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Transcript
```Answer Key
4con
Section 4-3
Interaction Forces
Table 1
Force
Magnitude
Direction
Fbook 1 on book 2
40 N
down
Fbook 2 on book 1
40 N
Fbook2 on desktop
50 N
down
Fdesktop onbook2
50 N
UP
Fbooksland2 on desktop
90 N
down
1esktopon books 1 and 2
90 N
up
2.. Terminal velocity is when the drag force on
a falling object is equal to the gravitational
force on that object.
3. Objects that are in free fall are considered
weightless.
= 12,000.0 N
12,010.0 N = —10.0 N,
or 10.0 N acting to the South
F
550N
5. a.. m=—=
56’cg
2
9.80 rn/s
g
F
590N
=1.OX1O 1 rn/s
m
56kg
F
510N
=9.lm/s
m
56kg
—
1.
2.
3,
4,
5.
6.
7,
8.
9,
10,
11,
12.
false
true
true
false
true
false
true
true
true
true
false
true
Section 4-1 Quiz
1. Force is a vector quantity, made up of both
magnitude and direction.
2. fhe forces acting on an object must be com
bined using vector addition to find the net
force on the object
3. Lquilibrium is when an object has no net
forces acting upon it,
F
270N
4. a-—=——----=7.7m/s
35kg
m
5. Ffl 240 N + 120 N 360 N East
F
970N
6. m=—=
—=220kg
a
44m/s
Section 4-2 Quiz
1. The weight of an object depends on the
mass of the object and the acceleration due
to gravity.
204
Chapters 1—5 Resources
Section 4-3 Quiz
1.. All forces are the result of the interaction
between objects, specifically an agent and a
system.
2. Tension is the specific name for the force
exerted by a string or a rope.
3.. Normal force is the support force resulting
from the contact of two objects. It is per
pendicular to the plane of contact between
two objects.
4 From the interaction pair, the force the safe
exerts on Earth is the same as what Earth
exerts on the safe, namely its weight.
F = mg = (15,000 kg)(9.80 m/s
) =
2
150,000 N
F
150,000 N
—
—
—
5. T
=
mg + ma
2.5X10 rn/s
20
2
(200.0 kg) (9.80 m/s
) +
2
(200.0 kg)(1.2 m/s
) = 2200 N
2
Chapter 4 Reinforcement
1. The fruit moved down to the base of the
tines on the fork.
2.. The fruit and the fork moved together
because of a downward force exerted by the
right arm. The left fist then exerted an
upward force that stopped the motion of the
hand and the fork. However, no such force
was stopping the motion of the fruit, so the
fruit continued moving down the tines. This
situation is an example of Newton’s first law,
Physics: Principles and Problems
Chapter4 continued
The motion of the fruit was stopped when
the fork exerted a force that stopped the
fruit. The fruit was moving downward, and
the force of the fork was in an upward direc
tion. This upward unbalanced force caused
an upward acceleration of the fruit, stopping
its downward motion. This situation is an
example of Newton’s second law.
3 The fruit flew off the fork into the sink.
4. The explanation is the same as before, with
one exception. In the first exercise, the fork
exerted a force on the fruit, stopping its
motion. In the second exercise, the fork did
not exert a force on the fruit, so the fruit
kept moving until it struck the sink.
Chapter 4 Enrichment
Procedure
1. The faster an object moves through the
water, the higher the drag force will be.
2. Attach a length of string to a small object
and attach the other end to a small spring
scale. Fill the basin with water. Submerge the
object at one end of the basin. Hold the oth
er end of the scale at the far end of the
basin. Pull the scale steadily away from the
basin a distance of 30 cm in 4 s. Observe the
spring scale as you do this and record the
force, Repeat several times. Then repeat sev
eral times using a time of 2 s.
3. An irregularly shaped object will create a
higher drag force than a streamlined one.
4. Attach a length of string to a small irregular
object. Attach the other end to a small
spring scale, Fill the basin with water. Sub
merge the object at one end of the basin,
Flold the other end of the scale at the far
end of the basin. Pull the scale steadily away
from the basin a distance of 30 cm in 4 s.
