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)