13. (a) From the force diagram, we write Sum(F) = ma: y-component: F_{N}- mg cosq = 0; x-component: - F - m_{k}F_{N}+ mg sinq = 0. Thus we have F = - m_{k}F_{N}+ mg sinq = - m_{k}mg cosq + mg sinq = mg (sinq - m_{k}cosq) = (380 kg)(9.80 m/s^{2})(sin 27° - 0.40 cos 27°) = 3.6x10^{2}N. (b) Because the piano is sliding down the incline, we have W_{F}= F d cos 180° = (3.6x10^{2}J)(3.5 m)(- 1) = - 1.3x10^{3}J. (c) For the friction force, we have W_{fr}= m_{k}mg cosq d cos 180° = (0.40)(380 kg)(9.80 m/s^{2})(cos 27°)(3.5 m)(- 1) = - 4.6x10^{3}J. (d) For the force of gravity, we have W_{grav}= mg d cos 63° = (380 kg)(9.80 m/s^{2})(3.5 m)(cos 63°) = 5.9x10^{3}J. (e) Because the normal force does no work, we have W_{net}= W_{grav}+ W_{F}+ W_{fr}+ W_{N}= 5.9x10^{3}J - 1.3 x10^{3}J - 4.6x10^{3}J + 0 = 0.

21. (a) A · (B + C) = (7.0 i - 8.5 j) · [(- 8.0 + 6.8) i + (8.1 - 7.2) j + 4.2 k] = (7.0)(- 1.2) + (- 8.5)(0.9) = - 16.1. (b) (A + C) · B = [(7.0 + 6.8) i + (- 8.5 - 7.2) j + 0 k] · (- 8.0 i + 8.1 j + 4.2 k) = (13.8)(- 8.0) + (- 15.7)(8.1) + 0 = - 238. (c) (B + A) · C = [(- 8.0 + 7.0) i + (8.1 - 8.5) j + (4.2 + 0) k] · (6.8 i - 7.2 j) = (- 1.0)(6.8) + (- 0.4)(- 7.2) + (4.2)(0) = - 3.9.

23. Because C lies in the xy-plane and is perpendicular to B, we have B · C = BxCx + ByCy = 9.6Cx + 6.7Cy = 0. For the scalar product of A and C we have A · C = AxCx + AyCy = - 4.8Cx + 7.8Cy = 20.0. When we solve these two equations for the two unknowns, we get Cx = -1.25, and Cy = 1.79. Thus C = - 1.3i + 1.8j.

34. The work done in moving the object is the area under the Fx vs. x graph. (a) For the motion from 0.0 m to 10.0 m, we find the area of two triangles and one rectangle: W = 1/2(400 N)(3.0 m - 0.0 m) + (400 N)(7.0 m - 3.0 m) + 1/2(400 N)(10.0 m - 7.0 m) = 2.8x10^{3}J. (b) For the motion from 0.0 m to 15.0 m, we add the negative area of two triangles and one rectangle: W = 2.8x10^{3}J - 1/2(200 N)(11.7 m - 10.0 m) - (200 N)(13.7 m - 11.7 m) - 1/2(200 N)(15.0 m - 13.7 m) = 2.1x10^{3}J.

52. Because the force F holds the spring a distance x, the spring constant is k = F/x. The work done by the spring changes the kinetic energy of the mass: W_{spring}= DK; - (1/2kx_{f}^{2}- 1/2kx_{i}^{2}) = 1/2mv_{f}^{2}- 1/2mv_{i}^{2}. (a) When the spring returns to xf = 0, we have - (0 - 1/2kx^{2}) = 1/2mv_{f}^{2}- 0, or v_{f}= (k/m)^{1/2}x = (Fx/m)^{1/2}. (b) When the spring returns to x_{f}= 1/2x, we have - [1/2k(1/2x)^{2}- 1/2kx^{2}] = 1/2mv_{f}^{2}- 0, or v_{f}= (3k/4m)^{1/2}x = (3Fx/4m)^{1/2}.

55. (a) For the work done by the applied force we have W_{P}= F_{P}d cosq = (150 N) cos 30° (5.0 m) = 6.5x10^{2}J. (b) For the force of gravity, we have W_{grav}= (- mg sinq )d = - (20 kg)(9.80 m/s^{2})(5.0 m) sin 30° = - 4.9x10^{2}J. (c) Because the normal force is perpendicular to the displacement, it does no work: W_{N}= 0. (d) The net work done on the block increases its kinetic energy: W_{net}= W_{P}+ W_{grav}+ W_{N}= 1/2mv^{2}- 1/2mv_{0}^{2}; 6.5x10^{2}J - 4.9x10^{2}J + 0 = 1/2(20 kg)v^{2}- 0, which gives v = 4.0 m/s.

