Solucionario de raymond serway

(b) For the molecule ΣF = ma. Its weight is negligble.
Fwall on molecule = 4.68 × 10–26 kg (–4.47 × 1015 m/s2)
= –2.09 × 10–10 N
F –
molecule on wall = +2.09 × 10–10 N
5.11 (a) F = ma and v2f
= v2i
+ 2ax or a =
v2f
– v2i
2x
Therefore,
F = m
(v2f
– v2i
)
2x
F = (9.11 × 10–31 kg)
[(7.00 × 105 m/s2)2 – (3.00 × 105 m/s)2]
(2)(0.0500 m)
= 3.64 × 10–18 N
4 Chapter 5 Solutions
©2000 by Harcourt College Publishers. All rights reserved.
(b) The weight of the electron is
Fg = mg = (9.11 × 10–31 kg)(9.80 m/s2) = 8.93 × 10–30 N
The accelerating force is 4.08 × 1011 times the weight of the electron.
Goal Solution
G: We should expect that only a very small force is required to accelerate an electron, but this
force is probably much greater than the weight of the electron ifthe gravitational force can
be neglected.
O: Since this is simply a linear acceleration problem, we can use Newton’s second law to find the
force as long as the electron does not approach relativistic speeds (much less than 3 × 108 m/s),
which is certainly the case for this problem. We know the initial and final velocities, and
the distance involved, so from these we can find theacceleration needed to determine the
force.
A: From v2f
= v2i
+ 2ax and ΣF = ma we can solve for the acceleration and the force.
a =
(v2f
– v2i
)
2x and so ΣF =
m(v2f
– v2i
)
2x
(a) F =
(9.11 × 10–31 kg) [(7.00 × 105 m/s2) – (3.00 × 105 m/s)2]
(2)(0.0500 m) = 3.64 × 10–18 N
(b) The weight of the electron is
Fg = mg = (9.11 × 10–31 kg)(9.80 m/s2) = 8.93 × 10–30 N
The ratio of theaccelerating force to the weight is
F
Fg
= 4.08 × 1011
L: The force that causes the electron to accelerate is indeed a small fraction of a newton, but it is
much greater than the gravitational force. For this reason, it is quite reasonable to ignore the
weight of the electron in electric charge problems.
5.12 (a) Fg = mg = 120 lb = 

 
4.448
N
lb (120 lb) = 534 N
(b) m =
Fg
g =
534 N9.80 m/s2 = 54.5 kg
5.13 Fg = mg = 900 N
m =
900 N
9.80 m/s2 = 91.8 kg
(Fg)on Jupiter = (91.8 kg)(25.9 m/s2) = 2.38 kN
Chapter 5 Solutions 5
© 2000 by Harcourt College Publishers. All rights reserved.
*5.14 Imagine a quick trip by jet, on which you do not visit the rest room and your perspiration is just
canceled out by a glass of tomato juice. By subtraction, (Fg)p = mgp and (Fg)C = mgCgive
ΔFg = m(gP – gC). For a person whose mass is 88.7 kg, the change in weight is
ΔFg = (88.7)(9.8095 – 9.7808) = 2.55 N
A precise balance scale, as in a doctor's office, reads the same in different locations because it
compares you with the standard masses on its beams. A typical bathroom scale is not precise
enough to reveal this difference.
5.15 (a) ΣF = F1 + F2 = (20.0i + 15.0j) N
ΣF =ma, 20.0i + 15.0j = 5.00 a
a = (4.00i + 3.00 j) m/s2
or a = 5.00 m/s2 at θ = 36.9°
(b) F2x = 15.0 cos 60.0° = 7.50 N
F2y = 15.0 sin 60.0° = 13.0 N
F2 = (7.50i + 13.0j) N
ΣF = F1 + F2
= (27.5i + 13.0j) N = ma = 5.00a
a = (5.50i + 2.60j) m/s2
5.16 We find acceleration: r – ri = vit +
1
2 at2
4.20 m i – 3.30 m j = 0 +
1
2 a(1.20 s)2 = 0.720 s2 a
a = (5.83 i – 4.58j) m/s2
Now ΣF = mabecomes
F2 = 2.80 kg (5.83i – 4.58j) m/s2 + (2.80 kg)9.80 m/s2 j
F2 = (16.3i + 14.6j) N
5.17 (a) You and earth exert equal forces on each other:
myg = Meae
If your mass is 70.0 kg,
m m
F2
a b
F1 F1
F2
90.0° 60.0°
6 Chapter 5 Solutions
© 2000 by Harcourt College Publishers. All rights reserved.
ae =
(70.0 kg)(9.80 m/s2)
5.98 × 1024 kg = ~ 10–22 m/s2
Chapter 5 Solutions 7
© 2000 byHarcourt College Publishers. All rights reserved.
(b) You and planet move for equal times in x =
1
2 at2. If the seat is 50.0 cm high,
2xy
a y
=
2xe
a e
xe =
ae
a y
xy =
my
me
xy =
70.0 kg
5.98 × 1024 kg (0.500 m) ~ 10–23 m
5.18 F = (20.0)2 + (10.0 – 15.0)2 = 20.6 N
a =
F
m
a = 5.15 m/s2 at 14.0° S of E
*5.19 Choose the x-axis forward. Then
ΣFx = max
(2000 Ni) – (1800 Ni)...