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 for year n is in =
Find a10 and s10 .
1
.
5+n
4. You are
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
s
F cos
= 0.567 s
mg F sin
(a) a = 0 since s = 0.6
(
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 + Wnc
( K + U ) = E = E f Ei = Wnc
Ei = mgh
Ef =
(1)
12
m
SP-8
Chapter 8, 36, 41, 64, 79, 83, 94
_
_
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
downward (same direction as Fg ), so (with h = 5R) we obta
SP-9
Chapter 9
14, 20, 30, 37, 49, 68, 72
_
_
Problem 9-14
(My solution)
m1 = 5.00g,
m2 = 3.00 g,
v1 = (v1 ,0),
v 2 = (v 2 cos , v 2 sin gt )
v1 = 10.0 m/s,
v2 = 20 m/s
The relation between v1 and v2 are given by
cos =
v2 cos = v1 ,
or
v1 1
=
v2 2
= 600
SP-10
Chapter 10 17, 38, 52, 66, 79, 82, 93
_
_
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 2
2
(b)
=
f 2 i 2 = 0 i 2 = 2
or
i
1.52
=
= 4.48 10 3
SP-11
Chapter 11
11, 13, 17, 28, 57, 61, 93
_
_
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 where the ball lands, we need to know its speed as it le
SP-12
Chapter 12
10, 19, 24, 29, 37, 68, 74
_
_
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, measured from the dashed line, is
0.30 m
= tan 1
= 1.9.
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 + F2 x + F3 x = 0 . From the condition that
F1 y + F2 y
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
Then we have
Fi =
Ai
kx = 83.3 N
Af
(WileyPlus)
29. Eq
SP-15
Chapter 15
22, 26, 35, 45, 51, 60, 63
_
_
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)
m2
m2
2
v = x0 + v0 = 0.5m
k
k
x = A cos(0t + 0 )
dx
=
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 fusion)
1 cal = 4.1868 J
(WileyPlus)
29. Let the mass of
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 F1 and F2), we have
F1 + F2 cos = m ax
= 56.
_
Problem
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 = 6.0 s.
Then, Eq. 2-11 applies to motion along the x axi
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 N particles on the surface A (= surface are of a cube
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 with those features. Here we briefly sketch the method.
rr
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 = 5(m + a) = 6.15 N
T1 = m(a + g ) = 1.23 N
T2 = 2m(a +
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 = (3.00 N)(8.00 m) + (7.00 N)(6.00 m) + (7.00 N)(2.00 m) = 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 =
3.0 10 3
0
(6.0 106 ) t (2.0 109 )t 2 dt
1
1
= (6.0
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 measured from the center of M). At point A we see that
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. The disk of this problem rotates about a point that i
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 striking the wall over an area during the time t
The c
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 A = r2/2, where r is the radius. Therefore, the
volume
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 finding the difference between that location and the
oasis ()