8.99 10=9jej i = 11.2 N C ia2.00 mfN m C je3.00 10 C ji = 18.7 N C ia1.20 mf9N m 2 C 2 5.00 10 9 C22922E R = E1 + E 2 + E 3 = 24.2 N C in +x direction.(b)E2E3ke qbgej=r = b11. 2 N C ge + jjrk q=r = b5.81 N C ge 0.371i +0.928 jjrE1 =r2ke qr = 8.46 N C 0.243 i + 0.970 j2e2Ex = E1 x + E3 x = 4.21i N CER = 9.42 N CEy = E1 y + E2 y + E3 y = 8.43 j N C = 63.4 above x axisFIG. P23.52(b)1920Electric Fields*P23.53(a)Each ion moves in a quarter circle. The electric force causes the centripetal acceleration. F = maqE =mv 2RE=dFor the x-motion,22v xf = v xi + 2 a x x f x i0 = v 2 + 2ax R(b)ax = Ex = mv 2qRiv 2 Fx qE x==m2R mmv 2. Similarly for the y-motion,2 qRv 2 = 0 + 2ay Ray = +v 2 qE y=m2REy =mv 22 qRThe magnitude of the field is22Ex + Ey =P23.54mv 2at 135 counterclockwise from the x -axis .2 qRFrom the free-body diagram shown, Fy = 0 :T cos 15.0 = 1.96 10 2 N .SoT = 2.03 10 2 N .FromorP23.55(a) Fx = 0 , we haveq=qE = T sin 15.0ej2.03 10 2 N sin 15.0T sin 15.0== 5.25 10 6 C = 5.25 C .E1.00 10 3 N CFIG. P23.54Let us sum force components to find Fx = qEx T sin = 0 , and Fy = qEy + T cos mg = 0 .Combining these two equations, we getq=e1.00 10 ja9.80feE cot + E j a3.00 cot 37.0+5.00f 103mgx=y5= 1.09 10 8 C= 10.9 nC(b)From the two equations forT=Free Body DiagramFIG. P23.55 Fxand FyqEx= 5.44 10 3 N = 5.44 mN .sin 37.0we also findChapter 23P23.5621This is the general version of the preceding problem. The known quantities are A, B, m, g, and . Theunknowns are q and T.The approach to this problem should be the same as for the last problem, but withoutnumbers to substitute for the variables. Likewise, we can use the free body diagram given in thesolution to problem 55. Fx = T sin + qA = 0andqA, into Eq. (2),sin Substituting T =(a)(1) Fy = +T cos + qB mg = 0Again, Newtons second law:(2)qA cos + qB = mg .sin mgA cot + BIsolating q on the left,aSubstituting this value into Eq. (1),(b)q=T=a A cos + B sin fmgAf..If we had solved this general problem first, we would only need to substitute theappropriate values in the equations for q and T to find the numerical results needed forproblem 55. If you find this problem more difficult than problem 55, the little list at the firststep is useful. It shows what symbols to think of as known data, and what to considerunknown. The list is a guide for deciding what to solve for in the analysis step, and forrecognizing when we have an answer.P23.57F=k e q1 q 2r2tan =:15.060.0 = 14.0e8.99 10 je10.0 10 jF =a0.150fe8.99 10 je10.0 10 jF =a0.600fe8.99 10 je10.0 10 jF =a0.619f6 29126 2932= 2.50 N6 292= 40.0 N2= 2.35 NFx = F3 F2 cos 14.0 = 2.50 2.35 cos 14.0 = 4.78 NFy = F1 F2 sin 14.0 = 40.0 2.35 sin 14.0 = 40.6 NFnet = Fx2 + Fy2 =tan =FyFx = 263=40.64.78a4.78f + a40.6f22= 40.9 NFIG. P23.5722Electric FieldsP23.58d cos 30.0 = 15.0 cm,15.0 cmd=cos 30.0From Figure A:or = sin 1From Figure B: = sin 1FG d IJH 50.0 cm KF 15.0 cm I = 20.3GH 50.0 cmacos 30.0f JKFigure AFqor= tan mgFq = mg tan 20.3From Figure C:Fq = 2 F cos 30.0(1)LM k q OP cos 30.0MN a0.300 mf PQCombining equations (1) and (2),L k q OP cos 30.0 = mg tan 20.32MMN a0.300 mf PQmg a0.300 mf tan 20.3q =2eFq = 2e2(2)Figure B22222 k e cos 30.0e2.00 10 kg je9.80 m s ja0.300 mf tan 20.3=2e8.99 10 N m C j cos 30.03q222922Figure Cq = 4.20 10 14 C 2 = 2.05 10 7 C = 0.205 CP23.59ChargeFIG. P23.58Qresides on each block, which repel as point charges:2F=Q=Solving for Q,*P23.60P23.61b gb g = kbL L g .Lk bL L g.2Lke Q 2 Q 2i2ikeIf we place one more charge q at the 29th vertex, the total force on the central charge will add up tok qQk e qQtoward vertex 29 .F28 charges =zero: F28 charges + e 2 away from vertex 29 = 0aa2According to the result of Example 23.7, the left-hand rod createsthis field at a distance d from its right-hand end:k eQE=d 2a + dadF =F=F=fk eQQdx2a d d + 2aak eQ2a2+ k eQ4a2FIG. P23.61fIJz a f FGHKFk Q I FFG ln 2a + b + ln b IJ = k Q ln bH bK 4a ab 2 afab + 2af = GH 4a JK lnGH bb 2abk Q2dx12a + x= e lnx x + 2ax2a2ax=b 2a2e22bb2a2e222b2 4a 2IJK23Chapter 23P23.62bgAt equilibrium, the distance between the charges is r = 2 0.100 m sin 10.0 = 3.47 10 2 mNow consider the forces on the sphere with charge +q , and use Fy = 0 : Fx = 0 : Fy = 0 :mgcos 10.0= F2 F1 = T sin 10.0T cos 10.0 = mg , or T =(1)Fnet L(2)q+qrFnet is the net electrical force on the charged sphere. Eliminate T from (2) by use of (1).mg sin 10.0Fnet == mg tan 10.0 = 2.00 10 3 kg 9.80 m s 2 tan 10.0 = 3.46 10 3 Ncos 10.0ejejFnet is the resultant of two forces, F1 and F2 . F1 is the attractive force on +q exertedby q , and F2 is the force exerted on +q by the external electric field.FIG. P23.62Fnet = F2 F1 or F2 = Fnet + F1e5.00 10 Cje5.00 10 Cj = 1.87 10C je3.47 10 mj8e9F1 = 8.99 10 N m282322NThus, F2 = Fnet + F1 yields F2 = 3.46 10 3 N + 1.87 10 2 N = 2.21 10 2 Nand F2 = qE , or E =P23.63zQ = d =zF2 2. 21 10 2 N== 4. 43 10 5 N C = 443 kN C .q5.00 10 8 C90 .0 0 cos Rd = 0 R sin 90 .0 90 .0 90.0 a f= 0 R 1 1 = 2 0 R = 10.0 C msob ga fF b3.00 Cge cos Rd j I1 F b3.00 C gbd g IdF =JJGH RJK cos = 41 GGH4 RKe3.00 10 Cje10.0 10 C mj cos dF = z e8.99 10 N m C ja0.600 mf8.99a30.0fF =e10 Nj z FGH 1 + 1 cos 2 IJK d0.6002 2F1 1I = 0.707 N Downward.F = a0.450 N fG + sin 2JKH2 4Q = 12.0 C = 2 0 0.600 m = 12.0 C020y2090.0 y2066922290 .0 y3 2 2cos01 2 2y1100360cos20360FIG. P23.63Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0 .P23.64At an equilibrium position, the net force on the charge Q is zero. The equilibrium position can belocated by determining the angle corresponding to equilibrium.In terms of lengths s,attractive force1a 3 , and r, shown in Figure P23.64, the charge at the origin exerts an2es +continued on next pagek eQq12a 3j224Electric FieldsThe other two charges exert equal repulsive forces of magnitudeof the two repulsive forces add, balancing the attractive force,Fnet = k eQq12k eQqr2. The horizontal componentsLM 2 cosMM r es +N2a12r=The equilibrium condition, in terms of , isFnet =Thus the equilibrium value of satisfies2 cos sin 2 sin 2as=From Figure P23.64OPP=03j PQ11a cot 2FG 4 IJ k QqFG 2 cos sin H a K GHe22ee3 + cot j2IJJ = 0 .3 + cot j K12=1.One method for solving for is to tabulate the left side. To three significant figures a value of corresponding to equilibrium is 81.7.The distance from the vertical side of the triangle to the equilibrium position is1s = a cot 81.7 = 0.072 9 a .22 cos sin 2 60e3 + cot 24702.654801.226900811.09181.51.02481.7FIG. P23.64j0.997A second zero-field point is on the negative side of the x-axis, where = 9.16 and s = 3.10 a .P23.65Fe T2 sin 2 = 0 and mg T2 cos 2 = 0 .FeT sin 2= 2= tan 2 .mg T2 cos 2From the Q charge we haveFe = T1 sin 1 = 0 and mg T1 cos 1 = 0 .Combining these we find(b)From the 2Q charge we haveCombining these we find(a)FeT sin 1= 1= tan 1 or 2 = 1 .mg T1 cos 1Fe =k e 2QQr2=2 k eQFIG. P23.652r2If we assume is small thentan r 2.Substitute expressions for Fe and tan into either equation found in part (a) and solve for r.F IGH JKFGHFe2k Q 2 14k eQ 2r= tan then e 2and solving for r we find r mg2mgmgrIJK13.25Chapter 23P23.66(a)The distance from each corner to the center of the square isFG L IJ + FG L IJH 2K H 2K22=L2z+q.+qL/2L/2+qThe distance from each positive charge to Q is thenz2 +xQ+qL2. Each positive charge exerts a force directed2FIG. P23.66k eQqalong the line joining q and Q , of magnitudez 2 + L2 2.zThe line of force makes an angle with the z-axis whose cosine is2z + L2 2The four charges together exert forces whose x and y components add to zero, while the4k e QqzF= kz-components add to3 22z + L2 2e(b)For z >> L , the magnitude of this force is Fz = 4k eQqzeL 2 j232F 4a2f k Qq I z = maGH L JK3 2=3ejzTherefore, the objects vertical acceleration is of the form a z = 2 zwith 2 =af4232k eQq3mL=k eQq 128mL3.Since the acceleration of the object is always oppositely directed to its excursion fromequilibrium and in magnitude proportional to it, the object will execute simple harmonicmotion with a period given byT=P23.67(a)2=2a128f14mL3=k e QqmL3.k eQqa8 f14FGHIJKqE,mwhich is constant and directed downward. Therefore, it behaves like a simple pendulum inthe presence of a modified uniform gravitational field with a period given by:The total non-contact force on the cork ball is: F = qE + mg = m g +T = 2L0.500 m= 2g + qE m9.80 m s 2 + 2.00 10 6 C 1.00 10 5 N C 1.00 10 3 kgejej= 0.307 s(b)Yes . Without gravity in part (a), we get T = 2T = 20.500 me2.00 10 Cje1.00 1065jLqE mN C 1.00 10 3 kg= 0.314 s (a 2.28% difference).26Electric FieldsP23.68The bowl exerts a normal force on each bead, directed along theradius line or at 60.0 above the horizontal. Consider the free-bodydiagram of the bead on the left: Fy = n sin 60.0 mg = 0 ,mg.sin 60.0orn=Also, Fx = Fe + n cos 60.0 = 0 ,keq 2orR2F mg IRGH k 3 JKq=Thus,= n cos 60.0 =nmgmg=.tan 60.03Femg12.FIG. P23.68eP23.69(a)60.0There are 7 terms which contribute:3 are s away (along sides)3 are1 is12s away (face diagonals) and sin =3s away (body diagonal) and sin == cos 213.FIG. P23.69The component in each direction is the same by symmetry.F=ke q2s2LM1 + 2 + 1 OPei + j + kj =N 2 2 3 3Qke q2keq 2s2a1.90fei + j + kj(b)P23.70F = Fx2 + Fy2 + Fz2 = 3. 29(a)Zero contribution from the same face due to symmetry, oppositeface contributesFG k q sin IJ whereHr K4e2sin =srr=s2away from the originFG s IJ + FG s IJH 2K H 2KE=42k e qsr3=2+ s 2 = 1.5 s = 1.22s4a1.22f3keqs2= 2.18keqs2FIG. P23.70(b)The direction is the k direction.27Chapter 23P23.71k e xQE = Ex =The field on the axis of the ring is calculated in Example 23.8,The force experienced by a charge q placed along the axis of the ring isex + a jLM xF = k QqMM ex + a jNF k Qq IJ xF = GHa Ke2 3 22eand when x << a , this becomes2 3 22OPPPQ3This expression for the force is in the form of Hookes law, with ank=effective spring constant ofk, we havemSince = 2 f =P23.72dE =FG+ a0.150 mf GHk e dqx2zE=dE = k e 2z0. 400 mx=0all chargef=I k e xi + 0.150 mjjdxJ=x + a0.150 mf JK x + a0.150 mfe xi + 0.150 mjjdxx + a0.150 mf x i + 0.150 m j222.2FIG. P23.720. 400 m22202ma 32 3 229k eQq122 3220. 400 m2a3eLMOPa0.150 mf jx+iE = k M+PPx + a0.150 mfa0.150 mf x + a0.150 mfMNQE = e8.99 10 N m C je35.0 10 C mj ia 2.34 6.67f m + ja6.24 0 f mE = e 1.36 i + 1.96 jj 10 N C = e 1.36 i + 1.96 jj kN Cek eQq09113P23.73The electrostatic forces exerted on the two charges result in a net torque = 2 Fa sin = 2Eqa sin .For small , sin and using p = 2 qa , we have = Ep .The torque produces an angular acceleration given by = I = ICombining these two expressions for torque, we haveThis equation can be written in the formd 2dt 2.FG IJH KEpd 2 =0.+2Idtd 2dt2FIG. P23.73= 2 where 2 =Ep.IThis is the same form as Equation 15.5 and the frequency of oscillation is found by comparison withEquation 15.11, orf=12pE1=I22 qaE.I28Electric FieldsANSWERS TO EVEN PROBLEMSP23.2P23.4P23.36(a) 200 pC; (b) 141 pC; (c) 58.9 pCP23.38see the solutionP23.40(a) 2.62 10 24 ; (b) 2.38 electrons for every10 9 present(a) 57.5 N91; (b) q1 is negative and q 2 is positive3P23.62.51 10P23.8514 kNP23.42electron: 4.39 Mm s ; proton: 2.39 km sP23.10x = 0.634d . The equilibrium is stable if thethird bead has positive charge.P23.44(a) 57.6 i Tm s 2 ; (b) 2.84i Mm s; (c) 49.3 nsP23.46(a) down; (b) 3.43 CP23.48The particle strikes the negative plate aftermoving in a parabola 0.181 mm high and0.961 mm.P23.50Possible only with +51.3 C atx = 16.0 cmP23.12md 3where m is the massk e qQ(a) period =2k e qQof the object with charge Q ; (b) 4amd 3P23.141.49 gP23.16720 kN C downP23.52(a) 24.2 N C at 0; (b) 9. 42 N C at 117P23.18(a) 18.0 i 218 j kN C ;P23.545.25 CP23.56(a)P23.580.205 Cej(b) 36.0 i 436 j mNmgmgA; (b)A cot + BA cos + B sin P23.20(a) 12.9 j kN C ; (b) 38.6 j mNP23.22see the solutionP23.24 2 ke qi6a2P23.62443 i kN CP23.26ke 0ix0P23.640.072 9 aP23.28ke 0i2x0P23.66see the solution; the period isP23.68RP23.30P23.32P23.34P23.60e je j(a) 383 MN C away; (b) 324 MN C away;(c) 80.7 MN C away; (d) 6.68 MN C awayLMej ead + hf + R jN2 k Qi L(b)R h MNh + ed + R j ead + hf + R jk eQ id 2 + R2he22a2toward the 29th vertexF mg IGH k 3 JK12esee the solution(a)k e qQ1 22 122221 22OP;QOPQ12P23.70(a) see the solution; (b) kP23.72e1.36i + 1.96 jj kN C81 4mL3k eQq24Gausss LawCHAPTER OUTLINE24.124.224.3Electric FluxGausss LawApplication of Gausss Lawto Various ChargeDistributionsConductors in ElectrostaticEquilibriumFormal Derivation ofGausss LawANSWERS TO QUESTIONS24.5The luminous flux on a given area is less when the sun is low inthe sky, because the angle between the rays of the sun and thelocal area vector, dA, is greater than zero. The cosine of thisangle is reduced. The decreased flux results, on the average, incolder weather.Q24.2If the region is just a point, line, or plane, no. Consider twoprotons in otherwise empty space. The electric field is zero atthe midpoint of the line joining the protons. If the field-freeregion is three-dimensional, then it can contain no charges, butit might be surrounded by electric charge. Consider the interiorof a metal sphere carrying static charge.Q24.324.4Q24.1The surface must enclose a positive total charge.Q24.4The net flux through any gaussian surface is zero. We can argue it two ways. Any surface containszero charge so Gausss law says the total flux is zero. The field is uniform, so the field lines enteringone side of the closed surface come out the other side and the net flux is zero.Q24.5Gausss law cannot tell the different values of the electric field at different points on the surface.When E is an unknown number, then we can say E cos dA = E cos dA . When E x , y , z is anzunknown function, then there is no such simplification.zbgQ24.6The electric flux through a sphere around a point charge is independent of the size of the sphere. Asphere of larger radius has a larger area, but a smaller field at its surface, so that the product of fieldstrength and area is independent of radius. If the surface is not spherical, some parts are closer to thecharge than others. In this case as well, smaller projected areas go with stronger fields, so that thenet flux is unaffected.Q24.7Faradays visualization of electric field lines lends insight to this question. Consider a section of a1field lines pointing out from it horizontally tovertical sheet carrying charge +1 coulomb. It has0the right and left, all uniformly spaced. The lines have the same uniform spacing close to the sheetand far away, showing that the field has the same value at all distances.2930Gausss LawQ24.8Consider any point, zone, or object where electric field lines begin. Surround it with a close-fittinggaussian surface. The lines will go outward through the surface to constitute positive net flux. ThenGausss law asserts that positive net charge must be inside the surface: it is where the lines begin.Similarly, any place where electric field lines end must be just inside a gaussian surface passing netnegative flux, and must be a negative charge.Q24.9Inject some charge at arbitrary places within a conducting object. Every bit of the charge repelsevery other bit, so each bit runs away as far as it can, stopping only when it reaches the outer surfaceof the conductor.Q24.10If the person is uncharged, the electric field inside the sphere is zero. The interior wall of the shellcarries no charge. The person is not harmed by touching this wall. If the person carries a (small)charge q, the electric field inside the sphere is no longer zero. Charge q is induced on the inner wallof the sphere. The person will get a (small) shock when touching the sphere, as all the charge on hisbody jumps to the metal.Q24.11The electric fields outside are identical. The electric fields inside are very different. We have E = 0everywhere inside the conducting sphere while E decreases gradually as you go below the surface ofthe sphere with uniform volume charge density.Q24.12There is zero force. The huge charged sheet creates a uniform field. The field can polarize theneutral sheet, creating in effect a film of opposite charge on the near face and a film with an equalamount of like charge on the far face of the neutral sheet. Since the field is uniform, the films ofcharge feel equal-magnitude forces of attraction and repulsion to the charged sheet. The forces addto zero.Q24.13Gausss law predicts, as described in section 24.4, that excess charge on a conductor will reside onthe surface of the conductor. If a car is left charged by a lightning strike, then that charge will remainon the outside of the car, not harming the occupants. It turns out that during the lightning strike, thecurrent also remains on the outside of the conductor. Note that it is not necessarily safe to be in afiberglass car or a convertible during a thunderstorm.SOLUTIONS TO PROBLEMSSection 24.1Electric Fluxjaef(a) E = EA cos = 3.50 10 3 0.350 0.700 cos 0 = 858 N m 2 C(b) = 90.0(c)P24.1 E = 3.50 10 3 0.350 0.700 cos 40.0 = 657 N m 2 CE = 0jaeefjejP24.2 E = EA cos = 2.00 10 4 N C 18.0 m 2 cos 10.0 = 355 kN m 2 CP24.3 E = EA cos aaA = r 2 = 0.200f5.20 10 5 = E 0.126 cos 0f2= 0.126 m 2E = 4.14 10 6 N C = 4.14 MN CChapter 24P24.4(a)afafA = 10.0 cm 30.0 cm30.0 cm2A = 300 cm = 0.030 0 m 2 E , A = EA cos jbe31g E , A = 7.80 10 4 0.030 0 cos 18010.0 cm60.0 E , A = 2.34 kN m 2 C(b)eja fF 10.0 cm IJ = 600 cm = 0.060 0 mA = a30.0 cmfa wf = a30.0 cmfGH cos 60.0 K= e7.80 10 jb0.060 0g cos 60.0 = +2.34 kN m C2224E, A(c)FIG. P24.4 E , A = EA cos = 7.80 10 4 A cos 60.0The bottom and the two triangular sides all lie parallel to E, so E = 0 for each of these. Thus, E, total = 2.34 kN m 2 C + 2.34 kN m 2 C + 0 + 0 + 0 = 0 .ej E = E A = a i + b j A i = aA E = a i + b j A j = bA(c)P24.6(a)(b)P24.5 E = a i + b j Ak = 0jejOnly the charge inside radius R contributes to the total flux.E =P24.7eq0 E = EA cos through the basea fa f E = 52.0 36.0 cos 180 = 1.87 kN m 2 C .Note the same number of electric field lines go through the base as go through thepyramids surface (not counting the base).FIG. P24.7For the slanting surfaces, E = +1.87 kN m 2 C .P24.8The flux entering the closed surface equals the flux exiting the surface. The flux entering the left sideof the cone is E = E dA = ERh . This is the same as the flux that exits the right side of the cone.zNote that for a uniform field only the cross sectional area matters, not shape.32Gausss LawSection 24.2P24.9(a)Gausss LawE =bg+5.00 C 9.00 C + 27.0 C 84.0 Cqin== 6.89 10 6 N m 2 C 208.85 10 12 C 2 N m 2 E = 6.89 MN m 2 C(b)P24.10(a)Since the net electric flux is negative, more lines enter than leave the surface.E=k eQr28.90 10:But Q is negative since E points inward.