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### Transcript of Solucionario Raymond Serway - Capitulo 6

2000 by Harcourt College Publishers. All rights reserved.Chapter 6 Solutions6.1 ( a ) Average speed = v = 200 m25.0 s = 8.00 m/ s (b) F = mv2r where r = 200 m2 = 31.8 mF = (1.50 kg)(8.00 m/ s)231.8 m = 3.02 N 6.2 ( a ) Fx = maxT = mv2r = 55.0 kg (4.00 m/ s)20.800 m = 1100 N (b) The tension is larger than her weight by1100 N(55.0 kg)(9.80 m/ s2) = 2.04 times 6.3 m = 3.00 kg: r = 0.800 m. The string will break if the tension exceeds the weight corresponding to25.0 kg, soTmax = Mg = 25.0 9.80 = 245 NWhen the 3.00 kg mass rotates in a horizontal circle, the tension provides the centripetal force,soT = mv2r = (3.00)v20.800 Then v2 = rTm = 0.800T3.00 (0.800Tmax)3.00 = 0.800 2453.00 = 65.3 m2/ s2and 0 < v < 65.3 or 0 < v < 8.08 m/ s 2000 by Harcourt College Publishers. All rights reserved.Goal Solution The string will break if the tension T exceeds the test weight it can support,Tmax = mg = (25.0 kg)(9.80 m/ s2) = 245 NAs the 3.00-kg mass rotates in a horizontal circle, the tension provides the central force.Chapter 6 Solutions 3 2000 by Harcourt College Publishers. All rights reserved.From F = ma, T = mv2r Then, v rTmaxm = (0.800 m)(245 N)(3.00 kg) = 8.08 m/ sSo the mass can have speeds between 0 and 8.08 m/ s. 6.4 ( a ) F = mv2r = (9.11 1031 kg)(2.20 106 m/ s)20.530 1010 m = 8.32 108 N inward(b) a = v2r = (2.20 106 m/ s)20.530 1010 m = 9.13 1022 m/ s2 inward6.5 Neglecting relativistic effects. F = mac = mv2r F = (2 1.661 1027 kg) (2.998 107 m/ s)2(0.480 m) = 6.22 1012 N 6.6 ( a ) We require that GmMer2 = mv2r but g = MeGR2e In this case r = 2Re, therefore, g4 = v22Re or v = gRe2 v = (9.80 m/ s2)(6.37 106 m)2 = 5.59 103 m/ s (b) T = 2rv = (2)(2)(6.37 106 m)5.59 103 m/ s = 239 min (c) F = GmMe(2Re)2 = mg4 = (300 kg)(9.80 m/ s2)4 = 735 N 6.7 The orbit radius is r = 1.70 106 m + 100 km = 1.80 106 m. Now using the information inExample 6.6,GMmmsr2 = ms222r2rT2 = msa( a ) a = GMmr2 = (6.67 1011)(7.40 1022)(1.80 106 m)2 = 1.52 m/ s2 (b) a = v2r , v = (1.52 m/ s2)(1.80 106 m) = 1.66 km/ s (c) v = 2rT , T = 2(1.80 106)1.66 103 = 6820 s 4 Chapter 6 Solutions 2000 by Harcourt College Publishers. All rights reserved.6.8 ( a ) Speed = distance/ time. If the radius of the hand of the clock is r thenv = 2rT vT = 2rrm = rs Tmvm = Tsvswhere vm = 1.75 103 m/ s, Tm = (60.0 60.0)s and Ts = 60.0 s.vs = ,_TmTs vm = ,_3.60 103 s60.0 s (1.75 103 m/ s) = 0.105 m/ s (b) v = 2rT for the second hand, r = vT2 = (0.105 m/ s)(60.0 s)2 = 1.00 mThen ar = v2r = (0.105 m/ s)21.00 m = 1.10 102 m/ s2 6.9 ( a ) st at ic fr ict ion (b) ma i = f i + n j + mg(j)Fy = 0 = n mgthus n = mg and Fr = m v2r = f = n = mgThen = v2rg = (50.0 cm/ s)2(30.0 cm)(980 cm/ s2) = 0.0850 *6.10 a = v2r =

