Regent Physics Review

1
Regent Physics Review
2
Physics Units

• I. Physics Skills
• II. Mechanics
• III. Energy
• IV. Electricity and Magnetism
• V. Waves
• VI. Modern Physics

3
I. Physics Skills

• A. Scientific Notation
• B. Graphing
• C. Significant Figures
• D. Units
• E. Prefixes
• F. Estimation

4
A. Scientific Notation

• Use for very large or very small numbers
• Write number with one digit to the left of the
  decimal followed by an exponent (1.5 x 105)
• Examples 2.1 x 103 represents 2100 and 3.6 x
  10-4 represents 0.00036

5
Scientific Notation Problems

• 1. Write 365,000,000 in scientific notation
• 2. Write 0.000087 in scientific notation

6
Answers

• 1.) 3.65 x 108
• 2.) 8.7 x 10-5

7
B. Graphing

• Use graphs to make a picture of scientific data
• independent variable, the one you change in
  your experiment is graphed on the x axis and
  listed first in a table
• dependent variable, the one changed by your
  experiment is graphed on the y axis and listed
  second in a table

8

• Best fit line or curve is drawn once points
  are plotted. Does not have to go through all
  points. Just gives you the trend of the points
• The slope of the line is given as the change in
  the y value divided by the change in the x
  value

9
Types of Graphs

• 1. Direct Relationship means an increase/decrease
  in one variable causes an increase/decrease in
  the other
• Example below

10

• 2. Inverse(indirect) relationship means that an
  increase in one variable causes a decrease in the
  other variable and vice versa
• Examples

11

• 3. Constant proportion means that a change in one
  variable doesnt affect the other variable
• Example

12

• 4. If either variable is squared(whether the
  relationship is direct or indirect), the graph
  will curve more steeply.

13
C. Significant figures

• Uncertainty in measurements is expressed by using
  significant figures
• The more accurate or precise a measurement is,
  the more digits will be significant

14
Significant Figure Rules

• 1. Zeros that appear before a nonzero digit are
  not significant (examples 0.002 has 1
  significant figure and 0.13 has 2 significant
  figures)
• 2. Zeros that appear between nonzero digits are
  significant (examples 1002 has 4 significant
  figures and 0.405 has 3 significant figures)

15
Significant figures rules(cont.)

• 3. zeros that appear after a nonzero digit are
  significant only if they are followed by a
  decimal point (20. has 2 sig figs) or if they
  appear to the right of the decimal point (35.0
  has 3 sig figs)

16
Sig Fig problems

• 1. How many significant figures does 0.050900
  contain?
• 2. How many significant figures does 4800 contain?

17
Answers

• 1. 5 sig figs
• 2. 2 sig figs

18
D. Units

• 1. Fundamental units are units that cant be
  broken down
• 2. Derived units are made up of other units and
  then renamed
• 3. SI units are standardized units used by
  scientists worldwide

19
Fundamental Units

• Meter (m) length, distance, displacement,
  height, radius, elongation or compression of a
  spring, amplitude, wavelength
• Kilogram (kg) mass
• Second (s) time, period
• Ampere (A) electric current
• Degree (o) angle

20
Derived Units

• Meter per second (m/s) speed, velocity
• Meter per second squared (m/s2) acceleration
• Newton (N) force
• Kilogram times meter per second (kg.m/s)
  momentum
• Newton times second (N.s)-- impulse

21
Derived Units (cont.)

• Joule (J) work, all types of energy
• Watt (W) power
• Coulomb (C) electric charge
• Newton per Coulomb (N/C) electric field strength
  (intensity)
• Volt (V)- potential difference (voltage)
• Electronvolt (eV) energy (small amounts)

22
Derived Units (cont.)

• Ohm (O) resistance
• Ohm times meter (O.m) resistivity
• Weber (Wb) number of magnetic field (flux) lines
  
• Tesla (T) magnetic field (flux) density
• Hertz (Hz)-- frequency

23
E. Prefixes

• Adding prefixes to base units makes them smaller
  or larger by powers of ten
• The prefixes used in Regents Physics are tera,
  giga, mega, kilo, deci, centi, milli, micro, nano
  and pico