(b)P24.11E =2Q = 5.56 10 8 C = 55.6 nCThe negative charge has a spherically symmetric charge distribution.qin0Through S1E =2Q + QQ= 00Through S 2E =+Q Q= 00Through S3E =2Q + Q Q2Q= 00Through S 4P24.12e8.99 10 jQ=a0.750f92E = 0(a)One-half of the total flux created by the charge q goes through the plane. Thus,q11 q E , plane = E , total ==.22 02 0(b)The square looks like an infinite plane to a charge very close to the surface. Hence,q E , square E , plane =.2 0(c)FG IJH KThe plane and the square look the same to the charge.P24.13The flux through the curved surface is equal to the flux through the flat circle, E0 r 2 .P24.14(a) E , shell =(b) E, half shell =(c)qin 12.0 10 6== 1.36 10 6 N m 2 C = 1.36 MN m 2 C0 8.85 10 1211.36 10 6 N m 2 C = 6.78 10 5 N m 2 C = 678 kN m 2 C2ejNo, the same number of field lines will pass through each surface, no matter how theradius changes.Chapter 24P24.15With very small, all points on the hemisphere are nearly ata distance R from the charge, so the field everywhere on thekQcurved surface is e 2 radially outward (normal to theRsurface). Therefore, the flux is this field strength times thearea of half a sphere:(a)z 033Q curved = E dA = Elocal A hemisphereFGH curved = k e(b)QR2IJ FG 1 4 R IJ = 1 Qa2 f =K H 2 K 4 20FIG. P24.15The closed surface encloses zero charge so Gausss law gives curved + flat = 0*P24.16+Q2 0 flat = curved =orQ.2 0Consider as a gaussian surface a box with horizontal area A, lying between 500 and 600 m elevation.zE dA =b+120 N CgA + b100 N CgA = Aa100 mfb20 N Cge8.85 10 C N m j = 1.77 10=q:001222100 m12C m3The charge is positive , to produce the net outward flux of electric field.P24.17The total charge is Q 6 q . The total outward flux from the cube isQ6q0, of which one-sixth goesthrough each face:b g=b g=E one faceE one faceP24.18Q6q6 0Q6q6 0=a5.00 6.00f 1066 8.85 10C N m212C2= 18.8 kN m 2 C .The total charge is Q 6 q . The total outward flux from the cube isthrough each face:b gE one faceP24.19=Q6q6 0.If R d , the sphere encloses no charge and E =qin= 0 .0If R > d , the length of line falling within the sphere is 2 R 2 d 2soE =2 R 2 d 20.Q6q0, of which one-sixth goes34Gausss LawP24.20 E , hole = E A hole =FG k Q IJ e rHR Ke2F e8.99 10 N m C je10.0 10 Cj IJJ e1.00 10j GGa0.100 mfHK292=6223jm2 E, hole = 28.2 N m 2 CP24.21E =qin170 10 6 C== 1.92 10 7 N m 2 C0 8.85 10 12 C 2 N m 21.92 10 7 N m 2 C1E =66(a)b g(b) E = 19.2 MN m 2 C(c)E one face=b gE one face= 3.20 MN m 2 CThe answer to (a) would change because the flux through each face of the cube wouldnot be equal with an asymmetric charge distribution. The sides of the cube nearer thecharge would have more flux and the ones further away would have less. The answerto (b) would remain the same, since the overall flux would remain the same.P24.22No charge is inside the cube. The net flux through the cube is zero. Positive fluxcomes out through the three faces meeting at g. These three faces together fillsolid angle equal to one-eighth of a sphere as seen from q, and together pass1 q. Each face containing a intercepts equal flux going into the cube:flux8 0FG IJH K0 = E , net E , abcd =Section 24.3P24.23FIG. P24.22q= 3 E , abcd +8 0q24 0Application of Gausss Law to Various Charge DistributionsThe charge distributed through the nucleus creates a field at the surface equal to that of a pointk qcharge at its center: E = e2re8.99 10 Nm C je82 1.60 10E=a208f 1.20 10 m9213E = 2.33 10 21 N C21519Cj2away from the nucleusChapter 24(a)E=(b)P24.24E=k eQra3= 0e8.99 10 je26.0 10 ja0.100f = 365 kN Caa0.400fk Q e8.99 10 je 26.0 10 j== 1.46 MN CE=ra0.400fk eQr369=369e(c)E=(d)22k eQr2e8.99 10 je26.0 10 j ==a0.600f69649 kN C2The direction for each electric field is radially outward .*P24.25mg = qE = q0P24.26120.01 9.8Q 2 0 mg 2 8.85 10=== 2.48 C m 2Aq0.7 10 60jbe2 8.99 10 9 Q 2.402k e 4E=3.60 10 =0.190rQ = +9.13 10 7 C = +913 nC(a)gE= 0(b)*P24.27ja fa feFG IJ = qFG Q A IJH2 K H 2 KThe volume of the spherical shell isaf a0.20 mf4 0.25 m333= 3.19 10 2 m3 .Its charge isejejV = 1.33 10 6 C m 3 3.19 10 2 m3 = 4.25 10 8 C .The net charge inside a sphere containing the protons path as its equator is60 10 9 C 4. 25 10 8 C = 1.02 10 7 C .The electric field is radially inward with magnitudeke qr2=q0 4 r2=e8.99 10 9 Nm 2 1.02 10 7 C2afC 0.25 m2j = 1.47 104N C.For the proton F = maF eEr IJv=GHmKeE =12mv 2rF 1.60 10=GGH19ejC 1.47 10 4 N C 0.25 m1.67 10 27 kgIJJK12= 5.94 10 5 m s .3536Gausss LawP24.28e = 8.60 10 6 C cm 2E=jFGH 100mcm IJK2= 8.60 10 2 C m 28.60 10 2== 4.86 10 9 N C away from the wall2 0 2 8.85 10 12ejThe field is essentially uniform as long as the distance from the center of the wall to the field point ismuch less than the dimensions of the wall.P24.29If is positive, the field must be radially outward. Choose as thegaussian surface a cylinder of length L and radius r, contained insidethe charged rod. Its volume is r 2 L and it encloses charge r 2 L .Because the charge distribution is long, no electric flux passesthrough the circular end caps; E dA = EdA cos 90.0 = 0 . The curvedsurface has E dA = EdA cos 0 , and E must be the same strengtheverywhere over the curved surface.q r 2 LGausss law, E dA =,becomesEdA =.00CurvedzFIG. P24.29zSurfaceNow the lateral surface area of the cylinder is 2 rL :b gE 2 r L =*P24.30 r 2 L.0rradially away from the cylinder axis .2 0E=Thus,Let represent the charge density. For the field inside the sphere at r1 = 5 cm we haveE1 4 r12 =q inside 4 r13 =03 0E1 =ejer1 3 0j12C 2 86 10 3 N3 0 E1 3 8.85 10=== 4.57 10 5 C m 3 .r10.05 mNm 2CNow for the field outside at r3 = 15 cm34 r2 E3 4 r32 =3 0af e4.57 10 Cj = 8.99 10k e 4 0.10 mE3 = 2r3 335m39eNm 2 1.91 10 7 Ca0.15 mf C22j = 7.64 104NCE 3 = 76. 4 kN C radially inwardP24.31E= 0(a)E=(b)P24.32k eQr2e8.99 10 je32.0 10 j = 7.19 MN C=a0.200f69E = 7.19 MN C radially outward2The distance between centers is 2 5.90 10 15 m . Each produces a field as if it were a point chargeat its center, and each feels a force as if all its charge were a point at its center.F=k e q1 q 2r2ea46f e1.60 10 CjC je2 5.90 10 mj29= 8.99 10 N m21921522= 3.50 10 3 N = 3.50 kNChapter 24P24.33Consider two balloons of diameter 0.2 m, each with mass 1 g, hanging apart with a0.05 m separation on the ends of strings making angles of 10 with the vertical.(a)mg Fy = T cos 10mg = 0 T = cos 10 Fx = T sin 10 Fe = 0 Fe = T sin 10 , soFe =FG mg IJ sin 10 = mg tan 10 = b0.001 kgge9.8 m s j tan 10H cos 10 K2Fe 2 10 3 N ~ 10 3 N or 1 mN(b)Fe =keq 2r2e8.99 10 N mNa0.25 mf92 1032jC 2 q22q 1. 2 10 7 C ~ 10 7 C or 100 nCkeq(c)E=(d)*P24.3437E =r2e8.99 109jeN m 2 C 2 1.2 10 7 Ca0.25 mf2N C ~ 10 kN C4 3a 3=3Q4 a 3The flux is that created by the enclosed charge within radius r:E =(b)4q1.2 10 7 C= 1. 4 10 4 N m 2 C ~ 10 kN m 2 C0 8.85 10 12 C 2 N m 2The charge density is determined by Q =(a)j 1.7 10E =q in 4 r 3 4 r 3 3QQr 3===03 03 0 4 a 30 a 3Q. Note that the answers to parts (a) and (b) agree at r = a .0(c)EQ000aFIG. P24.34(c)rFIG. P24.3338Gausss LawP24.35(a)ejej9222.00 10 6 C 7.00 m2 k e 2 8.99 10 N m C=E=0.100 mrE = 51.4 kN C , radially outward(b)bg E = EA cos = E 2 r cos 0j aefbga f E = 5.14 10 4 N C 2 0.100 m 0.020 0 m 1.00 = 646 N m 2 CP24.36(a)=Q5.70 10 6== 2.13 10 2 C m 334 34a0.040 033bgFG 4 r IJ = e2.13 10H3 KF4 I= G r J = e 2.13 10H3 Kqin = (b)32qin324jFGH 3 IJK b0.020 0g4jFGH 3 IJK b0.040 0g3= 7.13 10 7 C = 713 nC3= 5.70 C9.00 10 6 C m 2== 508 kN C , upward2 0 2 8.85 10 12 C 2 N m 2P24.37E=P24.38Note that the electric field in each case is directed radially inward, toward the filament.ejejejejejejej(a)E=69222 k e 2 8.99 10 N m C 90.0 10 C m== 16.2 MN C0.100 mr(b)E=69222 k e 2 8.99 10 N m C 90.0 10 C m== 8.09 MN C0.200 mr(c)69222 k e 2 8.99 10 N m C 90.0 10 C m== 1.62 MN CE=1.00 mrSection 24.4P24.39zConductors in Electrostatic Equilibriumb gEdA = E 2 rl =qin0E=q in l=2 0 r 2 0 r(a)r = 3.00 cmE= 0(b)r = 10.0 cmE=(c)r = 100 cmE=e30.0 10 9ja2 8.85 10 12 0.100e30.0 10 9f==ja f2 8.85 10 12 1.005 400 N C , outward540 N C , outwardChapter 24P24.40=P24.41EA =From Gausss Law,jaQ039fQ=0 E = 8.85 10 12 130 = 1.15 10 9 C m 2 = 1.15 nC m 2Ae conductorfor the field outside the aluminum looks0The fields are equal. The Equation 24.9 E = insulatorfor the field around glass. But its charge will spread out to2 0Qcover both sides of the aluminum plate, so the density is conductor =. The glass carries charge2AQQonly on area A, with insulator = . The two fields arethe same in magnitude, and both areA2 A 0perpendicular to the plates, vertically upward if Q is positive.different from Equation 24.8 E =*P24.42(a)All of the charge sits on the surface of the copper sphere at radius 15 cm. The field inside iszero .(b)The charged sphere creates field at exterior points as if it were a point charge at the center:E=ke qr2away =e8.99 109jeNm 2 40 10 9 C2afC 0.17 m(c)(d)P24.43e8.99 10E=(a)E=j outward =Nm 2 40 10 9 C2afC 0.75 m02ej outward =1.24 10 4 N C outwardAll three answers would be the same.9je2je639 N C outwardj = 8.00 10 4 8.85 10 12 = 7.08 10 7 C m 2 = 708 nC m 2 , positive on one face and negative on the other.P24.44jafQ2Q = A = 7.08 10 7 0.500 CAQ = 1.77 10 7 C = 177 nC , positive on one face and negative on the other.e(b)=(a)E= 0k eQ(b)E=(c)e8.99 10 je8.00 10 j = 7.99 10=b0.030 0g7NCE = 79.9 MN C radially outwarde8.99 10 je4.00 10 j = 7.34 10=b0.070 0g6NCE = 7.34 MN C radially outward69E= 0(d)E=r2k eQr2269240Gausss LawP24.45The charge divides equally between the identical spheres, with chargelike point charges at their centers:F=P24.46b gb g = k QaL + R + Rf 4aL + 2Rfke Q 2 Q 22e22=e8.99 10 9 N m 2 60.0 10 6 Caf4 C 2 2.01 mjQon each. Then they repel222= 2.00 N .The electric field on the surface of a conductor varies inversely with the radius of curvature of thesurface. Thus, the field is most intense where the radius of curvature is smallest and vice-versa. Thelocal charge density and the electric field intensity are related byE=(a) =0 E .or0Where the radius of curvature is the greatest,ejej =0 Emin = 8.85 10 12 C 2 N m 2 2.80 10 4 N C = 248 nC m 2 .(b)Where the radius of curvature is the smallest,ejej =0 Emax = 8.85 10 12 C 2 N m 2 5.60 10 4 N C = 496 nC m 2 .P24.47(a)Inside surface: consider a cylindrical surface within the metal. Since E inside the conductingshell is zero, the total charge inside the gaussian surface must be zero, so the insidecharge/length = .qin0 = + qinOutside surface:P24.48(a)(b)P24.49(a)The total charge on the metal cylinder isqout = 2 + (b)soso the outside charge/length isE=E=erk eQr2r3 .3radially outward2 0 re8.99 10 je6.40 10 j ==a0.150f6922.56 MN C , radially inwardThe charge density on each of the surfaces (upper and lower) of the plate is:E=FG IJH KeE=j81 q1 4.00 10 C== 8.00 10 8 C m 2 = 80.0 nC m 2 .2 A2 0.500 m 2afFG IJ k = F 8.00 10 C m I k = b9.04 kN CgkH K GH 8.85 10 C N m JK80(c)2 = qin + qoutE=0=(b)b g = 6k =2 k e 3= b9.04 kN Cgk12222Chapter 24(a)The charge +q at the center induces charge q on the inner surface of the conductor,where its surface density is:qa =.4 a 2(b)P24.50The outer surface carries charge Q + q with densityb =P24.5141Q+q.4 b 2Use Gausss Law to evaluate the electric field in each region, recalling that the electric field is zeroeverywhere within conducting materials. The results are:E = 0 inside the sphere and within the material of the shellE = keE = ke2Qoutside the shell, directed radially outward .r2ChargeQ is on the outer surface of the sphere .Charge+Q is on the inner surface of the shell ,andP24.52Qbetween the sphere and shell, directed radially inwardr2+2Q is on the outer surface of the shell.An approximate sketch is given at the right. Note that the electric field linesshould be perpendicular to the conductor both inside and outside.FIG. P24.52Section 24.5P24.53(a)Formal Derivation of Gausss LawUniform E, pointing radially outward, so E = EA . The arc length is ds = Rd ,and the circumference is 2 r = 2 R sin zA = 2 rds =zbzga2 R sin Rd = 2 R 2 sin d = 2 R 2 cos 0f00afa1QQE = 2 R 2 1 cos =1 cos 4 0 R 22 0afb= 2 R 2 1 cos [independent of R!]f(b)For = 90.0 (hemisphere): E =QQ1 cos 90 =.2 02 0(c)For = 180 (entire sphere): E =QQ1 cos 180 =2 00af[Gausss Law].gFIG. P24.5342Gausss LawAdditional ProblemsIn general,E = ay i + cxkzze E = E dA =wz(b)y =hx=0jx2 E = ch xdx = ch2x=0(a)y=0yay i + cxk kdAwP24.55zE = ay i + bz j + cxkIn the xy plane, z = 0 andP24.54x=0x=wchw 2=2dA = hdxxFIG. P24.54qin = +3Q Q = +2QThe charge distribution is spherically symmetric and qin > 0 . Thus, the field is directedradially outward .k e qinfor r c .E=(d)Since all points within this region are located inside conducting material, E = 0 forr2=2 k eQ(c)r2b < r < c.z(e) E = E dA = 0 qin =0 E = 0(f)qin = +3Q(g)E=(h)qin = V =(i)E=(j)From part (d), E = 0 for b < r < c . Thus, for aspherical gaussian surface with b < r < c ,qin = +3Q + qinner = 0 where qinner is thecharge on the inner surface of theconducting shell. This yields qinner = 3Q .(k)Since the total charge on the conductingshell is q net = qouter + qinner = Q , we havek e qinr2k e qinr2=3 k eQr2(radially outward) for a r < b .F +3Q I FG 4 r IJ = +3Q rGH a JK H 3 KaIk F=GH +3Q ra JK = 3 k Q ar (radially outward) for 0 r a .r43e2333333e3b gqouter = Q qinner = Q 3Q = +2Q .(l)This is shown in the figure to the right.EabcFIG. P24.55(l)rChapter 2443P24.56The sphere with large charge creates a strong field to polarize the other sphere. That means itpushes the excess charge over to the far side, leaving charge of the opposite sign on the near side.This patch of opposite charge is smaller in amount but located in a stronger external field, so it canfeel a force of attraction that is larger than the repelling force felt by the larger charge in the weakerfield on the other side.P24.57(a)zejE dA = E 4 r 2 =qin0FG 4 r IJH3 K3For r < a ,soE=For a < r < b and c < r ,qin = Q .SoQE=.4 r 2 0For b r c ,(b)qin = E = 0 , since E = 0 inside a conductor.r.3 0FIG. P24.57Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside theconductor, the total charge enclosed by a spherical surface of radius b r c must be zero.q1 + Q = 0Therefore,and1 =q14 b2=Q.4 b 2Let q 2 = induced charge on the outside surface of the hollow sphere. Since the hollowsphere is uncharged, we requireqQ2 = 1 2 =.q1 + q 2 = 0and4 c4 c 2P24.58zejE dA = E 4 r 2 =(a)e3.60 103qin0j afN C 4 0.100 m2=Q8.85 1012C 2 N m2aa < r < bfQ = 4.00 10 9 C = 4.00 nC(b)We take Q to be the net charge on the hollow sphere. Outside c,Q + Q2+2.00 10 2 N C 4 0.500 m =r>c8.85 10 12 C 2 N m 2ej afa fQ + Q = +5.56 10 9 C , so Q = +9.56 10 9 C = +9.56 nC(c)For b < r < c : E = 0 and qin = Q + Q1 = 0 where Q1 is the total charge on the inner surface ofthe hollow sphere. Thus, Q1 = Q = +4.00 nC .Then, if Q 2 is the total charge on the outer surface of the hollow sphere,Q 2 = Q Q1 = 9.56 nC 4.0 nC = +5.56 nC .44Gausss Law*P24.59yThe vertical velocity component of the moving chargeincreases according todv ymdv y dx= qE y .mdx dt= Fydtv0qNowqEy dxmvzFIG. P24.59q Ey dx .mv 0vyxQzvyv y = dv y =vxddx= v x has the nearly constant value v. Sodtdv y =The radially outward compnent of the electric field varies along the x axis, but is described byzzEy dA =zSoEy dx =tan =P24.60vyvb gEy 2 d dx ==Q.0qQQand v y =. The angle of deflection is described by2 d 0mv 2 d 0qQ2 0 dmv = tan 12qQ2 0 dmv 2.First, consider the field at distance r < R from the center of a uniform sphere of positive chargeQ = + e with radius R.begj4 r 2 E =(a)qin V==00FGH+e4 33 RIJK43r30so E =F e I r directed outwardGH 4 R JK03The force exerted on a point charge q = e located at distance r from the center is thenF = qE = eF e I r = F e I r =GH 4 R JK GH 4 R JK20330 Kr .k e2e2= e334 0 RR(b)K=(c)Fr = m e a r = F k e I r , so aGH R JKe23r=F k e I r = rGH m R JKee232Thus, the motion is simple harmonic with frequency(d)f = 2.47 10151Hz =2e8.99 109jef=1=22N m 2 C 2 1.60 10 19 Ce9.11 1031jj2kg R 3which yields R 3 = 1.05 10 30 m3 , or R = 1.02 10 10 m = 102 pm .kee2me R3.Chapter 24P24.61The field direction is radially outward perpendicular to the axis. The field strength depends on r butnot on the other cylindrical coordinates or z. Choose a Gaussian cylinder of radius r and length L.If r < a ,E =E=qin0and2 r 0orbgE 2 rL =E=bE=gbE=ar < a f.erjL + r 2 a 2 L0e + r 2 a 22 r 0gE 2 rL =If r > b ,L02 r 0E 2 rL =If a < r < b ,P24.6245jreaa < r < bf .jL + b 2 a 2 L0e + b 2 a 22 r 0jrar > b f.Consider the field due to a single sheet and let E+ and Erepresent the fields due to the positive and negative sheets. Thefield at any distance from each sheet has a magnitude given byEquation 24.8:E+ = E =(a).2 0To the left of the positive sheet, E+ is directed toward theleft and E toward the right and the net field over thisregion is E = 0 .(b)In the region between the sheets, E+ and E are bothdirected toward the right and the net field isE=(c)0to the right .FIG. P24.62To the right of the negative sheet, E+ and E are again oppositely directed and E = 0 .46Gausss LawP24.63The magnitude of the field due to the each sheet given byEquation 24.8 isE=(a)directed perpendicular to the sheet.2 0In the region to the left of the pair of sheets, both fields aredirected toward the left and the net field isE=0FIG. P24.63to the left .(b)In the region between the sheets, the fields due to the individual sheets are oppositelydirected and the net field isE= 0 .(c)In the region to the right of the pair of sheets, both are fields are directed toward the rightand the net field isto the right .0E=P24.64The resultant field within the cavity is the superposition of twofields, one E + due to a uniform sphere of positive charge of radius2a, and the other E due to a sphere of negative charge of radius acentered within the cavity.FGHIJK4 r 3= 4 r 2 E+3 0FGHsoSince r = a + r1 ,rrr=3 03 0soIJK4 r13 = 4 r12 E3 0E+ =E = r1 r1 =r1 .3 03 0a f r aE =3 0E = E+ + E =Thus,Ey =FIG. P24.64rraaa+== 0i +j.3 0 3 0 3 0 3 03 0Ex = 0and*P24.65b ga3 0at all points within the cavity.Consider the charge distribution to be an unbroken charged spherical shell with uniform chargedensity and a circular disk with charge per area . The total field is that due to the whole sphere,4 R 2Q==outward plus the field of the disk =radially inward. The total2202 0 2 04 0 R4 0 Rfield isoutward .=0 2 02 0Chapter 24P24.66The electric field throughout the region is directed along x; therefore, E will beperpendicular to dA over the four faces of the surface which are perpendicularto the yz plane, and E will be parallel to dA over the two faces which are parallelto the yz plane. Therefore,e E = Exx=aj A + eEx x=a+cjA = e3 + 2 a jab + e3 + 2aa + cf jab = 2abca2 a + cf .22Substituting the given values for a, b, and c, we find E = 0.269 N m 2 C .FIG. P24.66Q =0 E = 2.38 10 12 C = 2.38 pCP24.67zejE dA = E 4 r 2 =qin0zRejAR 55j4 Ar 55qin = Ar 2 4 r 2 dr = 4For r > R ,(a)0AR 5.5 0 r 2andFor r < R ,(b)E=qin = Ar 2 4 r 2 dr =zre0P24.68Ar 3.5 0E=andThe total flux through a surface enclosing the charge Q isdisk isQ. The flux through the0z disk = E dAwhere the integration covers the area of the disk. We must evaluate this integral1Qto find how b and R are related. In the figure, take dA to beand set it equal to 40the area of an annular ring of radius s and width ds. The flux through dA isbFIG. P24.68gE dA = EdA cos = E 2 sds cos .The magnitude of the electric field has the same value at all points within the annular ring,E=1 Q1Q=4 0 r 2 4 0 s 2 + b 2andcos =bb=2rs + b2ej12.Integrate from s = 0 to s = R to get the flux through the entire disk. E , disk =Qb2 0zeR0sdss2 + b 2j=32The flux through the disk equalsThis is satisfied if R = 3 b .LM eNQb s2 + b 22 0j OPQ12R=0LMMM eNQb12 0R2 + b2Qbprovided that4 0R2 + b2ej12=OPj PQP121.24748Gausss LawP24.69zE dA =qin1=0 0zzr0a4 r 2 drr4 a r4 a r 2E 4 r 2 =rdr =0 00 2E=a= constant magnitude2 0(The direction is radially outward from center for positive a; radially inward for negative a.)P24.70use a gaussian surface which is a cylinder of radius r, lengthdistribution.(a)bshell of radius r, lengthbg FGH 2 r 2E 2 r =(b)00z FGHra0g00IJK, and thickness dr so that dV = 2 r dr .I FG a r IJ so inside the cylinder, E =JK H 2 3b KFGH 0r2ra2 03bz FGH a br IJK b2 r drg or outside the cylinder, E =FGHRIJK.0R22Ra2 0 r3b0IJKConsider a cylindrical shaped gaussian surface perpendicularto the yz plane with one end in the yz plane and the other endcontaining the point x:Use Gausss law:zzE dA =rdV . The element of volume is a cylindricalbWhen r > R , Gausss law becomesb(a)00E 2 r =P24.71gWhen r < R , this becomes E 2 r =z1dV . We0, and is coaxial with the chargeIn this case the charge density is not uniform, and Gausss law is written asE dA =.yqin0gaussiansurfaceBy symmetry, the electric field is zero in the yz plane and isperpendicular to dA over the wall of the gaussian cylinder.Therefore, the only contribution to the integral is over the endcap containing the point x :za f(b)z AxqE dA = in or EA =00so that at distance x from the mid-line of the slab, E =a=a fFGHxxx.0IJKe EeF==xmemem e 0FIG. P24.71The acceleration of the electron is of the forma = 2 x with =Thus, the motion is simple harmonic with frequencyf=1=22e.m e 0e.m e 0Chapter 24P24.72Consider the gaussian surface described in the solution to problem 71.