]1,_86.5 kmh ,_1 h3600 s ,_1000 m1 km261.0 m ,_1 g9.80 m/ s2 = 0.966g 6.11 n = mg since ay = 0The centripetal force is the frictional force f.From Newton's second lawf = mar = mv2r nmgfarChapter 6 Solutions 5 2000 by Harcourt College Publishers. All rights reserved.But the friction condition isf s ni.e., mv2r s mgv s rg = (0.600)(35.0 m)(9.80 m/ s2) v 14.3 m/ s 6.12 (b) v = 235 m36.0 s = 6.53 m/ s The radius is given by 14 2 r = 235 mr = 150 m( a ) ar = ,_v2r toward center= (6.53 m/ s)2150 m at 35.0 north of west= (0.285 m/ s2)(cos 35.0( i) + sin 35.0 j)= 0.233 m/ s2 i + 0.163 m/ s2 j (c)a = (vf vi)t = (6.53 m/ s j 6.53 m/ s i)36.0 s = 0.181 m/ s2 i + 0.181 m/ s2 j 6 Chapter 6 Solutions 2000 by Harcourt College Publishers. All rights reserved.6.13 T cos 5.00 = mg = (80.0 kg)(9.80 m/ s2)( a ) T = 787 NT = (68.6 N)i + (784 N)j (b) T sin 5.00 = marar = 0.857 m/ s2 6.14 ( a ) The reaction force n1 representsthe apparent weight of the womanF = mai.e., mg n1 = mv2r , so n1 = mg mv2r n1 = 600 ,_6009.80 (9.00)211.0 = 149 N (b) If n1 = 0, mg = mv2r This gives v = rg = (11.0 m)(9.80 m/ s2) = 10.4 m/ s xymgT5.00vn1amgChapter 6 Solutions 7 2000 by Harcourt College Publishers. All rights reserved.6.15 Let the tension at the lowest point be T.F = maT mg = mar = mv2r T = m ,_g + v2r = (85.0 kg)

]19.80 m/ s2 + (8.00 m/ s)210.0 m = 1.38 kN > 1000 NHe doesn' t make it across the river because the vine breaks. Tmgar6.16 ( a ) ar = v2r = (4.00 m/ s)212.0 m = 1.33 m/ s2 (b) a = a2r + a2T a = (1.33)2 + (1.20)2 = 1.79 m/ s2 at an angle = tan-1 ,_araT = 47.9 inward 6.17 M = 40.0 kg, R = 3.00 m, T = 350 N( a ) F = 2T Mg = Mv2R v2 = (2T Mg) ,_RM v2 = [700 (40.0)(9.80)] ,_3.0040.0 = 23.1(m2/ s2)v = 4.81 m/ s vr 12.0 mT TMgchild + seat8 Chapter 6 Solutions 2000 by Harcourt College Publishers. All rights reserved.(b) n Mg = F = Mv2R n = Mg + Mv2R = 40.0 ,_9.80 + 23.13.00 = 700 N Goal Solution G: If the tension in each chain is 350 N at the lowest point, then the force of the seat on the childshould just be twice this force or 700 N. The childs speed is not as easy to determine, butsomewhere between 0 and 10 m/ s would be reasonable for the situation described.O: We should first draw a free body diagram that shows the forces acting on the seat and applyNewtons laws to solve the problem.A: We can see from the diagram that the only forces acting on the system of child+seat are thetension in the two chains and the weight of the boy:F = 2T mg = ma where a = v2r is the centripetal accelerationF = Fnet = 2(350 N) (40.0 kg)(9.80 m/ s2) = 308 N upwardsv = Fmaxrm = (308 N)(3.00 m)40.0 kg = 4.81 m/ s The child feels a normal force exerted by the seat equal to the total tension in the chains.n = 2(350 N) = 700 N (upwards) L: Our answers agree with our predictions. It may seem strange that there is a net upward forceon the boy yet he does not move upwards. We must remember that a net force causes anacceleration, but not necessarily a motion in the direction of the force. In this case, theacceleration is due to a change in the direction of the motion. It is also interesting to note thatthe boy feels about twice as heavy as normal, so he is experiencing an acceleration of about2gs.6.18 ( a ) Consider the forces acting on the system consisting of the child and the seat:Fy = may 2T mg = m v2R v2 = R ,_2Tm g v = R ,_2Tm g nMgchildaloneChapter 6 Solutions 9 2000 by Harcourt College Publishers. All rights reserved.