AP MIDTERM SAMPLE

Name: ______________________ Class: _________________ Date: _________
ID: B
AP MIDTERM SAMPLE
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
____
____
____
1. The radius of Earth is 6 370 000 m. Express this measurement in km in scientific notation with the
correct number of significant digits.
a. 63.7 × 10 4 km
c. 6.37 × 10 3 km
b. 6.37 × 10 6 km
d. 637 × 10 3 km
2. Calculate the following, and express the answer in scientific notation with the correct number of
significant figures: 21.4 + 15 + 17.17 + 4.003
a. 5.8 × 10 1
c. 5.75 × 10 1
b. 5.7573 × 10 1
d. 5.757 × 10 1
3. Calculate the following, and express the answer in scientific notation with the correct number of
significant figures: 10.5 × 8.8 × 3.14
a. 2.9 × 10 2
c. 2.90 × 10 2
b. 290
d. 290.136
4. Calculate the following, and express the answer in scientific notation with the correct number of
significant figures: (0.82 + 0.042)(4.4 × 10 3 )
a. 3.78 × 10 3
c. 3.8 × 10 3
b. 3784
d. 3.784 × 10 3
____
5. Which expression has the same dimensions as an expression yielding a value for acceleration (m/s 2 )?
(Δv has units of m/s.)
a. (Δv) 2 ⁄ Δt
c. Δv ⁄ (Δx) 2
b. Δv ⁄ (Δt) 2
d. (Δv) 2 ⁄ Δx
____
6. Which of the following expressions gives units of kg•m 2 ⁄s 2 ?
a. m•(Δx) 2 ⁄(Δt) 2
c. (Δt) 2 ⁄ m•(Δx) 2
b. m 2 •Δx ⁄(Δt) 2
d. m•(Δx) 2 ⁄Δt
7. If the change in position Δx is related to velocity v (with units of m/s) in the equation Δx = Av, the
constant A has which dimension?
____
a.
b.
____
m2
m
c. s
d. m/s 2
8. If a is acceleration (m/s 2 ), Δv is change in velocity (m/s), Δx is change in position (m), and Δt is the
time interval (s), which equation is not dimensionally correct?
a. Δt 2 = 2Δx ⁄a
c. Δt = Δx ⁄v
b. Δv = a ⁄Δt
d. a = v 2 ⁄Δx
1
Name: ______________________
ID: B
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9. For the winter, a duck flies 10.0 m/s due south against a gust of wind with a speed of 2.5 m/s. What
is the resultant velocity of the duck?
a. 12.5 m/s south
c. –7.5 m/s south
b. –12.5 m/s south
d. 7.5 m/s south
____
10. An ant on a picnic table travels 3.0 × 10 1 cm eastward, then 25 cm northward, and finally 15 cm
westward. What is the magnitude of the ant’s displacement relative to its original position?
a. 52 cm
c. 70 cm
b. 29 cm
d. 57 cm
11. A late traveler rushes to catch a plane, pulling a suitcase with a force directed 30.0° above the
horizontal. If the horizontal component of the force on the suitcase is 60.6 N, what is the force
exerted on the handle?
a. 95.6 N
c. 70.0 N
b. 65.2 N
d. 53.0 N
12. A waitperson carrying a tray with a platter on it tips the tray at an angle of 12° below the horizontal. If
the gravitational force on the platter is 5.0 N, what is the magnitude of the force parallel to the tray that
tends to cause the platter to slide down the tray? (Disregard friction.)
a. 0.42 N
c. 5.0 N
b. 1.0 N
d. 4.9 N
13. A net force of 6.8 N accelerates a 31 kg scooter across a level parking lot. What is the magnitude of
the scooter’s acceleration?
a. 3.2 m/s 2
c. 0.22 m/s 2
b. 0.69 m/s 2
d. 4.6 m/s 2
____
____
____
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14. An airplane with a mass of 1.20 × 10 4 kg tows a glider with a mass of 0.60 × 10 4 kg. If the airplane
propellers provide a net forward thrust of 3.60 × 10 4 N, what is the acceleration of the glider?
(Disregard friction.)
a. 9.80 m/s 2
c. 3.00 m/s 2
b. 6.00 m/s 2
d. 2.00 m/s 2
____
15. A sled weighing 1.0 × 10 2 N is held in place on a frictionless 20.0° slope by a rope attached to a
stake at the top. The rope is parallel to the slope. What is the normal force of the slope acting on the
sled?
a. 94 N
c. 34 N
b. 37 N
d. 47 N
16. There are six books in a stack, and each book weighs 5 N. The coefficient of static friction between
the books is 0.2. With what horizontal force must one push to start sliding the top five books off the
bottom one?
a. 1 N
c. 5 N
b. 3 N
d. 7 N
17. A construction worker pushes a wheelbarrow 5.0 m with a horizontal force of 50.0 N. How much
work is done by the worker on the wheelbarrow?
a. 10 J
c. 250 J
b. 1250 J
d. 55 J
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2
Name: ______________________
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____
____
ID: B
18. A horizontal force of 200 N is applied to move a 55 kg television set across a 10 m level surface.
What is the work done by the 200 N force on the television set?
a. 550 J
c. 11000 J
b. 2000 J
d. 6000 J
19. A child pulls a balloon for 12 m with a force of 1.0 N at an angle 60° below horizontal. How much
work does the child do on the balloon?
