Giancoli 5th Ed. Hwk 6 (Due 3/6/23) 10 point hwk Ch. 7 P. 1,2,5,8,12,18,20,22,25,37,39,41,45,55,56,59,60,65 MisConQ. 1-13 (odd) MisConQ 1. d 3. e 5. d 7. c 9. b 11. b 13. d Problems: 1. W=mgdcos(theta) = 10,000 J 2. W=mgdcos(theta) -> d = 3.86 m 5. Wpush = Fpush d cos(theta) = u_k mg d(1) = 1900 J 8. Wp = work by forcd parallel to the plane Wp = F d cos(0) = mgd sin (theta) = 950*9.8*510*sin(9)=740,000 J 12. box moves distance d=x-x_0=v_0 t +at^2/2 = 64 m W = Fdcos(0) = m*a*d = 4(2.0)(64)=510 J 18. A dot B = AxBx + AyBy + AzBz = 22x^2 - 10x^2 = 12x^2 20. A dot B = ABcos(theta) so theta = cos^-1 ((A dot B) / AB) = cos^-1 (76.34/(9.145*11.607))=44 deg 22. V1 dot V2 = V1V2cos(theta) = (75)(48)cos(138) = -2700 25. a) A dot (B + C) = 7 b) (A + C) dot B = -230 c) (B + A) dot C = 28 37. The plot of the force exerted to stretch the spring, F_stretch, versus x will be a straight line with slope +k and y-intercept 0. The work from x1 to x2 is the area of a trapezoid = 1/2(H1+H2)(Width) W = 1/2 (kx_1 + kx_2)(x_2-x_1) or 1/2 k (x_2^2 - x_1^2) = 0.154 J 39. The plot of F vs x begins at (0,0), follows a straight line to (3,380), follows a horizontal line to (7,380), and then down to (12,0). W = 1/2(12.0m + 4.0 m)(380 N) = 3040 J 41. a) W = 2800 J for displacement from x=0 to 10 m b) W = 2100 J for displacement from x=0 to 15 m 45. W = int (A x^(-1/2) dx integrated from x=0 to 1 m. = 2 A x^(1/2) from x=0 to 1 m. = 2(3)(1.0)^(1/2) = 6.0 J 55. K = 1/2 mv^2 so v = sqrt(2K/m) and v ~ sqrt(k) So if the K is tripled, the v is multiplied by sqrt(3) ~ 1.7 56. Work done on the electron equals change in K.E., by the work-KE principle. W = 1/2 m (v_f^2 - v_i^2) = -5.51E-10 J 59. F_on glove = 74.24 J / 0.22 m = 337 N in the direction of the original velocity of the ball. 60. Fd = W = K_f - K_i = 1/2 m(v_f^2-v_i^2) v_2 = sqrt(2Fd/m + v_1^2) = 43 m/s 65. W = K_f - K_i = -1/2 m v_i^2 but also W = Fdcos(theta) = -mu_k mgd, so -1/2 m v_i^2 = -mu_k mgd. Solve for mu_k = 0.16. Please check Tot= 10 pts Prob. 20, 45 (1 pt, 1 pt) MiscQ. 1,3,5--- (0.5 pt each)