Observe the spring scale as you do this and
record the force, Repeat several times. Then
repeat several times with a streamlined
object using the same distance and time.
Physics: Principles and Problems
Results
1. The drag force (reading on the spring scale)
was higher at higher speeds. The hypothesis
was supported.
2. The drag force (reading on the spring scale)
was higher for the irregularly shaped object.
The hypothesis was supported.
3. The object may hit the bottom of the basin,
adding drag. It is difficult to pull the scale at
a constant speed and to read the scale while
it is moving. Attaching the string at different
points on the object may result in different
data.
4. Measure the drag force using very cold water.
Then do the same experiment except with
very hot water. (If students do this experi
ment they will probably not see significant
differences. They should not use excessively
hot water because of the danger of burns.)
5. Flot water is less dense than cold water, so
there should be less resistance. The differ
ence in density is very small, so any differ
ences in the drag force may not be measur
able with this experiment.
Transparency Worksheet 44
Combining Forces on an Object
1. a. FA represents the friction between the
ground and the sled.
b. F
8 represents the pull of the person on
the sled.
c. F represents the pull of the rope on the
sled.
d. D represents the pull of the rope on the
person.
e. FE represents the pull of the sled on the
person.
f. FE represents the friction between the
person and the ground.
2. sled on ground, ground on sled; rope on
sled, sled on rope; rope on person, person on
rope; person on ground, ground on person
3. The net force that moves the sled is the force
the person exerts on the ground minus the
friction between the sled and the ground.
Chapters 1 5 Resources
205
Chavter4 continued
Transparency Worksheet 42
Motion and Newton’s Second Law
1. asXv/t
0.0 km/h)/(0.75 h 0.0 h)
= (3.0 km/h
= (3.0 km/h)/(0.75 km/h)
(3000 m)/((3600 s)(2700 s))
=3.09X10 2
m/s
4
0 rn/s
2
—
a
3.0 km/h)/(2.5 h
(1 km/h)/(0.75 h)
(1000 m)/((3600 5) (2700 s))
10X10 4
m/s
4 urn/s
5 aF/m
F ma
(3.1X1O 4 m/s
)(120,000 kg)
2
—
(4.0 km/h
—
—
1.75 h)
37 N
6 I
ma
)(120,000 kg)
2
(0 m/s
=ON
Z I ma
(1.OX1O 2
m
4
)
(120,000kg)
/s
12 N
-
8,
rn/s
4
12X10
2
Transparency Worksheet 4-3
Newton’s Third Law: Interaction Pairs
the hand, the bowling ball, and Farth
2. hnd on bowling ball S the force that the
hand exerts upward on the bowling ball.
bowling ball on hand is the force that Earth
1
exerts downward on the howling ball.
‘bowling ball on Farth is the force that the
bowling ball exerts upward on Earth.
Fhand on bowling ball and bowling ball on hand’
1
on bowling ball
and
bowling ball on Earth’
1
‘they are interaction pairs because they are
of equal magnitude and opposite direction
arid act on different objects.
‘“bowling ball on hand acts only on the hand,
bow1ing ball on Earth acts only on Earth, and
hand on bowling ball and Fa on bowling ball
act only on the bowling ball.
5.. The movement of the ball is due to unba1
anced forces on it, not the balanced force of
interaction pairs that act on each object.
Transparency Worksheet 4-4
Weight and Normal Force
1. Weight is the force defined by the formula
Pg = mg, where weight is a force caused by
the acceleration due to gravity on a mass.
2. The normal force is a support force resulting
from the contact of two objects. It is always
perpendicular to the plane of contact
between the two objects.
3.. The weight of the box and the magnitude of
the normal force are equal in Figure a.
4.. The magnitude of the normal force is greater
than the weight of the box in Figure b.
5. External forces other than gravity and the
mass of the object may change the normal
force that an object exerts.
6. The box’s apparent weight is different from
the weight caused by its mass and gravity in
Figures b and c.
