Chapter 4 Work energy power. class 9

36. Work, Energy & Power
AP Physics 1

37. There are many different TYPES of Energy.
• Energy is expressed in
JOULES (J)
• 4.19 J = 1 calorie
• Energy can be
expressed more
specifically by using the
term WORK(W)
Work = The Scalar Dot Product between Force and Displacement.
If you apply a force on an object and it covers a displacement IN THE
DIRECTION OF THE FORCE you have supplied ENERGY to, or done
WORK on, that object.

38. Scalar Dot Product?
A product is obviously a result of multiplying 2
numbers. A scalar is a quantity with NO
DIRECTION. So basically Work is found by
multiplying the Force times the displacement
and result is ENERGY, which has no direction
associated with it.

cos
x
F
x
F
W








A dot product is basically a CONSTRAINT
on the formula. In this case it means that
F and x MUST be parallel. To ensure that
they are parallel we add the cosine on the
end.
W = Fx
Area = Base x Height

39. Work The VERTICAL component of the force DOES NOT
cause the block to move the right. The energy imparted to
the box is evident by its motion to the right. Therefore
ONLY the HORIZONTAL COMPONENT of the force
actually creates energy or WORK.
When the FORCE and DISPLACEMENT are in the SAME
DIRECTION you get a POSITIVE WORK VALUE. The
ANGLE between the force and displacement is ZERO
degrees. What happens when you put this in for the
COSINE?
When the FORCE and DISPLACEMENT are in the
OPPOSITE direction, yet still on the same axis, you get a
NEGATIVE WORK VALUE. This negative doesn't mean
the direction!!!! IT simply means that the force and
displacement oppose each other. The ANGLE between the
force and displacement in this case is 180 degrees. What
happens when you put this in for the COSINE?
When the FORCE and DISPLACEMENT are
PERPENDICULAR, you get NO WORK!!! The ANGLE
between the force and displacement in this case is 90
degrees. What happens when you put this in for the
COSINE?

40. The Work Energy Theorem
Up to this point we have learned Kinematics and Newton's
Laws. Let 's see what happens when we apply BOTH to our
new formula for WORK!
1. We will start by applying
Newton's second law!
2. Using Kinematic #3!
3. An interesting term appears
called KINETIC ENERGY or
the ENERGY OF MOTION!

41. The Work Energy Theorem
And so what we really have is called the
WORK-ENERGY THEOREM. It basically
means that if we impart work to an
object it will undergo a CHANGE in
speed and thus a change in KINETIC
ENERGY. Since both WORK and
KINETIC ENERGY are expressed in
JOULES, they are EQUIVALENT TERMS!
" The net WORK done on an object is equal to the change in kinetic
energy of the object."

42. Example W=Fxcos
A 70 kg base-runner begins to slide into second base when moving at a
speed of 4.0 m/s. The coefficient of kinetic friction between his clothes
and the earth is 0.70. He slides so that his speed is zero just as he
reaches the base (a) How much energy is lost due to friction acting on
the runner? (b) How far does he slide?
)
8
.
9
)(
70
)(
70
.
0
(


 mg
F
F n
f 

= 480.2 N







f
o
f
f
W
mv
W
K
W
a
2
2
)
4
)(
70
(
2
1
2
1
0
)
-560 J




x
x
x
F
W f
f
)
180
(cos
)
2
.
480
(
560
cos
1.17 m

43. Example
A 5.00 g bullet moving at 600 m/s penetrates a tree trunk to a depth of 4.00
cm. (a) Use the work-energy theorem, to determine the average frictional
force that stops the bullet.(b) Assuming that the frictional force is constant,
determine how much time elapses between the moment the bullet enters
the tree and the moment it stops moving
2
1
0 (0.005)(600)
2
friction
W K
W
W
 
 
 -900 J
 
cos
900 0.04
f f
f
f
W F x
F
F



 22,500 N
6
22,500 (0.005)
0 600 ( 4.5 10 )
f NET
o
F F ma a
a
v v at x t
t
  

    

