Name: ________________________ Class: ___________________ Date: __________ ID: A
1
Fall 2012 Physics Final Exam
Multiple Choice
Make no marks on these sheets. Identify the choice that best completes the statement or answers the question and record
them on the bubble sheet.
1. (1 point) The magnitude of the gravitational force acting on an object is
a. mass. b. frictional force. c. weight. d. inertia.
2. (1 point) Which of the following equations expresses Newton’s law of universal gravitation?
a. Fc =mv t
2
r b. g = G
mE
r2 c. Fg = G
m1m2
r2 d. Fg =
m1m2
r
3. (1 point) The Greek letter ∆ (delta) indicates a(n)
a. difference or change. b. inverse proportion. c. direct proportion. d. sum or total.
4. (1 point) Acceleration is defined as
a. a rate of displacement. b. the rate of change of displacement. c. the rate of change of velocity. d. the change in
velocity.
5. (1 point) A coin released at rest from the top of a tower hits the ground after falling 1.5 s. What is the speed of the coin as it
hits the ground? (Disregard air resistance. a = −g = −9.81 m/s2.)
a. 15 m/s2 b. 150 m/s c. 1.5 m/s d. 15 m/s
6. (1 point) What is the speed of an object at rest?
a. 1.0 m/s b. 9.81 m/s c. 9.8 m/s d. 0.0 m/s
7. (1 point) Which of the following is the tendency of an object to maintain its state of motion?
a. inertia b. force c. acceleration d. velocity
8. (1 point) If you want to open a swinging door with the least amount of force, where should you push on the door?
a. in the middle b. as far from the hinges as possible c. It does not matter where you push. d. close to the
hinges
9. (1 point) When an object is moving with uniform circular motion, the object’s tangential speed
a. is circular. b. is perpendicular to the plane of motion. c. is constant. d. is directed toward the center of
motion.
Name: ________________________ ID: A
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10. (1 point) How many displacement vectors shown in the figure above have components that lie along the y-axis and are
pointed in the –y direction?
a. 0 b. 5 c. 2 d. 3
11. (1 point) How many displacement vectors shown in the figure above have horizontal components?
a. 5 b. 2 c. 4 d. 3
12. (1 point) Which displacement vectors shown in the figure above have vertical components that are equal?
a. d1 and d2 b. d4 and d5 c. d1 and d3 d. d2 and d5
13. (1 point) A model rocket flies horizontally off the edge of a cliff at a velocity of 50.0 m/s. If the canyon below is 100.0
m deep, how far from the edge of the cliff does the model rocket land? (a y = −g = −9.81 m/s2)
a. 22.6 m b. 220 m c. 226 m d. 22.0 m
14. (1 point) Which are simultaneous equal but opposite forces resulting from the interaction of two objects?
a. action-reaction pairs b. field forces c. net external forces d. gravitational forces
15. (1 point) Which of the following energy forms is associated with an object due to its position?
a. total energy b. kinetic energy c. potential energy d. positional energy
16. (1 point) As an object falls toward Earth,
a. the object exerts a downward force on Earth. b. the upward acceleration of Earth is negligible because of its large
mass. c. the object does not exert a force on Earth. d. Newton’s third law does not apply.
17. (1 point) What causes a moving object to change direction?
a. velocity b. force c. acceleration d. inertia
18. (1 point) Which of the following equations expresses Newton’s law of universal gravitation?
a. g = GmE
r2 b. Fc =
mv t2
r c. Fg =
m1m2
r d. Fg = G
m1m2
r2
19. (1 point) If a is acceleration (m/s2), ∆v is change in velocity (m/s), ∆x is change in position (m), and ∆t is the time
interval (s), which equation is not dimensionally correct?
a. ∆v = a ⁄∆t b. ∆t2 = 2∆x ⁄a c. ∆t = ∆x ⁄v d. a = v2⁄∆x
20. (1 point) When a car makes a sharp left turn, what causes the passengers to move toward the right side of the car?
a. inertia b. centripetal acceleration c. centrifugal force d. centripetal force
Name: ________________________ ID: A
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21. (1 point) A car goes forward along a level road at constant velocity. The additional force needed to bring the car into
equilibrium is
a. zero. b. greater than the normal force times the coefficient of static friction. c. equal to the normal force times the
coefficient of static friction. d. the normal force times the coefficient of kinetic friction.
22. (1 point) 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. 55 J b. 1250 J c. 250 J d. 10 J
23. (1 point) What is the term for the net force directed toward the center of an object’s circular path?
a. centripetal force b. circular force c. orbital force d. centrifugal force
24. (1 point) Which of the following units is the SI unit of velocity?
a. meter•second b. second per meter c. meter d. meter per second
25. (1 point) In this text, which of the following symbols represents gravitational field strength?
a. Fc b. g c. G d. Fg
26. (1 point) There are 20 tumtum trees in the tulgey wood.