(a)d,21E dA =dq0zCA0EA =E=(b)3Cd24 0For E=(a)dq = dV = Adx = CAx 2 dxFor x >zP24.7349zd 2x 2 dx =0FG IJ F d IH K GH 8 JK31 CA3 0E=ordd<x<22zCd 3di for x > ;24 02E dA =Cx 3i for x > 0 ;3 0E=zE=Cd 3di for x < 24 02z1CA x 2CAx 3dq =x dx =3 000 0Cx 3i for x < 03 0A point mass m creates a gravitational accelerationg=zGmr2r at a distance r.Gmej4 r 2 = 4 Gm .r2Since the r has divided out, we can visualize the field as unbroken field lines. The same fluxwould go through any other closed surface around the mass. If there are several or nomasses inside a closed surface, each creates field to make its own contribution to the net fluxaccording toThe flux of this field through a sphere isz(b)g dA = g dA = 4 Gm in .Take a spherical gaussian surface of radius r. The field is inward sog dA = g 4 r 2 cos 180 = g 4 r 2zandThen,Or, since44 Gm in = 4 G r 3 .344 32 g 4 r = 4 G r and g = rG .33M EGrMEM EGr, g=or g =inward .= 4333RERE3 REANSWERS TO EVEN PROBLEMSP24.2355 kN m 2 CP24.10(a) 55.6 nC ; (b) The negative charge has aspherically symmetric distribution.P24.4(a) 2.34 kN m 2 C ; (b) +2.34 kN m 2 C ;(c) 0P24.12(a)P24.14(a) 1.36 MN m 2 C ; (b) 678 kN m 2 C ;(c) No; see the solution.P24.6q0P24.8ERhqq; (b); (c) Plane and square2 02 0both subtend a solid angle of a hemisphereat the charge.50Gausss LawP24.16P24.181.77 pC m3 positiveP24.46(a) 248 nC m 2 ; (b) 496 nC m 2Q6qP24.48(a) 2.56 MN C radially inward; (b) 0P24.50(a)P24.52see the solutionP24.54chw 22P24.56see the solutionP24.58(a) 4.00 nC; (b) +9.56 nC ; (c) +4.00 nCand +5.56 nCP24.60(a, b) see the solution; (c)6 0P24.2028. 2 N m 2 CP24.22q24 0P24.24(a) 0; (b) 365 kN C ; (c) 1.46 MN C;(d) 649 kN CP24.26(a) 913 nC ; (b) 0P24.284.86 GN C away from the wall. It isconstant close to the wallP24.3076.4 kN C radially inwardP24.32q4 a23.50 kN; (b)Q+q4 b 212ke e2me R3;(d) 102 pmP24.34P24.36P24.38(a)3QrQ; (c) see the solution; (b)00 a 3713 nC ; (b) 5.70 C(a) 16.2 MN C toward the filament;(b) 8.09 MN C toward the filament;(c) 1.62 MN C toward the filament2P24.401.15 nC mP24.42(a) 0; (b) 12.4 kN C radially outward;(c) 639 N C radially outward; (d) Nothingwould change.P24.44(a) 0; (b) 79.9 MN C radially outward;(c) 0; (d) 7.34 MN C radially outwardto the right; (c) 00P24.62(a) 0; (b)P24.64see the solutionP24.660.269 N m 2 C ; 2.38 pCP24.68see the solutionP24.70(a)P24.72(a) E =FGHIJKFGH 0r R22r2R; (b) 0aa2 03b2 0 r3bIJKCd 3di for x > ;24 023CddE=i for x < ;24 023CxCx 3i for x > 0 ; E = i for x < 0(b) E =3 03 025Electric PotentialCHAPTER OUTLINE25.125.225.325.425.525.625.725.8Potential Difference andElectric PotentialPotential Difference in aUniform Electric FieldElectric Potential andPotential Energy Due toPoint ChargesObtaining the Value of theElectric Field from theElectric PotentialElectric Potential Due toContinuous ChargeDistributionsElectric Potential Due to aCharged ConductorThe Milliken Oil DropExperimentApplication of ElectrostatisticsANSWERS TO QUESTIONSQ25.1When one object B with electric charge is immersed in theelectric field of another charge or charges A, the systempossesses electric potential energy. The energy can be measuredby seeing how much work the field does on the charge B as itmoves to a reference location. We choose not to visualize Aseffect on B as an action-at-a-distance, but as the result of a twostep process: Charge A creates electric potential throughout thesurrounding space. Then the potential acts on B to inject thesystem with energy.Q25.2The potential energy increases. When an outside agent makes itmove in the direction of the field, the charge moves to a regionof lower electric potential. Then the product of its negativecharge with a lower number of volts gives a higher number ofjoules. Keep in mind that a negative charge feels an electric forcein the opposite direction to the field, while the potential is thework done on the charge to move it in a field per unit charge.Q25.3To move like charges together from an infinite separation, at which the potential energy of thesystem of two charges is zero, requires work to be done on the system by an outside agent. Henceenergy is stored, and potential energy is positive. As charges with opposite signs move togetherfrom an infinite separation, energy is released, and the potential energy of the set of chargesbecomes negative.Q25.4The charge can be moved along any path parallel to the y-z plane, namely perpendicular to the field.Q25.5The electric field always points in the direction of the greatest change in electric potential. This isVVV, Ey = and Ez = .implied by the relationships Ex = xyzQ25.6(a)The equipotential surfaces are nesting coaxial cylinders around an infinite line of charge.(b)The equipotential surfaces are nesting concentric spheres around a uniformly chargedsphere.Q25.7If there were a potential difference between two points on the conductor, the free electrons in theconductor would move until the potential difference disappears.5152Electric PotentialQ25.8No. The uniformly charged sphere, whether hollow or solid metal, is an equipotential volume. Sincethere is no electric field, this means that there is no change in electrical potential. The potential atevery point inside is the same as the value of the potential at the surface.Q25.9Infinitely far away from a line of charge, the line will not look like a point. In fact, without anydistinguishing features, it is not possible to tell the distance from an infinitely long line of charge.Another way of stating the answer: The potential would diverge to infinity at any finite distance, if itwere zero infinitely far away.Q25.10The smaller sphere will. In the solution to the example referred to, equation 1 states that each willqhave the same ratio of charge to radius, . In this case, the charge density is a surface chargerq, so the smaller-radius sphere will have the greater charge density.density,4 r 2Q25.11The main factor is the radius of the dome. One often overlooked aspect is also the humidity of theairdrier air has a larger dielectric breakdown strength, resulting in a higher attainable electricpotential. If other grounded objects are nearby, the maximum potential might be reduced.Q25.12The intenseoften oscillatingelectric fields around high voltage lines is large enough to ionize theair surrounding the cables. When the molecules recapture their electrons, they release that energy inthe form of light.Q25.13A sharp point in a charged conductor would imply a large electric field in that region. An electricdischarge could most easily take place at that sharp point.Q25.14Use a conductive box to shield the equipment. Any stray electric field will cause charges on the outersurface of the conductor to rearrange and cancel the stray field inside the volume it encloses.Q25.15No charge stays on the inner sphere in equilibrium. If there were any, it would create an electricfield in the wire to push more charge to the outer sphere. All of the charge is on the outer sphere.Therefore, zero charge is on the inner sphere and 10.0 C is on the outer sphere.Q25.16The grounding wire can be touched equally well to any point on the sphere. Electrons will drainaway into the ground and the sphere will be left positively charged. The ground, wire, and sphereare all conducting. They together form an equipotential volume at zero volts during the contact.However close the grounding wire is to the negative charge, electrons have no difficulty in movingwithin the metal through the grounding wire to ground. The ground can act as an infinite source orsink of electrons. In this case, it is an electron sink.SOLUTIONS TO PROBLEMSSection 25.1P25.1Potential Difference and Electric PotentialV = 14.0 VV =W,QejejjbgandQ = N A e = 6.02 10 23 1.60 10 19 = 9.63 10 4 CsoW = QV = 9.63 10 4 C 14.0 J C = 1.35 MJeChapter 25P25.253a f7.37 10 17 = q 115K = q Vq = 6. 41 10 19 CP25.3(a)Energy of the proton-field system is conserved as the proton moves from high to lowpotential, which can be defined for this problem as moving from 120 V down to 0 V.0 + qV + 0 =12mv p + 02e1.60 10K i + Ui + Emech = K f + U fC 120 V19ja1fFGH 1 V J C IJK = 1 e1.67 10227j2kg v pv p = 1.52 10 5 m s(b)The electron will gain speed in moving the other way,from Vi = 0 to V f = 120 V :K i + Ui + Emech = K f + U f0+0+0=0=12mv e + qV2jb129.11 10 31 kg v e + 1.60 10 19 C 120 J C2ej egv e = 6.49 10 6 m sP25.4W = K = qV019.11 10 31 kg 4.20 10 5 m s2ejej2ej= 1.60 10 19 C VFrom which, V = 0.502 V .Section 25.2P25.5(a)Potential Difference in a Uniform Electric FieldWe follow the path from (0, 0) to (20.0 cm, 0) to (20.0 cm, 50.0 cm).U = (work done)U = (work from origin to (20.0 cm, 0)) (work from (20.0 cm, 0) to (20.0 cm, 50.0 cm))Note that the last term is equal to 0 because the force is perpendicular to the displacement.b gejbgafU = qEx x = 12.0 10 6 C 250 V m 0.200 m = 6.00 10 4 J(b)P25.6E=V =U6.00 10 4 J== 50.0 J C = 50.0 Vq12.0 10 6 CV 25.0 10 3 J C== 1.67 10 6 N C = 1.67 MN Cd1.50 10 2 m54Electric PotentialP25.7U = jLNMe1.40 10 m sj e3.70 10= e1.60 10 jV11m v 2 vi2 = 9.11 10 31 kgf22eje+6.23 10 18U = qV :526msj OQP = 6.23 10218J19V = 38.9 V. The origin is at highest potential.jbeg(a)V = Ed = 5.90 10 3 V m 0.010 0 m = 59.0 V(b)P25.81mv 2 = qV :f2ja f19.11 10 31 v 2 = 1.60 10 19 59.0f2ejev f = 4.55 10 6 m szzzBCBAP25.9ACVB VA = E ds = E ds E dsaf z dy aE cos 90.0f= a325fa0.800 f = +260 VzVB VA0 .5000. 4000.300VB VA = E cos 1800 . 200dxFIG. P25.9*P25.10Assume the opposite. Then at some point A on some equipotential surface the electric field has anonzero component Ep in the plane of the surface. Let a test charge start from point A and movezBsome distance on the surface in the direction of the field component. Then V = E ds is nonzero.AThe electric potential charges across the surface and it is not an equipotential surface. Thecontradiction shows that our assumption is false, that Ep = 0 , and that the field is perpendicular tothe equipotential surface.P25.11(a)Arbitrarily choose V = 0 at 0. Then at other pointsV = ExandU e = QV = QEx .Between the endpoints of the motion,bK + Us+ Ueg = bK + Uis+ Uegf1 22QE0 + 0 + 0 = 0 + kx max QEx max so x max =.2k(b)At equilibrium, Fx = Fs + Fe = 0 orkx = QE .So the equilibrium position is at x =continued on next pageQE.kFIG. P25.11Chapter 25(c)d2x Fx = kx + QE = m dt 2The blocks equation of motion is.QEQE,, or x = x +kkso the equation of motion becomes:d 2 x + QE kQEd 2 xk=k x ++ QE = mx ., or2kmdtdt 2x = x LetFGHbIJKgFG IJH KThis is the equation for simple harmonic motion a x = 2 x withThe period of the motion is then(d)T=bK + Ugb+ U e i + Emech = K + U s + U es0 + 0 + 0 k mgx max = 0 +b2 QE k mgx max =P25.12k.m=gg2= 2m.kf1 2kx max QEx max2k1ayt 22y f yi = v yi t +For the entire motion,1ayt 222mvi mg qE = tm 2 vigE=q t0 0 = vi t + Fy = ma y :FGHdsoandIJK22v yf = v yi + 2 a y y f yiFor the upward flight:FGH0 = vi2 + 2 2 vitIJ byKay = E=andy max =imax0gFGIJFGIJ FG 1 v tIJzHKHKH 4 KI2.00 kg F 2b 20.1 m sgV =GH 4.10 s 9.80 m s JK LMN 1 b20.1 m sga4.10 sfOPQ =45.00 10 CV = FGHIJKm 2 vi g j.q t1vi t4yymaxE dy = +0maxm 2 vim 2 vig y=gq tq t0i26P25.132 vit40.2 kVArbitrarily take V = 0 at the initial point. Then at distance d downfield, where L is the rod length,V = Ed and U e = LEd .(a)aK + U f = a K + U fi0+0=v=(b)f1Lv 2 LEd22 EdThe same.=ejbgaf=2 40.0 10 6 C m 100 N C 2.00 mb0.100 kg mg0.400 m s5556Electric PotentialP25.14Arbitrarily take V = 0 at point P. Then (from Equation 25.8) the potential at the original position ofthe charge is E s = EL cos . At the final point a, V = EL . Suppose the table is frictionless:K +U i = K +U faf af0 qEL cos =v=Section 25.31mv 2 qEL2af2 qEL 1 cos =mjbefaga2 2.00 10 6 C 300 N C 1.50 m 1 cos 60.00.010 0 kgf=0.300 m sElectric Potential and Potential Energy Due to Point Chargesejejejej8.99 10 9 N m 2 C 2 1.60 10 19 Cq== 1.44 10 7 V .r1.00 10 2 m(a)The potential at 1.00 cm is V1 = k e(b)P25.158.99 10 9 N m 2 C 2 1.60 10 19 CqThe potential at 2.00 cm is V2 = k e == 0.719 10 7 V .r2.00 10 2 mThus, the difference in potential between the two points is V = V2 V1 = 7.19 10 8 V .(c)The approach is the same as above except the charge is 1.60 10 19 C . This changes thesign of each answer, with its magnitude remaining the same.That is, the potential at 1.00 cm is 1.44 10 7 V .The potential at 2.00 cm is 0.719 10 7 V , so V = V2 V1 = 7.19 10 8 V .(a)Since the charges are equal and placed symmetrically, F = 0 .(b)Since F = qE = 0 , E = 0 .(c)P25.16qV = 2 k e = 2 8.99 10 9 N m 2 C 2rejFGH 2.00 10m C IJK0.8006V = 4.50 10 4 V = 45.0 kVP25.17(a)E=Q4 0 r 2V=r=(b)Q4 0 rV3 000 V== 6.00 mE 500 V mV = 3 000 V =Q=Q4 0 6.00 ma3 000 Ve8.99 109V m Cfja6.00 mf =2.00 CFIG. P25.16Chapter 25P25.18Ex =(a)k e q1x2+ke q2ax 2.00f2=0Ex = k ebecomesaDividing by k e ,2 qx 2 = q x 2.00Therefore E = 0f2F + q + 2 q I = 0 .GH x ax 2.00f JK22x 2 + 4.00 x 4.00 = 0 .whenx=4.00 16.0 + 16.0= 4.83 m .2(Note that the positive root does not correspond to a physically valid situation.)V=(b)k e q1k q+ e 2 =0x2.00 xFG + q 2 q IJ = 0 .H x 2.00 x K2 qx = qa 2.00 xf .orV = kewhenx = 0.667 mFor x < 0x = 2.00 m .Again solving for x,For 0 x 2.00 V = 0andP25.19V = ki2 qq=.x2xqiriejeV = 8.99 10 9 7.00 10 6111jLMN 0.010 0 0.010 0 + 0.038 7 OPQV = 1.10 10 7 V = 11.0 MVFIG. P25.19P25.20(a)ejejefj5.00 10 9 C 3.00 10 9 C 8.99 10 9 V m CqQU=== 3.86 10 7 J4 0 r0.350 maThe minus sign means it takes 3.86 10 7 J to pull the two charges apart from 35 cm to amuch larger separation.(b)V=Q1Q2+4 0 r1 4 0 r2e5.00 10 Cje8.99 10=90.175 mV = 103 V9V m C57j + e3.00 10 Cje8.99 1090.175 m9V m Cj58Electric PotentialP25.21U e = q 4V1 + q 4V2 + q 4V3 = q 4eU e = 10.0 10 6 Cje2FG 1 IJ FG qH 4 K H r01+18.99 10 9 N m 2 C 2q 2 q3+r2 r3IJKF 11jGG 0.600 m + 0.150 m + 0.600 m 1+ 0.150 maf afH22IJJKU e = 8.95 JP25.22V=(a)FG IJH Kk e q1 k e q 2k q+=2 er1r2rF e8.99 10 N m C je2.00 10 Cj IGGJJa1.00 mf + a0.500 mfHK9V=226222V = 3.22 10 4 V = 32.2 kVejejU = qV = 3.00 10 6 C 3.22 10 4 J C = 9.65 10 2 J(b)P25.23FIG. P25.22U = U1 + U 2 + U 3 + U 4bg bU = 0 + U12 + U 13 + U 23 + U 14 + U 24 + U 34U =0+U=k eQ 2 k e Q 2+ssk eQs2FG 4 + 2 IJ =H 2KFG 1 + 1IJ + k Q FG 1 +H 2 K s He5.41k eQs2g12IJK+12FIG. P25.23FGHAn alternate way to get the term 4 +IJ is to recognize that there are 4 side pairs and 2 face2K2diagonal pairs.P25.24Each charge creates equal potential at the center. The total potential is:V =5P25.25(a)LM k b qg OP =NM R QPe5keq.REach charge separately creates positive potential everywhere. The total potential producedby the three charges together is then the sum of three positive terms. There is no pointlocated at a finite distance from the charges, at which this total potential is zero.(b)V=2k e qke q keq+=aaaChapter 25P25.2659Consider the two spheres as a system.e jm1 v1m2Conservation of momentum:0 = m1 v1 i + m 2 v 2 i or v 2 =By conservation of energy,0=and(a)2 2k e q1 q 2 k e q1 q 2 11 m 1 v12= m1 v1 +22 m2r1 + r2dv1 =d=b gk e q1 q 21122m1 v1 + m 2 v 2 +r1 + r222FG 1 1 IJbg H r + r dK2b0.700 kg ge8.99 10 N m C je 2 10 C je3 10 C j FGH 8 101b0.100 kg gb0.800 kg g2 m 2 k e q1 q 2m1 m1 + m 2129v1 =b gk e q1 q 226263m11.00 mIJK= 10.8 m sv2 =(b)bgm1 v1 0.100 kg 10.8 m s== 1.55 m s0.700 kgm2If the spheres are metal, electrons will move around on them with negligible energy loss toplace the centers of excess charge on the insides of the spheres. Then just before they touch,the effective distance between charges will be less than r1 + r2 and the spheres will really bemoving faster than calculated in (a) .P25.27Consider the two spheres as a system.e jConservation of momentum:0 = m 1 v1 i + m 2 v 2 iorv2 =By conservation of energy,0=and(a)2 2k e q1 q 2 k e q1 q 2 11 m 1 v12= m1 v1 +.22 m2r1 + r2db gk e q1 q 2v1 =v2 =(b)m1 v1.m2d=b gk e q1 q 21122m1 v1 + m 2 v 2 +r1 + r222IJK2m k q q F 1Gm bm + m g H r + r2 m 2 k e q1 q 2m 1 m1 + m 2bFG m IJ vHm K121=FG 1gHr +r121d1 e 1 2212121dIJKIf the spheres are metal, electrons will move around on them with negligible energy loss toplace the centers of excess charge on the insides of the spheres. Then just before they touch,the effective distance between charges will be less than r1 + r2 and the spheres will really bemoving faster than calculated in (a) .60Electric Potential*P25.28(a)In an empty universe, the 20-nC charge can be placed at its location with no energyinvestment. At a distance of 4 cm, it creates a potentialV1 =ejej8.99 10 9 N m 2 C 2 20 10 9 Ck e q1== 4.50 kV .0.04 mrTo place the 10-nC charge there we must put in energyejejU12 = q 2 V1 = 10 10 9 C 4.5 10 3 V = 4.50 10 5 J .Next, to bring up the 20-nC charge requires energybU 23 + U 13 = q3 V2 + q3 V1 = q 3 V2 + V1ge= 20 10 9 C 8.99 10 9 N m 2 C 2 10 10jFGH 100.04 m C + 200.08 m C IJK99= 4.50 10 5 J 4.50 10 5 JThe total energy of the three charges isU12 + U 23 + U13 = 4.50 10 5 J .(b)The three fixed charges create this potential at the location where the fourth is released:eV = V1 + V2 + V3 = 8.99 10 9 N m 2 C 2FjGH20 10 90.04 2 + 0.03 2+10 10 9 20 10 90.030.05I CmJKV = 3.00 10 3 VEnergy of the system of four charged objects is conserved as the fourth charge flies away:FG 1 mv + qV IJ = FG 1 mv + qV IJH2K H2K10 + e 40 10 C je3.00 10 V j = e 2.00 1022e1.20 10 Jj= 3.46 10 m sv=22if9313jkg v 2 + 0442 10 13 kg*P25.29The original electrical potential energy isU e = qV = qke q.dIn the final configuration we have mechanical equilibrium. The spring and electrostatic forces onk q2k qeach charge are k 2d + q e 2 = 0 . Then k = e 3 . In the final configuration the total potential18d3da f a f1 k q+ qV =a 2d f2 18dkeq 4 keq2=. The missing energy must have become internal33d 9 dk q 2 4k q 2energy, as the system is isolated: e = e + Eintd9denergy isEint =1 2kx25 keq2.9 de22+qChapter 25P25.30afV x =(a)b gk +Qk e Q1 k e Q 2+= e+r1r2x2 + a2af2 k eQkQ= e22ax +aV x =af2b k Q ag = b x agF 2GGH b x ag2b g+ a afk e +Qx2IJJ+1K2V xe2+1FIG. P25.30(a)bgV y =(b)b gb gk e Q 1 k e Q 2 k e +Q k e Q+=+r1r2yay+ab g k aQ FGH y a1 1 y a1+ 1 IJKF 1 1 IV b yg= Gbk Q ag H y a 1 y a + 1 JKeV y =eFIG. P25.30(b)P25.31V=ejej8.99 10 9 N m 2 C 2 8.00 10 9 CkQk eQ72.0 V m=so r = e =.VVVrFor V = 100 V , 50.0 V, and 25.0 V, r = 0.720 m, 1.44 m, and 2.88 m .The radii are inversely proportional to the potential.P25.32Using conservation of energy for the alpha particle-nucleus system,we haveK f + U f = K i + Ui .ButUi =andri .Thus,Ui = 0 .AlsoK f = 0 ( v f = 0 at turning point),soU f = Kirik e q qgoldorrmin =k e q qgoldrmin2 k e q qgold2m v==12m v2eja fa fekg je 2.00 102 8.99 10 9 N m 2 C 2 2 79 1.60 10 19 Ce6.64 10277msj2j2= 2.74 10 14 m = 27.4 fm .6162Electric PotentialP25.33Using conservation of energyk e eQ k e qQ 1=+ mv 2we have:2r1r22 k e eQ 1 1mr1 r2FGHIJKa2fe8.99 10N m 2 C 2 1.60 10 19 C 10 9 Cwhich gives:orv=Thus,P25.34v=v = 7.26 10 6 m s .k e qi q j9je9.11 10e, summed over all pairs of i , j where i j .rij222FIG. P25.346 2Each charge moves off on its diagonal line. All charges have equal speeds. K +U i = K +U fa0+fa224k e q2k q+ eL2LfF 1 I 4k q= 4G mv J +H 2 K 2L2FG 2 + 1 IJ k q = 2mvH 2K LF 1 IJ k qv = G1 +H 8 K mLe2ee2+2k e q 22 2L22A cube has 12 edges and 6 faces. Consequently, there are 12 edge pairs separated by s, 2 6 = 12 facediagonal pairs separated by 2s and 4 interior diagonal pairs separated 3s .U=Section 25.4P25.37229P25.36kgjF 1 1 I .GH 0.030 0 m 0.020 0 m JKb gL qb2 qg + b2 qgb3qg + b2 qgb3qg + qb2 qg + qb3qg + 2 qb2 qg OPU=k MabaPa +ba +b QNM bL 2 6 + 6 + 2 + 3 4 OPU=k q MN 0.400 0.200 0.400 0.200 0.447 0.447 QL 4 4 1 OP = 3.96 JU = e8.99 10 je6.00 10 j MN 0.400 0.200 0.447 QU=eP25.35je31LMNOPQkeq2k q2124+= 22.8 e12 +ss23Obtaining the Value of the Electric Field from the Electric PotentialbgV = a + bx = 10.0 V + 7.00 V m xV = 10.0 VV = 11.0 VAt x = 6.00 m ,(b)At x = 0 ,At x = 3.00 m ,(a)V = 32.0 VE=bgdV= b = 7.00 V m = 7.00 N C in the + x directiondxChapter 25k eQRdV= 0Er = dr(a)For r < RV=(b)P25.38For r RV=k eQrkQkQdV= e2 = e2Er = drrrFGHP25.39IJKV = 5 x 3 x 2 y + 2 yz 2bEvaluate E at 1, 0 , 2ga fa faf a f4yz = 4a0fa 2f = 0+ E = a 5 f + a 5 f + 0 =V= 5 + 6 xy = 5 + 6 1 0 = 5xV22= +3 x 2 2 z 2 = 3 1 2 2 = 5Ey = yEx = Ez = V=z22E = Ex + EyP25.4022z27.07 N CVsE A > EB since E =(a)2a f62 VV== 200 N C down2 cms(b)EB = (c)The figure is shown to the right, with sample field linessketched in.FIG. P25.40P25.41Ey = Ey =Section 25.5P25.42LMMNMV k eQ=lnyyLMMNk eQ1yFGGH2+yy22+ y2 ++ y22+ y2OPPQ =I OPJJ PK QPk eQy2+ y2Electric Potential Due to Continuous Charge DistributionsV = V2 R V0 =k eQa fR2 + 2R2k eQ k eQ=RRFG 1 1IJ =H 5 K0.553k eQR6364Electric PotentialLM OP = C FG 1 IJ =N x Q m H mK(a) =(b)P25.43V = kezCm2zLMNzFGHLdqdxxdxL= ke= k e= k e L d ln 1 +rrd+xd0IJ OPKQFIG. P25.43P25.44V=zk e dq= kerzxdxbb2 + L 2 xg2Lx.2Let z =Then x =V = k eL z , and dx = dz2zbga f = k LL 2 z dz2b +ze22zdz2b +z2+ k ezzdz2b +z2=FHLMF L I F L I OPGH 2 xJK + GH 2 xJK + b P + k FGH L xIJK + b2MNQLO L FL Ik L M L 2 L + b L 2 g + b PF LIV=ln MPP + k MMN GH 2 LJK + b GH 2 JK2MN L 2 + bL 2g + b QLOk L M b + eL 4j L 2 PV= ln M2MN b + eL 4j + L 2 PPQL2k LV = e ln22L22e0022e222e2+ b2OPPQ2222IKk eLln z + z 2 + b 2 + k e z 2 + b 222eP25.45z14 0V = dV =zdqrAll bits of charge are at the same distance from O.So V =P25.46dV =FG IJ eH K1Q= 8.99 10 9 N m 2 C 24 0 Rk e dqzba61.51 MV .where dq = dA = 2 rdrr 2 + x2V = 2k e.50 jFGH 70.140 10 C IJK =mrdr2r + x2= 2 k eLMNx2 + b2 x2 + a2OPQFIG. P25.46Chapter 25P25.47V = kezza fV = k e ln xV = k e lnSection 25.6P25.48zz3RRdqdxdsdx= ke+ ke+ kexrRxRall chargesemicircle3 RR3 R+ke3R R + k e ln x RRa3R+ k e + k e ln 3 = k e + 2 ln 3RfElectric Potential Due to a Charged ConductorSubstituting given values into V =ke qr7.50 10 3 V =e8.99 109jN m2 C 2 q0.300 m.Substituting q = 2.50 10 7 C ,N=P25.49(a)2.50 10 7 C= 1.56 10 12 electrons .1.60 10 19 C e E= 0 ;ejej8.99 10 9 26.0 10 6ke q== 1.67 MVV=0.140R(b)e8.99 10 je26.0 10 j = 5.84 MN Cra0.200fk q e8.99 10 je 26.0 10 j== 1.17 MVV=E=keq269=2away69eR(c)E=V=0.200keqe8.99 10 je26.0 10 j ==a0.140fR269ke q= 1.67 MVR211.9 MN C away6566Electric Potential*P25.50(a)Both spheres must be at the same potential according towhere alsoq1 + q 2 = 1.20 10 6 C .Thenq1 =k e q1 k e q 2=r1r2q 2 r1r2q 2 r1+ q 2 = 1.20 10 6 Cr21.20 10 6 C= 0.300 10 6 C on the smaller sphere1 + 6 cm 2 cmq2 =q1 = 1.20 10 6 C 0.300 10 6 C = 0.900 10 6 CV=(b)ejej8.99 10 9 N m 2 C 2 0.900 10 6 Ck e q1== 1.35 10 5 Vr16 10 2 mOutside the larger sphere,E1 =k e q1r12r=V11.35 10 5 Vr=r = 2. 