Chapter 4 Assessment
Forces
Understanding Physics Concepts
1. b
2. a
3.. c
4.. b
5.. a
6.. a
8.. c
9.. c
10.. c
11.. force
12.. magnitude
13.. away from
Physics: Principles and Problems
__crIccy
t
c
1
cL
i
nue
o
14.
15.
16.
17.
18.
19.
20.
21.
22.
vector
equilibrium
gravity
weightlessness
velocity (or motion), surface area (or shape)
interactions
magnitude, opposite
rope (or string, cable, wire, etc.)
normal
4.
-
Interaction pairs are forces that act on difft’r
ent objects; they are equal in magnitude but
opposite in direction. The drawing shows an
interaction pair in the force of gravity from
the weight of the boat balanced by the normal force of the boat on the ground.
Thinking Critically
1. The weight of any object is equal to the
product of the mass of the object and the
acceleration due to gravity. Mass does not
change, but weight is dependent upon the
gravitational force. The force of gravity on
Mars is different from the force of gravity on
Earth, so objects would have the same mass
on Mars but a different weight.
2. As the elevator slows, your acceleration is in
the direction opposite to your velocity. The
direction of the acceleration of the elevator
is down. Thus the net force on you is down
ward. Your apparent weight is equal to an
upward force equal to your weight plus the
net force acting on you. In this case, the net
force is downward, so your apparent weight
would be Fg
would decrease.
3. No, there is no net force on the boat
because it has neither vertical nor horizontal
acceleration. There is, however, constant forward motion because the force of tension is
greater than the force of friction,
Fnormai
jrm
5. The forces exerted by your arm muscles and
the force exerted by the rope are acting on
6. Have the pilot take the jump plane to a
higher altitude.
(a)
Fgravty > Fdrag
Physics: Principles and Problems
F
m
T
T> Ffrjctjon
Fd rag
Fgravjty
i
—
avIty
—
Fdrag I
Fgravity < Fdrag
Applying Physics Knowledge
ma
g
(6.0 kg)(2.() m/s
)
2
9.80 m/s
12 N
3.06 kg
mg
g and Fg
T
l’herefore,
= mg = (4.2 kg)(9.80 m/s
)
2
41 N
4. A force of —125 N acting in the opposite
direction will produce equilibrium.
. F
1
5. F
11 + Pg
= ma + mg
nz(a + g)
= (1.10X i0
3 kg)(0.45 m/s
2 + 9.80 m/s
)
2
= l.1x10
4 N
3.
Fgravity
(c)
Fd rag
Fgravty
1.
‘
(b)
Fd rag
Fgravty
2.
Ftrcton
T
Farm
1
F
Chapters 1 —5 Resources
207
Chapter4 continued
6. a. in
=
=
g
thmst
Ii. a=
m
2.OX io N
9.80 rn/s
2
—
—
=
N
6
25X10
2.OXlO
k
6
g
2.OX 106 kg
The Coefficient of Friction
—
2
13rn/s
—
25 N
7. a. 2
2
=
1
m
.6kg
=,
47N
4.8 kg
2
g = 9.80 rn/s
P
25 N
= 3.4 2
rn/s
+ 2
m
2.6 kg + 4.8 kg
2
rn
b.
cord
8.
=
tota1
=
a = (2.6 kg)(3.4 m/s
2
m
)
2
rng = (102 kg)(9.80 m/s
)
2
N
3
1.OOX1O
=
(lflweIder + lflequjpment)g
(102 kg ± 14 kg) (9.80 m/s
)
2
=
=
8.8 N
1100 N
Chapter 5
Static Friction Force, FN)
FN (N)
Trial 1
Trial 2
Trial 3
Average
1.5
1.7
1.4
1.5
2.1
Data Table 2
Kinetic Friction Force, F{N)
Trial 1
Trial 2
Trial 3
Average
0.85
0.65
0.75
0.75
2.10
Data Table 3
FN (N)
F(N)
(N)
1
F
ILs
2.10
1.53
0.75
0.73
0.36
Data Table 4, Angle, 0, when sliding begins on
an incline
Mini Lab
Analyze and Conclude
2. The weight of the 500-g object is 4.9 N.
While the object is being pulled up the
incline it is less—approximately 3.5 N. The
measurement on the inclined plane should
be less because the board partially supports
the object’s weight.