In each tulgey wood is one frumious Bandersnatch.
There are 5 slithy toves in 2 borogoves.
There are 2 mome raths per Jabberwock.
There are 2 Jubjub birds in 200 tumtum trees.
There are 200 mome raths in each borogove.
There are 5 Jubjub birds per slithy tove.
If there are 5 frumious Bandersnatches, how many Jabberwocks should there be?
a. 8.5 b. 18 c. 6 d. 8
27. (1 point) A worker does 25 J of work lifting a bucket, then sets the bucket back down in the same place. What is the
total net work done on the bucket?
a. –25 J b. 50 J c. 0 J d. 25 J
28. (1 point) When calculating the gravitational force between two extended bodies, you should measure the distance
a. from the most distant points on each body. b. from the center of each body. c. from the closest points on
each body. d. from the center of one body to the closest point on the other body.
29. (1 point) In a coordinate system, a vector is oriented at angle θ with respect to the x-axis. The x component of the vector equals
the vector’s magnitude multiplied by which trigonometric function?
a. tan θ b. cos θ c. cot θ d. sin θ
30. (1 point) The SI base unit used to measure mass is the
a. liter. b. second. c. kilogram. d. meter.
A child rides a bicycle in a circular path with a radius of 2.0 m. The tangential speed of the bicycle is 2.0 m/s. The
combined mass of the bicycle and the child is 43 kg.
31. (1 point) What kind of force provides the centripetal force on the bicycle?
a. friction b. air resistance c. gravitational force d. normal force
32. (1 point) What is the magnitude of the centripetal force on the bicycle?
a. 4.0 N b. 43 N c. 86 N d. 3.7 kN
Name: ________________________ ID: A
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33. (1 point) What is the magnitude of the bicycle’s centripetal acceleration?
a. 2.0 m/s2 b. 4.0 m/s2 c. 1.0 m/s2 d. 8.0 m/s2
34. (1 point) In which of the following scenarios is work done?
a. A student holds a spring in a compressed position. b. A car decelerates while traveling on a flat stretch of road.
c. A weightlifter holds a barbell overhead for 2.5 s. d. A construction worker carries a heavy beam while walking
at constant speed along a flat surface.
35. (1 point) A car on a roller coaster loaded with passengers has a mass of 2.0 × 103 kg. At the lowest point of the track,
the radius of curvature of the track is 24 m and the roller car has a tangential speed of 17 m/s. What centripetal force
is exerted on the roller car at the lowest point?
a. 2.4 × 104 N b. 2.4 × 101 N c. 2.4 × 103 N d. 2.4 × 102 N
36. (1 point) Which of the following energy forms is associated with an object in motion?
a. potential energy b. kinetic energy c. nonmechanical energy d. elastic potential energy
37. (1 point) A 35 kg child moves with uniform circular motion while riding a horse on a carousel. The horse is 3.2 m from
the carousel’s axis of rotation and has a tangential speed of 2.6 m/s. What is the centripetal force on the child?
a. 0.74 N b. 740 N c. 7.4 N d. 74 N
38. (1 point) The symbol mm represents a
a. micrometer. b. millimeter. c. megameter. d. manometer.
39. (1 point) 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. 11000 J b. 550 J c. 6000 J d. 2000 J
40. (1 point) 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/s2)
a. 280 N b. 20 N c. 28 N d. 2.8 N
41. (1 point) A hiker uses a compass to navigate through the woods. Identify the area of physics that this involves.
a. relativity b. thermodynamics c. electromagnetism d. quantum mechanics
42. (1 point) Which of the following statements does not describe force?
a. Force causes objects to start moving. b. Force causes objects to stop moving. c. Force causes objects to change
direction. d. Force causes objects at rest to remain stationary.
43. (1 point) While following directions on a treasure map, a person walks 45.0 m south, then turns and walks 7.50 m east. Which
single straight-line displacement could the person have walked to reach the same spot?
a. 4.56 m @ 80.5 deg S of E b. 45.6 m @ 80.5 deg E of S c. 45.6 m @ 18.5 deg S of E d. 45.6 m @ 80.5
deg S of E
44. (1 point) A race car accelerates from 0.0 m/s to 30.0 m/s with a displacement of 45.0 m. What is the car’s acceleration?
a. 9.0 m/s2 b. 19.0 m/s
2 c. 10.0 m/s
2 d. 100.0 m/s
2
45. (1 point) The gravitational force between two masses is 36 N. What is the gravitational force if the distance between
them is tripled? (G = 6.673 × 10−11 N•m2/kg2)
a. 18 N b. 4.0 N c. 27 N d. 9.0 N
Name: ________________________ ID: A
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46. (1 point) The sun is composed mostly of hydrogen. The mass of the sun is 2.0 × 1030 kg, and the mass of a hydrogen
atom is 1.67 × 10−27 kg. Estimate the number of atoms in the sun.
a. 1030 b. 103 c. 1075 d. 1057
47. (1 point) In this text, which of the following symbols represents the constant of universal gravitation?
a. Fg b. g c. Fc d. G
48. (1 point) What is the speed of an object at rest?
a. 0.0 m/s b. 9.8 m/s c. 1.0 m/s d. 9.81 m/s
49. (1 point) A force does work on an object if a component of the force
a. is parallel to the displacement of the object. b. parallel to the displacement of the object moves the object along
a path that returns the object to its starting position. c. perpendicular to the displacement of the object moves the
object along a path that returns the object to its starting position. d. is perpendicular to the displacement of the
object.
