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### Transcript of Solucionario Raymond Serway - Capitulo 7

2000 by Harcourt College Publishers. All rights reserved.Chapter 7 Solutions*7.1 W = Fd = (5000 N)(3.00 km) = 15.0 MJ *7.2 The component of force along the direction of motion isF cos = (35.0 N) cos 25.0 = 31.7 NThe work done by this force isW = (F cos )d = (31.7 N)(50.0 m) = 1.59 103 J 7.3 ( a ) W = mgh = (3.35 105)(9.80)(100) J = 3.28 102 J (b) Since R = mg, Wair resistance = 3.28 102 J 7.4 ( a ) Fy = F sin + n mg = 0n = mg F sin Fx = F cos kn = 0n = F cos k mg F sin = F cos k F = kmgk sin + cos F = (0.500)(18.0)(9.80)0.500 sin 20.0 + cos 20.0 = 79.4 N (b) WF = Fd cos = (79.4 N)(20.0 m) cos 20.0 = 1.49 kJ (c) fk = F cos = 74.6 NWf = fk d cos = (74.6 N)(20.0 m) cos 180 = 1.49 kJ 7.5 ( a ) W = Fd cos = (16.0 N)(2.20 m) cos 25.0 = 31.9 J (b) and (c) The normal force and the weight are both at 90 to the motion. Both do 0 work.(d ) W = 31.9 J + 0 + 0 = 31.9 J 20.0 mgnFd 20.0 m18.0 kgfR = kn 2 Chapter 7 Solutions 2000 by Harcourt College Publishers. All rights reserved.7.6 Fy = mayn + (70.0 N) sin 20.0 147 N = 0n = 123 Nfk = kn = 0.300 (123 N) = 36.9 N( a ) W = Fd cos = (70.0 N)(5.00 m) cos 20.0 = 329 J (b) W = Fd cos = (123 N)(5.00 m) cos 90.0 = 0 J (c) W = Fd cos = (147 N)(5.00 m) cos 90.0 = 0 (d ) W = Fd cos = (36.9 N)(5.00 m) cos 180 = 185 J (e) K = Kf Ki = W = 329 J 185 J = +144 J 7.7 W = mg(y) = mg(l l cos )= (80.0 kg)(9.80 m/ s2)(12.0 m)(1 cos 60.0) = 4.70 kJ 60ly7.8 A = 5.00; B = 9.00; = 50.0A B = AB cos = (5.00)(9.00) cos 50.0 = 28.9 7.9 A B = AB cos = 7.00(4.00) cos (130 70.0) = 14.0 7.10 A B = (Axi + Ay j + Azk) (Bxi + By j + Bzk)A B = AxBx (i i) + AxBy (i j) + AxBz (i k) +AyBx (j i) + AyBy (j j) + AyBz (j k) +AzBx (k i) + AzBy (k j) + AzBz (k k)70.0 N sin 20.0 mg 147 Nnfk70.0 N cos 20.0d 5.00 m15.0 kgChapter 7 Solutions 3 2000 by Harcourt College Publishers. All rights reserved.A B = AxBx + AyBy + AzBz 4 Chapter 7 Solutions 2000 by Harcourt College Publishers. All rights reserved.7.11 ( a ) W = F d = Fxx + Fyy = (6.00)(3.00) N m + (2.00)(1.00) N m = 16.0 J (b) = cos1 F dFd = cos1 16[(6.00)2 + (2.00)2][(3.00)2 + (1.00)2] = 36.9 7.12 A B = (3.00i + j k) (i + 2.00j + 5.00k)A B = 4.00i j 6.00kC (A B) = (2.00j 3.00k) (4.00i j 6.00k) = 0 + (2.00) + (+18.0) = 16.0 7.13 ( a ) A = 3.00i 2.00j B = 4.00i 4.00j = cos1 A BA B = cos1 12.0 + 8.00(13.0)(32.0) = 11.3 (b) B = 3.00i 4.00j + 2.00k A = 2.00i + 4.00jcos = A BA B = 6.00 16.0(20.0)(29.0) = 156 (c) A = i 2.00j + 2.00k B = 3.00j + 4.00k = cos1 ,_A BA B = cos1 ,_ 6.00 + 8.009.00 25.0 = 82.3 *7.14 We must first find the angle between the two vectors. It is: = 360 118 90.0 132 = 20.0ThenF v = Fv cos = (32.8 N)(0.173 m/ s) cos 20.0orF v = 5.33 N ms = 5.33 Js = 5.33 W xy118132v 17.3 cm/ sF 32.8 NChapter 7 Solutions 5 2000 by Harcourt College Publishers. All rights reserved.7.15 W = if Fdx = area under curve from xi to xf( a ) xi = 0 xf = 8.00 mW = area of triangle ABC = ,_12 AC altitude,W08 = ,_12 8.00 m 6.00 N = 24.0 J (b) xi = 8.00 m xf = 10.0 mW = area of CDE = ,_12 CE altitude,W810 = ,_12 (2.00 m) (3 .00 N) = 3.00 J (c) W010 = W08 + W810 = 24.0 + (3.00) = 21.0 J *7.16 Fx = (8x 16) N( a )2010010203 2 1 4x (m)Fx (N)(3, 8)(b) Wnet = (2.00 m)(16.0 N)2 + (1.00 m)(8.00 N)2 = 12.0 J 7.17 W = Fx dx andW equals the area under the Force-Displacement Curve( a ) For the region 0 x 5.00 m,W = (3.00 N)(5.00 m)2 = 7.50 J (b) For the region 5.00 x 10.0,W = (3.00 N)(5.00 m) = 15.0 J 4202466 4 2 8 10 12x (m)Fx (N)ABC ED21036 4 2 0 8 10 12x (m)14 16Fx (N)6 Chapter 7 Solutions 2000 by Harcourt College Publishers. All rights reserved.(c) For the region 10.0 x 15.0,W = (3.00 N)(5.00 m)2 = 7.50 J (d ) For the region 0 x 15.0W = (7.50 + 7.50 + 15.0) J = 30.0 J 7.18 W = if F ds = 05 m (4xi + 3yj) N dxi05 m (4 N/ m) x dx + 0 = (4 N/ m)x2/ 2 5 m0 = 50.0 J *7.19 k = Fy = Mgy = (4.00)(9.80) N2.50 102 m = 1.57 103 N/ m( a ) For 1.50 kg mass y = mgk = (1.50)(9.80)1.57 103 = 0.938 cm (b) Work = 12 ky2Work = 12 (1.57 103 N m)(4.00 102 m) 2 = 1.25 J 7.20 ( a ) Spring constant is given by F = kxk = Fx = (230 N)(0.400 m) = 575 N/ m (b) Work = Favg x = 12 (230 N)(0.400 m) = 46.0 J 7.21 Compare an initial picture of the rolling car with a final picture with both springs compressedKi + W = Kf1500100050000 20 40 10 30 50d (cm)F (N)Chapter 7 Solutions 7 2000 by Harcourt College Publishers. All rights reserved.Use equation 7.11.Ki + 12 k1 (x21i x21f ) + 12 k2 (x22i x22f ) = Kf12 mv2i + 0 12 (1600 N/ m)(0.500 m) 2 + 0 12 (3400 N/ m)(0.200 m) 2 = 012 (6000 kg) v2i 200 J 68.0 J = 0vi = 2 268 J/ 6000 kg = 0.299 m/ s 7.22 ( a ) W = if F dsW = 00.600 m (15000 N + 10000 x N/ m 25000 x2 N/ m2) dx cos 0W = 15,000x + 10,000x22 25,000x33 0.6000 W = 9.00 kJ + 1.80 kJ 1.80 kJ = 9.00 kJ (b) Similar ly,W = (15.0 kN)(1.00 m) + (10.0 kN/ m)(1.00 m)22 (25.0 kN/ m2)(1.00 m)33 W = 11.7 kJ , larger by 29.6%7.23 4.00 J = 12 k(0.100 m)2 k = 800 N/ mand to stretch the spring to 0.200 m requiresW = 12 (800)(0.200) 2 4.00 J = 12.0 J Goal Solution G: We know that the force required to stretch a spring is proportional to the distance the springis stretched, and since the work required is proportional to the force and to the distance, thenW x2. This means if the extension of the spring is doubled, the work will increase by afactor of 4, so that for x = 20 cm, W = 16 J, requiring 12 J of additional work.O: Lets confirm our answer using Hookes law and the definition of work.8 Chapter 7 Solutions 2000 by Harcourt College Publishers. All rights reserved.A: The linear spring force relation is given by Hookes law: Fs = kxIntegrating with respect to x, we find the work done by the spring is:Ws = xxxy Fs dx = xxxy (kx)dx = 12 k (x2f x2i )However, we want the work done on the spring, which is W = Ws = 12 k(x2f x2i )We know the work for the first 10 cm, so we can find the force constant:k = 2W010x2010 = 2(4.00 J)(0.100 m)2 = 800 N/ mSubstituting for k, xi and xf, the extra work for the next step of extension isW = ,_12 (800 N/ m) [(0.200 m)2 (0.100 m)2] = 12.0 JL: Our calculated answer agrees with our prediction. It is helpful to remember that the forcerequired to stretch a spring is proportional to the distance the spring is extended, but the workis proportional to the square of the extension.7.24 W = 12 kd 2 k = 2Wd 2 W = 12 k(2d)2 12 kd2W = 32 kd2 = 3W 7.25 ( a ) The radius to the mass makes angle with the horizontal, so its weightmakes angle with the negative side ofthe x-axis, when we take the x-axis inthe direction of motion tangent to thecylinder.Fx = maxF mg cos = 0F = mg cos xmgnFRChapter 7 Solutions 9 2000 by Harcourt College Publishers. All rights reserved.