a. 6.0 J
c. –10 J
b. 12 J
d. –6.0 J
20. What is the kinetic energy of a 0.135 kg baseball thrown at 40.0 m/s?
a. 87.0 J
c. 216 J
b. 54.0 J
d. 108 J
21. What is the potential energy of a 1.0 kg mass 1.0 m above the ground?
a. 9.8 J
c. 10 J
b. 1.0 J
d. 96 J
22. How much elastic potential energy is stored in a bungee cord with a spring constant of 10.0 N/m
when the cord is stretched 2.00 m?
a. 40.0 J
c. 20.0 J
b. 200 J
d. 10.0 J
23. A 3.00 kg toy falls from a height of 1.00 m. What will the kinetic energy of the toy be just before the
toy hits the ground? (Assume no air resistance and that g = 9.81 m/s 2 .)
a. 29.4 J
c. 9.8 J
b. 0.98 J
d. 294 J
24. How much power is required to lift a 2.0 kg mass at a speed of 2.0 m/s?
a. 2.0 J
c. 4.0 J
b. 9.8 J
d. 39 J
25. What is the average power supplied by a 60.0 kg person running up a flight of stairs a vertical
distance of 4.0 m in 4.2 s?
a. 57 W
c. 560 W
b. 670 W
d. 240 W
26. Which of the following has the greatest power output?
a.
a crane that lifts a 2.5 × 10 4 N beam at a speed of 1.2 m/s
b.
a mechanic’s lift that raises a 1.2 × 10 4 N car 2.1 m in 12 s
c.
d.
a car engine that does 1.2 × 10 3 J of work in 5.0 s
a weightlifter who lifts a 250 N weight 2.1 m in 3.0 s
Short Answer
27. Convert 92 × 10 3 km to decimeters using scientific notation.
28. Convert 1 µm to meters using scientific notation.
29. Convert 5.52 × 10 8 g to kilograms using scientific notation.
30. Convert 8.66 × 10 −9 m to millimeters using scientific notation.
3
Name: ______________________
ID: B
31. Construct a graph of position versus time of a jogger, using the data in the table above. Explain how
the graph indicates that the jogger’s speed is constant.
32. Construct a graph of position versus time of a jogger, using the data in the table above. What is the
average velocity of the jogger?
33. A motorized scooter starts from rest and accelerates for 4 s at 2 m/s 2 . It continues at a constant speed
for 6 s. Graph the scooter’s velocity versus time. What is the scooter’s average velocity for the
interval 0–4 s?
34. What is the kinetic energy of a 1.5 × 10 3 kg car traveling at 25 m/s?
Problem
35. Calculate the following, expressing the answer in scientific notation with the correct number of
Ê
ˆ
significant figures: ÁÁÁ 8.86 + 1.0 × 10 −3 ˜˜˜ ÷ 3.610 × 10 −3
Ë
¯
4
Name: ______________________
ID: B
36. A horse trots past a fencepost located 12 m to the left of a gatepost. It then passes another fencepost
located 24 m to the right of the gatepost 11 s later. What is the average velocity of the horse?
37. The graph above shows displacement versus time. What is the average velocity for line A?
38. The graph above shows displacement versus time. What is the average velocity for line B?
39. The graph above shows displacement versus time. What is the average velocity for line C?
40. The graph above shows the motion of a cat. What is the cat’s average velocity during the time interval
0.0–20.0 s?
41. A shopping cart is given an initial velocity of 2.0 m/s and undergoes a constant acceleration of 3.0
m/s 2 . What is the magnitude of the cart’s displacement after the first 4.0 s of its motion?
5
Name: ______________________
ID: B
42. A skater glides off a frozen pond onto a patch of ground at a speed of 1.8 m/s. Here she is slowed at
a constant rate of 3.00 m/s 2 . How fast is the skater moving when she has slid 0.37 m across the
ground?
43. A sports car traveling at 27.8 m/s slows at a constant rate to a stop in 8.00 s. What is the displacement
of the sports car in this time interval?
44. Human reaction time is usually about 0.20 s. If your lab partner holds a ruler between your finger
and thumb and releases it from rest without warning, how far can you expect the ruler to fall before
you catch it? (Disregard air resistance. a = −g = −9.81 m/s 2 .)
45. A rock is thrown downward from the top of a cliff with an initial speed of 12 m/s. If the rock hits the
ground after 2.0 s, what is the height of the cliff? (Disregard air resistance. a = −g = −9.81 m/s 2 .)
46. A rock is thrown straight upward with an initial velocity of 24.5 m/s where the acceleration due to
gravity has a magnitude of 9.81 m/s 2 . What is the rock’s displacement after 1.00 s?
47. A lightning bug flies at a velocity of 0.25 m/s due east toward another lightning bug seen off in the
distance. A light easterly breeze blows on the bug at a velocity of 0.25 m/s. What is the resultant
velocity of the lightning bug?
48. A jogger runs 10.0 blocks due east, 5.0 blocks due south, and another 2.0 blocks due east. Assume
all blocks are of equal size. Use the graphical method to find the magnitude of the jogger’s net
displacement.
49. A cave explorer travels 3.0 m eastward, then 2.5 m northward, and finally 15.0 m westward. Use the
graphical method to find the magnitude of the net displacement.