4.5x106 m/s/s
1.33x10-4 s

44. Lifting mass at a constant speed
Suppose you lift a mass upward at a constant speed,
v = 0 & K=0. What does the work equal now?
Since you are lifting at a constant
speed, your APPLIED FORCE
equals the WEIGHT of the object
you are lifting.
Since you are lifting you are raising
the object a certain “y”
displacement or height above the
ground.
When you lift an object above the ground it is said to have POTENTIAL ENERGY

45. Suppose you throw a ball upward
What does work while it is
flying through the air?
Is the CHANGE in kinetic
energy POSITIVE or
NEGATIVE?
Is the CHANGE in potential
energy POSITIVE or
NEGATIVE?
W K U
   
GRAVITY
NEGATIVE
POSITIVE
( )
o o
o o
o o
BEFORE AFTER
K U
K K U U
K K U U
U K U K
Energy Energy
  
   
   
  

Note KE = K, PE = U; these symbols are
used interchangeably.

46. ENERGY IS CONSERVED
The law of conservation of mechanical energy states:
Energy cannot be created or destroyed, only
transformed!
Energy Before Energy After
Am I moving? If yes,
KEo
Am I above the
ground? If yes, PEo
Am I moving? If yes,
KE
Am I above the
ground? If yes, PE

47. Energy consistently changes forms

48. Energy consistently changes forms
Position m v U K ME
1 60 kg 8 m/s
Am I above the ground?
Am I moving?
NO, h = 0, U = 0 J
0 J
Yes, v = 8 m/s, m = 60 kg
2 2
1 1 (60)(8)
2 2
1920
K mv
K J
 

1920 J
(= U+K)
1920 J

49. Energy consistently changes forms
Position m v U K ME
1 60 kg 8 m/s 0 J 1920 J 1920 J
2 60 kg
Energy Before = Energy After
KO = U + K
1920= (60)(9.8)(1) + (.5)(60)v2
1920= 588 + 30v2
588 J
1332 = 30v2
44.4 = v2
v = 6.66 m/s
6.66 m/s 1920 J
1332 J

50. Energy consistently changes forms
Position m v U K ME
1 60 kg 8 m/s 0 J 1920 J 1920 J
2 60 kg 6.66 m/s 588 J 1332 J 1920 J
3 60 kg 1920 J
Am I moving at the top? No, v = 0 m/s
0 m/s 0 J
1920 J
EB = EA
Using position 1
Ko = U
1920 = mgh
1920 =(60)(9.8)h
h = 3.27 m

51. Example
A 2.0 m pendulum is released from rest when the support
string is at an angle of 25 degrees with the vertical. What
is the speed of the bob at the bottom of the string?
L
 Lcos
h
h = L – Lcos
h = 2-2cos
h = 0.187 m
EB = EA
UO = K
mgho = 1/2mv2
gho = 1/2v2
2(1.83) = v2
1.94 m/s = v

52. Springs – Hooke’s Law
Hooke's Law describes the force
needed to stretch an elastic object.
This is primarily in reference to
SPRINGS.
kx
or
kx
F
k
k
x
F
s
s




N/m)
:
nit
Constant(U
Spring
ality
Proportion
of
Constant

The negative sign only
tells us that “F” is what is
called a RESTORING
FORCE, in that it works in
the OPPOSITE direction
of the displacement.