25 10 6 V m away .0.06 mr1Outside the smaller sphere,1.35 10 5 Vr = 6.74 10 6 V m away .0.02 mE2 =The smaller sphere carries less charge but creates a much stronger electric field than thelarger sphere.Section 25.7The Milliken Oil Drop ExperimentSection 25.8Application of ElectrostatisticsP25.51(a)Emax = 3.00 10 6 V m =6k eQar2=fFG IJHKFG IJHKk eQ 11= Vmaxr rrVmax = Emax r = 3.00 10 0.150 = 450 kV(b)P25.52V=k eQmaxr2= EmaxRor k QS rTemax= VmaxUVWke qk qVand E = e2 . Since E = ,rrr(b)r=6.00 10 5 VV== 0.200 m andE 3.00 10 6 V m(a)q=Vr= 13.3 CkeQ max =a6Emax r 2 3.00 10 0.150=ke8.99 10 9f2= 7.51 CChapter 2567Additional ProblemsP25.53q qU = qV = k e 1 2 = 8.99 10 9r12P25.54(a)ea38fa54f 1.60 10j a5.50 +e6.20f 10 j19 215= 4.04 10 11 J = 253 MeVTo make a spark 5 mm long in dry air between flat metal plates requires potential differenceejejV = Ed = 3 10 6 V m 5 10 3 m = 1.5 10 4 V ~ 10 4 V .The area of your skin is perhaps 1.5 m 2 , so model your body as a sphere with this surfacearea. Its radius is given by 1.5 m 2 = 4 r 2 , r = 0.35 m . We require that you are at thepotential found in part (a):(b)ke qrV=q=af FGH1.5 10 4 V 0.35 mVrJ=k e 8.99 10 9 N m 2 C 2 V CIJ FG N m IJKH J Kq = 5.8 10 7 C ~ 10 6 C .ejej2ejej2199k e q1 q 2 8.99 10 1.60 10U==r2 2 0.052 9 10 9(c)P25.56(a)(b)P25.55199k e q1 q 2 8.99 10 1.60 10=U=r0.052 9 10 9U=ej= 4.35 10 18 J = 27.2 eV= 6.80 eVk e q1 q 2 k e e 2== 0rFrom Example 25.5, the potential created by the ring at the electrons starting point isVi =k eQx i2+a2=bgxi22k e 2a+awhile at the center, it is V f = 2 k e . From conservation of energy,b gv2f1m e v 2 + eV ff2iIek F2e=dV V i = 4m GG1 x a+ a JJmHK4 e1.60 10 je8.99 10 je1.00 10 j F0.200GG1 =9.11 10H a0.100f + a0.200f0 + eVi =efdei2ie19v2fv f = 1.45 10 7 m s9312722IJJK68Electric Potential*P25.57b g = 2VV0 V00.ddAssume the ball swings a small distance x to the right. It moves to a place where the voltage created2Vby the plates is lower by Ex = 0 x . Its ground connection maintains it at V = 0 by allowingd2V x k q2V xR. Then the ballcharge q to flow from ground onto the ball, where 0 + e = 0q= 0dRkedThe plates create uniform electric field to the right in the picture, with magnitudefeels electric force F = qE =4V02 xRked 2to the right. For equilibrium this must be balanced by thehorizontal component of string tension according to T cos = mgtan =4V02 xR=2k e d mgFGHk d 2 mgxfor small x. Then V0 = eL4RLIJKT sin =124V02 xRked 2.If V0 is less than this value, the only equilibrium position of the ball is hanging straight down. If V0exceeds this value the ball will swing over to one plate or the other.P25.58(a)Take the origin at the point where we will find the potential. One ring, of width dx, hasQdxchargeand, according to Example 25.5, creates potentialhk eQdxdV =.h x2 + R2The whole stack of rings creates potentialzV=dV =all charge(b)zd+hdk eQdxh x2 + R2=FHk eQln x + x 2 + R 2hA disk of thickness dx has chargeIKd+h=dFGGHa f2d + h + d + h + R2k eQlnhd + d2 + R2QdxQdxand charge-per-area. According toh R2hExample 25.6, it creates potentialQdx R2hIntegrating,dV = 2 k eV=2 k eQ2zW = Vdq0where V =ke q.RTherefore, W =22k eQ 2.2R2e22e2QP25.59IKx2 + R2 x .z R h FH x + R dx xdxIK = 2Rk Q LMN 12 x x + RhLkQMad + hf ad + hf + R d d + R 2dh hR hMNMd+hdV=FH2222+FHIK OPQF d + h + ad + hf + R I OPlnGGH d + d + R JJK PQPR2x2ln x + x 2 + R 2 2222+R22d+hd22IJJK.Chapter 25P25.60The positive plate by itself creates a field E =6936.0 10 9 C m 2= 2.03 kN C away=2 0 2 8.85 10 12 C 2 N m 2ejfrom the + plate. The negative plate by itself creates the same size field and between the plates it isin the same direction. Together the plates create a uniform field 4.07 kN C in the space between.(a)Take V = 0 at the negative plate. The potential at the positive plate is thenV 0=zb12.0 cmg4.07 kN C dx .0ejafThe potential difference between the plates is V = 4.07 10 3 N C 0.120 m = 488 V .(b)FG 1 mv + qV IJ = FG 1 mv + qV IJH2K H2K1qV = e1.60 10C ja 488 V f = mv222if19(c)v 2 = vi2 + 2 a x f xif= 7.81 10 17 Jv f = 306 km s(d)2fde3.06 105msj2iaf= 0 + 2 a 0.120 ma = 3.90 10 11 m s 2(e)(f)P25.61 F = ma = e1.67 10 27jejE=(a)VB VA = E ds and the field at distance r from a uniformlykg 3.90 10 11 m s 2 = 6.51 10 16 NF 6.51 10 16 N== 4.07 kN Cq 1.60 10 19 CzBAcharged rod (where r > radius of charged rod) isE=2 0 r2ke.r=In this case, the field between the central wire and the coaxialcylinder is directed perpendicular to the line of charge so thatVB VA = zrbraorFG IJH Kr2kedr = 2 k e ln a ,rrbFG r IJHr KV = 2 k e lncontinued on next pageab.FIG. P25.6170Electric Potential(b)From part (a), when the outer cylinder is considered to be at zero potential, the potential at adistance r from the axis isV = 2 k e lnFG r IJ .HrKaThe field at r is given byE=FGHrV= 2 k e rarbV F 1 IE=GJlnbr r g H r KaP25.62(a)a2eV.ln ra rbBut, from part (a), 2k e =Therefore,IJ FG r IJ = 2k .KH r K rg.bFrom Problem 61,E=V 1.ln ra rb rbgWe require just outside the central wire5.50 10 6 V m =50.0 10 3 Vln 0.850 m rbbFG 1 IJgHr Kbe110 m jr lnFGH 0.850 m IJK = 1 .r1orbbWe solve by homing in on the required valuea fjr lnFGH 0.850 m IJKrrb me110 m1b0.01000.001000.001500.001450.001430.001424.890.7401.051.0171.0050.999bThus, to three significant figures,rb = 1.42 mm .(b)At ra ,E=P25.63zr2bV2 V1 = E dr = r1V2 V1 =FG 1 IJ =g H 0.850 m K50.0 kVln 0.850 m 0.001 42 mzr2r12 0 rFG IJH Krln 2r12 0dr9.20 kV m .71Chapter 25*P25.64Take the illustration presented with the problem as an initial picture.No external horizontal forces act on the set of four balls, so its center ofmass stays fixed at the location of the center of the square. As thecharged balls 1 and 2 swing out and away from each other, balls 3 and 4move up with equal y-components of velocity. The maximum-kineticenergy point is illustrated. System energy is conserved:v1+P25.65CM3v2+4vFIG. P25.64keq2 keq 2 1111=+ mv 2 + mv 2 + mv 2 + mv 23a2222a2k e q 2= 2mv 23avkeq23 amv=g k rbqg + k br2 qgbeV x , y, z =For the given charge distribution,e12ax + Rf + y + z and r = x + y + z .V b x , y , zg = 0F1 2Ik qG J = 0 , or 2r = r .Hr r K4a x + R f + 4y + 4z = x + y + zF8 IF4 Ix + y + z + G RJ x + a0fy + a0fz + G R J = 0 .H3 KH3 K2r1 =whereThe surface on whichis given bye12which may be written in the form:222222122This gives:222222222b[1]gThe general equation for a sphere of radius a centered at x 0 , y 0 , z0 is:bx x g + by y g + bz z g a = 0x + y + z + b 2 x gx + b 2 y gy + b 2 z gz + e xor222020200220020j22+ y 0 + z0 a 2 = 0 .[2]Comparing equations [1] and [2], it is seen that the equipotential surface for which V = 0 is indeed asphere and that:2 x 0 =Thus, x 0 = 48222R ; 2 y 0 = 0 ; 2 z 0 = 0 ; x 0 + y 0 + z 0 a 2 = R 2 .33FGHIJK416 4 2 4 2R , y 0 = z 0 = 0 , and a 2 = R = R .39 39The equipotential surface is therefore a sphere centered atFG 4 R, 0 , 0IJH 3K, having a radius2R .372Electric PotentialP25.66(a)E A = 0 (no charge within)From Gausss law,EB = k e1.00 10ej e r j = FGH 89.9 IJK V mrrbq + q g = e8.99 10 j e5.00 10 j = FG 45.0 IJ V mH r Krr8qA= 8.99 10 92229EC = k eVC = k e(b)AbqAB+ qBr At r2 , V = 92g = e8.99 10 j e5.00 10 j = FG 45.0 IJ VH r Kr9945.0= 150 V0.300Inside r2 , VB = 150 V + At r1 , V = 450 +P25.672FGHzrIJ FG 450 + 89.9 IJ VK Hr K89.911=dr = 150 + 89.9 2r 0.300r2 r89.9= +150 V so VA = +150 V .0.150From Example 25.5, the potential at the center of the ring iskQVi = e and the potential at an infinite distance from the ring isRV f = 0 . Thus, the initial and final potential energies of the pointcharge-ring system are:U i = QVi =k eQ 2RFIG. P25.67U f = QV f = 0 .andFrom conservation of energy,K f + U f = K i + UiorgivingP25.68k Q21Mv 2 + 0 = 0 + efR2vf =V = keza+Ladxx2 + b22 k eQ 2MR.= k ln L x + e xMNe2+b2j OPQ=aL a + L + aa + Lfk ln MMMN a + a + b2a +Le22+ b2OPPPQChapter 25*P25.69(a)V=keq ke q keq=r2 r1r1r2r1 r2bgFrom the figure, for r >> a ,r2 r1 2 a cos .Then(b)73VEr = ke qk p cos 2 a cos e 2.r1 r2r2 k e p cos V=rr3FG IJH K1 In spherical coordinates, the component of the gradient is.r E = Therefore,FIG. P25.69FG IJH Kk p sin 1 V.= e 3r ra f 2rk pE a90f = 0 ,E a0f = 0k pE a90f =.rFor r >> aE r 0 =ande3rande3These results are reasonable for r >> a . Their directions are as shown in Figure 25.13 (c).afHowever, for r 0 , E 0 . This is unreasonable, since r is not much greater than a if itis 0.(c)V=andexk e py2+ y2jEx = 3 23 k e pxyV=522xx + y2eje22V k e p 2 y x=Ey = 5 2yx2 + y2ejj74Electric PotentialP25.70Inside the sphere, Ex = Ey = Ez = 0 .j IKLF 3ISoE = M0 + 0 + E a z G J e x + y + z ja2xfOPQ = 3E a xzexH 2KNVI= F V E z + E a ze x + y + z jE =Kyy HF 3IE = E a zG J e x + y + z j2 y = 3E a yze x + y + z jH 2KVF 3IE == E E a zG J e x + y + z ja 2 zf E a e x + y + z jH 2KzE = E + E a e 2 z x y je x + y + z jEx = Outside,FHV=V0 E0 z + E0 a 3 z x 2 + y 2 + z 2xxx000y30z0zP25.71203002ye30232322220332+ y2 + z2j5 22 5 22 5 2203222 3 22 5 22k e dqFor an element of area which is a ring of radius r and width dr, dV =b22 3 20222 5 2222 5 23 2.r 2 + x2gdq = dA = Cr 2 rdr andbV = C 2 k egzR0P25.72Lb gMMNr 2 dr= C k e R R 2 + x 2 + x 2 lnr 2 + x2dU = Vdq where the potential V =FGH R +xR2 + x2I OPJK PQ.ke q.rejThe element of charge in a shell is dq = (volume element) or dq = 4 r 2 dr and the charge q in asphere of radius r iszrq = 4 r 2 dr = 0F 4 r I .GH 3 JK3Substituting this into the expression for dU, we haveFG k q IJ dq = k FG 4 r IJ FG 1 IJ e4 r dr j = k FG 16 IJ r drHrKH 3 KH 3 KH r KF 16 I r dr = k F 16 I RU = z dU = k GGH 15 JKH 3 JK zdU =3ee2e2R224e2e22 4503 k eQ 24But the total charge, Q = R 3 . Therefore, U =.5 R3Chapter 25*P25.73(a)The whole charge on the cube is3q = 100 10 6 C m3 0.1 m = 10 7 C . Divide up the cube intojaef64 or more elements. The little cube labeled a creates at Pke q. The others in thepotential264 6.25 + 1.25 2 + 1.25 2 10 2 mhorizontal row behind it contributeekeq64 10 2FGmj H18.75 2 + 3.125+111.25 2 + 3.1251+13.75 2I.J+ 3.125 KdcbaPThe little cubes in the rows containing b and c adde2k e q64 102Le6.25 + 1.25 + 3.75 j + e8.75 + 15.625jNmj MOP+ e11.25 + 15.625j+ e13.75 + 15.625 jQ22 1 221 222ekeq64 10 2LMe6.25mj N2+ 28.125The whole potential at P isj1 21 21.25 cm1 22and the bits in row d make potential at Pe+ + 13.75 2 + 28.125j OQP .1 28.987 6 10 9 Nm 2 10 7 CeFIG. P25.73jC 2 64 10 2 mb1.580 190g4 =8 876 V . If we usemore subdivisions of the large cube, we get the same answer to four digits.(b)75A sphere centered at the same point would create potentialk e q 8.987 6 10 9 Nm 2 10 7 C== 8 988 V , larger by 112 V .rC210 1 mANSWERS TO EVEN PROBLEMSP25.26.41 10 19 CP25.40.502 VP25.6(a) 59.0 V ; (b) 4.55 Mm sP25.10see the solutionP25.1240.2 kVP25.140.300 m sP25.16(a) 0; (b) 0; (c) 45.0 kVP25.18(a) 4.83 m ; (b) 0.667 m and 2.00 mP25.20(a) 386 nJ ; (b) 103 V(a) 32.2 kV ; (b) 96.5 mJP25.24P25.26(a) 10.8 m s and 1.55 m s ; (b) greaterP25.28(a) 45.0 J ; (b) 34.6 km sP25.30see the solutionP25.3227.4 fmP25.343.96 JP25.3622.81.67 MN CP25.8P25.225k e qRkeq 2s76Electric Potentialk eQP25.60P25.38(a) 0; (b)P25.40(a) larger at A; (b) 200 N C down;(c) see the solutionP25.420.553P25.44Lk L Mln M2MNrradially outward2(d) 390 Gm s 2 toward the negative plate;(e) 6.51 10 16 N toward the negative plate;(f) 4.07 kN C toward the negative platek eQReLMNP25.62b2b2(a) 1.42 mm ; (b) 9. 20 kV mP25.64+ e L 4j L 2 OPP+ e L 4j + L 2 PQ2Fk q IGH 3am JK2x2 + b 2 x2 + a2P25.66OPQP25.462 k eP25.481.56 10 12 electronsP25.50(a) 135 kV ; (b) 2.25 MV m away from theP25.54(a) ~ 10 V ; (b) ~ 10P25.566P25.702e222B22222OPI PPJJ PK PQ2ej ;E = 3E a yze x + y + z j;E a e2z x y jE =E +outside andex + y + z jz0320E = 0 insideP25.722Ex = 3E0 a 3 xz x 2 + y 2 + z 2022222eyd + h + d + h + R2k eQlnhd + d 2 + R22FG 89.9 IJ V m radiallyHr KF 45.0 IJ V m radiallyoutward; E = G H r Koutward;89.9 IF(b) V = 150 V ; V = G 450 +J V;Hr KF 45.0 IJ VV = GH r KL a + L + aa + Lf + b OPk ln MMNM a + a + b PQP(a) E A = 0 ; E B =CCFa f IJ ;GGJKHLMad + hf ad + hf + R d d + RkQ MF d + h + a d + hf + R(b)R h M2dh h + R lnGMMGH d + d + RN(a)12AP25.6814.5 Mm sP25.5822(a) 13.3 C ; (b) 0.200 m4eClarge sphere and 6.74 MV m away fromthe small sphereP25.52(a) 488 V ; (b) 7.81 10 17 J ; (c) 306 km s ;3 k eQ 25 R5 22 5 22322222 5 2226Capacitance and DielectricsCHAPTER OUTLINE26.126.226.3ANSWERS TO QUESTIONSDefinition of CapacitanceCalculating CapacitanceCombinations ofCapacitorsEnergy Stored in a ChargedCapacitorCapacitors with DielectricsElectric Dipole in an ElectricFieldAn Atomic Description ofDielectrics26.526.626.7Q26.4Nothing happens to the charge if the wires are disconnected. Ifthe wires are connected to each other, charges in the singleconductor which now exists move between the wires and theplates until the entire conductor is at a single potential and thecapacitor is discharged.Q26.2336 km. The plate area would need to beQ26.326.4Q26.1The parallel-connected capacitors store more energy, since theyhave higher equivalent capacitance.1m2 .0Seventeen combinations:IndividualC1 , C 2 , C 3ParallelC 1 + C 2 + C 3 , C 1 + C 2 , C1 + C 3 , C 2 + C 3FG 1 + 1 IJ + C , FG 1 + 1 IJ + C , FG 1 + 1 IJ + CHC C KHC C KHC C KFG 1 + 1 IJ , FG 1 + 1 IJ , FG 1 + 1 IJHC +C C K HC +C C K HC +C C KFG 1 + 1 + 1 IJ , FG 1 + 1 IJ , FG 1 + 1 IJ , FG 1 + 1 IJHC C C K HC C K HC C K HC C K1Series-Parallel11321123112312313113121Series2231121123113Q26.5This arrangement would decrease the potential difference between the plates of any individualcapacitor by a factor of 2, thus decreasing the possibility of dielectric breakdown. Depending on theapplication, this could be the difference between the life or death of some other (most likely moreexpensive) electrical component connected to the capacitors.Q26.6Nonot just using rules about capacitors in series or in parallel. See Problem 72 for an example. Ifconnections can be made to a combination of capacitors at more than two points, the combinationmay be irreducible.7778Capacitance and DielectricsQ26.7A capacitor stores energy in the electric field between the plates. This is most easily seen when usinga dissectable capacitor. If the capacitor is charged, carefully pull it apart into its component pieces.One will find that very little residual charge remains on each plate. When reassembled, the capacitoris suddenly rechargedby inductiondue to the electric field set up and stored in thedielectric. This proves to be an instructive classroom demonstration, especially when you ask astudent to reconstruct the capacitor without supplying him/her with any rubber gloves or otherinsulating material. (Of course, this is after they sign a liability waiver).Q26.8The work you do to pull the plates apart becomes additional electric potential energy stored in thecapacitor. The charge is constant and the capacitance decreases but the potential difference increases1to drive up the potential energy QV . The electric field between the plates is constant in strength2but fills more volume as you pull the plates apart.Q26.9A capacitor stores energy in the electric field inside the dielectric. Once the external voltage source isremovedprovided that there is no external resistance through which the capacitor candischargethe capacitor can hold onto this energy for a very long time. To make the capacitor safeto handle, you can discharge the capacitor through a conductor, such as a screwdriver, provided thatyou only touch the insulating handle. If the capacitor is a large one, it is best to use an externalresistor to discharge the capacitor more slowly to prevent damage to the dielectric, or welding of thescrewdriver to the terminals of the capacitor.Q26.10The work done, W = QV , is the work done by an external agent, like a battery, to move a chargethrough a potential difference, V . To determine the energy in a charged capacitor, we must addthe work done to move bits of charge from one plate to the other. Initially, there is no potentialdifference between the plates of an uncharged capacitor. As more charge is transferred from oneplate to the other, the potential difference increases as shown in Figure 26.12, meaning that morework is needed to transfer each additional bit of charge. The total work is the area under the curve of1Figure 26.12, and thus W = QV .2Q26.11Energy is proportional to voltage squared. It gets four times larger.Q26.12Let C = the capacitance of an individual capacitor, and C s represent the equivalent capacitance of thegroup in series. While being charged in parallel, each capacitor receives chargeejafQ = CVcharge = 500 10 4 F 800 V = 0.400 C .While being discharged in series,(or 10 times the original voltage).Vdischarge =QQ0. 