24
Prefix Examples

• A terameter is 1012 meters, so 4 Tm would be 4
  000 000 000 000 meters
• A gigagram is 109 grams, so 9 Gg would be 9 000
  000 000 grams
• A megawatt is 106 watts, so 100 MW would be 100
  000 000 watts
• A kilometer is 103 meters, so 45 km would be 45
  000 meters

25
Prefix examples (cont.)

• A decigram is 10-1 gram, so 15 dg would be 1.5
  grams
• A centiwatt is 10-2 watt, so 2 dW would be 0.02
  Watt
• A millisecond is 10-3 second, so 42 ms would be
  0.042 second

26
Prefix examples (cont.)

• A microvolt is 10-6 volt, so 8 µV would be 0.000
  008 volt
• A nanojoule is 10-9 joule, so 530 nJ would be
  0.000 000 530 joule
• A picometer is 10-12 meter, so 677 pm would be
  0.000 000 000 677 meter

27
Prefix Problems

• 1.) 16 terameters would be how many meters?
• 2.) 2500 milligrams would be how many grams?
• 3.) 1596 volts would be how many gigavolts?
• 4.) 687 amperes would be how many nanoamperes?

28
Answers

• 1.) 16 000 000 000 000 meters
• 2.) 2.500 grams
• 3.) 1596 000 000 000 gigavolts
• 4.) 0.000 000 687 amperes

29
F. Estimation

• You can estimate an answer to a problem by
  rounding the known information
• You also should have an idea of how large common
  units are

30
Estimation (cont.)

• 2 cans of Progresso soup are just about the mass
  of 1 kilogram
• 1 medium apple weighs 1 newton
• The length of an average Physics students leg is
  1 meter

31
Estimation Problems

• 1.) Which object weighs approximately one newton?
  Dime, paper clip, student, golf ball
• 2.) How high is an average doorknob from the
  floor? 101m, 100m, 101m, 10-2m

32
Answers

• 1.) golf ball
• 2.) 100m

33
II. Mechanics

• A. Kinematics vectors, velocity, acceleration
• B. Kinematics freefall
• C. Statics
• D. Dynamics
• E. 2-dimensional motion
• F. Uniform Circular motion
• G. Mass, Weight, Gravity
• H. Friction
• I. Momentum and Impulse

34
Kinematics vectors, velocity, acceleration

• In physics, quantities can be vector or scalar
• VECTOR quantities have a magnitude (a number), a
  unit and a direction
• Example 22m(south)

35

• SCALAR quantities only have a magnitude and a
  unit
• Example 22m

36

• VECTOR quantities displacement, velocity,
  acceleration, force, weight, momentum, impulse,
  electric field strength
• SCALAR quantities distance, mass, time, speed,
  work(energy), power

37
Distance vs. Displacement

• Distance is the entire pathway an object travels
• Displacement is the shortest pathway from the
  beginning to the end

38
Distance/Displacement Problems

• 1.) A student walks 12m due north and then 5m due
  east. What is the students resultant
  displacement? Distance?
• 2.) A student walks 50m due north and then walks
  30m due south. What is the students resultant
  displacement? Distance?

39
Answers

• 1.) 13m (NE) for displacement
• 17 m for distance
• 2.) 20m (N) for displacement 80 m for distance

40
Speed vs. Velocity

• Speed is the distance an object moves in a unit
  of time
• Velocity is the displacement of an object in a
  unit of time

41
Average Speed/Velocity Equations
42
Symbols
43
Speed/Velocity Problems

• 1.) A boy is coasting down a hill on a
  skateboard. At 1.0s he is traveling at 4.0m/s and
  at 4.0s he is traveling at 10.0m/s. What distance
  did he travel during that time period? (In all
  problems given in Regents Physics, assume
  acceleration is constant)

44
Answers

• 1.) You must first find the boys average speed
  before you are able to find the distance

45
Answers (cont.)
46
Acceleration

• The time rate change of velocity is acceleration
  (how much you speed up or slow down in a unit of
  time)
• We will only be dealing with constant (uniform)
  acceleration

47
Symbols (cont.)
48
Constant Acceleration Equations
49
Constant Acceleration Problems

• 1.) A car initially travels at 20m/s on a
  straight, horizontal road. The driver applies the
  brakes, causing the car to slow down at a
  constant rate of 2m/s2 until it comes to a stop.
  What was the cars stopping distance? (Use two
  different methods to solve the problem)