50. An airplane flying at 120 km/h due west moves into a region where the wind is blowing at 40 km/h
due east. If the plane’s original vector velocity is vplane, what is the expression for the plane’s
resulting velocity in terms of vplane?
51. A dog walks 28 steps north and then walks 55 steps west to bury a bone. If the dog walks back to the
starting point in a straight line, how many steps will the dog take? Use the graphical method to find
the magnitude of the net displacement.
52. A quarterback takes the ball from the line of scrimmage and runs backward for 1.0 × 10 1 m. He then
runs sideways, parallel to the line of scrimmage, for 15 m. Next, he throws the ball forward 5.0 × 10 1
m, perpendicular to the line of scrimmage. The receiver is tackled immediately. How far is the football
displaced from its original position?
53. A string attached to an airborne kite was maintained at an angle of 40.0° with the ground. If 120 m of
string was reeled in to return the kite back to the ground, what was the horizontal displacement of the
kite? (Assume the kite string did not sag.)
54. An athlete runs 110 m across a level field at an angle of 30.0° north of east. What is the north
component of this displacement?
55. A small airplane flies at a velocity of 145 km/h toward the south as observed by a person on the
ground. The airplane pilot measures an air velocity of 172 km/h south. What is the velocity of the
wind that affects the plane?
6
Name: ______________________
ID: B
56. In a game of tug-of-war, a rope is pulled by a force of 102 N to the right and by a force of 75 N to
the left. What is the magnitude and direction of the net horizontal force on the rope?
57. Two ice-hockey players simultaneously strike a puck with their sticks. The stick of one player exerts
an eastward force on the puck of 12 N. The other player’s stick exerts a northward force of 15 N on
the puck. Assuming that there is no frictional force between the puck and the ice, what is the
magnitude of the net horizontal force on the puck?
58. A package of meteorological instruments is held aloft by a balloon that exerts an upward force of 634
N on the package. The gravitational force acting on the package is 365 N. What is the magnitude and
direction of the force that a scientist must exert on a rope attached to package to keep it from rising?
59. A wagon having a mass of 32 kg is accelerated across a level road at 0.50 m/s 2 . What net force acts
on the wagon horizontally?
60. A farmhand attaches a 25 kg bale of hay to one end of a rope passing over a frictionless pulley
connected to a beam in the hay barn. Another farmhand then pulls down on the opposite end of the
rope with a force of 277 N. Ignoring the mass of the rope, what will be the magnitude and direction of
the bale’s acceleration if the gravitational force acting on it is 245 N?
61. Basking in the sun, a 1.10 kg lizard lies on a flat rock tilted at an angle of 15.0° with respect to the
horizontal. What is the magnitude of the normal force exerted by the rock on the lizard?
62. A three-tiered birthday cake rests on a table. From bottom to top, the cake tiers weigh 16 N, 9 N, and
5 N, respectively. What is the magnitude and direction of the normal force acting on the second tier?
63. A row of six 1.0 N wooden blocks is being pushed across a tabletop at a constant speed by a toy
tractor that exerts a force of 1.3 N on the row. What is the coefficient of kinetic friction between the
wooden blocks and the tabletop?
64. A waitperson pushes the bottom of a glass tumbler full of water across a tabletop at constant speed.
The tumbler and its contents have a mass of 0.70 kg, and the coefficient of kinetic friction for the
surfaces in contact is 0.41. What force does the waitperson exert on the glass? g = 9.81 m/s 2 )
65. How much work is done on a bookshelf being pulled 5.00 m at an angle of 37.0° from the
horizontal? The magnitude of the component of the force that does the work is 43.0 N.
66. A skier with a mass of 88 kg hits a ramp of snow at 16 m/s and becomes airborne. At the highest
point of flight, the skier is 3.7 m above the ground. What is the skier’s gravitational potential energy
at this point?
67. An 80.0 kg climber climbs to the top of Mount Everest, which has a peak height of 8848 m above sea
level. What is the climber’s potential energy with respect to sea level?
68. A scale contains a spring with a spring constant of 275 N/m. Placing a mass on the scale causes the
spring to be compressed by 3.25 cm. Calculate the elastic potential energy stored in the spring.
69. A pendulum with a mass of 1 kg is released from a height of 1.5 cm above the height of its resting
position. How fast will the pendulum be moving when it passes through the lowest point of its swing?
70. What is the average power output of a weightlifter who can lift 250 kg a height of 2.0 m in 2.0 s?
7
ID: B
AP MIDTERM SAMPLE
Answer Section
MULTIPLE CHOICE
1. ANS: C
Solution
ÁÊ
˜ˆ
(6 370 000 m) ÁÁÁÁ 1 km ˜˜˜˜ = 6.37 × 10 3 km
Ë 1000 m ¯
PTS: 1
2. ANS: A
Solution
DIF:
IIIA
OBJ: 1-2.2
21.4
15.
17.17
+4.003
57.573
Answer rounds to 58 and is written as 5.8 × 10 1 in scientific notation.
PTS: 1
DIF: IIIA
3. ANS: A
Solution
(10.5) × (8.8) × (3.14) = 290.136
OBJ: 1-2.4
The answer rounds to 290 and is written as 2.9 × 10 2 in scientific notation.
PTS: 1
4. ANS: C
Solution
DIF:
IIIA
OBJ: 1-2.4
Ê
ˆ
Ê
ˆ
(0.82 + 0.042) ÁÁÁ 4.4 × 10 3 ˜˜˜ = (0.86) ÁÁÁ 4.4 × 10 3 ˜˜˜ = 3784
Ë
¯
Ë
¯
The answer rounds to 3800 and is written as 3.8 × 10 3 in scientific notation.
