Idea 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
Answer Key
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 Your apparent weight 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 your hand. 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 What’s Your Angle? 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 2 are added together. 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