50
Answers

• First Method
• vi20m/s
• vf0m/s
• a2m/s2
• Use vf2vi22ad to find d
• d100m

51
Answer (cont.)

• Second Method

52
B. Kinematics freefall

• In a vacuum (empty space), objects fall freely at
  the same rate
• The rate at which objects fall is known as g,
  the acceleration due to gravity
• On earth, the g is 9.81m/s2

53
Solving Freefall Problems

• To solve freefall problems use the constant
  acceleration equations
• Assume a freely falling object has a vi0m/s
• Assume a freely falling object has an a9.81m/s2

54
Freefall Problems

• 1.) How far will an object near Earths surface
  fall in 5s?
• 2.) How long does it take for a rock to fall 60m?
  How fast will it be going when it hits the ground
• 3.) In a vacuum, which will hit the ground first
  if dropped from 10m, a ball or a feather?

55
Answers

• 1.)

56
Answers (cont.)

• 2.)

57
Answers (cont.)

• 3.) Both hit at the same time because g the
  acceleration due to gravity is constant. It
  doesnt depend on mass of object because it is a
  ratio

58
Solving Anti??? Freefall Problems

• If you toss an object straight up, that is the
  opposite of freefall.
• So
• vf is now 0m/s
• a is -9.8lm/s2
• Because the object is slowing down not speeding up

59
Antifreefall Problems

• 1.) How fast do you have to toss a ball straight
  up if you want it reach a height of 20m?
• 2.) How long will the ball in problem 1 take to
  reach the 20m height?

60
Answers

• 1.)

61
Answers (cont.)

• 2.)

62
C. Statics

• The study of the effect of forces on objects at
  rest
• Force is a push or pull
• The unit of force is the newton(N) (a derived
  vector quantity)

63
Adding forces

• When adding concurrent (acting on the same object
  at the same time) forces follow three rules to
  find the resultant (the combined effect of the
  forces)
• 1.) forces at 00, add them
• 2.) forces at 1800, subtract them
• 3.) forces at 900, use Pythagorean Theorem

64
Force Diagrams

• Forces at 00
• Forces at 1800
• Forces at 900

65
Composition of Forces Problems

• 1.) Find the resultant of two 5.0N forces at 00?
  1800? And 900?

66
Answers

• 1.) 00
• 5N 5N 10N

67
Answers (cont.)

• 1.) 1800
• 5N 5N 0N

68
Answers (cont.)

• 1.) 900
• 5N
• 5N 7.1N

69
Resolution of forces

• The opposite of adding concurrent forces.
• Breaking a resultant force into its component
  forces
• Only need to know components(2 forces) at a 900
  angle to each other

70
Resolving forces using Graphical Method

• To find the component forces of the resultant
  force
• 1.) Draw x and y axes at the tail of the
  resultant force
• 2.) Draw lines from the head of the force to each
  of the axes
• 3.) From the tail of the resultant force to where
  the lines intersect the axes, are the lengths of
  the component forces

71
Resolution Diagram

• Black arrowresultant force
• Orange linesreference lines
• Green arrowscomponent forces
• y
• x

72
Resolving Forces Using Algebraic Method
73
Equilibrium

• Equilibrium occurs when the net force acting on
  an object is zero
• Zero net force means that when you take into
  account all the forces acting on an object, they
  cancel each other out

74
Equilibrium (cont.)

• An object in equilibrium can either be at rest or
  can be moving with constant (unchanging) velocity
• An equilibrant is a force equal and opposite to
  the resultant force that keeps an object in
  equilibrium

75
Equilibrium Diagram

• Black arrowscomponents
• Blue arrowresultant
• Red arrowequilibrant

76
Problems

• 1.) 10N, 8N and 6N forces act concurrently on an
  object that is in equilibrium. What is the
  equilibrant of the 10N and 6N forces? Explain.
• 2.) A person pushes a lawnmower with a force of
  300N at an angle of 600 to the ground. What are
  the vertical and horizontal components of the
  300N force?

77
Answers

• 1.) The 8N force is the equilibrant (which is
  also equal to and opposite the resultant) The 3
  forces keep the object in equilibrium, so the
  third force is always the equilibrant of the
  other two forces.