53. Hooke’s Law
Common formulas which are set equal to Hooke's law are N.S.L. and
weight

54. Example
A load of 50 N attached to a spring hanging vertically stretches the spring
5.0 cm. The spring is now placed horizontally on a table and stretched
11.0 cm. What force is required to stretch the spring this amount?



k
k
kx
Fs
)
05
.
0
(
50
1000 N/m



s
s
s
F
F
kx
F
)
11
.
0
)(
1000
(
110 N

55. Hooke’s Law from a Graphical Point of View
x(m) Force(N)
0 0
0.1 12
0.2 24
0.3 36
0.4 48
0.5 60
0.6 72
graph
x
vs.
F
a
of
Slope



k
x
F
k
kx
F
s
s
Suppose we had the following data:
Force vs. Displacement y = 120x + 1E-14
R2
= 1
0
10
20
30
40
50
60
70
80
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Displacement(Meters)
Force(Newtons)
k =120 N/m

56. We have seen F vs. x Before!!!!
Force vs. Displacement y = 120x + 1E-14
R2
= 1
0
10
20
30
40
50
60
70
80
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Displacement(Meters)
Force(Newtons)
Work or ENERGY = Fx
Since WORK or ENERGY
is the AREA, we must get
some type of energy when
we compress or elongate
the spring. This energy is
the AREA under the line!
Area = ELASTIC
POTENTIAL ENERGY
Since we STORE energy when the spring is compressed and
elongated it classifies itself as a “type” of POTENTIAL ENERGY, Us.
In this case, it is called ELASTIC POTENTIAL ENERGY (EPE).

57. Elastic Potential Energy
The graph of F vs.x for a
spring that is IDEAL in
nature will always
produce a line with a
positive linear slope.
Thus the area under the
line will always be
represented as a
triangle.
NOTE: Keep in mind that this can be applied to WORK or can be conserved
with any other type of energy.

58. Conservation of Energy in Springs

59. Example
A slingshot consists of a light leather cup, containing a stone, that is
pulled back against 2 rubber bands. It takes a force of 30 N to stretch
the bands 1.0 cm (a) What is the potential energy stored in the bands
when a 50.0 g stone is placed in the cup and pulled back 0.20 m from
the equilibrium position? (b) With what speed does it leave the
slingshot?











v
v
mv
U
K
U
E
E
c
k
kx
U
b
k
k
kx
F
a
s
s
A
B
s
s
2
2
2
)
050
.
0
(
2
1
2
1
)
)
20
)(.
(
5
.
0
2
1
)
)
01
.
0
(
30
) 3000 N/m
300 J
109.54 m/s

60. Power
One useful application of Energy is to
determine the RATE at which we
store or use it. We call this
application POWER!
As we use this new application, we
have to keep in mind all the
different kinds of substitutions we
can make.
Unit = WATT or Horsepower

62. Work and Energy
Physics 100 Chapt 5

63. Physicist’s definition of “work”
dist
Work = F x dist∥
dist∥

64. Atlas holds up the Earth
But he doesn’t move,
dist∥ = 0
Work= Fx dist∥ = 0
He doesn’t do any work!

65. Garcon does work when
he picks up the tray
but not while he
carries it around
the room
dist is not zero,
but dist∥ is 0

66. Why this definition?
Newton’s 2nd law: F=m a
Definition of work
+ a little calculus
Work= change in ½mv2
This scalar quantity is given
a special name: kinetic energy

67. Work = change in KE
This is called:
the Work-Energy Theorem

68. Units again…
Kinetic Energy = ½mv2
kg m2
s2
work = F x dist∥
N m =kg m
s2
m
=1Joule
same!

69. Work done by gravity
start
end
dist dist∥
W=mg
Work = F x dist∥
= -mg x change in height
= -change in mg h
change in
vertical height

70. Gravitational Potential Energy
Workgrav = -change in mgh
This is called:
“Gravitational Potential
Energy” (or PEgrav)
Workgrav = -change in PEgrav
change in PEgrav = -Workgrav

71. If gravity is the only force doing work….
-change in mgh = change in ½ mv2
0 = change in mgh + change in ½ mv2
change in (mgh + ½ mv2) = 0
mgh + ½ mv2 = constant
Work-energy theorem:

72. Conservation of energy
mgh + ½ mv2 = constant
Gravitational
Potential energy
Kinetic energy
If gravity is the only force that does work:
PE + KE = constant
Energy is conserved