400 C=== 8.00 kVC s C 10 5.00 10 5 FQ26.13Put a material with higher dielectric strength between the plates, or evacuate the space between theplates. At very high voltages, you may want to cool off the plates or choose to make them of adifferent chemically stable material, because atoms in the plates themselves can ionize, showingthermionic emission under high electric fields.Q26.14The potential difference must decrease. Since there is no external power supply, the charge on thecapacitor, Q, will remain constantthat is assuming that the resistance of the meter is sufficientlylarge. Adding a dielectric increases the capacitance, which must therefore decrease the potentialdifference between the plates.Q26.15Each polar molecule acts like an electric compass needle, aligning itself with the external electricfield set up by the charged plates. The contribution of these electric dipoles pointing in the samedirection reduces the net electric field. As each dipole falls into a configuration of lower potentialenergy it can contribute to increasing the internal energy of the material.Chapter 2679Q26.16The material of the dielectric may be able to support a larger electric field than air, without breakingdown to pass a spark between the capacitor plates.Q26.17The dielectric strength is a measure of the potential difference per unit length that a dielectric canwithstand without having individual molecules ionized, leaving in its wake a conducting path fromplate to plate. For example, dry air has a dielectric strength of about 3 MV/m. The dielectric constantin effect describes the contribution of the electric dipoles of the polar molecules in the dielectric tothe electric field once aligned.Q26.18In water, the oxygen atom and one hydrogen atom considered alone have an electric dipole momentthat points from the hydrogen to the oxygen. The other O-H pair has its own dipole moment thatpoints again toward the oxygen. Due to the geometry of the molecule, these dipole moments add tohave a non-zero component along the axis of symmetry and pointing toward the oxygen.A non-polarized molecule could either have no intrinsic dipole moments, or have dipolemoments that add to zero. An example of the latter case is CO 2 . The molecule is structured so thateach CO pair has a dipole moment, but since both dipole moments have the same magnitude andopposite directiondue to the linear geometry of the moleculethe entire molecule has no dipolemoment.Q26.19Heating a dielectric will decrease its dielectric constant, decreasing the capacitance of a capacitor.When you heat a material, the average kinetic energy per molecule increases. If you refer back to theanswer to Question 26.15, each polar molecule will no longer be nicely aligned with the appliedelectric field, but will begin to ditherrock back and fortheffectively decreasing its contributionto the overall field.Q26.20The primary choice would be the dielectric. You would want to chose a dielectric that has a largedielectric constant and dielectric strength, such as strontium titanate, where 233 (Table 26.1). Aconvenient choice could be thick plastic or mylar. Secondly, geometry would be a factor. Tomaximize capacitance, one would want the individual plates as close as possible, since thecapacitance is proportional to the inverse of the plate separationhence the need for a dielectricwith a high dielectric strength. Also, one would want to build, instead of a single parallel platecapacitor, several capacitors in parallel. This could be achieved through stacking the plates of thecapacitor. For example, you can alternately lay down sheets of a conducting material, such asaluminum foil, sandwiched between your sheets of insulating dielectric. Making sure that none ofthe conducting sheets are in contact with their next neighbors, connect every other plate together.Figure Q26.20 illustrates this idea.DielectricConductorFIG. Q26.20This technique is often used when home-brewing signal capacitors for radio applications, asthey can withstand huge potential differences without flashover (without either discharge betweenplates around the dielectric or dielectric breakdown). One variation on this technique is to sandwichtogether flexible materials such as aluminum roof flashing and thick plastic, so the whole productcan be rolled up into a capacitor burrito and placed in an insulating tube, such as a PVC pipe, andthen filled with motor oil (again to prevent flashover).80Capacitance and DielectricsSOLUTIONS TO PROBLEMSSection 26.1Definition of CapacitanceejafejafQ = CV = 4.00 10 6 F 12.0 V = 4.80 10 5 C = 48.0 CQ = CV = 4.00 10 6 F 1.50 V = 6.00 10 6 C = 6.00 C(a)C=(b)P26.2(a)(b)P26.1V =Section 26.2P26.3E=Q 10.0 10 6 C== 1.00 10 6 F = 1.00 FV10.0 VQ 100 10 6 C== 100 VC 1.00 10 6 FCalculating Capacitancekeqr2e4.90 10 N Cja0.210 mfe8.99 10 N m C j4q=:2922= 0.240 Cq 0.240 10 6== 1.33 C m 2A 4 0.120 2(a)(b)P26.4=C = 4 0 r = 4 8.85 10 12 0.120 = 13.3 pF(a)C = 4 0 RCR== k eC = 8.99 10 9 N m 2 C 2 1.00 10 12 F = 8.99 mm4 0afjaefejeejejj=4 8.85 10 12 C 2 2.00 10 3 m(b)(c)P26.5C = 4 0 R =Q = CV = 2.22 10 13 F 100 V = 2.22 10 11 C(a)Q1 R1=Q 2 R2jaeFGHQ1 + Q 2 = 1 +Q 2 = 2.00 C(b)N mV1 = V2 =20. 222 pFfIJKR1Q 2 = 3.50Q 2 = 7.00 CR2Q1 = 5.00 CQ1 Q 25.00 C=== 8.99 10 4 V = 89.9 kV19C1 C 28.99 10 m F0.500 mej afChapter 26P26.6a fejej1232 0 A 1.00 8.85 10 C 1.00 10 mC==dN m 2 800 maf2= 11.1 nFThe potential between ground and cloud isjaefV = Ed = 3.00 10 6 N C 800 m = 2.40 10 9 Va f ejejQ = C V = 11.1 10 9 C V 2.40 10 9 V = 26.6 CP26.7V = Ed(a)E=E=(b)20.0 V1.80 10 3 m= 11.1 kV m0ejej = 1.11 10 4 N C 8.85 10 12 C 2 N m 2 = 98.3 nC m 2ejbjeg(c)8.85 10 12 C 2 N m 2 7.60 cm 2 1.00 m 100 cm0 A=C=d1.80 10 3 m(d)V =QCafejQ = 20.0 V 3.74 10 12 F = 74.7 pCP26.8C= 0 A= 60.0 10 15 Fda feje1221.0 10 12 0 A 1 8.85 10=C60.0 10 15d = 3.10 10 9 m = 3.10 nmd=P26.9Q=d=a f0 AVda f=0 Vja f VQ= = 0Ade8.85 10e30.0 10912jaC 2 N m 2 150 VjefC cm 2 1.00 10 4 cm 2 m 2j= 4.42 m2= 3.74 pF8182Capacitance and DielectricsP26.10With = , the plates are out of mesh and the overlap area is zero. With R2. By proportion, the2 R2effective area of a single sheet of charge is2When there are two plates in each comb, the number of adjoiningsheets of positive and negative charge is 3, as shown in the sketch. Whenthere are N plates on each comb, the number of parallel capacitors is 2 N 1and the total capacitance is = 0 , the overlap area is that of a semi-circle,aAf distance = a2 N 1f d a2 f RaC = 2N 1P26.11(a)(b)C=00effectivef250.0c h 2e8.99 10 j lnc h =F bIV = 2 k lnG JMethod 1:H aK2 k e lnba=97. 272.582a2 N 1f a fR0=FIG. P26.102.d2.68 nFe8.10 10 6 C= 1.62 10 7 C m50.0 m7.27= 3.02 kVV = 2 8.99 10 9 1.62 10 7 ln2.58=q=eMethod 2:P26.12V =j FGHjeIJKQ 8.10 10 6== 3.02 kVC 2.68 10 9Let the radii be b and a with b = 2 a . Put charge Q on the inner conductor and Q on the outer.Electric field exists only in the volume between them. The potential of the inner sphere is Va =that of the outer is Vb =k eQ. ThenbFGHVa Vb =k e Q k eQQba=ab4 0 abHere C =IJ and C = QKV V4 0 2 a 2= 8 0 aaaa=4 0 ab.baC.8 0Volume =The intervening volume isb=Volume =FGHbgafe7 20.0 10 63842e8.85 10(a)C=0.070 0 0.140ab== 15.6 pFke b a8.99 10 9 0.140 0.070 0(b)C=QVa f eV =jbIJ FG IJK H KC N mj= 2.13 10C N m jC34 3 4 3447C 3 b a = 7 a3 = 7 3 3 3 =33338 0 384 2 30The outer sphere is 360 km in diameter.P26.13k eQ;agQ 4.00 10 6 C== 256 kVC 15.6 10 12 F123222 316m3 .Chapter 26T cos mg = 0T sin Eq = 0Dividing,tan =soE=andP26.15 Fy = 0 : Fx = 0 :P26.14V = Ed =mgtan qmgd tan .qejejC = 4 0 R = 4 8.85 10 12 C N m 2 6.37 10 6 m = 7.08 10 4 FSection 26.3P26.16Eqmg(a)Combinations of CapacitorsCapacitors in parallel add. Thus, the equivalent capacitor has a value ofC eq = C1 + C 2 = 5.00 F + 12.0 F = 17.0 F .(b)The potential difference across each branch is the same and equal to the voltage of thebattery.V = 9.00 V(c)bfgaQ5 = CV = 5.00 F 9.00 V = 45.0 Cbgafand Q12 = CV = 12.0 F 9.00 V = 108 CP26.17(a)In series capacitors add as11111=+=+C eq C 1 C 2 5.00 F 12.0 Fand(c)C eq = 3.53 F .The charge on the equivalent capacitor isQ eq = C eq V = 3.53 F 9.00 V = 31.8 C .bgafEach of the series capacitors has this same charge on it.So(b)Q1 = Q 2 = 31.8 C .The potential difference across each isV1 =Q1 31.8 C== 6.35 VC1 5.00 FandV2 =Q 2 31.8 C== 2.65 V .C 2 12.0 F8384Capacitance and DielectricsP26.18The circuit reduces first according to the rule for capacitors inseries, as shown in the figure, then according to the rule forcapacitors in parallel, shown below.FGHC eq = C 1 +IJK1 111+= C = 1.83C2 36FIG. P26.18C p = C1 + C 2111=+C s C1 C 2Substitute C 2 = C p C1C p C1 + C1111=+=.C s C1 C p C1 C1 C p C1Simplifying,P26.192C 1 C 1C p + C p C s = 0 .C1 =2C p C p 4C p C s2e=j11 2Cp C p C pCs24We choose arbitrarily the + sign. (This choice can be arbitrary, since with the case of the minus sign,we would get the same two answers with their names interchanged.)C1 =C 2 = C p C1 =P26.20bgb11 211Cp +C p C p C s = 9.00 pF +9.00 pF2424bg b9.00 pFgb2.00 pFg =26.00 pFg11 21Cp C p C p C s = 9.00 pF 1.50 pF = 3.00 pF242C p = C1 + C 2and111=+.C s C1 C 2SubstituteC 2 = C p C1 :Simplifying,andC p C1 + C1111=+=.C s C1 C p C1 C1 C p C1ej2C 1 C 1C p + C p C s = 0C1 =2C p C p 4C pC s2=11 2Cp +C p C pCs24where the positive sign was arbitrarily chosen (choosing the negative sign gives the same values forthe capacitances, with the names reversed).Then, fromC 2 = C p C1C2 =11 2Cp C p C pC s .24Chapter 26P26.21(a)85111=+C s 15.0 3.00C s = 2.50 FC p = 2.50 + 6.00 = 8.50 FC eq =(b)FG 1 + 1 IJH 8.50 F 20.0 F Kb1ga= 5.96 FfQ = CV = 5.96 F 15.0 V = 89.5 C on 20.0 FQ 89.5 C== 4.47 VC 20.0 F15.0 4.47 = 10.53 VV =bfgaQ = CV = 6.00 F 10.53 V = 63.2 C on 6.00 F89.5 63.2 = 26.3 C on 15.0 F and 3.00 F*P26.22FIG. P26.21(a)Capacitors 2 and 3 are in parallel and present equivalent capacitance 6C. This is in series111= 2C .with capacitor 1, so the battery sees capacitance+3C 6C(b)If they were initially unchanged, C 1 stores the same charge as C 2 and C 3 together. WithLMNOPQgreater capacitance, C 3 stores more charge than C 2 . Then Q1 > Q3 > Q 2 .bg(c)The C 2| C 3 equivalent capacitor stores the same charge as C 1 . Since it has greater|Qimplies that it has smaller potential difference across it than C 1 . Incapacitance, V =Cparallel with each other, C 2 and C 3 have equal voltages: V1 > V2 = V3 .(d)If C 3 is increased, the overall equivalent capacitance increases. More charge moves throughthe battery and Q increases. As V1 increases, V2 must decrease so Q 2 decreases. ThenQ 3 must increase even more: Q3 and Q1 increase; Q 2 decreases .P26.23QsoVQ = 120 Cand6.00 10 6 =C=andorQ1 = 120 C Q 2QV = :C120 Q 2Q= 26.003.00a3.00fb120 Q g = a6.00fQ2360= 40.0 CQ2 =9.00Q20.0120 Q 2 Q 2=C1C22Q1 = 120 C 40.0 C = 80.0 CFIG. P26.2386Capacitance and DielectricsP26.24In series , to reduce the effective capacitance:(a)111=+F 34.8 F C s32.01= 398 FCs =2.51 10 3 FIn parallel , to increase the total capacitance:(b)29.8 F + C p = 32.0 FC p = 2.20 FP26.25nC =1C1001+ +C+1C=100nCn capacitorsnC =*P26.26100Cso n 2 = 100 and n = 10nFor C 1 connected by itself, C 1 V = 30.8 C where V is the battery voltage: V =For C 1 and C 2 in series:FGH 1 C30.8 C.C1IJK1V = 23.1 C1 + 1 C2substituting,30.8 C 23.1 C 23.1 C=+C1C1C2C 1 = 0.333C 2 .For C 1 and C 3 in series:FGH 1 CIJK1V = 25.2 C+ 1 C3130.8 C 25. 2 C 25.2 C=+C1C1C3For all three:Q=FGH 1 CC 1 = 0.222C 3 .IJKC 1 V130.8 CV === 19.8 C .1 + C 1 C 2 + C1 C 3 1 + 0.333 + 0.222+ 1 C 2 + 1 C31This is the charge on each one of the three.P26.27FG 1 + 1 IJ = 3.33 FH 5.00 10.0 K= 2a3.33f + 2.00 = 8.66 F= 2a10.0f = 20.0 FF 1 + 1 IJ = 6.04 F=GH 8.66 20.0 KCs =C p1C p211C eqFIG. P26.27Chapter 26P26.28a f ejafQ eq = C eq V = 6.04 10 6 F 60.0 V = 3.62 10 4 CQ eqQ p1 = Q eq , so Vp1 =eC p1=3.62 10 4 C= 41.8 V8.66 10 6 Fjaj efQ 3 = C 3 Vp1 = 2.00 10 6 F 41.8 V = 83.6 CP26.29Cs =FG 1 + 1 IJH 5.00 7.00 K1= 2.92 FC p = 2.92 + 4.00 + 6.00 = 12.9 FFIG. P26.29*P26.30According to the suggestion, the combination ofcapacitors shown is equivalent toThen1111 C + C0 + C0 + C + C0=++=C C0 C + C0 C0C0 C + C0b2+ C0C 0Cg2= 2C + 3C 0 C222C + 2C 0 C C 0 = 0C=FIG. P26.30e j222C 0 4C 0 + 4 2C 04Only the positive root is physicalC=Section 26.4C02ej3 1Energy Stored in a Charged Capacitora f2=13.00 F 12.0 V2a f2=13.00 F 6.00 V2P26.32(a)U=1C V2(b)P26.31U=1C V2U=bgaf2= 216 Jbgaf2= 54.0 J1CV 22V =2U=Ca2 300 Jf30 10 6 C V= 4. 47 10 3 V8788Capacitance and DielectricsP26.33U=a f1C V22The circuit diagram is shown at the right.C p = C1 + C 2 = 25.0 F + 5.00 F = 30.0 F(a)U=ja f130.0 10 6 1002e2= 0.150 JF 1 + 1 IJ = FG 1 + 1 IJC =GH C C K H 25.0 F 5.00 F K1U = C a V f22a0.150f2U== 268 VV =1(b)s11= 4.17 F224.17 10 6CP26.34Use U = A1 Q2and C = 0 .2 CdIf d 2 = 2d1 , C 2 =1C 1 . Therefore, the stored energy doubles .2ejej(a)Q = CV = 150 10 12 F 10 10 3 V = 1.50 10 6 C(b)*P26.35U=a f1C V22U=CV =P26.36u=2e2 250 10 6 J150 1012Fj=U 1= 0 E 2V 21.00 10 7 1= 8.85 10 12 3 000V2jbeg2eV = 2.51 10 3 m 3 = 2.51 10 3 m 3P26.371.83 10 3 VzjFGH 1 000 L IJK =m3W = U = Fdxso F =F IGH JKFGHIJKdU d Q 2dQ2xQ2===2 0 Adx dx 2Cdx 2 0 A2.51 LFIG. P26.33Chapter 26P26.38With switch closed, distance d = 0.500d and capacitance C =a fa f eja0 A 2 0 A== 2C .ddf(a)Q = C V = 2C V = 2 2.00 10 6 F 100 V = 400 C(b)The force stretching out one spring isa f4C 2 VQ2F==2 0 A2 0 Aa f = 2CaV f=d A d gdb22C 2 V2P26.39a f FG 4 IJ = 8CaV fH dK dF 2C V=xd22=2H ET =The energy transferred is.0One spring stretches by distance x =k=2d, so4jae8 2.00 10 6 F 100 Ve8.00 10 3 maj2fef2= 2.50 kN m .11QV = 50.0 C 1.00 10 8 V = 2.50 10 9 J22jand 1% of this (or Eint = 2.50 10 7 J ) is absorbed by the tree. If m is the amount of water boiledaway,bf egajthenEint = m 4 186 J kg C 100 C 30.0 C + m 2.26 10 6 J kg = 2.50 10 7 Jgivingm = 9.79 kg .a f1C V22+a f1C V22a f= C V2(a)U=(b)*P26.40The altered capacitor has capacitance C =a f a f a f C a V f2C V + C V = C V +FGH14 VU = C23(c)(d)P26.41IJK2FGH114V+C223IJK2=C. The total charge is the same as before:2V =a V f4C4 V.323The extra energy comes from work put into the system by the agent pulling the capacitorplates apart.a fU=1C V2U=1 R2 keFGH2where C = 4 0 R =IJ FG k Q IJKH R Ke2=k eQ 22RkQkQRand V = e 0 = eRRke8990Capacitance and Dielectrics*P26.42(a)bg2222q11 q1 1 q 2 11 Q q1The total energy is U = U 1 + U 2 =+=+.2 C 1 2 C 2 2 4 0 R1 2 4 0 R 2dU= 0:dq1For a minimum we setbga f1 2 q11 2 Q q1+1 = 02 4 0 R1 2 4 0 R 2R 2 q1 = R1Q R1 q1Then q 2 = Q q1 =q1 =R1QR1 + R 2R 2Q= q2 .R1 + R 2V1 =k e q1k e R1Qk eQ==R1R1 R1 + R 2R1 + R 2V2 =(b)ke q2k e R2 Qk eQ==R2R 2 R1 + R 2R1 + R 2bgbgand V1 V2 = 0 .Section 26.5Capacitors with DielectricsejejP26.44(a)(b)P26.4312F m 1.75 10 4 m 2 0 A 2.10 8.85 10C=== 8.13 10 11 F = 81.3 pFd4.00 10 5 mVmax = Emax d = 60.0 10 6 V m 4.00 10 5 m = 2.40 kVejejQ max = CVmax ,Vmax = Emax d .but 0 A.dAlso,C=Thus,Q max =(a) 0 AEmax d = 0 AEmax .dbgWith air between the plates, = 1.00andEmax = 3.00 10 6 V m .Therefore,ejejejQ max = 0 AEmax = 8.85 10 12 F m 5.00 10 4 m 2 3.00 10 6 V m = 13.3 nC .(b)With polystyrene between the plates, = 2.56 and Emax = 24.0 10 6 V m .ejejejQ max = 0 AEmax = 2.56 8.85 10 12 F m 5.00 10 4 m 2 24.0 10 6 V m = 272 nCChapter 26P26.45C=or 0 Ad95.0 109=jbe3.70 8.85 10 12 0.070 0g0.025 0 10 3= 1.04 mP26.46Consider two sheets of aluminum foil, each 40 cm by 100 cm, with one sheet of plastic between2.54 cm. Then,them. Suppose the plastic has 3 , Emax ~ 10 7 V m and thickness 1 mil =1 000ejej12C 2 N m 2 0. 4 m 2 0 A 3 8.85 10~ 10 6 F~C=5d2.54 10 mejejVmax = Emax d ~ 10 7 V m 2.54 10 5 m ~ 10 2 VP26.47C=Originally,(a)a fe8.85 1012jea f0 A VijaC 2 N m 2 25.0 10 4 m 2 250 Ve1.50 102.djmf=369 pCFinally,Cf = 0 AQ=dVa fCf =e2f=Qd 0 Aa fOriginally,Ui =1C V2Finally,Uf =1C f V2e8.85 10U = 2i12C20a f2fi 0 Ad=U = U f U i =So,jee1.50 10 mj Aa V f d a V f==80.0 8.85 10 12 C 2 N m 2 25.0 10 4 m 2fa V f(c).iThe charge is the same before and after immersion, with value Q =Q=(b)0 AQ=Vda f0 A V2i2d=2=a f 0 A V2d 22i=a f0 A V2 d2i.a f a 1f 0 A V2 d2118 pF250 V= 3.12 V .80.0.2ije25.0 10 m ja250 V f a79.0f =2e1.50 10 mja80.0fNm42ij=245.5 nJ .9192Capacitance and Dielectricsa fejej121.00 10 4 m 2 0 A 173 8.85 10== 1.53 nFd0.100 10 3 m(a)C = C 0 =(b)P26.48The battery delivers the free chargea f ejafQ = C V = 1.53 10 9 F 12.0 V = 18.4 nC .(c)The surface density of free charge isQ18.4 10 9 C= == 1.84 10 4 C m 2 .A 1.00 10 4 m 2The surface density of polarization charge isFGHp = 1(d)We have E =E=P26.49IJ FGK HIJK11= 1= 1.83 10 4 C m 2 .173E0and E0 =V; hence,dV12.0 V== 694 V m .d173 1.00 10 4 ma fejThe given combination of capacitors is equivalent to the circuit diagramshown to the right.Put charge Q on point A. Then,bgbgbgQ = 40.0 F VAB = 10.0 F VBC = 40.0 F VCD .FIG. P26.49So, VBC = 4VAB = 4VCD , and the center capacitor will break down first, at VBC = 15.0 V . Whenthis occurs,VAB = VCD =bg1VBC = 3.75 V4and VAD = VAB + VBC + VCD = 3.75 V + 15.0 V + 3.75 V = 22.5 V .Section 26.6P26.50(a)Electric Dipole in an Electric FieldThe displacement from negative to positive charge isejejej2 a = 1.20 i + 1.10 j mm 1.40 i 1.30 j mm = 2.60 i + 2.40 j 10 3 m.The electric dipole moment isejejp = 2aq = 3.50 10 9 C 2.60 i + 2.40 j 10 3 m =(b)eje9.10 i + 8.40 jj 10ej = p E = 9.10 i + 8.40 j 10 12 C m 7.80 i 4.90 j 10 3 N Cej = +44.6k 65.5k 10 9 N m = 2.09 10 8 N mkcontinued on next page12Cm .Chapter 26(c)eje93jU = p E = 9.10 i + 8.40 j 10 12 C m 7.80 i 4.90 j 10 3 N CafU = 71.0 + 41.2 10 9 J = 112 nJa9.10f + a8.40fa7.80f + a4.90f2E=2 10 12 C m = 12.4 10 12 C m2p=(d)2 10 3 N C = 9.21 10 3 N CU max = p E = 114 nJ,U min = 114 nJU max U min = 228 nJP26.51(a)Let x represent the coordinate of the negative charge.Then x + 2 a cos is the coordinate of the positivecharge. The force on the negative charge isF = qE x i . The force on the positive charge isF+(b)F-afdE= + qEa x + 2 a cos fi qEa xfi + qa2 a cos fi .dxThe force on the dipole is altogetherThus,a f2 k e qdE=.dxx3a feke qx2jea fEFIG. P26.51(a)F = F + F+ = qThe balloon creates field along the x-axis ofF+pafdEdE2 a cos i = pcos i .dxdxi.j2 8.99 10 9 2.00 10 6dE== 8.78 MN C m3dx0.160At x = 16.0 cm ,ejejF = 6.30 10 9 C m 8.78 10 6 N C m cos 0 i = 55.3 i mNSection 26.7P26.52An Atomic Description of Dielectrics2 r E =soqin0E=2 r 0zzr2r2r1r1V = E d r = maxFG IJH Krdr =ln 1r22 r 02 0= Emax rinner2 0ejej FGH 025.0 IJK. 200V = 1. 20 10 6 V m 0.100 10 3 m lnVmax = 579 VFIG. P26.5294Capacitance and DielectricsP26.53(a)Consider a gaussian surface in the form of a cylindrical pillbox with ends of area A << Aparallel to the sheet. The side wall of the cylinder passes no flux of electric field since thissurface is everywhere parallel to the field. Gausss law becomesEA + EA =(b)Qaway from the positive and2 Atoward the negative sheet. Together, they create a field ofQ.AAssume that the field is in the positive x-direction. Then, the potential of the positive platerelative to the negative plate isz+ plateV = z+ plateE ds = plate(d)directed away from the positive sheet.In the space between the sheets, each creates fieldE=(c)QQA , so E =2 AAQQdi idx = +.AA plateeCapacitance is defined by: C =jQQ A 0 A===.ddV Qd AAdditional ProblemsP26.54(a)C=(c)Q acLM 1 + 1 OP + LM 1 + 1 OP = 3.33 FN 3.00 6.00 Q N 2.00 4.00 Q= C b V g = b 2.00 Fga90.0 V f = 180 C1ac1acTherefore, Q3 = Q6 = 180 Ci bdgafQ df = C df Vdf = 1.33 F 90.0 V = 120 C(b)(d)Q3C3QV6 = 6C6QV2 = 2C2QV4 = 4C4V3 =UT =180 C=3.00 F180 C==6.00 F120 C==2.00 F120 C==4.00 F=a f1C eq V22=60.0 V30.0 V60.0 V30.0 Vja13.33 10 6 90.0 V2ef2= 13.4 mJFIG. P26.54Chapter 26*P26.55Each face of P2 carries charge, so the three-plate system is equivalent to(a)P2P3P2P1Each capacitor by itself has capacitanceC=ej12422 0 A 1 8.85 10 C 7.5 10 m== 5.58 pF .dN m 2 1.19 10 3 mThen equivalent capacitance = 5.58 + 5.58 = 11.2 pF .af(b)Q = CV + CV = 11.2 10 12 F 12 V = 134 pC(c)Now P3 has charge on two surfaces and in effect three capacitors are in parallel:bgC = 3 5.58 pF = 16.7 pF .Only one face of P4 carries charge:(d)a fQ = CV = 5.58 10 12 F 12 V = 66.9 pC .*P26.56From the example about a cylindrical capacitor,Vb Va = 2 k e lnbae= 2a8.99fe1.4 10jejVb 345 kV = 2 8.99 10 9 Nm 2 C 2 1.40 10 6 C m ln3j12 m0.024 mJ C ln 5005= 1.564 3 10 VVb = 3. 45 10 5 V 1.56 10 5 V = 1.89 10 5 V*P26.57Imagine the center plate is split along its midplane and pulled apart.We have two capacitors in parallel, supporting the same V and Aand thecarrying total charge Q. The upper has capacitance C 1 = 0d Alower C 2 = 0 . Charge flows from ground onto each of the outside2dV1 = V2 = V .plates so that Q1 + Q 2 = QThenQ 1 Q 2 Q1 d Q 2 2 d===C 1 C 2 0 A 0 AQ1 = 2Q 22Q 2 + Q 2 = Q .