78
Answers (cont.)

• 2.)

79
C. Dynamics

• The study of how forces affect the motion of an
  object
• Use Newtons Three Laws of Motion to describe
  Dynamics

80
Newtons 1st Law of Motion

• Also called the law of inertia
• Inertia is the property of an object to resist
  change. Inertia is directly proportional to the
  objects mass
• An object will remain in equilibrium (at rest or
  moving with constant speed) unless acted upon by
  an unbalanced force

81
Newtons Second Law of Motion

• When an unbalanced (net) force acts on an
  object, that object accelerates in the direction
  of the force
• How much an object accelerates depends on the
  force exerted on it and the objects mass (See
  equation)

82
Symbols

• Fnetthe net force exerted on an object (the
  resultant of all forces on an object) in newtons
  (N)
• mmass in kilograms (kg)
• aacceleration in m/s2

83
Newtons Third Law of Motion

• Also called law of action-reaction
• When an object exerts a force on another object,
  the second object exerts a force equal and
  opposite to the first force
• Masses of each of the objects dont affect the
  size of the forces (will affect the results of
  the forces)

84
Free-Body Diagrams

• A drawing (can be to scale) that shows all
  concurrent forces acting on an object
• Typical forces are the force of gravity, the
  normal force, the force of friction, the force of
  acceleration, the force of tension, etc.

85
Free-body Forces

• Fg is the force of gravity or weight of an object
  (always straight down)
• FN is the normal force (the force of a surface
  pushing up against an object)
• Ff is the force of friction which is always
  opposite the motion
• Fa is the force of acceleration caused by a push
  or pull

86
Free-Body Diagrams

• If object is moving with constant speed to the
  right.
• Black arrowFf
• Green arrowFg
• Yellow arrowFN
• Blue arrowFa

87
Free-Body Diagrams on a Slope

• When an object is at rest or moving with constant
  speed on a slope, some things about the forces
  change and some dont
• 1.) Fg is still straight down
• 2.) Ff is still opposite motion
• 3.) FN is no longer equal and opposite to Fg
• 4.) Ff is still opposite motion
• More

88
Free-Body Diagrams on a Slope

• FaFfAxAcos?
• FNAyAsin?
• Fgmg (still straight down)
• On a horizontal surface, force of gravity and
  normal force are equal and opposite
• On a slope, the normal force is equal and
  opposite to the y component of the force of
  gravity

89
Free-Body Diagram on a Slope

• Green arrowFN
• Red arrowFg
• Black arrowsFa and Ff
• Orange dashesAy
• Purple dashesAx

90
Dynamics Problems

• 1.) Which has more inertia a 0.75kg pile of
  feathers or a 0.50kg pile of lead marbles?
• 2.) An unbalanced force of 10.0N acts on a 20.0kg
  mass for 5.0s. What is the acceleration of the
  mass?

91
Answers

• 1.) The 0.75kg pile of feathers has more inertia
  because it has more mass. Inertia is dependent on
  the mass of the object

92
Answers (cont.)

• 2.)

93
More Problems

• 3.) A 10N book rests on a horizontal tabletop.
  What is the force of the tabletop on the book?
• 4.) How much force would it take to accelerate a
  2.0kg object 5m/s2? How much would that same
  force accelerate a 1.0kg object?

94
Answers

• 1.) The force of the tabletop on the book is also
  10N (action/reaction)

95
Answers (cont.)

• 2.)

96
E. 2-Dimensional Motion

• To describe an object moving 2-dimensionally, the
  motion must be separated into a horizontal
  component and a vertical component (neither has
  an effect on the other)
• Assume the motion occurs in a perfect physics
  world a vacuum with no friction

97
Types of 2-D Motion

• 1.) Projectiles fired horizontally
• an example would be a baseball tossed straight
  horizontally away from you

98
Projectile Fired Horizontally

• Use the table below to solve these type of 2-D
  problems

Quantities Horizontal Vertical
vi Same as vf 0m/s
vf Same as vi
a 0m/s2 9.81m/s2
d
t Same as vertical time Same as horizontal time
99
Types of 2-D Motion

• 2.) Projectiles fired at an angle
• an example would be a soccer ball lofted into
  the air and then falling back onto the ground

100
Projectile Fired at an Angle

• Use the table below to solve these type of 2-D
  problems
• AxAcos? and AyAsin?

