Homework parts I and II

Pages 130-136

100 points for part I and 100 points for part II

 

4.1 and 4.2: Force/Net Force. Inertia and 1st Law

1-7, 10,12,14

Helpful Tips:

6: Try this. We will try it in class with low friction carts and a bubble level

12: Find x and y components of F1 and F2 and add them together. F3 will be opposite in magnitude to these components.

14: for part b) figure out a force, call it F5. When you add F5 to F1-F4 the total force should be zero Newtons.

 

4.3 Newton’s Second Law

15, 17, 20, 21, 25 – 28, 30, 34

Helpful Tips:

17: “Soft” means catching the ball and allowing your arms to continue in the path of the ball. This increases contact time with the ball and thereby reduces the force.

20: You are given Force, you are given acc. solve for the mass of the crate

26: Look on your conversion sheet for pounds. Pounds is a FORCE, not a mass!

 

4.4 Newton’s 3rd Law

37-39, 43,44

Helpful tips: 37: This point is really important to understand!

43: Realize the force he pushes on her is equal to the force she pushes on him. So, since she has a lower mass, her acceleration will be greater than his. This problem we will do in class as the “skateboard lab”. FUN!

44: For part b) Xjohn = 1/2ajohnt2 and Xjane = 1/2ajanet2 ALSO, Xjohn + Xjane = 10m. Therefore, if you find a ration between the accelerations of these two, you will find a ratio between their displacement as well. (You should find a ratio of 1.2ajohn = 1ajane). So, Xjohn + 1.2Xjohn = 10 m. Solve for X which will give you displacement relative to John.

 

------------------------------End of Part I---------------------------------------------------

 

4.5 Free-Body Diagrams

45,46,48,50, 51, 52, 53, 56, 58. Atwood machine problems: 60, 61, 64, 65

Helpful tips:

48: Part a) Net force = 2(600N Cos45). Part b) Net force = 2(100N Cos45). Remember the skier has a mass of 75.0 Kg, solve for a.

50: see example problem 4.11 p. 125

52: (the symbol “S” means the sum of)for part a) SFx = mgsinq = ma. Since the m’s will cancel, solve for a. For part b) V2 = Vo2 + 2ax

53: Just like 52. First solve for a. Then, when using the second equation, V = 0m/s

58: see example 4.3 p. 106

 

4.6 Friction

66-71, 74, 78, 82, 83, 85, 86, 88, 89, 90

Helpful Tips:

70: Try it to prove it to yourself – besides, your parents will thank you for mowing the lawn

74: see example 4.10 p. 124. Assume force is in the horizontal direction only

78: see table 4.1 coefficients of friction on p. 123

82: see example 4.12 on p. 126

90:  Re-draw the blocks so that they are in a horizontal line.  Notice from their original position that m2 will be pulling to the right while m1 will be pulling to the left.  Also, m3 will not be pulling either direction as it is being pulled by m1 and m2.

 

Even Answers: 

2)  c

4)  c

6)  a) fwd

10)  zero

12)  F3 = 2.5N at 36o above the +x axis; no, only a =0

14)  a.  no    b.  F5 = 4.1N at 13o above –x axis

20)  1.5 x 103 N

26)  668 N, 68.2 Kg

28)  a.  1.10 x 103 Kg    b.  2.43 x 103 lb

30)  7.0 m/s2

34)  2.7 x 102 N

38)  d

44)  a.  0.77 m/s2 toward Jane    b.  4.5 m from John’s original position

46)  a.  735 N   b.  735 N    c.  885 N

48)  a.  849 N    b.  1.88 m/s2

50)  a.  0.72 m/s2    b.  2.8 x 102N

52)  a.  5.9 m/s2   b.  21 m/s

56)  a.  2.5 x 102 N   b.  3.5 x 102 N

58)  a.  3.0 m/s2   b.  T1 = 3.0N,  T2 = 9.0N

60)  1.1 m/s2 up

64)  1.2 Kg; 1.2 Kg

66)  c

68)  c

70)  pull

74)  a.  0.802    b.  0.569

78)  a.  38 m    b.  53 m

82)  yes, tanq = mk

86)  0.15

88)  a.  6.0 kg    b.  1.2 m/s2

90)  a.  0.179 kg    b.  0.862 m/s2