(b)Q2 =QQ. On the lower plate the charge is .33Q1 =(a)2Q2Q. On the upper plate the charge is .33V =Q12Qd=C13 0 Ad2dFIG. P26.579596Capacitance and DielectricsP26.58(a)We use Equation 26.11 to find the potential energy of the capacitor. As we will see, thepotential difference V changes as the dielectric is withdrawn. The initial and finalenergies are U i =F IGH JK1 Q22 CiUf =andF IGH JK1 Q2.2 CfF IGH JK1 Q2But the initial capacitance (with the dielectric) is Ci = C f . Therefore, U f = .2CiSince the work done by the external force in removing the dielectric equals the change inF I F I F Ia fGH JK GH JK GH JK1 Q21 Q21 Q2= 1 .potential energy, we have W = U f U i = 22 Ci2 CiCib gTo express this relation in terms of potential difference Vi , we substitute Q = Ci Vi , andb ga f ejafaf1122C i Vi 1 = 2.00 10 9 F 100 V 5.00 1.00 = 4.00 10 5 J .22The positive result confirms that the final energy of the capacitor is greater than the initialenergy. The extra energy comes from the work done on the system by the external force thatpulled out the dielectric.evaluate: W =The final potential difference across the capacitor is V f =(b)CiQ.Cfab gfand Q = Ci Vi gives V f = Vi = 5.00 100 V = 500 V .Even though the capacitor is isolated and its charge remains constant, the potentialdifference across the plates does increase in this case.Substituting C f =P26.59 = 3.00 , Emax = 2.00 10 8 V m =ForC= 0 A= 0.250 10 6 FdVmaxdjbjeeeg0. 250 10 6 4 000CVmaxCdA== 0.188 m 2== 0 0 Emax 3.00 8.85 10 12 2.00 10 8*P26.60jThe original kinetic energy of the particle isK=11mv 2 = 2 10 16 kg 2 10 6 m s22ejej2= 4.00 10 4 J .Q 1 000 C== 100 V .10 FCFor the particle to reach the negative plate, the particle-capacitor system would need energyThe potential difference across the capacitor is V =jaefU = qV = 3 10 6 C 100 V = 3.00 10 4 J .Since its original kinetic energy is greater than this, the particle will reach the negative plate .As the particle moves, the system keeps constant total energyaK + U fat + platea= K +Ufat plate:jaef 1 e2 10 jv24.00 10 4 J + 3 10 6 C +100 V =vf =ej=2 1.00 10 4 J2 1016kg1.00 10 6 m s .162f+0Chapter 26P26.61(a)C1 =FG 1HC2 1 0 A 2 A2 A2; C2 = 2 0; C3 = 3 0dd 2d 2+1C3C = C1 +(b)*P26.6297IJK1FGIJHK A FGH 2 + + IJKd=FG 1HC2 A 2 3C 2C 3= 0C 2 + C3d 2 + 3+1C3IJK10=1223FIG. P26.613C total = 1.76 10 12 F = 1.76 pF .Using the given values we find:The initial charge on the larger capacitor isa fQ = CV = 10 F 15 V = 150 C .An additional charge q is pushed through the 50-V battery, giving the smaller capacitor charge q andthe larger charge 150 C + q .150 C + q5 F10 F500 C = 2 q + 150 C + q50 V =Thenq+q = 117 CSo across the 5- F capacitorq 117 C== 23.3 V .5 FCAcross the 10- F capacitorP26.63V =V =150 C + 117 C= 26.7 V .10 F(a)Put charge Q on the sphere of radius a and Q on the other sphere. Relative to V = 0 atinfinity,the potential at the surface of a isk eQ k eQadand the potential of b isVb = k e Q k eQ+.bdThe difference in potential isVa Vb =and(b)Va =C=As d ,C=k e Q k eQ k eQ k eQ+abddQ=Va VbF 4 IGH b1 ag + b1 bg b2 dg JK11becomes negligible compared to . Then,da4 0111+and =4 0 a 4 0 b1 a+1 bCas for two spheres in series.0.98Capacitance and Dielectricsa xf + x =(a)C=0d(b)U=1C V2(c)F=(d)P26.64F=a f2FG dU IJ i =H dx K2F a fIGHJK1 0 V2d=2a f+ x 12a f+ x 1a f a 1f to the left0 V2d2b2 000g e8.85 10 jb0.050 0ga4.50 1f =2e 2.00 10 j2P26.650d(out of the capacitor)1231.55 10 3 NThe portion of the capacitor nearly filled by metal hasa fcapacitance 0 xdand stored energyQ20.2CThe unfilled portion has0capacitancedQ=The charge on this portion is(a)a xf .a xfQ0.The stored energy isa fa fa f22 x Q0Q0 x dQ2=U==.x d2C 2 02 0 3(b)F=F=F a fIGHJK22Q0 ddUd Q0 x d==+dxdx2 0 32 0 32Q0 d2 03to the right (into the capacitor)2Q0F=d2 0(c)Stress =(d)11u = 0 E 2 = 00224FG IJH K2FGHQ01= 020 2IJK2=2Q02 04P26.661 gal.00b126 000 Btu galgb1 054 J BtugFGH 3.786.0010 m IJK FGH 1670 m IJK = 5.24 10kgb12.0 J Cgb100 C sgb3 600 sg = 2.70 10 J kg3Gasoline:33Chapter 26799J kg5Battery:16.0 kgCapacitor:12a0.100 Ffa12.0 Vf20.100 kg= 72.0 J kgGasoline has 194 times the specific energy content of the battery and 727 000 times that of thecapacitor.P26.67Call the unknown capacitance C ub g b gd iC d V ib10.0 Fga30.0 Vf ===bV g dV i a100 V 30.0 V fQ = C u Vi = Cu + C V fCufi*P26.684.29 FfShe can clip together a series combination of parallel combinations of two100- F capacitors. The equivalent capacitance is1= 100 F . When 90 V is connected across the11200 F + 200 Fbg bgcombination, only 45 V appears across each individual capacitor.P26.69(a)C0 =FIG. P26.680 A Q0=V0dWhen the dielectric is inserted at constant voltage,C = C 0 =U0 =U=andQ;V0b gC 0 V022b gC V02U= .U02=e jC 0 V022The extra energy comes from (part of the) electrical work done by the battery in separatingthe extra charge.(b)Q 0 = C 0 V0andsoQ = CV0 = C 0 V0Q= .Q0100P26.70Capacitance and DielectricsThe vertical orientation sets up two capacitors in parallel, with equivalent capacitanceCp =b g + b A 2g = FG + 1 IJ AH 2 K ddd0 A 200where A is the area of either plate and d is the separation of the plates. The horizontal orientationproduces two capacitors in series. If f is the fraction of the horizontal capacitor filled with dielectric,the equivalent capacitance isb g OP d , or C = LM OP A .MN f + b1 f g QP dQP A +1=gives, or a + 1f f + b1 f g = 2 .2f + b1 f gb g LMNM1 f df + 1 ffd1=+=0 AC s 0 ARequiring that C p = C ss00a fFor = 2.00 , this yields 3.00 2.00 1.00 f = 4.00 , with the solution f =P26.712.3Initially (capacitors charged in parallel),a f bga f= C a V f = b 2.00 F ga 250 V f = 500 C .q1 = C1 V = 6.00 F 250 V = 1 500 Cq22After reconnection (positive plate to negative plate),qtotal = q1 q 2 = 1 000 C and V =qtotal 1 000 C== 125 V .8.00 FC totalTherefore,a f bga fq = C a V f = b 2.00 F ga125 V f =q1 = C1 V = 6.00 F 125 V = 750 C2P26.722250 C .Assume a potential difference across a and b, and notice that the potential difference across the8.00 F capacitor must be zero by symmetry. Then the equivalent capacitance can be determinedfrom the following circuit:FIG. P26.72C ab = 3.00 F .Chapter 26P26.73Emax occurs at the inner conductors surface.Emax =2kefrom Equation 24.7.aV = 2 k e lnEmax =FG b IJ from Example 26.2H aKVa ln b ab gFG b IJ = e18.0 10 V mje0.800 10H aKF bIV = 2 lnG JH aKF bIa lnG JH aKLMlnFG b IJ + aF 1 I FG b IJ OP = 0NM H a K GH b a JK H a K QP6Vmax = Emax a lnP26.74E=2;aVmax = EmaxdVmax= EmaxdalnP26.753.00j FGH 03.800 IJK =m ln19.0 kV .2FG b IJ = 1 or b = eH aKa1so a =beBy symmetry, the potential difference across 3C is zero, so the circuit reduces toC eq =FG 1 + 1 IJH 2C 4C K1=84C=C .63FIG. P26.75101102P26.76Capacitance and DielectricsThe electric field due to the charge on the positive wire is perpendicular to the wire, radial, and ofmagnitudeE+ =.2 0 rThe potential difference between wires due to the presence of this charge isV1 = z+ wireE dr = wire2 0FGHzdIJKdrDdln.=rd2 0DdThe presence of the linear charge density on the negative wire makes an identical contributionto the potential difference between the wires. Therefore, the total potential difference isb gV = 2 V1 =FGHDdln 0dIJKand the capacitance of this system of two wires, each of lengthC= 0Q===.V V 0 ln D d d ln D d dbg afThe capacitance per unit length is:*P26.77, isaC=f 0ln D d daf.The condition that we are testing is that the capacitance increases by less than 10%, or,C< 1.10 .CSubstituting the expressions for C and C from Example 26.2, we have,c h = lnc h < 1.10 .lnchchb2 k e ln 1.10 aC=C2 k e ln babab1.10 aThis becomes,lnFG b IJ < 1.10 lnFG b IJ = 1.10 lnFG b IJ + 1.10 lnFG 1 IJ = 1.10 lnFG b IJ 1.10 lna1.10f .H aKH 1.10a KH aKH 1.10 KH aKWe can rewrite this as,FG b IJ < 1.10 lna1.10fH aKF bIlnG J > 11.0 lna1.10f = lna1.10fH aK0.10 ln11.0where we have reversed the direction of the inequality because we multiplied the whole expressionby 1 to remove the negative signs. Comparing the arguments of the logarithms on both sides of theinequality, we see that,a fb> 1.10a11.0= 2.85 .Thus, if b > 2.85 a , the increase in capacitance is less than 10% and it is more effective to increase.Chapter 26103ANSWERS TO EVEN PROBLEMSP26.2(a) 1.00 F ; (b) 100 VP26.4(a) 8.99 mm ; (b) 0.222 pF ; (c) 22.2 pCP26.611.1 nF ; 26.6 CP26.83.10 nmP26.10P26.400P26.22(a) 13.3 nC ; (b) 272 nCP26.46~ 10 6 F and ~ 10 2 V for two 40 cm by100 cm sheets of aluminum foilsandwiching a thin sheet of plastic.(a) 1.53 nF ; (b) 18.4 nC ; (c) 184 C m 2free; 183 C m 2 induced; (d) 694 V m(a) 17.0 F ; (b) 9.00 V ;(c) 45.0 C and 108 CP26.501.83C2+R1QR 2Qand q 2 =;R1 + R 2R1 + R 2(b) see the solution(a) q1 =P26.44mgd tan qCpej(a) 9.10 i + 8. 40 j pC m ;(b) 20.9 nN mk ; (c) 112 nJ ; (d) 228 nJ2Cp C p C s and4Cp22Cp4 C pC sP26.52579 VP26.54(a) 3.33 F ;(b) V3 = 60.0 V ; V6 = 30.0 V ;(a) 2C ; (b) Q1 > Q3 > Q 2 ;(c) V1 > V2 = V3 ;V2 = 60.0 V ; V4 = 30.0 V ;(c) Q 3 = Q6 = 180 C ; Q 2 = Q 4 = 120 C ;(d) 13.4 mJ(d) Q 3 and Q1 increase and Q 2 decreasesP26.24(a) 398 F in series; (b) 2.20 F in parallelP26.56189 kVP26.2619.8 CP26.58(a) 40.0 J ; (b) 500 VP26.2883.6 CP26.60yes; 1.00 Mm sP26.30eP26.6223.3 V ; 26.7 VP26.64(a)P26.32P26.343 1;(d) Positive work is done on the system bythe agent pulling the plates apart.2dP26.14P26.202P26.48a2 N 1f a fR2.13 10 16 m3P26.182P26.42P26.12P26.16a f ; (b) 4V ; (c) 4C aV f33(a) C Vj C204.47 kVenergy doublesP26.362.51 10 3 m3 = 2.51 LP26.38(a) 400 C ; (b) 2.50 kN m(b)(c)02a f+ x 1a f0 Vd22;a f;+ x 12da f a 1f to the left ;0 V22d(d) 1.55 mN left104Capacitance and DielectricsGasoline has 194 times the specific energycontent of the battery, and 727 000 timesthat of the capacitor.P26.723.00 FP26.74see the solutionP26.68see the solution; 45 VP26.76see the solutionP26.7023P26.6627Current and ResistanceCHAPTER OUTLINE27.127.227.3Electric CurrentResistanceA Model for ElectricalConductionResistance andTemperatureSuperconductorsElectric PowerANSWERS TO QUESTIONSQ27.2Voltage is a measure of potential difference, not of current.Surge implies a flowand only charge, in coulombs, can flowthrough a system. It would also be correct to say that the victimcarried a certain current, in amperes.Geometry and resistivity. In turn, the resistivity of the materialdepends on the temperature.Q27.427.527.6Individual vehiclescars, trucks and motorcycleswouldcorrespond to charge. The number of vehicles that pass acertain point in a given time would correspond to the current.Q27.327.4Q27.1Resistance is a physical property of the conductor based on thematerial of which it is made and its size and shape, includingthe locations where current is put in and taken out. Resistivityis a physical property only of the material of which the resistoris made.Q27.5The radius of wire B is 3 times the radius of wire A, to make its crosssectional area 3 times larger.Q27.6Not all conductors obey Ohms law at all times. For example, consider an experiment in which avariable potential difference is applied across an incandescent light bulb, and the current ismeasured. At very low voltages, the filament follows Ohms law nicely. But then long before theVbecomes non-linear, because the resistivity is temperaturefilament begins to glow, the plot ofIdependent.Q27.7A conductor is not in electrostatic equilibrium when it is carrying a current, duh! If charges areplaced on an isolated conductor, the electric fields established in the conductor by the charges willcause the charges to move until they are in positions such that there is zero electric field throughoutthe conductor. A conductor carrying a steady current is not an isolated conductorits ends must beconnected to a source of emf, such as a battery. The battery maintains a potential difference acrossthe conductor and, therefore, an electric field in the conductor. The steady current is due to theresponse of the electrons in the conductor due to this constant electric field.105106Current and ResistanceQ27.8The bottom of the rods on the Jacobs Ladder are close enough so that the supplied voltage issufficient to produce dielectric breakdown of the air. The initial spark at the bottom includes a tubeof ionized air molecules. Since this tube containing ions is warmer than the air around it, it is buoyedup by the surrounding air and begins to rise. The ions themselves significantly decrease theresistivity of the air. They significantly lower the dielectric strength of the air, marking longer sparkspossible. Internal resistance in the power supply will typically make its terminal voltage drop, sothat it cannot produce a spark across the bottom ends of the rods. A single continuous spark,therefore will rise up, becoming longer and longer, until the potential difference is not large enoughto sustain dielectric breakdown of the air. Once the initial spark stops, another one will form at thebottom, where again, the supplied potential difference is sufficient to break down the air.Q27.9The conductor does not follow Ohms law, and must have a resistivity that is current-dependent, ormore likely temperature-dependent.Q27.10A power supply would correspond to a water pump; a resistor corresponds to a pipe of a certaindiameter, and thus resistance to flow; charge corresponds to the water itself; potential differencecorresponds to difference in height between the ends of a pipe or the ports of a water pump.Q27.11The amplitude of atomic vibrations increases with temperature. Atoms can then scatter electronsmore efficiently.Q27.12In a metal, the conduction electrons are not strongly bound to individual ion cores. They can movein response to an applied electric field to constitute an electric current. Each metal ion in the latticeof a microcrystal exerts Coulomb forces on its neighbors. When one ion is vibrating rapidly, it can setits neighbors into vibration. This process represents energy moving though the material by heat.Q27.13The resistance of copper increases with temperature, while the resistance of silicon decreases withincreasing temperature. The conduction electrons are scattered more by vibrating atoms whencopper heats up. Silicons charge carrier density increases as temperature increases and more atomicelectrons are promoted to become conduction electrons.Q27.14A current will continue to exist in a superconductor without voltage because there is no resistanceloss.Q27.15Superconductors have no resistance when they are below a certain critical temperature. For mostsuperconducting materials, this critical temperature is close to absolute zero. It requires expensiverefrigeration, often using liquid helium. Liquid nitrogen at 77 K is much less expensive. Recentdiscoveries of materials that have higher critical temperatures suggest the possibility of developingsuperconductors that do not require expensive cooling systems.Q27.16In a normal metal, suppose that we could proceed to a limit of zero resistance by lengthening theaverage time between collisions. The classical model of conduction then suggests that a constantapplied voltage would cause constant acceleration of the free electrons, and a current steadilyincreasing in time.On the other hand, we can actually switch to zero resistance by substituting asuperconducting wire for the normal metal. In this case, the drift velocity of electrons is establishedby vibrations of atoms in the crystal lattice; the maximum current is limited; and it becomesimpossible to establish a potential difference across the superconductor.Q27.17Because there are so many electrons in a conductor (approximately 10 28 electrons m 3 ) the averagevelocity of charges is very slow. When you connect a wire to a potential difference, you establish anelectric field everywhere in the wire nearly instantaneously, to make electrons start driftingeverywhere all at once.Chapter 27107Q27.18Current moving through a wire is analogous to a longitudinal wave moving through the electrons ofthe atoms. The wave speed depends on the speed at which the disturbance in the electric field canbe communicated between neighboring atoms, not on the drift velocities of the electronsthemselves. If you leave a direct-current light bulb on for a reasonably short time, it is likely that nosingle electron will enter one end of the filament and leave at the other end.Q27.19More power is delivered to the resistor with the smaller resistance, since P =Q27.20The 25 W bulb has a higher resistance. The 100 W bulb carries more current.Q27.21One amperehour is 3 600 coulombs. The amperehour rating is the quantity of charge that thebattery can lift though its nominal potential difference.Q27.22Choose the voltage of the power supply you will use to drive the heater. Next calculate the requiredV 2resistance R as. Knowing the resistivity of the material, choose a combination of wire lengthPand crosssectional area to makeV 2.RFG IJ = FG R IJ . You will have to pay for less material if you make bothH AK H Kand A smaller, but if you go too far the wire will have too little surface area to radiate away theenergy; then the resistor will melt.SOLUTIONS TO PROBLEMSSection 27.1P27.1I=Electric CurrentQtN=P27.2jaefQ = It = 30.0 10 6 A 40.0 s = 1.20 10 3 C1.20 10 3 CQ== 7.50 10 15 electronse 1.60 10 19 C electronThe molar mass of silver = 107.9 g mole and the volume V isa faf ejejV = area thickness = 700 10 4 m 2 0.133 10 3 m = 9.31 10 6 m 3 .ejejThe mass of silver deposited is m Ag = V = 10.5 10 3 kg m3 9.31 10 6 m 3 = 9.78 10 2 kg .And the number of silver atoms deposited iseN = 9.78 10 2 kgI=10 atomsjFGH 6.02 107.9 g IJK FGH 1 1000 g IJK = 5.45 10kg23V 12.0 V== 6.67 A = 6.67 C s1.80 Rt =ejej23atoms5.45 10 23 1.60 10 19 CQ Ne=== 1.31 10 4 s = 3.64 hII6.67 C s108P27.3Current and ResistanceafzteQ t = Idt = I 0 1 e t 0afjj a0.632fI e(a)(b)Q 10 = I 0 1 e 10 =(c)Q = I 0 1 e = I 0afe(a)Usingkee2(b)P27.4Q = I 0 1 e 1 =The time for the electron to revolve around the proton once is:a fj b0.999 95gI er200jkee2= 2.19 10 6 m s .mrmv 2, we get: v =r=ej11m2 r 2 5.29 10t=== 1.52 10 16 s .v2.19 10 6 m sejThe total charge flow in this time is 1.60 10 19 C , so the current isI=P27.51.60 10 19 C1.52 10 16 sThe period of revolution for the sphere is T =revolving charge is I =P27.62, and the average current represented by thisqq=.2Tq = 4t 3 + 5t + 6eA = 2.00 cm 2aJ=(b)I=1.00 mjFGH 100 cmIJKfI 1.00 s =(a)P27.7= 1.05 10 3 A = 1.05 mA .dqdt2= 2.00 10 4 m 2e= 12t 2 + 5t = 1.00 sjt =1.00 s= 17.0 AI17.0 A== 85.0 kA m 2A 2.00 10 4 m 2dqdta100 Af sinFGH 120s t IJK dt100100 C L F Iq=MNcosGH 2 JK cos 0OPQ = +120C = 0.265 C120z zq = dq = Idt =z1 240 s0Chapter 275.00 AI=A 4.00 10 3 m= 99.5 kA m 2(a)J=(b)J2 =1I1 I=J1 ;4A 2 4 A1A1 =P27.81A 2 so 4.00 10 34ej2eej2=1 r224jr2 = 2 4.00 10 3 = 8.00 10 3 m = 8.00 mm8.00 10 6 AI=A 1.00 10 3 m= 2.55 A m 2(a)J=(b)From J = nev d , we haven=(c)P27.9Q, we haveFrom I =t6.02 10 23 1.60 10 19 CQ N A e=== 1. 20 10 10 s .t =II8.00 10 6 Aej22.55 A m 2J== 5.31 10 10 m 3 .ev d1.60 10 19 C 3.00 10 8 m sejeejjej(This is about 382 years!)P27.10(a)K=The speed of each deuteron is given bye2.00 10 je1.60 10 Jj = 1 e2 1.67 10261927jkg v 2 and1mv 22v = 1.38 10 7 m s .qtThe time between deuterons passing a stationary point is t inI=10.0 10 6 C s = 1.60 10 19 C t ort = 1.60 10 14 s .ejejSo the distance between them is vt = 1.38 10 7 m s 1.60 10 14 s = 2.21 10 7 m .