Quantities Horizontal Vertical
vi Ax Ay
vf Ax 0m/s
a 0m/s2 - 9.81m/s2
d
t twice vertical time
101
2-D Motion Problems

• 1.) A girl tossed a ball horizontally with a
  speed of 10.0m/s from a bridge 7.0m above a
  river. How long did the ball take to hit the
  river? How far from the bottom of the bridge did
  the ball hit the river?

102
Answers

• 1.) In this problem you are asked to find time
  and horizontal distance (see table on the next
  page)

103
Answers (cont.)
Quantities Horizontal Vertical
vi 10.0m/s 0.0m/s
vf 10.0m/s Dont need
a 0.0m/s 9.81m/s2
d ? 7.0m
t ? ?
104
Answers (cont.)

• Use

105
More 2-D Motion Problems

• 2.) A soccer ball is kicked at an angle of 600
  from the ground with an angular velocity of
  10.0m/s. How high does the soccer ball go? How
  far away from where it was kicked does it land?
  How long does its flight take?

106
Answers

• 2.) In this problem you are asked to find
  vertical distance, horizontal distance and
  horizontal time. Finding vertical time is usually
  the best way to start. (See table on next page)

107
Answers (cont.)
Quantities Horizontal Vertical
vi Ax5.0m/s Ay8.7m/s
vf Ax5.0m/s 0.0m/s
a 0.0m/s - 9.81m/s2
d ? ?
t ? Need to find
108
Answers (cont.)

• Find vertical t first using
• vfviat with.
• vf0.0m/s
• vi8.7m/s
• a-9.81m/s2
• Sovertical t0.89s and horizontal t is
  twice that
• and equals 1.77s

109
Answers (cont.)

• Find horizontal d using

110
Answers (cont.)

• Find vertical d by using

111
F. Uniform Circular Motion

• When an object moves with constant speed in a
  circular path
• The force (centripetal) will be constant towards
  the center
• Acceleration (centripetal) will only be a
  direction change towards the center
• Velocity will be tangent to the circle in the
  direction of movement

112
Uniform Circular Motion Symbols

• Fccentripetal force, (N)
• vconstant velocity (m/s)
• accentripetal acceleration (m/s2)
• rradius of the circular pathway (m)
• mmass of the object in motion (kg)

113
Uniform Circular Motion Diagram

Fc
a
v
114
Uniform Circular Motion Equations
115
Uniform Circular Motion Problems

• 1.) A 5kg cart travels in a circle of radius 2m
  with a constant velocity of 4m/s. What is the
  centripetal force exerted on the cart that keeps
  it on its circular pathway?

116
Answers

• 1.)

117
G. Mass, Weight and Gravity

• Mass is the amount of matter in an object
• Weight is the force of gravity pulling down on an
  object
• Gravity is a force of attraction between objects

118
Mass

• Mass is measured in kilograms (kg)
• Mass doesnt change with location (for example,
  if you travel to the moon your mass doesnt
  change)

119
Weight

• Weight is measured in newtons (N)
• Weight does change with location because it is
  dependent on the pull of gravity
• Weight is equal to mass times the acceleration
  due to gravity

120
Weight/Force of Gravity Equations
121
Gravitational Field Strength

• gacceleration due to gravity but it is also
  equal to gravitational field strength
• The units of acceleration due to gravity are m/s2
• The units of gravitational field strength are
  N/kg
• Both quantities are found from the equation

122
Mass, Weight, Gravity Problems

• 1.) If the distance between two masses is
  doubled, what happens to the gravitational force
  between them?
• 2.) If the distance between two objects is halved
  and the mass of one of the objects is doubled,
  what happens to the gravitational force between
  them?

123
Answers

• 1.) Distance has an inverse squared relationship
  with the force of gravity.
• Sosince r is multiplied by 2 in the problem,
  square 2so.224, then take the inverse of that
  square which equals ¼.so.the answer is ¼ the
  original Fg

124
Answers (cont.)

• 2.) Mass has a direct relationship with Fg and
  distance has an inverse squared relationship with
  Fg.
• Firstsince m is doubled so is Fg and since r is
  halved, square ½ , which is ¼ and then take the
  inverse which is 4.
• Thencombine 2x48
• Soanswer is 8 times Fg

125
More Problems

• 3.) Determine the force of gravity between a 2kg
  and a 3kg object that are 5m apart.
• 4.) An object with a mass of 10kg has a weight of
  4N on Planet X. What is the acceleration due to
  gravity on Planet X? What is the gravitational
  field strength on Planet X?

