MATH 346 HW 4 (Due Oct 7, Wed)
1. Find 200 + 203 + 206 + 209 + . . . + 299.
2. The force of interest of a fund is t =
2t
,
8+t2
t 0. Find an and sn .
3. Suppose that the annual eective interest rate f
SP-6
Chapter 6
14, 25, 33, 50, 60, 65, 95
_
_
Problem 6-14
(My solution)
F = 0.5 mg,
= 20
Suppose that the acceleration a of the block of mass m is equal to zero.
N + F sin = mg
F cos f = ma
f s N
or
SP-7
Chapter 7
23, 32, 48, 65, 71, 77, 80
_
Problem 7-23
= 50,
Fr = 50 N,
K = 80 J.
K = Wnc + Wc
or
(Note)
Wc = mgd sin = mgh
Wnc = Fr d
K = W = Wc + Wnc = mgd sin Fr d
U = Wc = mgd sin
or
K = U +
SP-8
Chapter 8, 36, 41, 64, 79, 83, 94
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_
Problem 8-8
(WileyPlus)
8. We use Eq. 7-12 for Wg and Eq. 8-9 for U.
(a) The displacement between the initial point and Q has a vertical component of h R
dow
SP-10
Chapter 10 17, 38, 52, 66, 79, 82, 93
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_
Problem 10-17
(My solution)
i = 1.5rad / s
f = 0
= 40rev = 40 2 = 80
We use the following equations.
f i = 2 ( f i ) = 2
2
2
f = i t
1
f i = = it t
SP-11
Chapter 11
11, 13, 17, 28, 57, 61, 93
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_
Problem 11-11
(My solution)
h=2m
,
H = 6 m.
Energy conservation law
Then we have
or
where
Equations of motion:
When y = 0,
and
(WileyPlus)
11. To find w
SP-12
Chapter 12
10, 19, 24, 29, 37, 68, 74
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_
Problem 12-10
(My solution)
F
y
= F 2T sin = 0
tan =
0.3
9
or
F
= 8.25 103 N
2 sin
= 1.91
T=
(WileyPlus)
10. The angle of each half of the rope, measu
SP-13
Chapter 13
11, 16, 23, 39, 74, 94, 99
_
Problem 13-11
(My solution)
2r sin(60) = a
or
Gmm4
r2
GMm4
F3 =
r2
F1 = F2 =
r=
a
3
(a)
Ftot = F1 + F2 + F3 = 0
From the symmetry, it is evident that F1x
SP-14
Chapter 14
29, 36, 43, 65, 67, 70, 77
_
_
Problem 14-29
(My solution)
k = 3.0x104 N/m,
x = 0.05 m,
Ao =18 Ai
Pascals principle
F
Pi = i
Ai
Po =
Fo
Ao
Pi = Pf
For the force on the spring,
Fo = kx
SP-15
Chapter 15
22, 26, 35, 45, 51, 60, 63
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_
Problem 15-22
(My solution)
m = 2.0 kg,
k
m
v0 = 3.415 m/s at t = 1.0 s.
k = 100 N/m,
x0 = 0.129 m,
0 =
(a)
E=
121212
mv + kx = kA
2
2
2
or
A = x2 +
(b)
SP-18
Chapter 18
29, 42, 56, 61, 73, 82, 95
_
Problem 18-29
(Solution)
Ls = 2256 x 103 J/kg (latent heat of evaporation)
Cw = 4180 J/kg K (heat capacity of water)
LF = 333 x 103 J/kg (latent heat of f
SP-5
Chapter 5
12, 35, 48, 54, 59, 66, 99
_
_
Problem 5-12
(WileyPlus)
From the slope of the graph we find ax = 3.0 m/s2. Applying Newtons second law to the
x axis (and taking to be the angle between
SP-4
Chapter 4
15, 31, 46, 66, 70, 73, 76
_
_
Problem 4-15
(WileyPlus)
We find t by applying Eq. 2-11 to motion along the y axis (with vy = 0 characterizing y =
ymax ):
0 = (12 m/s) + (2.0 m/s2)t t =
Homework solutions
HW-A
Chapter 1
Chapter 2
28,40,45
21, 40, 50
_
_
Problem 1-28
(My solution)
The radius and density of the sand are given by
r = 50 m = 50 106 m
= 2600kg / m3
Suppose that there are
HW-B
Chapter-3:
Chapter-4:
35, 40, 43
26, 38, 56
_
_
Problem 3-35
(WileyPlus)
We apply Eq. 3-30 and Eq.3-23. If a vector-capable calculator is used, this makes a good
exercise for getting familiar wit
HW-C
Chapter 5
Chapter 6
45, 51, 55
29, 45, 55
_
Problem 5-45
(My solution)
m = 0.1 kg, a = 2.50 m/s2.
T1 mg = ma
T2 T1 mg = ma
T3 T2 mg = ma
T4 T3 mg = ma
F T4 mg = ma
From these equations we have
F
HW-D
Chapter 7
51, 57, 60
Chapter 8
21, 63, 72
_
Problem 7-51
(WileyPlus)
51. (a) The objects displacement is
rrr
d = d f di = (8.00 m) + (6.00 m) + (2.00 m) k .
i
j
Thus, Eq. 7-8 gives
rr
W = F = (3.
HW-E
Chapter 9
Chapter 10
35, 45, 54
48, 63, 69
_
Problem 9-35
(WileyPlus)
35. (a) We take the force to be in the positive direction, at least for earlier times. Then the
impulse is
J =
3.0 10 3
0
Fdt
HW-G
Chapter 13
Chapter 14
24, 41, 65
31, 41, 62
_
_
Problem 13-24
(WileyPlus)
24. (a) What contributes to the GmM/r2 force on m is the (spherically distributed) mass M
contained within r (where r is
HW-H
Chapter 15
Chapter 18
43, 49, 53
34, 58, 65
_
Problem 15-43
(WileyPlus)
2
43. (a) A uniform disk pivoted at its center has a rotational inertia of 1 Mr , where M is
2
its mass and r is its radius
HW-I
Chapter 19
Chapter 20
23, 41, 76
18, 28, 32
_
_
Problem 19-23
(My solution)
A = 2.0 cm2 = 2.0 x 10-4 m2
m = 3.3 x 10-27 kg,
N0 = 1023 moles/s,
v = 1.0 x 103 m/s =
N 0 t is the number of particles
Selected problems
SP-1
7, 13, 23, 25, 29, 30,44
_
_
Problem 1-7
(WileyPlus)
The volume of ice is given by the product of the semicircular surface area and the
thickness. The area of the semicircle is
SP-3
Chapter 3
23, 28, 30, 39, 41, 44, 55
_
Problem 3-23
(WileyPlus)
The strategy is to find where the camel is () by adding the two consecutive displacements
described in the problem, and then findin