50. (1 point) When a car makes a sharp left turn, what causes the passengers to move toward the right side of the car?
a. centripetal acceleration b. centripetal force c. inertia d. centrifugal force
51. (1 point) Which of the following quantities measures the ability of a force to rotate or accelerate an object around an
axis?
a. torque b. lever arm c. axis of rotation d. tangential force
52. (1 point) In a coordinate system, the magnitude of the x component of a vector and θ, the angle between the vector and x-axis,
are known. The magnitude of the vector equals the x component
a. divided by the cosine of θ. b. multiplied by the cosine of θ. c. multiplied by the sine of θ. d. divided by the sine of
θ.
53. (1 point) Ball A has triple the mass and speed of ball B. What is the ratio of the kinetic energy of ball A to ball B.
a. 27 b. 3 c. 9 d. 6
54. (1 point) In a coordinate system, a vector is oriented at angle θ with respect to the x-axis. The y component of the vector equals
the vector’s magnitude multiplied by which trigonometric function?
a. tan θ b. cot θ c. cos θ d. sin θ
55. (1 point) If the change in position ∆x is related to velocity v (with units of m/s) in the equation ∆x = Av, the constant A
has which dimension?
a. s b. m/s2 c. m d. m2
56. (1 point) What is the path of a projectile (in the absence of friction)?
a. a parabola b. a wavy line c. a hyperbola d. Projectiles do not follow a predictable path.
57. (1 point) A 35 kg child moves with uniform circular motion while riding a horse on a carousel. The horse is 3.2 m from
the carousel’s axis of rotation and has a tangential speed of 2.6 m/s. What is the child’s centripetal acceleration?
a. 2.1 m/s^2 b. 210 m/s^2 c. 0.21 m/s^2 d. 21 m/s^2
58. (1 point) Gravitational potential energy is always measured in relation to
a. a zero level. b. total potential energy. c. kinetic energy. d. mechanical energy.
59. (1 point) The most appropriate SI unit for measuring the length of an automobile is the
a. kilometer. b. meter. c. nanometer. d. micron.
Name: ________________________ ID: A
6
60. (1 point) According to Newton’s second law, when the same force is applied to two objects of different masses,
a. the object with greater mass will experience a greater acceleration, and the object with less mass will experience
a smaller acceleration. b. the object with greater mass will experience a great acceleration, and the object with less
mass will experience an even greater acceleration. c. the object with greater mass will experience a small
acceleration, and the object with less mass will experience an even smaller acceleration. d. the object with greater
mass will experience a smaller acceleration, and the object with less mass will experience a greater acceleration.
61. (1 point) In general, Fnet equals
a. Ff. b. Fn. c. ΣF. d. Fg.
62. (1 point) Which of the following is a physical quantity that has both magnitude and direction?
a. frame of reference b. scalar c. resultant d. vector
63. (1 point) Which of the following forces arises from direct physical contact between two objects?
a. field force b. fundamental force c. contact force d. gravitational force
64. (1 point) When calculating the gravitational force between two extended bodies, you should measure the distance
a. from the closest points on each body. b. from the most distant points on each body. c. from the center of
each body. d. from the center of one body to the closest point on the other body.
65. (1 point) 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?
a. 1.4 N, downward b. 14 N, downward c. 1.4 N, upward d. 14 N, upward
66. (1 point) A newton is equivalent to which of the following quantities?
a. kg•m/s2 b. kg•(m/s)
2 c. kg•m/s d. kg
67. (1 point) 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.)
a. 96 m b. 92 m c. 94 m d. 90 m
68. (1 point) What is the potential energy of a 1.0 kg mass 1.0 m above the ground?
a. 96 J b. 9.8 J c. 1.0 J d. 10 J
69. (1 point) The length of a force vector represents the
a. direction of the force. b. cause of the force. c. type of force. d. magnitude of the force.
70. (1 point) Which of the following can be a centripetal force?
a. gravity b. tension c. friction d. all of the above
71. (1 point) If a nonzero net force is acting on an object, then the object is definitely
a. losing mass. b. being accelerated. c. moving with a constant velocity. d. at rest.
