Homework parts
I and II
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.
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