(b)One nucleus will put its nearest neighbor at potentialejej8.99 10 9 N m 2 C 2 1.60 10 19 Ckeq== 6.49 10 3 V .V=r2.21 10 7 mThis is very small compared to the 2 MV accelerating potential, so repulsion within thebeam is a small effect.109110P27.11Current and ResistanceWe use I = nqAv d n is the number of charge carriers per unit volume, and is identical to the numberof atoms per unit volume. We assume a contribution of 1 free electron per atom in the relationshipabove. For aluminum, which has a molar mass of 27, we know that Avogadros number of atoms,N A , has a mass of 27.0 g. Thus, the mass per atom is27.0 g27.0 g== 4.49 10 23 g atom .NA6.02 10 23n=Thus,density of aluminum2.70 g cm 3=mass per atom4.49 10 23 g atomn = 6.02 10 22 atoms cm3 = 6.02 10 28 atoms m3 .5.00 AI== 1.30 10 4 m s283nqA6.02 10 m1.60 10 19 C 4.00 10 6 m 2Therefore,vd =or,v d = 0.130 mm s .Section 27.2*P27.12I=P27.14jejE=0.740 V m2.44 10 8 mFG 1 A IJ =H 1V K3.03 10 7 A m 2V 120 V== 0.500 A = 500 mAR240 (a)Applying its definition, we find the resistance of the rod,R=(b)V15.0 V== 3 750 = 3.75 k .I4.00 10 3 A. Solving forAand substituting numerical values for R, A, and the value of given for carbon in Table 27.1,we obtainThe length of the rod is determined from the definition of resistivity: R ==P27.15jeResistanceJ = E =P27.13eRAe3.75 10 je5.00 10=e3.50 10 mj635m2j=536 m .V = IRandR=:AIV =:AF 1.00 m I = 6.00 10A = a0.600 mmf GH 1 000 mm JKVA a0.900 V fe6.00 10 m jI==e5.60 10 mja1.50 mf2278I = 6.43 A27m2111Chapter 27P27.16J=Ir2= E =af = 55.3 mb3.00 Ag 0.012 0 m1=b= 120 N C21= 0.018 1 mGiven(a)M = dV = d Awe obtain:P27.17A=V=(b)Mdwherer d d mass density,Taking r resistivity,.MR=Thus,gR=e1.00 10 ja0.500fe1.70 10 je8.92 10 j3=8M,d3Mr= d=e1.00 10 3 8.92 103ja1.82fdiameter = 280 m .The volume of the gram of gold is given by =V=m=10 3 kgmVej= 5.18 10 8 m3 = A 2.40 10 3 m19.3 10 3 kg m3A = 2.16 10 11 m 2R=P27.19(a)ej2. 44 10 8 m 2.4 10 3 m== 2.71 10 6 A2.16 10 11 m 2Suppose the rubber is 10 cm long and 1 mm in diameter.ejej4 10 13 m 10 1 m4R=~== ~ 10 18 223A d 10 meejjej(b)4 1.7 10 8 m 10 3 m4R=~~ 10 7 22 d2 2 10 m(c)I=eI~V10 2 V~ 18~ 10 16 AR10 10 2 V10 7 ~ 10 9 AjM.dr = 1.40 10 4 m .The diameter is twice this distance:*P27.18= 1.82 m . r2 =orThus,rr == r dAM dM2.112P27.20Current and ResistanceF 90.0 g I=GH 10.5 g cm JKThe distance between opposite faces of the cube isR=(a) 1.59 10 8 m= 2 = == 7.77 10 7 = 777 n2A2.05 10 mV 1.00 10 5 V== 12.9 AR7.77 10 7 10.5 g cm 36.02 10 23 electrons moln=107.87 g moleen = 5.86 10 22 electrons cm 3I = nqvA and v =P27.22Originally, R = Alb g rAl2= CuP27.23J = EP27.24R==363e28m3jbjeFinally, R f =b 3g = 3A9A=g2= 3.28 m sR.92=so13A m2J 6.00 10== 6.00 10 15 mE100 V ma1 1 2 2 1 1 + 2+=A1A2d2e4.00 10R=P27.2510jFGH 1.00100 m cm IJK = 5.86 10.12.9 C sI=nqA5.86 10 28 m3 1.60 10 19 C 0.020 5 m.Ab g rCuj Al2.82 10 8== 1.29 Cu1.70 10 8rAl=rCuSection 27.3= 2.05 cm .3I=(b)P27.21133jaf12f ejaf=jjej m 0.250 m + 6.00 10 3 m 0.400 me3.00 103jm2378 A Model for Electrical Conductionmnq 2 soeje9.11 10 31m=== 2.47 10 14 snq 21.70 10 8 8. 49 10 28 1.60 10 19evd =qEme1.60 10 jEe2.47 10 j19so7.84 10 4 =Therefore,E = 0.181 V m .9.11 10 3114Chapter 27P27.26113n is unaffected(a)J =(b)IIAso it doubles .J = nev d(c)so v d=(d)doubles .mis unchanged as long as does not change due to a temperature change in thenq 2conductor.P27.27From Equation 27.17,=me9.11 10 31=e8.49 10 je1.60 10 j e1.70 10 j= v = e8.60 10 m sje 2.47 10sj = 2.12 102nq 288145Section 27.419 2= 2. 47 10 14 s8m = 21.2 nmResistance and TemperatureAt the low temperature TC we writeRC =V= R0 1 + TC T0ICAt the high temperature Th ,P27.28Rh =V V== R0 1 + Th T0 .Ih1Abgwhere T0 = 20.0 C .baV f a1.00 Af = 1 + e3.90 10aV f I 1 + e3.90 10F 1.15 IJ = 1.98 AI = a1.00 A fGH 0.579 Kgja38.0fja108f3Then3CandP27.29Ca fR = R 0 1 + TgivesSolving,And, the final temperature isaf ej.140 = 19.0 1 + 4.50 10 3 C T .T = 1.42 10 3 C = T 20.0 C .T = 1.44 10 3 C .114P27.30Current and ResistancebgbR = R c + Rn = R c 1 + c T T0 + Rn 1 + n T T0bgbg0 = Rc c T T0 + Rn n T T0 so R c = RnR = Rnn+ RncF IJF IJ= RG 1 R = RG 1 H KH KL e0.400 10 Cj OP= 10.0 k M1 MN e0.500 10 Cj PQ1Rnn1n13Rnnccccg3Rn = 5.56 kR c = 4.44 kandbg eajf(a) = 0 1 + T T0 = 2.82 10 8 m 1 + 3.90 10 3 30.0 = 3.15 10 8 m(b)J=(c)F d I = 6.35 10I = JA = J GH 4 JK e(d)P27.31n=E=0.200 V m3.15 10 8 m2= 6.35 10 6 A m 2LM e1.00 10A m jMM4N266.02 10 23 electronse626.98 g 2.70 10 g m3je4j OP =PPQm249.9 mA= 6.02 10 28 electrons m 3j6.35 10 6 A m 2J== 659 m svd =ne6.02 10 28 electrons m3 1.60 10 19 Ce(e)P27.32bjejfgaV = E = 0.200 V m 2.00 m = 0.400 VFor aluminum, E = 3.90 10 3 C 1 = 24.0 10 6 C 1R=bga(Table 27.1)(Table 19.1)fbag affFGHIJK 0 1 + E T 1 + T1 + E T1.39== R0= 1.71 = 1.234 2A1 + T1.002 4A 1 + TafChapter 27P27.33R = R 0 1 + TR R0 = R0TR R0= T = 5.00 10 3 25.0 = 0.125R0eP27.34jaAssuming linear change of resistance with temperature, R = R0 1 + Tajaf effR77 K = 1.00 1 + 3.92 10 3 216 C = 0.153 .af1FG HWIJK = 0 1 + T orTW =Require that W = 4 0 Cu so thatTW =Therefore,P27.35TW = 47.6 C + T0 = 67.6 C .Section 27.5W10WF 1GH 4.50 10I FG 4e1.70 10 j 1IJ = 47.6 C .JK C J G 5.60 10KH83SuperconductorsProblem 48 in Chapter 43 can be assigned with this section.Section 27.6P27.36I=Electric PowerPV=and R =*P27.37P27.38600 W= 5.00 A120 VV 120 V== 24.0 .I5.00 AebgbgP = 0.800 1 500 hp 746 W hp = 8.95 10 5 Wb8.95 10 5 = I 2 000P = IVP27.39jP = IV = 500 10 6 A 15 10 3 V = 7.50 WgI = 448 AThe heat that must be added to the water isbgbfgaQ = mcT = 1.50 kg 4 186 J kg C 40.0 C = 2.51 10 5 J .Thus, the power supplied by the heater isP=WQ 2.51 10 5 J=== 419 Wt t600 saV f = a110 V f2and the resistance is R =P2419 W= 28.9 .8115116*P27.40Current and ResistanceThe battery takes in energy by electric transmissiona fa feFG 3 600 s IJ = 469 J .H 1h KjPt = V I t = 2.3 J C 13.5 10 3 C s 4.2 hIt puts out energy by electric transmissionaV fIatf = 1.6 J C e18 103FG 3 600 s IJ = 249 J.H 1h KjC s 2. 4 huseful output 249 J== 0.530total input469 J(a)efficiency =(b)The only place for the missing energy to go is into internal energy:469 J = 249 J + EintEint = 221 J(c)We imagine toasting the battery over a fire with 221 J of heat input:Q = mcTT =P27.41221 JQ=mc 0.015 kgkg C= 15.1 C975 Ja f R = FG V IJ = FG 140 IJ = 1.361b g R H V K H 120 KF P P IJ a100%f = FG P 1IJ a100%f = a1.361 1f100% =% = GH P KHP KVP=P0V0222200036.1%0a f aV f = 500 WRa110 V f = 24.2 R=a500 Wf2P27.42P = I V =2a24.2 f e2.50 10=4mj2(a)R=A(b)R = R0 1 + T = 24.2 1 + 0.400 10 3 1 180 = 35.6 =soRAeaV f = a110f2P=R35.62= 340 W1.50 10 6 mjbg= 3.17 mChapter 27P27.43R=e117j1.50 10 6 m 25.0 m== 298 2A 0.200 10 3 mejafafV = IR = 0.500 A 298 = 149 VV149 V= 5.97 V m25.0 m(a)E==(b)P = V I = 149 V 0.500 A = 74.6 W(c)R = R0 1 + T T0 = 298 1 + 0.400 10 3 C 320 C = 337 a f abI=fagaafffe149 VV== 0.443 AR337 a f afajfP = V I = 149 V 0.443 A = 66.1 Wa f a f afaP27.45(a)U = q V = It V = 55.0 A h 12.0 V(b)P27.44Cost = 0.660 kWh0.660 kWh3.96a f V = IRaV f = a10.0f = 0.833 WP=P = I V2R*P27.46FG $0.060 0 IJ =H 1 kWh K1fFGH 11AC s IJK FGH 1 V J C IJK FGH 1 WJ s IJK = 660 W h =1(a)2120The resistance of 1 m of 12-gauge copper wire is=R=A d 2b g2=4d2=ej4 1.7 10 8 m 1 me 0.205 3 102jm2= 5.14 10 3 .af2The rate of internal energy production is P = IV = I 2 R = 20 A 5.14 10 3 = 2.05 W .(b)PAl = I 2 R =PAl Al=PCu CuI 2 4 Al d2PAl =2.82 10 8 m2.05 W = 3.41 W1.7 10 8 mAluminum of the same diameter will get hotter than copper.118*P27.47Current and ResistanceThe energy taken in by electric transmission for the fluorescent lamp isafFGH 3 600 s IJK = 3.96 10 J1hF $0.08 IJ FG k IJ FG W s IJ FG h IJ = $0.088JGH kWh K H 1 000 K H J K H 3 600 s K6Pt = 11 J s 100 hcost = 3.96 10 6For the incandescent bulb,afFGH 3 600 s IJK = 1.44 101hF $0.08 IJ = $0.32JGH 3.6 10 J KPt = 40 W 100 hcost = 1.44 10 77J6saving = $0.32 $0.088 = $0.232P27.48The total clock power ise270 10jFGH6clocks 2.50JsclockIJ FG 3 600 s IJ = 2.43 10KH 1 h K12J h.Wout, the power input to the generating plants must be:QinFrom e =Q in Wout t 2.43 10 12 J h=== 9.72 10 12 J hte0.250and the rate of coal consumption iseRate = 9.72 10 12 J hP27.49a f afa.00coaljFGH 133.0 kg10 J IJK = 2.95 1065kg coal h = 295 metric ton h .fP = I V = 1.70 A 110 V = 187 WafafEnergy used in a 24-hour day = 0.187 kW 24.0 h = 4.49 kWhFG $0.060 0 IJ = $0.269 = 26.9H kWh KP = IV = a 2.00 A fa120 V f = 240 WE = b0.500 kg gb 4 186 J kg C ga77.0 C f = 161 kJ cost = 4.49 kWhP27.50intt =EintP=1.61 10 5 J= 672 s240 WChapter 27P27.51At operating temperature,afaf(a)P = IV = 1.53 A 120 V = 184 W(b)Use the change in resistance to find the final operating temperature of the toaster.aR = R0 1 + Tf120 1201 + 0.400 10 3 T=1.53 1.80eT = 441 C*P27.52jT = 20.0 C + 441 C = 461 CYou pay the electric company for energy transferred in the amount E = P t(a)(b)(c)P27.531197d1fFGH 1 week IJK FGH 861400 s IJK FGH 1 WJ s IJK = 48.4 MJdF 7 d IJ FG 24 h IJ FG k IJ = 13.4 kWhP t = 40 Wa 2 weeksfGH 1 week K H 1 d K H 1 000 KF 7 d IJ FG 24 h IJ FG k IJ FG 0.12 $ IJ = $1.61P t = 40 Wa 2 weeksfGH 1 week K H 1 d K H 1 000 K H kWh KaP t = 40 W 2 weekshk.12fFGH 601min IJK FGH 1 000 IJK FGH 0kWh$ IJK = $0.005 82F 1 h IJ FG k IJ FG 0.12 $ IJ = $0.416P t = 5 200 Wa 40 minfGH 60 min K H 1 000 K H kWh KaP t = 970 W 3 min= 0.582Consider a 400-W blow dryer used for ten minutes daily for a year. The energy transferred to thedryer isbgbFG 1 kWh IJ 20 kWh .H 3.6 10 J KfgaP t = 400 J s 600 s d 365 d 9 10 7 J6We suppose that electrically transmitted energy costs on the order of ten cents per kilowatt-hour.Then the cost of using the dryer for a year is on the order ofafbgCost 20 kWh $0.10 kWh = $2 ~ $1 .120Current and ResistanceAdditional ProblemsP27.54(a)I=R=(b)(c)(d)VRVa V fRaV f = a120 V fR=2P = IV =soa f = a120 V f2225.0 WP2= 576 and100 WP2= 144 P25.0 WQ 1.00 C== 0.208 A ==V120 Vtt1.00 Ct == 4.80 s0.208 AThe bulb takes in charge at high potential and puts out the same amount of charge at lowpotential.I=U 1.00 J1.00 J=t == 0.040 0 stt25.0 WThe bulb takes in energy by electrical transmission and puts out the same amount of energyby heat and light.P = 25.0 W =bgbfgaU = Pt = 25.0 J s 86 400 s d 30.0 d = 64.8 10 8 JThe electric company sells energy .FG $0.070 0 IJ FG k IJ FG W s IJ FG h IJ = $1.26H kWh K H 1 000 K H J K H 3 600 s K$0.070 0 FkWhCost per joule =GH 3.60 10 J IJK = $1.94 10 JkWhCost = 64.8 10 6 J86*P27.55The original stored energy is Ui =11 Q2.QVi =22 C(a)When the switch is closed, charge Q distributes itself over the plates of C and 3C in parallel,Qpresenting equivalent capacitance 4C. Then the final potential difference is V f =for4Cboth.(b)The smaller capacitor then carries charge CV f =charge 3C(c)Q3Q.=4C4The smaller capacitor stores final energycapacitor possesses energy(d)QQC=. The larger capacitor carries4C4FG IJH K1Q3C24C2=1C V f2d i2=FG IJH K1QC24C2=Q2. The larger32C3Q 2.32CQ 2 3Q 2 Q 2+=. The loss of potential energy is the energy32C 32C 8C3Q 2Q2 Q2appearing as internal energy in the resistor:=+ EintEint =.8C2C 8CThe total final energy isChapter 27P27.56I = nqv d A = nqv d r 2We find the drift velocity fromvd =v=Inq rxt2t==1 000 A8.49 1028m3e1.60 1019jeC 10j2m1 d. dTWe begin with the differential equation=z zSeparating variables,lnd0FG IJ = bT T g andH KT= dTT0 = 0 e bT T0 g .00afFrom the series expansion e x 1 + x , x << 1 ,(b)b 0 1 + T T0P27.58= 2.34 10 4 m s200 10 3 mx== 8.54 10 8 s = 27.0 yrv 2.34 10 4 m s(a)P27.572gThe resistance of one wire is.FG 0.500 IJ a100 mif = 50.0 .H mi KThe whole wire is at nominal 700 kV away from ground potential, but the potential differencebetween its two ends isbfgaIR = 1 000 A 50.0 = 50.0 kV .a f ejbgThen it radiates as heat power P = V I = 50.0 10 3 V 1 000 A = 50.0 MW .a V f ARA=IR ( )10.4 ( m)1.41 10 621.11.50 10 61.543=(m)0.5401.028P27.5931.81.50 10 6 = 1.47 10 6 m (in agreement with tabulated value of 1.50 10 6 m in Table 27.1)P27.602 wires = 100 mR=af0.108 100 m = 0.036 0 300 m(a)aV f(b)P = I V = 110 A 116 V = 12.8 kW(c)Pwires = I 2 R = 110 Ahomea f= Va f alinea fbfaag IR = 120 110 0.036 0 = 116 Vff b0.036 0 g =2436 W121122Current and Resistanceaaff0 4.00 VdVi== 8.00 i V m0.500 0 mdx4.00 10 8 m 0.500 m== 0.637 R=2A 1.00 10 4 mI=4.00 VV== 6.28 AR0.637 (d)J=I6. 28 Ai=A 1.00 10 4 m(e)J = 4.00 10 8 m 2.00 10 8 i A m 2 = 8.00 i V m = E(a)E=(b)R=4 L=A d2(c)I=V d 2V=R4 L(d)J=IVi=iLA(e)P27.63E=(c)P27.62(a)(b)P27.61J=jeje2= 2.00 10 8 i A m 2 = 200 i MA m 2jejafdV xVi=idxLVi= ELbgsoI0,10soR = R0 1 + T T0R=feIn this case, I =P27.64jaeLM OPLM OPN QN Q19T = T + a9f = 20+= 2 020 C0.004 50 CT = T0 +1 R1 I01 . 1 = T0 + R0 I06.00V 12.0==thus 12.0 I 36.0 = 6.00 I and I = 6.00 A .III 3.00Therefore, R =af12.0 V= 2.00 .6.00 A.Chapter 27P27.65(a)P = IVso I =(b)123P8.00 10 3 W== 667 A .V12.0 VUt =P=2.00 10 7 J38.00 10 W= 2.50 10 3 sbgejand x = vt = 20.0 m s 2.50 10 3 s = 50.0 km .(a)We begin withR=which reduces toP27.66bgbb g1 + bT T g 1 + bT T g1 + 2 bT T gg 0 1 + T T0 0 1 + T T0=,AA0 1 + 2 T T0R=R000.0(b) 0 = 1.70 10 8 m , = 3.90 10 3 C 1 , and = 17.0 10 6 C 1For copper:R0 =ja fje1.70 10 8 2.000 0== 1.08 .2A0 0.100 10 3eThe simple formula for R gives:ajaf efR = 1.08 1 + 3.90 10 3 C 1 100 C 20.0 C = 1.420 while the more complicated formula gives:R=P27.67a1.08 f 1 + e3.90 103jajaf e C ja80.0 C f C 1 80.0 C 1 + 17.0 10 6 C 1 80.0 Ce1 + 2 17.0 10 6f= 1.418 .1Let be the temperature coefficient at 20.0C, and be the temperature coefficient at 0 C. Thenafa = 0 1 + T 20.0 C , and = 1 + T 0 Ctemperature T. That is, we must have:fmust both give the correct resistivity at anyaaff 0 1 + T 20.0 C = 1 + T 0 C .a f= 1 + a 20.0 C f .Setting T = 0 in equation (1) yields: = 0 1 20.0 C ,and setting T = 20.0 C in equation (1) gives:0Put from the first of these results into the second to obtain:afaf 0 = 0 1 20.0 C 1 + 20.0 C .continued on next page(1)124Current and ResistanceafTherefore1 + 20.0 C =which simplifies to =11 20.0 Caf.1 20.0 CafFrom this, the temperature coefficient, based on a reference temperature of 0C, may be computedfor any material. For example, using this, Table 27.1 becomes at 0 C :MaterialSilverTemp Coefficients at 0C4.1 10 3 CCopper4.2 10 3 CGold3.6 10 3 CAluminum4.2 10 3 CTungsten4.9 10 3 CIron5.6 10 3 CPlatinum4.25 10 3 CLead4.2 10 3 CNichrome0.4 10 3 CCarbon0.5 10 3 CGermaniumSiliconP27.6824 10 3 C30 10 3 C(a)A thin cylindrical shell of radius r, thickness dr, and length L contributes resistancedR =Fb g GHIJKddr dr==.A2 L r2 r LThe resistance of the whole annulus is the series summation of the contributions of the thinshells:zrFG IJH Kr b dr=R=ln bra2 L r r2 La(b)In this equationwe solve for.FG IJH KrV=ln bIra2 L=2 LVI ln rb rabg.Chapter 27P27.69125Each speaker receives 60.0 W of power. Using P = I 2 R , we then haveP60.0 W== 3.87 A .4.00 RI=The system is not adequately protected since the fuse should be set to melt at 3.87 A, or less .P27.70V = E or dV = E dxV = IR = E I=dq E AAdVdV=== AE = E = AdtRdxdxCurrent flows in the direction of decreasing voltage. Energy flows as heat in the direction ofdecreasing temperature.P27.71R=zzy y1dxdx=where y = y1 + 2xAwyLzLMg Ny y1LLdx=R=ln y1 + 2xw 0 y1 + y 2 y 1 L x w y 2 y 1LR=P27.72bbgF Ig GH JKOPQL0FIG. P27.71yLln 2y1w y 2 y1bFrom the geometry of the longitudinal section of the resistor shown in the figure,we see thatbrba=.yha f a fFrom this, the radius at a distance y from the base isFor a disk-shaped element of volume dR =Using the integral formula*P27.73zaduau + bf2=dy: r2R=1,a au + bafR=zah0dyfb gab y h +b h. abd=.A A 0 A.The capacitance of the capacitor is C =dd 0 A 0=Then RC =is a characteristic of the material only.A d(a)The resistance of the dielectric block is R =(b)R=a fFIG. P27.72a fy +b .hr = ab 0 0 75 10 16 m 3.78 8.85 10 12 C 2=== 1.79 10 15 CC14 10 9 FN m22.126P27.74Current and ResistanceLM F eV I 1OP and R = VINM GH k T JK QPI = I 0 expBwith I 0 = 1.00 10 9 A , e = 1.60 10 19 C , and k B = 1.38 10 23 J K .The following includes a partial table of calculated values and a graph for each of the specifiedtemperatures.(i)For T = 280 K :afV Vafa fRI A0. 4000.015 6 25.60. 4400.081 85.380. 4800.4291.122.250.2320.5200.56011.80.047 60.60061.60.009 7FIG. P27.74(i)(ii)For T = 300 K :afV VafI AafR0. 4000.00577.30. 4400.02418.10. 4800.1144.220.5200.5340.9730.5602.510.2230.60011.80.051FIG. P27.74(ii)(iii)For T = 320 K :a f IaAfafV VR0. 4000.002 0 2030. 4400.008 452.50. 4800.035 713.40.5200.1523.420.5600.6480.8640.6002.760.217FIG. P27.74(iii)Chapter 27*P27.75(a)Think of the device as two capacitors in parallel. The one on the left has 1 = 1 ,A1 =FG + xIJH2 K. The equivalent capacitance is 1 0 A1 2 0 A 2 0+=ddd(b)127FG + xIJ + FG xIJ =H2 K d H2 K002daf+ 2 x + 2x .The charge on the capacitor is Q = CV VQ= 0+ 2 x + 2x .2dThe current is VvdQ dQ dx 0 V== 1 .I=0 + 2 + 0 2 v = 0dtdx dtd2dThe negative value indicates that the current drains charge from the capacitor. Positive Vvcurrent is clockwise 0 1 .dafafa fa fANSWERS TO EVEN PROBLEMSP27.23.64 hP27.321.71 P27.4(a) see the solution; (b) 1.05 mAP27.340.153 P27.6(a) 17.0 A ; (b) 85.0 kA m 2P27.365.00 A , 24.0 P27.8(a) 99.5 kA m 2 ; (b) 8.00 mmP27.38448 AP27.10(a) 221 nm ; (b) no; see the solutionP27.40(a) 0.530; (b) 221 J; (c) 15.1CP27.1230.3 MA m 2P27.42(a) 3.17 m ; (b) 340 WP27.14(a) 3.75 k ; (b) 536 mP27.44(a) 0.660 kWh ; (b) 3.96P27.160.018 1 mP27.46(a) 2.05 W; (b) 3.41 W; noP27.182.71 MP27.48295 metric ton hP27.20(a) 777 n ; (b) 3. 28 m sP27.50672 sP27.22rAl= 1.29rCuP27.52(a) $1.61; (b) $0.005 82; (c) $0.416P27.54(a) 576 and 144 ;(b) 4.80 s; The charge is the same. Thecharge-field system is in a lower-energyconfiguration.