PTS: 1
DIF: IIIA
5. ANS: D
Solution
2
2
(Δv) 2
(m/s) 2
=
= m /s = m/s 2
Δx
m
m
PTS:
1
DIF:
OBJ: 1-2.4
IIIA
OBJ: 1-3.3
1
ID: B
6. ANS: A
Solution
2
m
(Δx)
(Δt) 2
ÊÁÁ 2 ˆ˜˜
ÁË m ˜¯
Ê
ˆ
Á
˜
= Ë kg ¯ × Ê ˆ = kgm 2 /s 2
ÁÁÁ s 2 ˜˜˜
Ë ¯
PTS: 1
7. ANS: C
Solution
Δx = Av
DIF:
IIIA
OBJ: 1-3.3
Rearrange the equation to solve for A and substitute units.
A = Δx = m = s
v
m/s
PTS: 1
8. ANS: B
Solution
DIF:
IIIA
OBJ: 1-3.3
Does Δv = a ?
Δt
Substitute units into the right side of the equation.
a = m/s 2 =
Δt
s
ÊÁÁ m
ÁÁÁ
ÁË s 2
ˆ˜˜ ÁÊ 1 ˜ˆ
˜˜˜ ÁÁÁ ˜˜˜
˜¯ ÁË s ˜¯
m
ÊÁÁ 1 ˆ˜˜ = s 3
(s) ÁÁÁ ˜˜˜
Ës¯
Substituting units into both sides of the equation yields the following result:
m≠ m
s
s3
PTS: 1
DIF:
9. ANS: D
Given
v 1 = 10.0 m/s south
v 2 = 2.5 m/s north
IIIA
OBJ: 1-3.3
Solution
v R = v 1 − v 2 = 10.0 m/s − 2.5 m/s = 7.5 m/s
vR = 7.5 m/s south
PTS:
1
DIF:
IIIA
OBJ: 3-1.2
2
ID: B
10. ANS: B
Given
Δx 1 = 3.0 × 10 1 cm
Δy 1 = 25 cm
Δx 2 = –15 cm
Solution
Δx tot = Δx 1 + Δx 2 = (3.0 × 10 1 cm) + ( − 15 cm) = 15 cm
Δy tot = Δy 1 = 25 cm
d 2 = (Δx tot ) 2 + (Δy tot ) 2
(Δx tot ) 2 + (Δy tot ) 2 =
d=
(15 cm) 2 + (25 cm) 2
d = 29 cm
PTS: 1
11. ANS: C
DIF:
IIIA
OBJ: 3-2.2
Given
F y = 60.0 N
θ = 30.0°
Solution
Fy
cos θ =
F
F=
Fy
cos θ
= 60.6 N = 70.0 N
cos 30.0°
PTS: 1
12. ANS: B
DIF:
IIIA
OBJ: 4-2.2
Given
F g = 5.0 N
α = 12°
Solution
θ = 90° − 12° = 78°
ΣF x = F g, x = F g cos θ = (5.0 N)(cos 78°) = 1.0 N
PTS:
1
DIF:
IIIA
OBJ: 4-2.2
3
ID: B
13. ANS: C
Given
F applied = 6.8 N
m = 31 kg
Solution
F net = ∑ F x = F applied = ma x
F applied
ax =
= 6.8 N = 0.22 m/s 2
m
31 kg
PTS: 1
14. ANS: D
DIF:
IIIA
OBJ: 4-3.2
Given
Fnet = 3.60 × 10 4 N, forward
m 1 = 1.20 × 10 4 kg
m 2 = 0.60 × 10 4 kg
Solution
F net = ma = (m 1 + m 2 )a
a=
F net
m1 + m2
=
3.60 × 10 4 N
= 2.00 m/s 2
4
4
(1.20 × 10 kg) + (0.60 × 10 kg)
PTS: 1
15. ANS: A
DIF:
IIIA
OBJ: 4-3.2
Given
F g = 1.0 × 10 2 N
θ = 20.0°
Solution
ΣF y = F n − F g, y = 0
F n = F g, y = F g cos θ = (1.0 × 10 2 N)(cos 20.0°) = 94 N
PTS:
1
DIF:
IIIA
OBJ: 4-4.2
4
ID: B
16. ANS: C
Given
F g, book = 5 N
µ s = 0.2
Solution
ΣF x = F applied − F s,max = 0
F applied = F s,max = µ s F n = µ s F g
F g = (5Ν + 5Ν + 5Ν + 5Ν + 5Ν) = 25Ν
F applied = (0.2) (25Ν) = 5Ν
PTS: 1
17. ANS: C
2.5 × 10 2 J
DIF:
IIIA
OBJ: 4-4.4
Given
F = 50.0 N
d = 5.0 m
Solution
W = Fd = (50.0 N) (5.0 m) = 2.5 × 10 2 J
PTS: 1
18. ANS: B
DIF:
IIIA
OBJ: 5-1.4
Given
F = 200 N
d = 10 m
Solution