(b) Consider the forces acting on the child alone:Fy = may n = m ,_g + v2R and from above, v2 = R ,_2Tm g , son = m ,_g + 2Tm g = 2T 6.19 Fy = mv2r = mg + nBut n = 0 at this minimum speed condition, somv2r = mg v = gr = (9.80 m/ s2)(1.00 m) = 3.13 m/ s 6.20 At the top of the vertical circle,T = m v2R mgor T = (0.400) (4.00)20.500 (0.400)(9.80) = 8.88 N 6.21 ( a ) v = 20.0 m/ s, n = force of track on roller coaster, and R = 10.0 m.F = Mv2R = n MgFrom this we findn = Mg + Mv2R = (500 kg)(9.80 m/ s2) + (500 kg)(20.0 m/ s2)10.0 m n = 4900 N + 20,000 N = 2.49 104 N ABCr10 m15 m10 Chapter 6 Solutions 2000 by Harcourt College Publishers. All rights reserved.(b) At B, n Mg = Mv2R The max speed at B corresponds ton = 0Mg = Mv2maxR vmax = Rg = 15.0(9.80) = 12.1 m/ s 6.22 ( a ) ar = v2r r = v2ar = (13.0 m/ s)22(9.80 m/ s2) = 8.62 m (b) Let n be the force exerted by rail.Newton's law gives Mg + n = Mv2r n = M ,_v2r g = M(2g g) = Mg, downward (c) ar = v2r = (13.0 m/ s)220.0 m = 8.45 m/ s2 If the force by the rail is n1, thenn1 + Mg = Mv2r = Marn1 = M(ar g) which is < 0,since ar = 8.45 m/ s2Thus, the normal force would have to point away from the center of the curve. Unlessthey have belts, the riders will fall from the cars. To be safe we must require n1 to bepositive. Then ar > g. We needv2r > g or v > rg = (20.0 m)(9.80 m/ s2) v > 14.0 m/ s6.23 v = 2rT = 2(3.00 m)(12.0 s) = 1.57 m/ s( a ) a = v2r = (1.57 m/ s)2(3.00 m) = 0.822 m/ s2 Chapter 6 Solutions 11 2000 by Harcourt College Publishers. All rights reserved.(b) For no sliding motion,ff = ma = (45.0 kg)(0.822 m/ s2) = 37.0 N (c) ff = mg, = 37.0 N(45.0 kg)(9.80 m/ s2) = 0.0839 6.24 ( a ) Fx = Ma, a = TM = 18.0 N5.00 kg = 3.60 m/ s2 to the right.(b) If v = const, a = 0, so T = 0 (This is also an equilibrium situation.) (c) Someone in the car (noninertial observer) claims that the forces on the mass along x are Tand a fictitious force (Ma). Someone at rest outside the car (inertial observer) claimsthat T is the only force on M in the x-direction.5.00 kg6.25 Fx = T sin = max(1)Fx = T cos mg = 0or T cos = mg (2)( a ) Dividing (1) by (2) gives tan = axg = tan1 ,_axg = tan1 ,_3.009.80 = 17.0 (b) From (1), T = maxsin = (0.500 kg)(3.00 m/ s2)sin 17.0 = 5.12 N 12 Chapter 6 Solutions 2000 by Harcourt College Publishers. All rights reserved.Goal Solution G: If the horizontal acceleration were zero, then the angle would be 0, and if a = g, then theangle would be 45, but since the acceleration is 3.00 m/ s2, a reasonable estimate of the angleis about 20. Similarly, the tension in the string should be slightly more than the weight ofthe object, which is about 5 N.O: We will apply Newtons second law to solve the problem.A: The only forces acting on the suspended object are the force of gravity mg and the force oftension T, as shown in the free-body diagram. Applying Newton's second law in the x and ydirections,mgT cos T sinFx = T sin = ma(1)Fy = T cos mg = 0or T cos = mg (2)( a ) Dividing equation (1) by (2) givestan = ag = 3.00 m/ s29.80 m/ s2 = 0.306Solving for , = 17.0(b) From Equation (1),T = m asin = (0.500 kg)(3.00 m/ s2)sin (17.0) = 5.12 NL: Our answers agree with our original estimates. This problem is very similar to Prob. 5.30, sothe same concept seems to a