15. Because the motion is in the x-direction, we see that the weight and normal forces do no work: W_{F}_{N}= W_{mg}= 0. From the force diagram, we write Sum(F) = ma: x-component: F_{P}cosq - F_{fr}= 0, or F_{fr}= F_{P}cosq. For the work by these two forces, we have W_{FP}= F_{P}Dx cosq = (14 N)(15 m) cos 20° = 2.0x10^{2}J. W_{fr}= F_{P}cosq Dx cos 180°= (14 N) cos 20° (15 m)(- 1) = - 2.0x10^{2}J. As expected, the total work is zero: W_{FP}= - W_{fr}= 2.0x10^{2}J.

24. For the sum and difference vectors, we have C_{sum}= A + B, C_{diff}= A - B. If we form the scalar product, we get C_{sum}· C_{diff}= (A + B) · (A - B) = A · A + B · A - A · B - B · B = A^{2}- B^{2}= 0, because A and B have the same magnitude. Because the scalar product is zero, the sum and difference vectors are perpendicular.

51. On the level the normal force is F_{N}= mg, so the friction force is F_{fr}= m_{k}mg. The normal force and the weight do no work. The net work increases the kinetic energy of the mass: W_{net}= DK = 1/2mv_{f}^{2}- 1/2mv_{i}^{2}; F(L_{1}+ L_{2}) - m_{k}mgL_{2}= 1/2mv_{f}^{2}- 0; (225 N)(11.0 m + 10.0 m) - (0.20)(66.0 kg)(9.80 m/s^{2})(10.0 m) = 1/2(66.0 kg)v_{f}^{2}, which gives v_{f}= 10.2 m/s.

53. There will be an additional (negative) work done by the friction force. The net work increases the kinetic energy of the mass: - (1/2kx_{f}^{2}- 1/2kx_{i}^{2}) - m_{k}mg Dx = 1/2mv_{f}^{2}- 1/2mv_{i}^{2}; - (0 - 1/2kx^{2}) - m_{k}mgx = 0 - 0, which gives m_{k}= kx/2mg = F/2mg.

56. (a) The work done by the applied force is the same: W_{P}= 6.5x10^{2}J. (b) The work done by gravity is the same: W_{grav}= - 4.9x10^{2}J. (c) The work done by the normal force is the same: W_{N}= 0. (d) From the force diagram, we write Sum(F) = ma: y-component: F_{N}- mg cosq - F_{P}sinq = 0, or F_{N}= mg cosq + F_{P}sinq = (20 kg)(9.80 m/s^{2}) cos 30° + (150 N) sin 30° = 245 N. The friction force is F_{fr}= m_{k}F_{N}, so the work done by friction is W_{fr}= - m_{k}F_{N}d = - (0.10)(245 N)(5.0 m) = - 1.2x10^{2}J. The net work done on the block increases its kinetic energy: W_{net}= W_{P}+ W_{grav}+ W_{N}+ W_{fr}= 1/2mv^{2}- 1/2mv_{0}^{2}; 6.5x10^{2}J - 4.9x10^{2}J + 0 - 1.2x10^{2}J = 1/2(20 kg)v^{2}- 0, which gives v = 1.9 m/s.

67. The work done by the force is W = F · d = Fx dx + Fy dy + Fz dz = (10.0 kN)(5.0 m) + (9.0 kN)(4.0 m) + (12.0 kN)(0) = 86 kJ. If we use the other expression for the scalar product, we have W = F · d = Fd cosq; 86 kJ = [(10.0 kN)^{2}+ (9.0 kN)^{2}+ (12.0 kN)^{2}]^{1/2}[(5.0 m)^{2}+ (4.0 m)^{2}]^{1/2}cosq. which gives cosq = 0.746, q = 42°.

38. If we let y be the length of chain hanging over the edge and use m for linear weight density, the weight hanging is w = my. As the next differential length dy comes over the edge, the weight w will fall a distance dy, during which the force of gravity will do work dW = w dy. If L_{1}is the initial hanging length and L is the total length of the chain, we integrate this from L_{1}to L:

74. (a) The net work decreases the kinetic energy: W_{snow}+ W_{grav}= W_{snow}+ mgd = DK = 1/2mv_{f}^{2}- 1/2mv_{i}^{2}; W_{snow}+ (80 kg)(9.80 m/s^{2})(1.1 m) = 1/2(80 kg)[0 - (50 m/s)^{2}] = - 1.0x10^{5}J. (b) We find the average force from F = W/d = (- 1.0x10^{5}J)/(1.1 m) = - 9.1x10^{4}N. (c) With air resistance during the fall we have W_{air}+ W_{grav}= W_{air}+ mgh = DK= (1/2mv_{f}^{2}- 1/2mv_{i}^{2}) W_{air}+ (80 kg)(9.80 m/s^{2})(370 m) = 1/2(80 kg)[(50 m/s)^{2}- 0], which gives W_{air}= - 1.9x10^{5}J.