W = Fd = (200 N) (10 m) = 2 × 10 3 J
PTS:
1
DIF:
IIIA
OBJ: 5-1.4
5
ID: B
19. ANS: A
Given
F = 1.0 N
d = 12 m
θ = −60°
Solution
È
˘
W = Fdcos θ = (1.0 N) (12 m) ÍÍÍÎ cos (−60°) ˙˙˙˚ = 6.0 J
PTS: 1
20. ANS: D
DIF:
IIIA
OBJ: 5-1.4
Given
m = 0.135 kg
v = 40.0 m/s
Solution
KE = 12 mv 2 =
PTS: 1
21. ANS: A
1
2
(0.135kg)(40.0m / s) 2 = 108J
DIF:
IIIA
OBJ: 5-2.2
Given
m = 1.0 kg
h = 1.0 m
g = 9.81 m/s 2
Solution
Ê
ˆ
PE = mgh = ÊÁË 1.0 kg ˆ˜¯ ÁÁÁ 9.81 m/s 2 ˜˜˜ (1.0 m) = 9.8 J
Ë
¯
PTS:
1
DIF:
IIIA
OBJ: 5-2.6
6
ID: B
22. ANS: C
Given
k = 10.0 N/m
x = 2.00 m
Solution
PE = 12 kx 2 =
PTS: 1
23. ANS: A
1
2
(10.0 N/m) (2.00 m) 2 = 20.0 J
DIF:
IIIA
OBJ: 5-2.6
Given
m = 3.00 kg
h = 1.00 m
g = 9.81 m/s 2
Solution
Ê
ˆ
KE f = PE g, i = mgh = ÊÁË 3.00 kg ˆ˜¯ ÁÁÁ 9.81 m/s 2 ˜˜˜ (1.00 m) = 29.4 J
Ë
¯
PTS: 1
24. ANS: D
DIF:
IIIA
OBJ: 5-3.3
Given
m = 2.0 kg
v = 2.0 m/s
g = 9.81 m/s 2
Solution
Ê
ˆ
P = Fv = mgv = ÊÁË 2.0 kg ˜ˆ¯ ÁÁÁ 9.81 m/s 2 ˜˜˜ (2.0 m/s) = 39 J/s
Ë
¯
PTS:
1
DIF:
IIIA
OBJ: 5-4.2
7
ID: B
25. ANS: C
Given
m = 60.0 kg
d = 4.0 m
Δt = 4.2 s
g = 9.81 m/s 2
Solution
ÊÁ 60.0 kg ˆ˜ ÁÊÁÁ 9.81 m/s 2 ˜ˆ˜˜ (4.0 m)
Ë
¯Ë
mgd
¯
P = W = Fd =
=
= 5.6 × 10 2 W
Δt
Δt
Δt
4.2 s
PTS:
1
DIF:
IIIA
OBJ: 5-4.2
8
ID: B
26. ANS: A
Given
a: F = 250 N
d = 2.1 m
Δt = 3.0 s
b: F = 1.2 × 10 4 N
d = 2.1 m
Δt = 12 s
c: W = 1.2 × 10 3 J
Δt = 3.0 s
d: F = 2.5 × 10 4 N
v = 1.2 m/s
Solution
P = W = Fd = Fv
Δt
Δt
(250 N) (2.1 m)
P a = Fd =
= 2.6 × 10 2 J
Δt
2.0 s
ÊÁÁ
ˆ
4
ÁË 1.2 × 10 N ˜˜˜¯ (2.1 m)
P b = Fd =
= 2.1 × 10 3 J
Δt
12 s
ÊÁÁ
3 ˆ
ÁË 1.2 × 10 J ˜˜˜¯
Pc = W =
= 2.4 × 10 2 J
Δt
5.0 s
Ê
ˆ
P d = Fv = ÁÁÁ 2.5 × 10 4 N ˜˜˜ (1.2 m/s) = 3.0 × 10 4 J
Ë
¯
Pd > Pb > Pa > Pc
PTS:
1
DIF:
IIIA
OBJ: 5-4.2
9
ID: B
SHORT ANSWER
27. ANS:
9.2 × 10 8 dm
Solution
Ê
ˆ
ÊÁÁ
ˆ˜˜ ÁÁÁÁ 10 4 dm ˜˜˜˜
3
7
8
ÁË 92 × 10 km ˜¯ ÁÁÁ 1 km ˜˜˜ = 92 × 10 dm = 9.2 × 10 dm
ÁË
˜¯
PTS: 1
28. ANS:
1 × 10 −6 m
DIF:
IIIA
OBJ: 1-2.2
Solution
ÁÊÁ −6 ˜ˆ˜
ÊÁ 1 µm ˆ˜ ÁÁÁÁ 10 m ˜˜˜˜ = 1 × 10 −6 m
Ë
¯ ÁÁÁ 1 µm ˜˜˜
ÁË
˜¯
PTS: 1
29. ANS:
5.52 × 10 5 kg
DIF:
IIIA
OBJ: 1-2.2
Solution
Ê
ˆ
ÊÁÁ
ˆ˜˜ ÁÁÁÁ 1 kg ˜˜˜˜
8
5
ÁË 5.52 × 10 g ˜¯ ÁÁÁ 3 ˜˜˜ = 5.52 × 10 kg
ÁË 10 g ˜¯
PTS: 1
30. ANS:
8.66 × 10 −6 mm
DIF:
IIIA
OBJ: 1-2.2
Solution
Ê
ˆ
ÊÁÁ
ˆ˜˜ ÁÁÁÁ 10 3 mm ˜˜˜˜
−9
−6
ÁË 8.66 × 10 m ˜¯ ÁÁÁ 1 m ˜˜˜ = 8.66 × 10 mm
ÁË
˜¯
PTS:
1
DIF:
IIIA
OBJ: 1-2.2
10
ID: B
31. ANS:
The jogger is moving at a constant speed because the position versus time graph is a straight line with
a positive slope.