72. (1 point) Which of the following is the motion of objects moving in two dimensions under the influence of gravity?
a. directrix b. projectile motion c. vertical velocity d. horizontal velocity
73. (1 point) A professional skier starts from rest and reaches a speed of 56 m/s on a ski slope angled 30.0° above the
horizontal. Using the work-kinetic energy theorem and disregarding friction, find the minimum distance along the
slope the skier would have to travel in order to reach this speed.
a. 32 m b. 320 m c. 30 m d. 300 m
Name: ________________________ ID: A
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74. (1 point) In which of the following scenarios is no net work done?
a. A car decelerates as it travels up a hill. b. A car accelerates down a hill. c. A car decelerates on a flat road.
d. A car travels at constant speed on a flat road.
75. (1 point) In a coordinate system, a vector is oriented at angle θ with respect to the x-axis. The x component of the vector equals
the vector’s magnitude multiplied by which trigonometric function?
a. cot θ b. tan θ c. sin θ d. cos θ
76. (1 point) What is the SI unit of acceleration?
a. m2/s b. m/s c. m•s
2 d. m/s
2
77. (1 point) Which of the following equations expresses the work-kinetic energy theorem?
a. ∆W = ∆KE b. Wnet = ∆PE c. Wnet = ∆KE d. MEi = ME f
78. (1 point) A worker pushes a box with a horizontal force of 50.0 N over a level distance of 5.0 m. If a frictional force of
43 N acts on the box in a direction opposite to that of the worker, what net work is done on the box?
a. 3.2 J b. 32 J c. 3.5 J d. 35 J
79. (1 point) The SI base unit for time is
a. 1 hour. b. 1 second. c. 1 minute. d. 1 day.
80. (1 point) When an object is moving with uniform circular motion, the centripetal acceleration of the object
a. is zero. b. is circular. c. is perpendicular to the plane of motion. d. is directed toward the center of motion.
81. (1 point) A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s2. What is the final
velocity after 6.0 s?
a. 17 m/s b. 15 m/s c. 18 m/s d. 16 m/s
82. (1 point) Which of the following is the equation for average velocity?
a. vavg =∆t∆x
b. vavg = ∆x∆t c. vavg =v i − v f
2 d. vavg =
∆x∆t
83. (1 point) The centripetal force on an object in circular motion is
a. in the direction opposite the centripetal acceleration. b. in the direction opposite the tangential speed. c. in the
same direction as the tangential speed. d. in the same direction as the centripetal acceleration.
84. (1 point) Which of the following is a physical quantity that has a magnitude but no direction?
a. frame of reference b. vector c. scalar d. resultant
85. (1 point) A hammer drives a nail into a piece of wood. Identify an action-reaction pair in this situation.
a. The hammer exerts a force on the nail; the wood exerts a force on the nail. b. The hammer exerts a force on the nail;
the nail exerts a force on the hammer. c. The nail exerts a force on the hammer; the hammer exerts a force on the wood.
d. The hammer exerts a force on the nail; the hammer exerts a force on the wood.
86. (1 point) A car on a roller coaster loaded with passengers has a mass of 2.0 × 103 kg. At the lowest point of the track,
the radius of curvature of the track is 24 m and the roller car has a tangential speed of 17 m/s. What is the
centripetal acceleration of the roller car at the lowest point on the track?
a. 10 m/s2 b. 14 m/s2 c. 11 m/s2 d. 12 m/s2
87. (1 point) 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.
a. 20.5 J b. 21.5 J c. 215 J d. 205 J
Name: ________________________ ID: A
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88. (1 point) Objects that are falling toward Earth in free fall move
a. at a constant velocity. b. slower then faster. c. faster and faster. d. slower and slower.
89. (1 point) 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. 0.69 m/s2 b. 4.6 m/s
2 c. 0.22 m/s
2 d. 3.2 m/s
2
90. (1 point) A lack of precision in scientific measurements typically arises from
a. lack of calibration. b. human error. c. limitations of the measuring instrument. d. too many significant
figures.
91. (1 point) What is the common formula for work? Assume that W is the work, F is a constant force, ∆v is the change in
velocity, and d is the displacement.
a. W = F2d b. W = Fd2 c. W = Fd d. W = F∆v
92. (1 point) The main difference between kinetic energy and potential energy is that
a. kinetic energy involves motion, and potential energy involves position. b. although both energies involve
position, only potential energy involves motion. c. although both energies involve motion, only kinetic energy
involves position. d. kinetic energy involves position, and potential energy involves motion.
93. (1 point) A pebble falls from the bottom of the basket of a hot-air balloon that is rising at 1.0 m/s. After 3.0 seconds, how far
below the basket is the pebble? (Assume no air resistance and a = −g = −9.81 m/s2.)
a. 46 m b. 4 m c. 4.4 m d. 44 m
94. (1 point) What two dimensions, in addition to mass, are commonly used by physicists to derive additional
measurements?
a. area and mass b. velocity and time c. length and width d. length and time
95. (1 point) Which of the following is the cause of an acceleration?
a. inertia b. velocity c. force d. speed
96. (1 point) Which would hit the ground first if dropped from the same height in a vacuum—a feather or a metal bolt?
a. the metal bolt b. They would be suspended in a vacuum. c. the feather d. They would hit the ground at the same
time.