126
Answers

• 3.)
• 4.)

127
H. Friction

• The force that opposes motion measured in newtons
  (N)
• Always opposite direction of motion
• Static Friction is the force that opposes the
  start of motion
• Kinetic Friction is the force of friction
  between objects in contact that are in motion

128
Coefficient of Friction

• The ratio of the force of friction to the normal
  force (no unit, since newtons cancel out)
• Equation
• Ff µFN
• µcoefficient of friction
• Ffforce of friction
• FNnormal force

129
Coefficient of Friction

• The smaller the coefficient, the easier the
  surfaces slide over one another
• The larger the coefficient, the harder it is to
  slide the surfaces over one another
• Use the table in the reference tables

130
Coefficient of Friction Problems

• 1.) A horizontal force is used to pull a 2.0kg
  cart at constant speed of 5.0m/s across a
  tabletop. The force of friction between the cart
  and the tabletop is 10N. What is the coefficient
  of friction between the cart and the tabletop? Is
  the friction force kinetic or static? Why?

131
Answers

• 1.)
• The friction force is kinetic because the cart is
  moving over the tabletop

132
I. Momentum and Impulse

• Momentum is a vector quantity that is the product
  of mass and velocity (unit is kg.m/s)
• Impulse is the product of the force applied to an
  object and time (unit is N.s)

133
Momentum and Impulse Symbols

• pmomentum
• ?pchange in momentum (usually) m(vf-vi)
• Jimpulse

134
Momentum and Impulse Equations

• pmv
• JFt
• J?p
• pbeforepafter

135
Momentum and Impulse Problems

• 1.) A 5.0kg cart at rest experiences a 10N.s (E)
  impulse. What is the carts velocity after the
  impulse?
• 2.) A 1.0kg cart at rest is hit by a 0.2kg cart
  moving to the right at 10.0m/s. The collision
  causes the 1.0kg cart to move to the right at
  3.0m/s. What is the velocity of the 0.2kg cart
  after the collision?

136
Answer

• 1.) Use J?p so
• J10N.s(E)?p10kg.m/s(E)
• and since original p was 0kg.m/s and
  ?p10kg.m/s(E),
• new p10kg.m/s(E)
• then use.. pmv so..
• 10kg.m/s(E)5.0kg x v so.
• v2m/s(E)

137
Answers (cont.)

• 2.) Use pbeforepafter
• Pbefore0kg.m/s 2kg.m/s(right)
• 2kg.m/s(right)
• Pafter2kg.m/s(right)3kg.m/s
• P(0.2kg cart) so.p of 0.2kg cart must be
  -1kg.m/s or 1kg.m/s(left)
• more..

138
Answers (cont.)

• So if p after collision of 0.2kg cart is
  1kg.m/s(left) and
• pmv
• 1kg.m/s(left)0.2kg x v
• And v5m/s(left)

139
III. Energy

• A. Work and Power
• B. Potential and Kinetic Energy
• C. Conservation of Energy
• D. Energy of a Spring

140
A. Work and Power

• Work is using energy to move an object a distance
• Work is scalar
• The unit of work is the Joule (J)
• Work and energy are manifestations of the same
  thing, that is why they have the same unit of
  Joules

141
Work and Power (cont.)

• Power is the rate at which work is done so there
  is a time factor in power but not in work
• Power and time are inversely proportional the
  less time it takes to do work the more power is
  developed
• The unit of power is the watt (W)
• Power is scalar

142
Work and Power Symbols

• Wwork in Joules (J)
• Fforce in newtons (N)
• ddistance in meters (m)
• ?ETchange in total energy in Joules (J)
• Ppower in watts (W)
• ttime in seconds (s)

143
Work and Power Equations

• WFd?ET
• When work is done vertically, F can be the
  weight of the object Fgmg

144
Work and Power Problems

• 1.) A 2.5kg object is moved 2.0m in 2.0s after
  receiving a horizontal push of 10.0N. How much
  work is done on the object? How much power is
  developed? How much would the objects total
  energy change?
• 2.) A horizontal 40.0N force causes a box to move
  at a constant rate of 5.0m/s. How much power is
  developed? How much work is done against friction
  in 4.0s?