(b) W = if F ds We use radian measure to express the next bit of displacement as ds = r d in terms of thenext bit of angle moved through:W = 0/ 2 mg cos Rd = mgR sin / 20 W = mgR (1 0) = mgR *7.26 [k] =

]1Fx = Nm = kg m/ s2m = kgs2 7.27 ( a ) KA = 12 (0.600 kg)(2.00 m/ s) 2 = 1.20 J (b)12 mv2B = KBvB = 2KBm = (2)(7.50)0.600 = 5.00 m/ s (c) W = K = KB KA = 12 m(v2B v2A )= 7.50 J 1.20 J = 6.30 J *7.28 ( a ) K = 12 mv2 = 12 (0.300 kg)(15 .0 m/ s) 2 = 33.8 J (b) K = 12 (0.300)(30.0) 2 = 12 (0.300)(15.0) 2 (4) = 4(33.8) = 135 J 7.29 vi = (6.00i 2.00j) m/ s( a ) vi = v2i x + v2iy = 40.0 m/ sKi = 12 mv2i = 12 (3.00 kg)(40.0 m2/ s2) = 60.0 J (b) v = 8.00i + 4.00jv2 = v v = 64.0 + 16.0 = 80.0 m2/ s2K = K Ki = 12 m(v2 v2i ) = 3.002 (80.0) 60.0 = 60.0 J 10 Chapter 7 Solutions 2000 by Harcourt College Publishers. All rights reserved.7.30 ( a ) K = W12 (2500 kg) v2 = 5000 Jv = 2.00 m/ s (b) W = F d5000 J = F(25.0 m)F = 200 N 7.31 ( a ) K = 12 mv2 0 = W, sov2 = 2W/ m and v = 2W/ m (b) W = F d = Fxd Fx = W/ d 7.32 ( a ) K = Kf Ki = 12 mv2f 0 = W = (area under curve from x = 0 to x = 5.00 m)vf = 2(ar ea)m = 2(7.50 J)4.00 kg = 1.94 m/ s (b) K = Kf Ki = 12 mv2f 0 = W = (area under curve from x = 0 to x = 10.0 m)vf = 2(ar ea)m = 2(22.5 J)4.00 kg = 3.35 m/ s (c) K = Kf Ki = 12 mv2f 0 = W = (area under curve from x = 0 to x = 15.0 m)vf = 2(ar ea)m = 2(30.0 J)4.00 kg = 3.87 m/ s *7.33 Fy = mayn 392 N = 0 n = 392 Nfk = kn = 0.300(392 N) = 118 N( a ) WF = Fd cos = (130)(5.00) cos 0 = 650 J (b) Wfk = fk d cos = (118)(5.00) cos 180 = 588 J mg 392 NnfkF 130 Nd 5.00 m40.0 kgChapter 7 Solutions 11 2000 by Harcourt College Publishers. All rights reserved.(c) Wn = nd cos = (392)(5.00) cos 90 = 0 (d ) Wg = mg cos = (392)(5.00) cos (90) = 0 (e) K = Kf Ki = W12 mv2f 0 = 650 J 588 J + 0 + 0 = 62.0 J ( f ) vf = 2Kfm = 2(62.0 J)40.0 kg = 1.76 m/ s 7.34 ( a ) Ki + W = Kf = 12 mv2f 0 + W = 12 (15.0 103 kg)(780 m/ s) 2 = 4.56 kJ (b) F = Wd cos = 4.56 103 J(0.720 m) cos 0 = 6.34 kN (c) a = v2f v2i2x = (780 m/ s)2 02(0.720 m) = 422 km/ s2 (d ) F = ma = (15 103 kg)(422 103 m/ s2) = 6.34 kN 7.35 ( a ) Wg = mgl cos (90.0 + ) = (10.0 kg)(9.80 m/ s2)(5.00 m) cos 110 = 168 J (b) fk = kn = kmg cos Wf = lfk = lkmg cos cos 180Wf = (5.00 m)(0.400)(10.0)(9.80) cos 20.0 = 184 J (c) WF = Fl = (100)(5.00) = 500 J (d ) K = W = WF + Wf + Wg = 148 J (e) K = 12 mv2f 12 mv2i vf = 2(K)m + v2i = 2(148)10.0 + (1.50)2 = 5.65 m/ s F 100 Nvil12 Chapter 7 Solutions 2000 by Harcourt College Publishers. All rights reserved.7.36 W = K = 00L mg sin 35.0 dl 0d kx dx = 0mg sin 35.0 (L) = 12 kd2d = 2 mg sin 35.0(L)k = 2(12.0 kg)(9.80 m/ s2) sin 35.0 (3.00 m)3.00 104 N/ m = 0.116 m 7.37 vi = 2.00 m/ s k = 0.100W = K fk x = 0 12 mv2i kmgx = 12 mv2i x = v2i2k g = (2.00 m/ s)2