97. (1 point) A ball is whirled on a string, then the string breaks. What causes the ball to move off in a straight line?
a. centripetal acceleration b. centrifugal force c. centripetal force d. inertia
98. (1 point) Find the resultant of these two vectors: 2.00 × 102 units due east and 4.00 × 10
2 units 30.0° north of west.
a. 300 units, 29.8° north of west b. 546 units, 59.3° north of west c. 581 units, 20.1° north of east d. 248 units, 53.9°
north of west
99. (1 point) A measure of the quantity of matter is
a. mass. b. force. c. weight. d. density.
100. (1 point) The statement by Newton that for every action there is an equal but opposite reaction is which of his laws of motion?
a. third b. second c. fourth d. first
Name: ________________________ ID: A
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Short Answer. In the space below the box, write your answers to Short Answer and show your work for the
Problems.
101. (12 points) In increasing order, list the four metric prefixes, symbol and their respective multiplier that you are expected
to memorize for Physics.
102. (21 points) For the seven SI base units, list the unit’s names, the of measure, and symbol.
Problems.
103. (1 point) A coin released at rest from the top of a tower hits the ground after falling 1.5 s. What is the speed of the coin as it
hits the ground? (Disregard air resistance. a = −g = −9.81 m/s2.)
104. (1 point) 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.)
105. (1 point) 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?
106. (1 point) A bucket filled with water has a mass of 23 kg and is attached to a rope wound around a cylinder with a
radius of 0.050 m at the top of a well. What torque does the weight of the water and bucket produce on the cylinder?
(g = 9.81 m/s2)
107. (1 point) How much energy would be required to do 675 J of work with a machine that was 35% efficient?
108. (1 point) 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?
109. (1 point) A 40.0 N crate starting at rest slides down a rough 6.0 m long ramp inclined at 30.0° with the horizontal. The
force of friction between the crate and ramp is 6.0 N. Using the work-kinetic energy theorem, find the velocity of the
crate at the bottom of the incline.
110. (1 point) 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.
ID: A
1
Fall 2012 Physics Final Exam
Answer Section
MULTIPLE CHOICE
1. ANS: C PTS: 1 DIF: I OBJ: 4-4.1
2. ANS: C PTS: 1 DIF: I OBJ: 7-2.2
3. ANS: A PTS: 1 DIF: I OBJ: 1-3.2
4. ANS: C PTS: 1 DIF: I OBJ: 2-2.1
5. ANS: D PTS: 1
6. ANS: D PTS: 1 DIF: I OBJ: 2-1.1
7. ANS: A PTS: 1 DIF: I OBJ: 4-2.1
8. ANS: B PTS: 1 DIF: I OBJ: 7-4.1
9. ANS: C PTS: 1 DIF: I OBJ: 7-1.1
10. ANS: C PTS: 1 DIF: I OBJ: 3-2.3
11. ANS: C PTS: 1 DIF: I OBJ: 3-2.3
12. ANS: C PTS: 1 DIF: I OBJ: 3-2.3
13. ANS: C PTS: 1
14. ANS: A PTS: 1 DIF: I OBJ: 4-3.3
15. ANS: C PTS: 1 DIF: I OBJ: 5-2.1
16. ANS: B PTS: 1 DIF: II OBJ: 4-3.3
17. ANS: B PTS: 1 DIF: I OBJ: 4-1.1
18. ANS: D PTS: 1 DIF: I OBJ: 7-2.2
19. ANS: A
Solution
Does ∆v =a∆t
?
Substitute units into the right side of the equation.