145
Answers

• 1.) to find work use WFd
• SoW10.0N x 2.0m20.0J
• To find power use PW/t
• SoP20.0J/2.0s10.0W
• To find total energy change its the same as work
  done so
• ?ETW20.0J

146
Answers (cont.)

• 2.) To find power use
• So P40.0N x 5.0m/s200W
• To find work use PW/t so200WW/4.0s
• So.W800J

147
More problems

• 3.) A 2.0kg object is raised vertically 0.25m.
  What is the work done raising it?
• 4.) A lift hoists a 5000N object vertically, 5.0
  meters in the air. How much work was done lifting
  it?

148
Answers

• 3.) to find work use WFd with F equal to the
  weight of the object
• So..Wmg x d
• So...W2.0kgx9.81m/s2x0.25m
• SoW4.9J

149
Answers (cont.)

• 4.) to find work use WFd
• Even though it is vertical motion, you dont have
  to multiply by g because weight is already
  given in newtons
• SoWFd5000N x 5.0m
• And W25000J

150
B. Potential and Kinetic Energy

• Gravitational Potential Energy is energy of
  position above the earth
• Elastic Potential Energy is energy due to
  compression or elongation of a spring
• Kinetic Energy is energy due to motion
• The unit for all types of energy is the same as
  for work the Joule (J). All energy is scalar

151
Gravitational Potential Energy Symbols and
Equation

• ?PEchange in gravitational potential energy in
  Joules (J)
• mmass in kilograms (kg)
• gacceleration due to gravity in (m/s2)
• ?hchange in height in meters (m)
• Equation ?PEmg?h
• Gravitational PE only changes if there is a
  change in vertical position

152
Gravitational PE Problems

• 1.) How much potential energy is gained by a
  5.2kg object lifted from the floor to the top of
  a 0.80m high table?
• 2.) How much work is done in the example above?

153
Answers

• 1.) Use ?PEmg?h to find potential energy gained
  so ?PE5.2kgx9.81m/s2x0.80m
• So?PE40.81J
• 2.) W?ET so..W is also 40.81J

154
Kinetic Energy Symbols and Equation

• KEkinetic energy in Joules (J)
• mmass in kilograms (kg)
• vvelocity or speed in (m/s)

155
Kinetic Energy Problems

• 1.) If the speed of a car is doubled, what
  happens to its kinetic energy?
• 2.) A 6.0kg cart possesses 75J of kinetic energy.
  How fast is it going?

156
Answers

• 1.) Using KE1/2mv2 if v is doubled, because v if
  squared KE will be quadrupled.
• 2.) Use KE1/2mv2 so..
• 75J1/2 x 6.0kg x v
• And..v5m/s

157
C. Conservation of Energy

• In a closed system the total amount of energy is
  conserved
• Total energy includes potential energy, kinetic
  energy and internal energy
• Energy within a system can be transferred among
  different types of energy but it cant be
  destroyed

158
Conservation of Energy in a Perfect Physics World

• In a perfect physics world since there is no
  friction there will be no change in internal
  energy so you dont have to take that into
  account
• In a perfect physics world energy will transfer
  between PE and KE

159
In the Real World

• In the real world there is friction so the
  internal energy of an object will be affected by
  the friction (such as air resistance)

160
Conservation of Energy Symbols

• ETtotal energy of a system
• PEpotential energy
• KEkinetic energy
• Qinternal energy
• all units are Joules (J)

161
Conservation of Energy Equations

• In a real world situation, ETKEPEQ because
  friction exists and may cause an increase in the
  internal energy of an object
• In a perfect physics world ETKEPE with
  KEPE equal to the total mechanical energy of
  the system object

162
Conservation of Energy Examples (perfect physics
world)

• position 1
• position 2
• position 3 more..