(c) 0.040 0 s; The energy enters by electrictransmission and exits by heat andelectromagnetic radiation;(d) $1.26; energy; 1.94 10 8 $ JP27.24378 P27.26(a) nothing; (b) doubles; (c) doubles;(d) nothingP27.281.98 AP27.30carbon, 4.44 k ; nichrome, 5.56 k128Current and ResistanceP27.5627.0 yrP27.66P27.5850.0 MWP27.68P27.60(a) 116 V ; (b) 12.8 kW ; (c) 436 WP27.62(a) E =V dVi4 L;; (c) I =; (b) R =2L4 Ld(d) J =Vi; (e) see the solutionLP27.642.00 2(a) see the solution;(b) 1.418 nearly agrees with 1.420 (a) R =P27.70see the solutionP27.72see the solutionP27.74see the solution2 LVrln b ; (b) =I ln rb ra2 L rabg28Direct Current CircuitsCHAPTER OUTLINE28.128.228.328.428.528.6Electromotive ForceResistors in Series andParallelKirchhoffs RulesRC CircuitsElectrical MetersHousehold Wiring andElectrical SafetyANSWERS TO QUESTIONSQ28.1The load resistance in a circuit is the effective resistance of all ofthe circuit elements excluding the emf source. In energy terms,it can be used to determine the energy delivered to the load byelectrical transmission and there appearing as internal energyto raise the temperature of the resistor. The internal resistanceof a battery represents the limitation on the efficiency of thechemical reaction that takes place in the battery to supplycurrent to the load. The emf of the battery represents itsconversion of chemical energy into energy which it puts out byelectric transmission; the battery also creates internal energywithin itself, in an amount that can be computed from itsinternal resistance. We model the internal resistance asconstant for a given battery, but it may increase greatly as thebattery ages. It may increase somewhat with increasing currentdemand by the load. For a load described by Ohms law, theload resistance is a precisely fixed value.Q28.2The potential difference between the terminals of a battery will equal the emf of the battery whenthere is no current in the battery. At this time, the current though, and hence the potential dropacross the internal resistance is zero. This only happens when there is no load placed on thebatterythat includes measuring the potential difference with a voltmeter! The terminal voltagewill exceed the emf of the battery when current is driven backward through the battery, in at itspositive terminal and out at its negative terminal.Q28.3No. If there is one battery in a circuit, the current inside it will be from its negative terminal to itspositive terminal. Whenever a battery is delivering energy to a circuit, it will carry current in thisdirection. On the other hand, when another source of emf is charging the battery in question, it willhave a current pushed through it from its positive terminal to its negative terminal.Q28.4Connect the resistors in series. Resistors of 5.0 k, 7.5 k and 2.2 k connected in series presentequivalent resistance 14.7 k.Q28.5Connect the resistors in parallel. Resistors of 5.0 k, 7.5 k and 2.2 k connected in parallel presentequivalent resistance 1.3 k.129130Direct Current CircuitsQ28.6Q28.7In series, the current is the same through each resistor. Without knowing individual resistances,nothing can be determined about potential differences or power.Q28.8In parallel, the potential difference is the same across each resistor. Without knowing individualresistances, nothing can be determined about current or power.Q28.9In this configuration, the power delivered to one individual resistor is significantly less than if onlyone equivalent resistor were used. This decreases the possibility of component failure, and possibleelectrical disaster to some more expensive circuit component than a resistor.Q28.10Each of the two conductors in the extension cord itself has a small resistance. The longer theextension cord, the larger the resistance. Taken into account in the circuit, the extension cord willreduce the current from the power supply, and also will absorb energy itself in the form of internalenergy, leaving less power available to the light bulb.Q28.11The whole wire is very nearly at one uniform potential. There is essentially zero difference inpotential between the birds feet. Then negligible current goes through the bird. The resistancethrough the birds body between its feet is much larger than the resistance through the wirebetween the same two points.Q28.12The potential difference across a resistor is positive when it is measured against the direction of thecurrent in the resistor.Q28.13The bulb will light up for a while immediately after the switch is closed. As the capacitor charges, thebulb gets progressively dimmer. When the capacitor is fully charged the current in the circuit is zeroand the bulb does not glow at all. If the value of RC is small, this whole process might occupy a veryshort time interval.Q28.14An ideal ammeter has zero resistance. An ideal voltmeter has infinite resistance. Real meters cannotattain these values, but do approach these values to the degree that they do not alter the current orpotential difference that is being measured within the accuracy of the meter. Hooray forexperimental uncertainty!Q28.15A short circuit can develop when the last bit of insulation frays away between the two conductors ina lamp cord. Then the two conductors touch each other, opening a low-resistance branch in parallelwith the lamp. The lamp will immediately go out, carrying no current and presenting no danger. Avery large current exists in the power supply, the house wiring, and the rest of the lamp cord up tothe contact point. Before it blows the fuse or pops the circuit breaker, the large current can quicklyraise the temperature in the short-circuit path.Chapter 28131Q28.16A wire or cable in a transmission line is thick and made of material with very low resistivity. Onlywhen its length is very large does its resistance become significant. To transmit power over a longdistance it is most efficient to use low current at high voltage, minimizing the I 2 R power loss in thetransmission line. Alternating current, as opposed to the direct current we study first, can be steppedup in voltage and then down again, with high-efficiency transformers at both ends of the powerline.Q28.17Car headlights are in parallel. If they were in series, both would go out when the filament of onefailed. An important safety factor would be lost.Q28.18Kirchhoffs junction rule expresses conservation of electric charge. If the total current into a pointwere different from the total current out, then charge would be continuously created or annihilatedat that point.Kirchhoffs loop rule expresses conservation of energy. For a single-loop circle with tworesistors, the loop rule reads + IR1 IR 2 = 0 . This is algebraically equivalent to q = qIR1 + qIR 2 ,where q = It is the charge passing a point in the loop in the time interval t . The equivalentequation states that the power supply injects energy into the circuit equal in amount to that whichthe resistors degrade into internal energy.Q28.19V 2, the bulbs present resistancesR222120 V120 V120 VV 2= 190 , and= 72 . The nominal 60 W lampR=== 240 , andP60 W75 W200 Whas greatest resistance. When they are connected in series, they all carry the same small current.Here the highest-resistance bulb glows most brightly and the one with lowest resistance is faintest.This is just the reverse of their order of intensity if they were connected in parallel, as they aredesigned to be.At their normal operating temperatures, from P =afafafQ28.20Answer their question with a challenge. If the student is just looking at a diagram, provide thematerials to build the circuit. If you are looking at a circuit where the second bulb really is fainter, getthe student to unscrew them both and interchange them. But check that the studentsunderstanding of potential has not been impaired: if you patch past the first bulb to short it out, thesecond gets brighter.Q28.21Series, because the circuit breaker trips and opens the circuit when the current in that circuit loopexceeds a certain preset value. The circuit breaker must be in series to sense the appropriate current(see Fig. 28.30).Q28.22The hospital maintenance worker is right. A hospital room is full of electrical grounds, including thebed frame. If your grandmother touched the faulty knob and the bed frame at the same time, shecould receive quite a jolt, as there would be a potential difference of 120 V across her. If the 120 V isDC, the shock could send her into ventricular fibrillation, and the hospital staff could use thedefibrillator you read about in Section 26.4. If the 120 V is AC, which is most likely, the current couldproduce external and internal burns along the path of conduction. Likely no one got a shock fromthe radio back at home because her bedroom contained no electrical groundsno conductorsconnected to zero volts. Just like the bird in Question 28.11, granny could touch the hot knobwithout getting a shock so long as there was no path to ground to supply a potential differenceacross her. A new appliance in the bedroom or a flood could make the radio lethal. Repair it ordiscard it. Enjoy the news from Lake Wobegon on the new plastic radio.132Direct Current CircuitsQ28.23So long as you only grab one wire, and you do not touch anything that is grounded, you are safe(see Question 28.11). If the wire breaks, let go! If you continue to hold on to the wire, there will be alargeand rather lethalpotential difference between the wire and your feet when you hit theground. Since your body can have a resistance of about 10 k, the current in you would be sufficientto ruin your day.Q28.24Both 120-V and 240-V lines can deliver injurious or lethal shocks, but there is a somewhat bettersafety factor with the lower voltage. To say it a different way, the insulation on a 120-V line can bethinner. On the other hand, a 240-V device carries less current to operate a device with the samepower, so the conductor itself can be thinner. Finally, as we will see in Chapter 33, the last stepdown transformer can also be somewhat smaller if it has to go down only to 240 volts from the highvoltage of the main power line.Q28.25As Luigi Galvani showed with his experiment with frogs legs, muscles contract when electriccurrent exists in them. If an electrician contacts a live wire, the muscles in his hands and fingerswill contract, making his hand clench. If he touches the wire with the front of his hand, his hand willclench around the wire, and he may not be able to let go. Also, the back of his hand may be drierthan his palm, so an actual shock may be much weaker.Q28.26Grab an insulator, like a stick or baseball bat, and bat for a home run. Hit the wire away from theperson or hit them away from the wire. If you grab the person, you will learn very quickly aboutelectrical circuits by becoming part of one.Q28.27A high voltage can lead to a high current when placed in a circuit. A device cannot supply a highcurrentor any currentunless connected to a load. A more accurate sign saying potentially highcurrent would just confuse the poor physics student who already has problems distinguishingbetween electrical potential and current.Q28.28The two greatest factors are the potential difference between the wire and your feet, and theconductivity of the kite string. This is why Ben Franklins experiment with lightning and flying akite was so dangerous. Several scientists died trying to reproduce Franklins results.Q28.29Suppose = 12 V and each lamp has R = 2 . Before the switch is closed the current is12 V= 2 A.6The potential difference across each lamp is 2 A 2 = 4 V . The power of each lamp is2 A 4 V = 8 W , totaling 24 W for the circuit. Closing the switch makes the switch and the wiresconnected to it a zero-resistance branch. All of the current through A and B will go through theswitch and (b) lamp C goes out, with zero voltage across it. With less total resistance, the (c) current12 Vin the battery= 3 A becomes larger than before and (a) lamps A and B get brighter. (d) The4voltage across each of A and B is 3 A 2 = 6 V , larger than before. Each converts power3 A 6 V = 18 W , totaling 36 W, which is (e) an increase.a fa fa fa fa fa fQ28.30a fa fThe starter motor draws a significant amount of current from the battery while it is starting the car.This, coupled with the internal resistance of the battery, decreases the output voltage of the batterybelow its the nominal 12 V emf. Then the current in the headlights decreases.Chapter 28Q28.31. Three runs in parallel:Two runs in series:two runs:. Junction of one lift and.Gustav Robert Kirchhoff, Professor of Physics at Heidelberg and Berlin, was master of theobvious. A junction rule: The number of skiers coming into any junction must be equal to thenumber of skiers leaving. A loop rule: the total change in altitude must be zero for any skiercompleting a closed path.SOLUTIONS TO PROBLEMSSection 28.1P28.1(a)Electromotive ForceP=aV f2Ra11.6 V fbecomesso(b)20.0 W =2R = 6.73 .RFIG. P28.1V = IRafso11.6 V = I 6.73 andI = 1.72 A = IR + Irso15.0 V = 11.6 V + 1.72 A rafr = 1.97 .P28.2(a)Vterm = IRabecomesso(b)10.0 V = I 5.60 fI = 1.79 A .Vterm = IrafabecomessoP28.310.0 V = 1.79 A 0.200 f = 10.4 V .The total resistance is R =3.00 V= 5.00 .0.600 A(a)Rlamp = R rbatteries = 5.00 0.408 = 4.59 (b)0.408 I 2Pbatteries== 0.081 6 = 8.16%Ptotal5.00 I 2aaff133FIG. P28.3134Direct Current Circuitsaf12.6 V(a)Here = I R + r , so I =(b)P28.4Let I 1 and I 2 be the currents flowing through the battery and theheadlights, respectively.Then,I 1 = I 2 + 35.0 A , and I 1 r I 2 r = 0b5.00 + 0.080 0 g = 2.48 A .Then, V = IR = a 2.48 A fa5.00 f = 12.4 V .bR+r=g agbfsoI 2 = 1.93 A .Thus,P28.5 = I 2 + 35.0 A 0.080 0 + I 2 5.00 = 12.6 VgivingSection 28.2V2 = 1.93 A 5.00 = 9.65 V .afafResistors in Series and Parallelafg a fbgTherefore, a 2.00 A fR = a1.60 A fb R + 3.00 g or R = 12.0 .bV = I 1 R1 = 2.00 A R1 and V = I 2 R1 + R 2 = 1.60 A R1 + 3.00 1P28.6FIG. P28.4(a)Rp =111b1 7.00 g + b1 10.0 g = 4.12 Rs = R1 + R 2 + R3 = 4.00 + 4.12 + 9.00 = 17.1 (b)V = IR34.0 V = I 17.1 afFIG. P28.6I = 1.99 A for 4.00 , 9.00 resistors.a1.99 Afa4.12 f = 8.18 V8.18 V = I a7.00 fApplying V = IR ,soI = 1.17 A for 7.00 resistora8.18 V = I 10.0 soP28.7fI = 0.818 A for 10.0 resistor.For the bulb in use as intended,P75.0 W== 0.625 AV120 VV120 V== 192 .R=0.625 AII=andFIG. P28.7Now, presuming the bulb resistance is unchanged,I=Across the bulb isso its power is120 V= 0.620 A .193.6 afP = IV = 0.620 Aa119 V f =V = IR = 192 0.620 A = 119 V73.8 W .Chapter 28P28.8120 V = IReq = IV2 =P28.9I=A2 A2FG HAe+11A1IJK++, or I =A 2 A3 A 4a120 V f+1A2+1A3+1A4jea120 Vf1A1+1A2+1A3+1A4135j= 29.5 VIf we turn the given diagram on its side, we find that it is the same as figure(a). The 20.0 and 5.00 resistors are in series, so the first reduction isshown in (b). In addition, since the 10.0 , 5.00 , and 25.0 resistors arethen in parallel, we can solve for their equivalent resistance as:1Req == 2.94 .11110.0 + 5.00 + 25 .0 chThis is shown in figure (c), which in turn reduces to the circuit shown infigure (d).VNext, we work backwards through the diagrams applying I =andRV = IR alternately to every resistor, real and equivalent. The 12.94 resistor is connected across 25.0 V, so the current through the battery inevery diagram isV25.0 V== 1.93 A .I=12.94 RIn figure (c), this 1.93 A goes through the 2.94 equivalent resistor to give apotential difference of:V = IR = 1.93 A 2.94 = 5.68 V .afafFrom figure (b), we see that this potential difference is the same across Vab ,the 10 resistor, and the 5.00 resistor.(b)(a)*P28.10Therefore, Vab = 5.68 V .Since the current through the 20.0 resistor is also the currentthrough the 25.0 line ab,Vab 5.68 V== 0.227 A = 227 mA .I=25.0 R abFIG. P28.9We assume that the metal wand makes low-resistance contact with the persons hand and that theresistance through the persons body is negligible compared to the resistance Rshoes of the shoe soles.The equivalent resistance seen by the power supply is 1.00 M + Rshoes . The current through both50.0 Vresistors is. The voltmeter displays1.00 M + Rshoes50.0 V 1.00 MV = I 1.00 M == V .1.00 M + Rshoesaafafaffb50.0 V 1.00 M = V 1.00 M + V Rshoes1.00 M 50.0 VRshoes =.V(a)We solve to obtain(b)With Rshoes 0 , the current through the persons body is50.0 V= 50.0 AThe current will never exceed 50 A .1.00 Mafg136P28.11Direct Current Circuits(a)Since all the current in the circuit must pass through the series100 resistor, P = I 2 R2Pmax = RI maxP25.0 W== 0.500 A100 RI max =soR eq = 100 +FG 1 + 1 IJH 100 100 K1FIG. P28.11 = 150 Vmax = R eq I max = 75.0 VafafP = IV = 0.500 A 75.0 V = 37.5 W total power(b)P1 = 25.0 WafaP2 = P3 = RI 2 100 0.250 AP28.122= 6.25 WUsing 2.00-, 3.00-, 4.00- resistors, there are 7 series, 4 parallel, and 6 mixed combinations:Series2.00 3.00 4.00 5.00 P28.13fParallel0.923 1.20 1.33 1.71 6.00 7.00 9.00 Mixed1.56 2.00 2.22 3.71 4.33 5.20 The resistors may be arranged in patterns:The potential difference is the same across either combination.V = IR = 3 I1+c11R+1500R=3500hFG 1 + 1 IJ = 3H R 500 KsoRandR = 1 000 = 1.00 k .FIG. P28.13*P28.14When S is open, R1 , R 2 , R3 are in series with the battery. Thus:6V(1)R1 + R 2 + R3 = 3 = 6 k .10 AWhen S is closed in position 1, the parallel combination of the two R 2 s is in series with R1 , R3 , and thebattery. Thus:16V= 5 k.(2)R1 + R 2 + R3 =21.2 10 3 AWhen S is closed in position 2, R1 and R 2 are in series with the battery. R3 is shorted. Thus:6V= 3 k.(3)R1 + R 2 =2 10 3 AFrom (1) and (3): R3 = 3 k.Subtract (2) from (1): R 2 = 2 k .From (3): R1 = 1 k.Answers: R1 = 1.00 k , R 2 = 2.00 k , R3 = 3.00 k .Chapter 28P28.15FG 1 + 1 IJ = 0.750 H 3.00 1.00 K= a 2.00 + 0.750 + 4.00f = 6.75 1Rp =Rs137V 18.0 V== 2.67 ARs 6.75 I battery =aP = I 2R:P2 = 2.67 Af a2.00 f2P2 = 14.2 W in 2.00 af a4.00 Af = 28.4 W= a 2.67 A fa 2.00 f = 5.33 V,= a 2.67 A fa 4.00 f = 10.67 VP4 = 2.67 AV2V42in 4.00 bVp = 18.0 V V2 V4 = 2.00 V = V3 = V1P3 =P1 =P28.16bV g = a2.00 V fbV g = a2.00 Vf3223.00 R31= 1.33 W in 3.00 FIG. P28.152= 4.00 W in 1.00 1.00 R1gDenoting the two resistors as x and y,x + y = 690, and11 1= +150 x yaf690 x + x111= +=150 x 690 xx 690 xaf2x 690 x + 103 500 = 0x=690 a690f 414 0002x = 470 *P28.172y = 220 a fA certain quantity of energy Eint = P time is required to raise the temperature of the water toaV f where aV f is a constant.RaV f t = aV f 2 t . Then R = 2 R .Thus comparing coils 1 and 2, we have for the energy2100C . For the power delivered to the heaters we have P = IV =22R1(a)2R2When connected in parallel, the coils present equivalent resistanceRp =a fa f22V t pV t2 R111===. Now1 R1 + 1 R 2 1 R1 + 1 2 R132 R1 3R1t p =a V f t = a V f t2(b)1For the series connection, Rs = R1 + R 2 = R1 + 2 R1 = 3 R1 andt s = 3 t .R123 R1s2 t.3138Direct Current Circuits(a)afaV = IR :33.0 V = I1 11.0 33.0 V = I 2 22.0 I 3 = 3.00 AI 2 = 1.50 AP = I 2R:P28.18P1 = 3.00 AP1 = 99.0 Waf a11.0 f2affa2P2 = 1.50 A 22.0 P2 = 49.5 WfThe 11.0- resistor uses more power.FIG. P28.18(a)a f a fa fP = I V = 4.50 33.0 = 148 W(b)P1 + P2 = 148 W(c)Rs = R1 + R 2 = 11.0 + 22.0 = 33.0 afV = IR :33.0 V = I 33.0 , so I = 1.00 AP = I 2R:P1 = 1.00 AP1 = 11.0 Waf a11.0 f2af a22.0 fP2 = 1.00 AP2 = 22.0 W2FIG. P28.18(c)The 22.0- resistor uses more power.(d)(e)*P28.19g a f a33.0 f =P = I a V f = a1.00 A fa33.0 V f = 33.0 WbP1 + P2 = I 2 R1 + R2 = 1.00 A233.0 WThe parallel configuration uses more power.(a)The resistors 2, 3, and 4 can be combined to a single 2R resistor. This is in series withresistor 1, with resistance R, so the equivalent resistance of the whole circuit is 3R. In series,1potential difference is shared in proportion to the resistance, so resistor 1 gets of the32battery voltage and the 2-3-4 parallel combination get of the battery voltage. This is the31potential difference across resistor 4, but resistors 2 and 3 must share this voltage. goes to322 and to 3. The ranking by potential difference is V4 > V3 > V1 > V2 .3(b)Based on the reasoning above the potential differences are242.V1 = , V2 =, V3 =, V4 =3993(c)All the current goes through resistor 1, so it gets the most. The current then splits at theparallel combination. Resistor 4 gets more than half, because the resistance in that branch isless than in the other branch. Resistors 2 and 3 have equal currents because they are inseries. The ranking by current is I 1 > I 4 > I 2 = I 3 .(d)Resistor 1 has a current of I. Because the resistance of 2 and 3 in series is twice that ofresistor 4, twice as much current goes through 4 as through 2 and 3. The current throughI2Ithe resistors are I 1 = I , I 2 = I 3 = , I 4 =.33continued on next pageChapter 28139(e)Increasing resistor 3 increases the equivalent resistance of the entire circuit. The current inthe circuit, which is the current through resistor 1, decreases. This decreases the potentialdifference across resistor 1, increasing the potential difference across the parallelcombination. With a larger potential difference the current through resistor 4 is increased.With more current through 4, and less in the circuit to start with, the current throughresistors 2 and 3 must decrease. To summarize, I 4 increases and I 1 , I 2 , and I 3 decrease .(f)If resistor 3 has an infinite resistance it blocks any current from passing through that branch,and the circuit effectively is just resistor 1 and resistor 4 in series with the battery. The circuit3now has an equivalent resistance of 4R. The current in the circuit drops to of the original44current because the resistance has increased by . All this current passes through resistors 133I3I, I2 = I3 = 0, I4 =.and 4, and none passes through 2 or 3. Therefore I 1 =44Section 28.3P28.20Kirchhoffs Rulesa f a fa f+15.0 7.00 I1 2.00 5.00 = 05.00 = 7.00 I 1soI 1 = 0.714 AsoI 2 = 1.29 AI 3 = I 1 + I 2 = 2.00 A0.714 + I 2 = 2.00a fa f+ 2.00 1.29 5.00 2.00 = 0P28.21FIG. P28.20 = 12.6 VWe name currents I 1 , I 2 , and I 3 as shown.From Kirchhoffs current rule, I 3 = I 1 + I 2 .Applying Kirchhoffs voltage rule to the loop containing I 2 and I 3 ,a f a f8.00 = a 4.00fI + a6.00fI12.0 V 4.00 I 3 6.00 I 2 4.00 V = 032Applying Kirchhoffs voltage rule to the loop containing I 1 and I 2 ,a fa f 6.00 I 2 4.00 V + 8.00 I1 = 0a8.00fI1a f= 4.00 + 6.00 I 2 .FIG. P28.21Solving the above linear system, we proceed to the pair of simultaneous equations:R8 = 4I + 4IS8I = 4 + 6 IT112+ 6I2or2R8 = 4I + 10ISI = 1.33I 0.667T1221and to the single equation 8 = 4I 1 + 13.3 I 1 6.67I1 =and14.7 V= 0.846 A .17.3 I 3 = I1 + I 2ThengiveafI 2 = 1.33 0.846 A 0.667I 1 = 846 mA, I 2 = 462 mA, I 3 = 1.31 A .All currents are in the directions indicated by the arrows in