PTS:
1
DIF:
IIIA
OBJ: 2-1.3
11
ID: B
32. ANS:
+0.40 m/s
Given
x i = 2.8 m
x f = 5.2 m
t i = 2.0 s
t f = 8.0 s
Solution
xf − xi
v avg = Δx =
= 5.2 m – 2.8 m = 2.4 m = 0.40 m/s
Δt
tf − ti
8.0 s – 2.0 s
6.0 s
PTS:
1
DIF:
IIIA
OBJ: 2-1.3
12
ID: B
33. ANS:
+2.0 m/s
Given
v i = 0.0 m/s
v f = 4.0 m/s
Solution
vi + vf
v avg =
= 0.0 m/s + 4.0 m/s = 2.0 m/s
2
2
PTS: 1
34. ANS:
DIF:
IIIA
OBJ: 2-2.2
4.7 × 10 3 J
Given
m = 1.5 × 10 3 kg
v = 25 m/s
Solution
KE =
PTS:
1
2
mv 2 =
1
1
2
ÊÁÁ
ˆ
3
2
5
ÁË 1.5 × 10 kg ˜˜˜¯ (25 m/s) = 4.7 × 10 J
DIF:
IIIA
OBJ: 5-2.2
13
ID: B
PROBLEM
35. ANS:
2.45 × 10 3
Solution
ÁÊÁÁ 8.86 + 1.0 × 10 −3 ˜ˆ˜˜
Ë
¯
(8.86)
3
=
ÊÁÁ
ˆ˜ = 2.45 × 10
−3 ˜
ÁÊÁÁ 3.610 × 10 −3 ˜ˆ˜˜
ÁË 3.610 × 10 ˜¯
Ë
¯
PTS: 1
36. ANS:
3.3 m/s, to the right
DIF:
IIIA
OBJ: 1-2.4
Given
x i = −12 m
x f = 24 m
Δt = 11 s
Solution
xf − xi
(24 m) − (−12 m)
v avg = Δx =
=
= 3.3 m/s,to the right
Δt
Δt
11 s
PTS: 1
37. ANS:
3.0 m/s
DIF:
IIIA
OBJ: 2-1.2
Given
x i = 0.0 m
x f = 9.0 m
t i = 0.0 s
t f = 3.0 s
Solution
xf − xi
v avg = Δx =
= 9.0 m – 0.0 m = 3.0 m/s
Δt
tf − ti
3.0 s – 0.0 s
PTS:
1
DIF:
IIIA
OBJ: 2-1.3
14
ID: B
38. ANS:
1.0 m/s
Given
x i = 0.0 m
x f = 3.0 m
t i = 0.0 s
t f = 3.0 s
Solution
xf − xi
v avg = Δx =
= 3.0 m – 0.0 m = 1.0 m/s
Δt
tf − ti
3.0 s – 0.0 s
PTS: 1
39. ANS:
0.33 m/s
DIF:
IIIA
OBJ: 2-1.3
Given
x i = 0.0 m
x f = 3.0 m
t i = 0.0 s
t f = 9.0 s
Solution
xf − xi
v avg = Δx =
= 3.0 m – 0.0 m = 0.33 m/s
Δt
tf − ti
9.0 s – 0.0 s
PTS:
1
DIF:
IIIA
OBJ: 2-1.3
15
ID: B
40. ANS:
7.5 × 10 −2 m/s, south
Given
x i = 2.0 m
x f = 0.5 m
t i = 0.0 s
t f = 20.0 s
Solution
xf − xi
v avg = Δx =
= 0.5 m – 2.0 m = –1.5 m = −7.5 × 10 −2 m/s
Δt
tf − ti
20.0 s – 0.0 s
20.0 s
v avg = 7.5 × 10 −2 m/s,south
PTS: 1
41. ANS:
32 m
DIF:
IIIA
OBJ: 2-1.3
Given
v i = 2.0 m/s
a = 3.0 m/s 2
Δt = 4.0 s
Solution
Δx = v i Δt + 12 a(Δt) 2
Δx = (2.0 m/s)(4.0 s) + 12 (3.0 m/s 2 )(4.0 s) 2 = 8.0 m + 24 m
Δx = 32 m
PTS:
1
DIF:
IIIA
OBJ: 2-2.3
16
ID: B
42. ANS:
1.0 m/s
Given
v i = 1.8 m/s
a = −3.00 m/s 2
Δx = 0.37 m
Solution
v f 2 = v i 2 + 2aΔx
vf =
v i 2 + 2aΔx =
(1.8 m/s) 2 + 2(−3.00 m/s 2 )(0.37 m)
vf =
3.2 m 2 /s 2 – 2.2 m 2 /s 2 =
1.0 m 2 /s 2
v f = 1.0 m/s
PTS: 1
43. ANS:
111 m
DIF:
IIIA
OBJ: 2-2.3
Given
v i = 27.8 m/s
v f = 0.0 m/s
Δt = 8.00 s
Solution
Δx = 12 (v i + v f )Δt =
1
2
PTS: 1
44. ANS:
at least 0.20 m
(27.8 m/s + 0.0 m/s)(8.00 s) = 111 m
DIF:
IIIA
OBJ: 2-2.3
Given
a = − g =− 9.81 m/s 2
Δt = 0.20 s
v i = 0.0 m/s
Solution
Δx = v i Δt + 12 a(Δt) 2
Δx = (0 m/s)(0.20 s) + 12 (−9.81 m/s 2 )(0.20 s) 2 = −0.20 m
PTS:
1
DIF:
IIIA
OBJ: 2-3.2
17
ID: B
45. ANS:
44 m
Given
a = − g = − 9.81 m/s 2
Δt = 2.0 s