a∆t
=m/s2
s=
m
s2
Ê
Ë
ÁÁÁÁÁÁÁÁ
ˆ
¯
˜̃̃˜̃̃˜̃
1s
Ê
Ë
ÁÁÁÁÁÁ
ˆ
¯
˜̃̃˜̃̃
s( )1s
Ê
Ë
ÁÁÁÁÁÁ
ˆ
¯
˜̃̃˜̃̃
=m
s3
Substituting units into both sides of the equation yields the following result:
ms
≠m
s3
PTS: 1 DIF: IIIA OBJ: 1-3.3
20. ANS: A PTS: 1 DIF: I OBJ: 7-1.3
21. ANS: A PTS: 1 DIF: I OBJ: 4-2.3
ID: A
2
22. ANS: C
2.5 × 102 J
Given
F = 50.0 N
d = 5.0 m
Solution
W = Fd = 50.0 N( ) 5.0 m( ) = 2.5 × 102 J
PTS: 1 DIF: IIIA OBJ: 5-1.4
23. ANS: A PTS: 1 DIF: I OBJ: 7-1.2
24. ANS: D PTS: 1 DIF: I OBJ: 2-1.1
25. ANS: B PTS: 1 DIF: I OBJ: 7-2.2
26. ANS: D PTS: 1
27. ANS: C PTS: 1 DIF: II OBJ: 5-1.3
28. ANS: B PTS: 1 DIF: I OBJ: 7-2.2
29. ANS: B PTS: 1 DIF: I OBJ: 3-2.3
30. ANS: C PTS: 1 DIF: I OBJ: 1-2.1
31. ANS: A PTS: 1 DIF: II OBJ: 7-1.2
32. ANS: C
Given
m = 43 kg
v t = 2.0 m/s
r = 2.0 m
Solution
Fc =mv t
2
r=
43 kgÊËÁÁ ˆ
¯˜̃ 2.0 m/s( )
2
2.0 m= 86 N
PTS: 1 DIF: IIIA OBJ: 7-1.2
33. ANS: A
Given
v t = 2.0 m/s
r = 2.0 m
Solution
ac =vt
2
r=
2.0 m/s( )2
2.0 m= 2.0 m/s2
PTS: 1 DIF: IIIA OBJ: 7-1.1
ID: A
3
34. ANS: B PTS: 1 DIF: I OBJ: 5-1.3
35. ANS: A
2.4 × 104 N
Given
m = 2.0 × 103 kg
v t = 17 m/s
r = 24 m
Solution
Fc =mv t
2
r=
2.0 × 103 kgÊ
ËÁÁÁÁ
ˆ
¯˜̃̃˜ 17 m/s( )
2
24 m= 2.4 × 104 N
PTS: 1
36. ANS: B PTS: 1 DIF: I OBJ: 5-2.1
37. ANS: D
74 N
Given
m = 35 kg
v t = 2.6 m/s
r = 3.2 m
Solution
Fc =mv t
2
r=
35 kgÊËÁÁ ˆ
¯˜̃ 2.6 m/s( )
2
3.2 m= 74 N
PTS: 1
38. ANS: B PTS: 1 DIF: I OBJ: 1-2.1
39. ANS: D
Given
F = 200 N
d = 10 m
Solution
W = Fd = 200 N( ) 10 m( ) = 2 × 103 J
PTS: 1 DIF: IIIA OBJ: 5-1.4
40. ANS: D PTS: 1
41. ANS: C PTS: 1 DIF: I OBJ: 1-1.1
42. ANS: D PTS: 1 DIF: I OBJ: 4-1.1
ID: A
4
43. ANS: D PTS: 1
44. ANS: C PTS: 1
45. ANS: B
Given
F1 = 36 N
r2 = 3r1
G = 6.673 × 10−11 N•m2/kg2
Solution
r2 = 3r1
F1 = Gm1m2
r12
= 36 N
F2 = Gm1m2
r22
= Gm1m2
3r1ÊËÁÁ ˆ
¯˜̃
2= G
m1m2
9r12
=1
9G
m1m2
r12
=1
9F1
F2 =1
936 N( ) = 4.0 N
PTS: 1 DIF: II OBJ: 7-2.2
46. ANS: D
Given
msun = 2.0 × 1030 kg
mH atom = 1.67 × 10−27 kg
Solution
Estimate the answer using an order-of-magnitude calculation.
1030
10−27= 1057
PTS: 1 DIF: IIIB OBJ: 1-3.4
47. ANS: D PTS: 1 DIF: I OBJ: 7-2.2
48. ANS: A PTS: 1 DIF: I OBJ: 2-1.1
49. ANS: A PTS: 1 DIF: I OBJ: 5-1.2
50. ANS: C PTS: 1 DIF: I OBJ: 7-1.3
51. ANS: A PTS: 1 DIF: I OBJ: 7-4.1
52. ANS: A PTS: 1 DIF: II OBJ: 3-2.3
53. ANS: A PTS: 1 DIF: II OBJ: 5-2.2
54. ANS: D PTS: 1 DIF: I OBJ: 3-2.3
ID: A
5
55. ANS: A
Solution
∆x = Av
Rearrange the equation to solve for A and substitute units.