3.
= rng sin 0 = (0.5 kg)(9.80 m/s
)(sin
2
3.5 N
4. Answers may vary, although the inclined
plane reading should be nearly the same as
the component from question 3.
450)
Chapters 1—5 Resources
Sample Data
Object material = wood
Surface material = wood
Data Table 1
FN (N)
Adding the torch and fuel tank to the weight
of the welder would exceed the 1100 N spec
ified for this stool,
208
Physics Lab
0
tan0
210
0.38
Analyze
1. Answers are in Tables 1 and 3.
2. Answers are in Tables 2 and 3.
3. Answer is in Table 3.
4. Answer is in Table 3.
5, Answer is in Table 4.
Conclude and Apply
1. Students should draw on their experience
moving objects and conclude that the force
necessary to start an object moving usually
is greater than the force needed to keep it
moving, so the value for ,u is larger than for
>
ILk. It is reasonable that
Physics: Principles and Problems
r- M!crJcY
16.
430
ChapterS continued
north of east, 641 m/s
19.
20.
21.
22.
NORTH
2
4
1
3
10O.O rn/s
Section 5.2
Friction
641.0 mIs
I
1.
2.
3.
4.
5.
6.
EAST
c
b
c
c
b.
Ffricition
Fwjflch
17.
In vector addition you are transforming two
vectors into one vector. In vector resolution
you are transforming one vector into two
vectors.
I 8. Add each x and ycomponent of the vectors
to obtain the answer resultant force,
P°5.0N 4.ON+ 1.ON=2.ON
F=3.0N+20N- 8.ON= 3.0N
R = V(2.0 N)
2 + (—3.0 N)
2
RA
sinO
sinx
=
0
=
=
R
Z net
Fwinch friction
2000 N
(0.2)(9800 N) = 40 N
=
40N
8,, a=
=0.04 rn/s
2
m
1000kg
—0 83
3.6N
—56°
2.0 N in the x-direction, —3.0 N in the
y-direction, resultant force is 3.6 N acting at
an angle of —56
=
—
—
3.6 N
—
sinO
Fgravty
Section 5.3
Force and Motion in
Two Dimensions
1. c
2. c
3. b
y
sinO
2N
r’”—-—
56
X
A
sin
4si4z
srn9O
4.1 N
-
=\/(9.8N)
A
B=VR
—
2
=
(4.IN)
8.8N
3N
210
Chapter
5 Resources
Physics: Principles and Problems
tr5conued
Enrichment
Transparency Worksheet 5-1
Forces in Two Dimensions
Vector Components
1.
Measured, approximately 97°, 41°, and 41°
15 m
10
2.
10 m
2 + B
A
2
0=cos 1
—
2
R
2AB
2(10.0 m)(l0.0 m)
=
=
97.2°
0=
1. The vectors are not perpendicular. The angle
between them is 110° (120.0°
10.0°).
2. First, a
1 and a
2 are decomposed into com
ponents. Second, the vertical components of
1 and a
a
2 are added. Then the horizontal
components of a
1 and a
Third, the sum of the horizontal compo
nents and the sum of the vertical compo
nents are added to find the resultant vector.
3. a
2 is negative because it points to the left.
4. a
2 has a larger vertical component.
5. a
1 has a larger horizontal component.
6. a
1 cos 10°
1 = sin 10°
a
1= a
7. a = a
2 cos 120°
y=a
2
a
2 sin 120°
—
anet
=
9.
a
=
(x+ety)
=
V(7.8
=
12km
2(10.0 m)(15,0 m)
41.4°
3. T=Tsin48.6° =0.750T
= Tcos 48.6° = 0.661T
4, 2(0,6614T) = 5,00X
N
5.