v i = −12 m/s
Solution
Δx = v i Δt + 12 a(Δt) 2
Δx = (−12 m/s)(2.0 s) + 12 (−9.81 m/s 2 )(2.0 s) 2 = −44 m
height of cliff = 44 m
PTS: 1
46. ANS:
19.6 m
DIF:
IIIA
OBJ: 2-3.2
Given
a = − g = − 9.81 m/s 2
v i = 24.5 m/s
Δt = 1.00 s
Solution
Δx = v i Δt + 12 a(Δt) 2
Δx = (24.5 m/s)(1.00 s) + 12 (−9.81 m/s 2 )(1.00 s) 2 = 19.6 m
PTS: 1
47. ANS:
0.00 m/s
DIF:
IIIA
OBJ: 2-3.2
Given
v 1 = 0.25 m/s east
v 2 = 0.25 m/s west
Solution
v R = v 1 − v 2 = 0.25 m/s − 0.25 m/s = 0.00 m/s
PTS:
1
DIF:
IIIA
OBJ: 3-1.2
18
ID: B
48. ANS:
13.0 blocks
Solution
Students should use graphical techniques. Their answers can be checked using the techniques
presented in Section 2.
d=
(12.0 blocks) 2 + (5.0 blocks) 2 = 13.0 blocks
PTS: 1
49. ANS:
12.2 m
DIF:
IIIA
OBJ: 3-1.2
Solution
Students should use graphical techniques. Their answers can be checked using the techniques
presented in Section 2.
d=
(12.0 m) 2 + (2.5 m) 2 = 12.2 m
PTS: 1
50. ANS:
2v
3 plane
DIF:
IIIA
OBJ: 3-1.2
Given
vplane = 120 km/h west = –120 km/h
vwind = 40 km/h east = +40 km/h
Solution
v R = v plane + v wind = −120 km/h + (40 km/h) = –80 km/h
vR
v plane
= –80 km/h = 2
–120 km/h 3
v R = 2 v plane
3
PTS:
1
DIF:
IIIA
OBJ: 3-1.3
19
ID: B
51. ANS:
62 steps
Solution
Students should use graphical techniques. Their answers can be checked using the techniques
presented in Section 2.
d=
(28 steps) 2 + (55 steps) 2 = 62 steps
PTS: 1
52. ANS:
43 m
DIF:
IIIA
OBJ: 3-2.2
Given
Δx 1 = −1.0 × 10 1 m
Δy 1 = 15 m
Δx 2 = +5.0 × 10 1 m
Solution
Δx tot = Δx 1 + Δx 2 = (−1.0 × 10 1 m) + (5.0 × 10 1 m) = 4.0 × 10 1 m
Δy tot = Δy 1 = 15 m
d 2 = (Δx tot ) 2 + (Δy tot ) 2
d=
(Δx tot ) 2 + (Δy tot ) 2 =
PTS: 1
53. ANS:
92 m
DIF:
(4.0 × 10 1 m) 2 + (1.5 × 10 1 m) 2 = 4.3 × 10 1 m
IIIA
OBJ: 3-2.2
Given
d = 120 m, θ = 40.0°
Solution
d x = dcos θ = (120 m)(cos 40.0°) = 92 m
PTS:
1
DIF:
IIIA
OBJ: 3-2.3
20
ID: B
54. ANS:
55 m
Given
d = 110 m
θ = 30.0°
Solution
d y = dsin θ = (110 m)(sin30.0°) = 55 m
PTS: 1
55. ANS:
27 km/h north
DIF:
IIIA
OBJ: 3-2.3
Given
vpg = velocity of plane to ground = 145 km/h south
vpa = velocity of plane to air = 172 km/h south
Solution
v pg = v pa + v ag
v ag = v pg − v pa
v ag = 145 km/h − 172 km/h = − 27 km/h
v ag = 27 km/h north
PTS: 1
56. ANS:
27 N, to the right
DIF:
IIIA
OBJ: 3-4.2
Given
F 1 = 102 N, to the right
F 2 = 75 N, to the left
Solution
F net = F 1 + F 2
F net = F 1 − F 2 = 102 N − 75 N = 27 N
F net = 27 N, to the right
PTS:
1
DIF:
IIIA
OBJ: 4-2.2
21
ID: B
57. ANS:
19 N
Given
Fx = 12 N, east
Fy = 15 N, north
Solution
Fx 2 + Fy 2 =
F net =
PTS: 1
58. ANS:
269 N, downward
(12 N) 2 + (15 N) 2 = 19 N
DIF:
IIIA
OBJ: 4-2.2
Given