A =∆xv
=m
m/s= s
PTS: 1 DIF: IIIA OBJ: 1-3.3
56. ANS: A PTS: 1 DIF: I OBJ: 3-3.2
57. ANS: A
2.1 m/s2
Given
v t = 2.6 m/s
r = 3.2 m
Solution
ac =vt
2
r=
2.6 m/s( )2
3.2 m= 2.1 m/s2
PTS: 1
58. ANS: A PTS: 1 DIF: I OBJ: 5-2.5
59. ANS: B PTS: 1 DIF: II OBJ: 1-2.1
60. ANS: D PTS: 1 DIF: II OBJ: 4-3.1
61. ANS: C PTS: 1 DIF: II OBJ: 4-3.1
62. ANS: D PTS: 1 DIF: I OBJ: 3-1.1
63. ANS: C PTS: 1 DIF: I OBJ: 4-1.1
64. ANS: C PTS: 1 DIF: I OBJ: 7-2.2
65. ANS: D PTS: 1
66. ANS: A PTS: 1 DIF: I OBJ: 4-1.1
67. ANS: B PTS: 1
68. ANS: B
Given
m = 1.0 kg
h = 1.0 m
g = 9.81 m/s2
Solution
PE = mgh = 1.0 kgÊËÁÁ ˆ
¯˜̃ 9.81 m/s2Ê
ËÁÁÁÁ
ˆ
¯˜̃̃˜ 1.0 m( ) = 9.8 J
PTS: 1 DIF: IIIA OBJ: 5-2.6
ID: A
6
69. ANS: D PTS: 1 DIF: I OBJ: 4-1.2
70. ANS: D PTS: 1 DIF: I OBJ: 7-1.2
71. ANS: B PTS: 1 DIF: I OBJ: 4-3.1
72. ANS: B PTS: 1 DIF: I OBJ: 3-3.1
73. ANS: B
320 m
Given
vi = 0 m/s
v f = 56 m/s
θ = 30.0°
g = 9.81 m/s2
Solution
Wnet = ∆KE
Wnet = Fd = (mg sinθ)d
∆KE = KE f − KEi =1
2mv f
2 − 0 =1
2mv f
2
mgdsinθ =1
2mv f
2
d =v f
2
2gsinθ
d =56 m/s( )
2
2( ) 9.81 m/s2Ê
ËÁÁÁÁ
ˆ
¯˜̃̃˜ sin30.0°( )
= 3.2 × 102 m
PTS: 1
74. ANS: D PTS: 1 DIF: I OBJ: 5-1.3
75. ANS: D PTS: 1 DIF: I OBJ: 3-2.3
76. ANS: D PTS: 1 DIF: I OBJ: 2-2.1
77. ANS: C PTS: 1 DIF: I OBJ: 5-2.3
ID: A
7
78. ANS: D
35 J
Given
Fw = 50.0 N
Fk = −43 N
d = 5.0 m
Solution
Wnet = Fnetd = Fw + FkÊËÁÁ ˆ
¯˜̃d = 50.0 N( ) + −43 N( )
ÈÎÍÍÍÍ
˘˚˙̇̇˙ 5.0 m( ) = 35 J
PTS: 1
79. ANS: B PTS: 1 DIF: I OBJ: 1-2.1
80. ANS: D PTS: 1 DIF: I OBJ: 7-1.1
81. ANS: A PTS: 1
82. ANS: D PTS: 1 DIF: I OBJ: 2-1.1
83. ANS: D PTS: 1 DIF: I OBJ: 7-1.2
84. ANS: C PTS: 1 DIF: I OBJ: 3-1.1
85. ANS: B PTS: 1 DIF: II OBJ: 4-3.3
86. ANS: D
12 m/s2
Given
vt = 17 m/s
r = 24 m
Solution
ac =vt
2
r=
17 m/s( )2
24 m= 12 m/s2
PTS: 1
87. ANS: C
215 J
Given
F = 43.0 N
d = 5.00 m
Solution
W = Fd = 43.0 N( ) 5.00 m( ) = 215 J
PTS: 1
88. ANS: C PTS: 1 DIF: I OBJ: 2-3.3
ID: A
8
89. ANS: C
Given
Fapplied = 6.8 N
m = 31 kg
Solution
Fnet = Fx = Fapplied = max∑
ax =Fapplied
m=
6.8 N31 kg
= 0.22 m/s2
PTS: 1 DIF: IIIA OBJ: 4-3.2
90. ANS: C PTS: 1 DIF: I OBJ: 1-2.3
91. ANS: C PTS: 1 DIF: I OBJ: 5-1.2
92. ANS: A PTS: 1 DIF: I OBJ: 5-2.4
93. ANS: D PTS: 1
94. ANS: D PTS: 1 DIF: I OBJ: 1-2.1
95. ANS: C PTS: 1 DIF: I OBJ: 4-1.1
96. ANS: D PTS: 1 DIF: I OBJ: 2-3.3
97. ANS: D PTS: 1 DIF: I OBJ: 7-1.3
ID: A
9
98. ANS: DGiven
d1 = 2.00 × 102 units east
d2 = 4.00 × 102 units 30.0° north of west
Solution
Measuring direction with respect to x = (east),
∆x1 = 2.00 × 102
units
∆y1 = 0