163
Conservation of Energy Examples (cont.)
Position 1
Position 2
Position 3
164
Conservation of Energy Examples (cont.)

• Position 1
• Position 2
• Position 3

165
Conservation of Energy (perfect physics world)

• Position 1 the ball/bob has not starting
  falling yet so the total energy is all in
  gravitational potential energy
• Position 2 the ball/bob is halfway down, so
  total energy is split evenly between PE and KE
• Position 3 the ball/bob is at the end of its
  fall so total energy is all in KE

166
Conservation of Energy Problems

• 1.) A 2.0kg block starts at rest and then slides
  along a frictionless track. What is the speed of
  the block at point B?
• A
• 7.0m
• B

167
Answer

• Since there is no friction, Q does not need to be
  included
• Souse ETPEKE
• At position B, the total energy is entirely KE
• Since you cannot find KE directly, instead find
  PE at the beginning of the slide and that will be
  equal to KE at the end of the slide more..

168
Answer (cont.)

• PE (at position A) mg?h2.0kgx9.81m/s2x7.0m
  137.3J
• KE (at position A) 0J because there is no speed
• So ET (at position A)137.3J
• At position B there is no height so the PE is 0J
• More.

169
Answer (cont.)

• At position B the total energy still has to be
  137.3J because energy is conserved and because
  there is no friction no energy was lost along
  the slide
• So.ET(position B)137.3J0JKE
• SoKE also equals 137.3J at position B
• More

170
Answer (cont.)

• Use KE1/2mv2
• SoKE137.3J1/2x2.0kgxv2
• So v (at position B)11.7m/s

171
More Conservation of Energy Problems

• 1.) From what height must you drop the 0.5kg ball
  so that the it will be traveling at 25m/s at
  position 3(bottom of the fall)?
• 2.)How fast will it be traveling at position 2
  (halfway down)?
• Assume no friction

• position 1
• position 2
• position 3

172
Answers

• 1.) At position 3, total energy will be all in
  KE because there is no height and no friction
• Souse ETKE1/2mv2
• KE1/2 x 0.5kg x (25m/s)2
• SoKE156.25JPE (at position1)
• So?PE156.25Jmg?h
• And ?h31.86m

173
Answers (cont.)

• 2.) Since position 2 is half way down total
  energy will be half in PE and half in KE
• SoKE at position 2 will be half that at
  position 3
• SoKE at position 2 is 78.125J
• Then use KE (at 2)78.125J 1/2 x 0.5kg x v2
• v17.68m/s at position 2

174
D. Energy of a Spring

• Energy stored in a spring is called elastic
  potential energy
• Energy is stored in a spring when the spring is
  stretched or compressed
• The work done to compress or stretch a spring
  becomes its elastic potential energy

175
Spring Symbols

• Fsforce applied to stretch or compress the
  spring in newtons (N)
• kspring constant in (N/m) specific for each
  type of spring
• xthe change in length in the spring from the
  equilibrium position in meters (m)

176
Spring Equations

• Fskx
• PEs1/2kx2

177
Spring Diagrams
x
178
Spring Problems

• 1.) What is the potential energy stored in a
  spring that stretches 0.25m from equilibrium when
  a 2kg mass is hung from it?
• 2.) 100J of energy are stored when a spring is
  compressed 0.1m from equilibrium. What force was
  needed to compress the spring?

179
Answers

• 1.) Using PEs1/2kx2 you know x but not k
• You can find k using Fskx
• With Fs equal to the weight of the hanging mass
• So FsFgmg2kgx9.81m/s2
• Fs19.62Nkxk x 0.25m
• k78.48N/m
• More

180
Answers (cont.)

• Now use PEs1/2kx2
• PEs1/2 x 78.48N/m x (0.25m)2
• So PEs2.45J

181
Answers (cont.)

• 2.) To find the force will use Fskx, but since
  you only know x you must find k also
• Use PEs1/2kx2 to find k
• PEs100J1/2k(0.1m)2
• k20 000N/m
• use Fskx20 000N/m x 0.1m
• Fs2000N

182
Examples of Forms of Energy

• 1.) Thermal Energy is heat energy which is the KE
  possessed by the particles making up an object
• 2.) Internal Energy is the total PE and KE of the
  particles making up an object
• 3.) Nuclear Energy is the energy released by
  nuclear fission or fusion
• 4.) Electromagnetic Energy is the energy
  associated with electric and magnetic fields