Fballoon,y = 634 N, upward
Fg = 365 N, downward
Solution
ΣF y = F balloon, y − F g − F applied, y = 0
F applied, y = F balloon, y − F g = 634 N – 365 N = 269 N
F applied, y = 269 N,downward
PTS: 1
59. ANS:
16 N
DIF:
IIIA
OBJ: 4-2.3
Given
m = 32 kg
ax = 0.50 m/s 2
Solution
ΣF x = ma x = (32 kg)(0.50 m/s 2 ) = 16 N
PTS:
1
DIF:
IIIA
OBJ: 4-3.2
22
ID: B
60. ANS:
1.3 m/s 2 , upward
Given
F applied, y = 277 N
F g = 245 N
m = 25 kg
Solution
ΣF y = F applied, y − F g = ma
a=
F applied, y − F g
m
= 277 N – 245 N = 1.3 m/s 2
25 kg
a = 1.3 m/s 2 , upward
PTS: 1
61. ANS:
10.4 N
DIF:
IIIA
OBJ: 4-3.2
Given
m = 1.10 kg
θ = 15.0°
g = 9.81 m/s 2
Solution
F net, y = ΣF y = F n − F g, y = 0
F n = F g, y = F g cos θ = mgcos θ
F n = (1.10 kg)(9.81 m/s 2 )(cos 15.0°) = 10.4 N
PTS:
1
DIF:
IIIA
OBJ: 4-4.2
23
ID: B
62. ANS:
14 N, upward
Given
F g, 1 = 5 N
F g, 2 = 9 N
F g, 3 = 16 N
Solution
F net, y = ΣF y = F n − F g, 1 − F g, 2 = 0
F n = F g, 1 + F g, 2 = 5 N + 9 N = 14 N
Fn = 14 N, upward
PTS: 1
63. ANS:
0.22
DIF:
IIIA
OBJ: 4-4.2
Given
F g = (6)(1.0 N) = 6.0 N
F applied = 1.3 N
Solution
F net, y = F n − F g = 0
Fn = Fg
F net, x = F applied − F f = 0
F applied = F f
µk =
PTS:
Ff
Fn
1
=
F applied
Fg
= 1.3 N = 0.22
6.0 N
DIF:
IIIA
OBJ: 4-4.4
24
ID: B
64. ANS:
2.8 N
Given
m = 0.70 kg
µ k = 0.41
g = 9.81 m/s 2
Solution
F net, y = F n − F g = 0
Fn = Fg
F net, x = F applied − F f = 0
F applied = F f
F f = µ k F n = µ k F g = µ k mg = (0.41)(0.70 kg)(9.81 m/s 2 ) = 2.8 N
F applied = F f = 2.8 N
PTS: 1
65. ANS:
215 J
DIF:
IIIA
OBJ: 4-4.4
Given
F = 43.0 N
d = 5.00 m
Solution
W = Fd = (43.0 N) (5.00 m) = 215 J
PTS:
1
DIF:
IIIA
OBJ: 5-1.4
25
ID: B
66. ANS:
3.2 × 10 3 J
Given
m = 88 kg
h = 3.7 m
g = 9.81 m/s 2
Solution
Ê
ˆ
PE = mgh = ÊÁË 88 kg ˆ˜¯ ÁÁÁ 9.81 m/s 2 ˜˜˜ (3.7 m) = 3.2 × 10 3 J
Ë
¯
PTS: 1
67. ANS:
DIF:
IIIA
OBJ: 5-2.6
6.94 × 10 6 J
Given
m = 80.0 kg
h = 8848 m
g = 9.81 m/s 2
Solution
Ê
ˆ
PE = mgh = ÊÁË 80.0 kg ˆ˜¯ ÁÁÁ 9.81 m/s 2 ˜˜˜ (8848 m) = 6.94 × 10 6 J
Ë
¯
PTS: 1
68. ANS:
0.145 J
DIF:
IIIA
OBJ: 5-2.6
Given
k = 275 N/m
x = 3.25 cm = 3.25 × 10 −2 m
Solution
PE =
PTS:
1
2
kx 2 =
1
1
2
Ê
ˆ2
(275 N/m) ÁÁÁ 3.25 × 10 −2 m ˜˜˜ = 0.145 J
Ë
¯
DIF:
IIIA
OBJ: 5-2.6
26
ID: B
69. ANS:
0.54 m/s
Given
h = 1.5 cm = 1.5 × 10 −2 m
g = 9.81 m/s 2
Solution
ME i = ME f
1
2
mgh i =
vf =
mv f 2
2gh =
PTS: 1
70. ANS:
2.5 kW
Ê
ˆÊ
ˆ
(2) ÁÁÁ 9.81 m/s 2 ˜˜˜ ÁÁÁ 1.5 × 10 −2 m ˜˜˜ = 0.54 m/s
Ë
¯Ë
¯
DIF:
IIIA
OBJ: 5-3.3
Given
m = 250 kg
d = 2.0 m
Δt = 2.0 s
g = 9.81 m/s 2
Solution
ÊÁ 250 kg ˆ˜ ÊÁÁÁ 9.81 m/s 2 ˆ˜˜˜ (2.0 m)
Ë
¯Ë
mgd
¯
P = W = Fd =
=
= 2.5 × 10 3 W = 2.5 kW
Δt
Δt
Δt
2.0 s
PTS:
1
DIF:
IIIA
OBJ: 5-4.2
27