∆x2 = d2 cos θ = (4.00 × 102
units)(cos 150.0°) = −3.46 × 102
units
∆y2 = d2 sinθ = (4.00 × 102
units)(sin 150.0°) = 2.00 × 102
units
∆x tot = ∆x1 + ∆x2 = (2.00 × 102 units) + (−3.46 × 102 units) = −1.46 × 102 units
∆y tot = ∆y1 + ∆y2 = 0 + (2.00 × 102
units) = 2.00 × 102
units
d2
= (∆x tot)2
+ (∆y tot)2
d = (∆x tot)2
+ (∆y tot )2
= (−1.46 × 102
units)2
+ (2.00 × 102
units)2
d = 2.48 × 102
units
θ = tan−1
∆y tot
∆x tot
Ê
Ë
ÁÁÁÁÁÁÁÁÁÁ
ˆ
¯
˜̃˜̃˜̃˜̃˜̃
= tan−1 2.00 × 10
2units
–1.46 × 102units
Ê
Ë
ÁÁÁÁÁÁÁÁÁÁ
ˆ
¯
˜̃˜̃˜̃˜̃˜̃
= −53.9°
d = 2.48 × 102
units, 53.9°north of west
PTS: 1 DIF: IIIB OBJ: 3-2.4
99. ANS: A PTS: 1 DIF: I OBJ: 4-4.1
100. ANS: A PTS: 1 DIF: I OBJ: 4-3.3
SHORT ANSWER
101. ANS:
micro- µ 10−6
milli- m 10−3
centi- c 10−2
kilo- k 103
PTS: 12
ID: A
10
102. ANS:
time seconds s
length meter m
mass (kilo)gram kg or g
temperature Kelvin K
luminosity candela cd
elec current amp, ampere A
amount of substance mole mol
PTS: 21
PROBLEM
103. ANS: 15 m/s
Given
a = −g = −9.81 m/s2
v i = 0.0 m/s
∆t = 1.5 s
Solution
v f = v i + a∆t = 0.0 m/s + (–9.81 m/s2)(1.5 s) = –15 m
speed = 15 m/s
PTS: 1 DIF: IIIB OBJ: 2-3.2
104. ANS: 92 m
Given
d = 120 m, θ = 40.0°
Solution
dx = d cos θ = (120 m)(cos 40.0°) = 92 m
PTS: 1 DIF: IIIA OBJ: 3-2.3
ID: A
11
105. ANS:
10.4 N
Given
m = 1.10 kg
θ = 15.0°
g = 9.81 m/s2
Solution
Fnet, y = ΣF y = Fn − Fg, y = 0
Fn = Fg, y = Fg cosθ = mgcosθ
Fn = (1.10 kg)(9.81 m/s2)(cos15.0°) = 10.4 N
PTS: 1 DIF: IIIA OBJ: 4-4.2
106. ANS:
11 N•m
Given
m = 23 kg
d = 0.050 m
g = 9.81 m/s2
Solution
τ = Fd = mgd = 23 kgÊËÁÁ ˆ
¯˜̃ 9.81 m/s2Ê
ËÁÁÁÁ
ˆ
¯˜̃̃˜ 0.050 m( ) = 11 Nm
PTS: 1 DIF: IIIA OBJ: 7-4.2
107. ANS:
1900 J
Given
Wout = 675 J
eff = 35% = 0.35
Solution
eff =Wout
Win
Win =Wout
eff=
670 J0.35
= 1.9 × 103 J
PTS: 1 DIF: IIIB OBJ: 7-4.4
ID: A
12
108. ANS:
6.94 × 106 J
Given
m = 80.0 kg
h = 8848 m
g = 9.81 m/s2
Solution
PE = mgh = 80.0 kgÊËÁÁ ˆ
¯˜̃ 9.81 m/s2Ê
ËÁÁÁÁ
ˆ
¯˜̃̃˜ 8848 m( ) = 6.94 × 106 J
PTS: 1 DIF: IIIA OBJ: 5-2.6
ID: A
13
109. ANS:
6.4 m/s
Given
vi = 0 m/s
Fg = 40.0 N
θ = 30.0°
d = 6.0 m
Fk = −6.0 N
g = 9.81 m/s2
Solution
Wnet = ∆KE
Wnet = Fnetd = Fg sinθ − Fk
ÊËÁÁÁ
ˆ¯˜̃̃d
∆KE = KE f − KEi =1
2mv f
2 − 0 =1
2mv f
2
Fg = mg
m =Fg
g
Fg sinθ + Fk
ÊËÁÁÁ
ˆ¯˜̃̃d =
1
2mv f
2 =1
2
Fg
g
Ê
Ë
ÁÁÁÁÁÁÁÁÁÁÁ
ˆ
¯
˜̃̃˜̃̃˜̃̃˜̃v f
2
v f =2dg Fg sinθ + Fk
ÊËÁÁÁ
ˆ¯˜̃̃
Fg
v f =2( ) 6.0 m( )(9.81 m/s2) 40.0 N( ) sin30.0°( ) + −6.0 N( )
ÈÎÍÍÍÍ
˘˚˙̇̇˙
40.0 N= 6.4 m/s
PTS: 1 DIF: IIIC OBJ: 5-2.3
110. ANS:
215 J
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