= 0.750T = (0,750)(5.00X
= 3750 N
=
6.
7.
8.
9.
10.
Phsics: Principles and Problems
x
+
0
)
1
anet
=
tan(
X
2 + (9.0 km)
km)
2
9.0 km’
0=tan -11I
\ 7.8 km
12 km at 49°
N)
5.OOXIO
S
2
N
3
OON
2 wires
To provide a horizontal force component
that prevents the sculpture from oscillating
like a pendulum.
Yes, but this would require a recalculation of
the equilibrium forces and the tension in
each wire. As long as the wires could handle
the maximum tension, a symmetrical sus
pension is not required.
No; the wire making the smaller angle with
the perpendicular would have the greater
tension.
Yes; the procedure is the same but with two
different equilibrium conditions, one for
each wire. The symmetrical suspension
merely simplifies the calculations.
V(a
8.
Transparency Worksheet 5-2
Surfaces and Friction
(9.80 m/s
)(2 kg)(0.50) = 10 N
2
(9.80 m/s
)(2 kg)(0.20) = 4 N
2
3. Yes, the force of static friction is 40 N X
0.90 = 36 N 40 N, less than the horizon
tal force.
4. The roughness of the surfaces makes than
hard to slide past each other.
5. The surface of the paper is not perfectly
smooth, but it also does not have sharp pro
jections that would give it a high coefficient
1.
2.
of friction.
6. The coefficients of static and kinetic friction
would be relatively high because of the
roughness of the surface of the sandpaper.
Chapters 1—5 Resources
213
5conti,
Transparency Worksheet 5-3
Static Friction
1..
2.
3..
4..
5.
6.
7..
Pg
=mg
)
2
= (20 kg)(9.80 m/s
= 200 N
The static friction force gradually increases.
The forces are equal. If they were not equal,
there would be a net force and the toboggan
would move.
F static =
= (0,20)(196 N)
= 39 N
Yes, the pulling force is greater than the
maximum static friction force.
Pf kinetic IL kEN
= (0.15)(196 N)
= 29 N
a=(P—Ff)/m
29.4 N)/20 kg
= (50 N
1 rn/s
2
)
2
= (35 kg)(9.80 m/s
= 343 N
Ff kinetic (0,20)(343 N)
= 69 N
No. The maximum static friction force
would be 68.6 N, which is greater than the
pulling force.
=
—
8..
7.. F would decrease and F, would increase.
8. FgyZPgcoSO
9 FgFgSflO
10. The inclined plane exerts an upward force,
on the trunk that is equal in magnitude
to F, and acts perpendicular to the surface
of the inclined plane.
Chapter Assessment
Forces in Two Dimensions
Understanding Physics Concepts
1. c
2.. c
3.
4..
5.
6..
c
a
a
c
7.b
8.. c
9.. b
10.. b
11.
NORTH
Transparency Worksheet 5-4
Forces on an Inclined Plane
1. The process is called vector decomposition.
2. F is a single force vector acting in a certain
direction. F is its horizontal component
and F is its vertical component.
+
= p
2
would point to the left, F, would be the
same, and F would lie between FX and F.Y
4 pV
5. If the angle is increased to 400, F, would
increase and F would decrease.
6. Earth’s gravity causes Fgi which is the weight
of the trunk. The vector points downward
because gravity acts towards Earth’s center.
214
Chapters 1—5 Resources
EAST
12. A
=
=
A
=
=
A cos 0
(500.0 N)(cos 30.0°)
433 N
A sin 0 = (500.0 N)(sin 30,0°)
2.50X 102 N
13. A=V=
\/(10.0 rn/s)
2 (2.00 rn/s)
2
2 +B
A
2
2AB cos 0
—
14. R
2
=
R = ‘s/(4.0 rn)
2 + (3.0 rn)
2
15.
o=
=
9.80 rn/s
—
—
2(4.0 rn)(3.0 rn) cos (142) =
((3.00 m)(sin 27.0k))
1
sin
4.00 m
=
6.6 m
20.0°
Physics: Principles and Problems
```