73. Free fall
(reminder)
V0 = 0
t = 0s
V1 = 10m/s
t = 1s
V2 = 20m/s
t = 2s
V3 = 30m/s
t = 3s
V4 = 40m/s
t = 4s
75m
60m
35m
0m
height
80m

74. m=1kg free falls from 80m
V0 = 0 h0=80m
t = 0s
V1 = 10m/s; h1=75m
t = 1s
V2 = 20m/s; h2=60m 600J 200J 800J
t = 2s
V3 = 30m/s; h3=35m 350J 450J 800J
t = 3s
V4 = 40m/s; h4=0 0 800J 800J
t = 4s
mgh ½ mv2 sum
800J 0 800J
750J 50J 800J

75. pendulum
W=mg
T
Two forces: T and W
T is always
┴ to the motion
(& does no work)

76. Pendulum conserves energy
hmax
E=mghmax
E=mghmax
E=1/2 m(vmax)2

77. Roller coaster

78. Work done by a spring
Relaxed
Position
F=0
F
x
I compress
the spring
(I do + work;
spring does
-work)
Work done by spring = - change in ½ kx2

79. Spring Potential Energy
Workspring = -change in ½ kx2
This is the:
“Spring’s Potential
Energy” (or PEspring)
Workspring = -change in PEspring
change in PEspring = -

80. If spring is the only force doing work….
-change in ½ kx2 = change in ½ mv2
0 = change in ½ kx2 + change in ½ mv2
change in ( ½ kx2 + ½ mv2) = 0
½ kx2 + ½ mv2 = constant
Work-energy theorem:

81. Conservation of energy
springs & gravity
mgh + ½ kx2 + ½ mv2 = constant
Gravitational
potential energy
Kinetic energy
If elastic force & gravity are the only force doing work:
PEgrav + PEspring + KE = constant
Energy is conserved
spring
potential energy

82. example
KineticE
Spring PE
grav PE

83. Two types of forces:
“Conservative” forces
forces that do + & – work
•Gravity
•Elastic (springs, etc)
•Electrical forces
•…
“Dissipative” forces
forces that only do – work
•Friction
•Viscosity
•….
-work 
change in PE
-work  heat
(no potential energy.)

84. (-)Work done by frictionheat

85. Thermal atomic motion
Heat energy= KE and PE associated with
the random thermal motion of atoms
Air solid

86. Work-energy theorem
(all forces)
Workfric = change in (PE+KE)
Work done
dissipative
Forces
(always -)
Kinetic
energy
-Workfric = change in heat energy
potential energy
From all
Conservative forces
-change in Heat Energy =
change in (PE+KE)
Workfric = -change in heat energy

87. Work – Energy Theorem
(all forces)
0 = change in Heat Energy +
change in (PE+KE)
0 = change in (Heat Energy+PE+KE)
Heat Energy + PE + KE = constant
Law of Conservation of Energy

88. Energy conversion while skiing
Friction: energy gets
converted to heat
Potential energy
Potential energykinetic energy

89. Units again
Heat units:
1 calorie = heat energy required to raise the
temp of 1 gram of H2O by 1o C
1 calorie= 4.18 Joules
Kg m2/s2

90. Food Calories
1 Calorie = 1000 calories = 1Kcalorie
1 Calorie= 4.18x103 Joules
The Calories you read on food labels
8 x 105 J
7 x 106 J
2 x 106 J

91. Power
Rate of using energy:
amout of energy
elapsed time
Units:
Joule
second
1 = 1 Watt
Power =
A 100 W light bulb
consumes 100 J of
electrical energy each
second to produce light

92. Other units
Over a full day, a work-horse can
have an average work output of
more than 750 Joules each second
1 Horsepower = 750 Watts

93. Kilowatt hours
energy
time
Power =  energy = power x time
 power unit x time unit = energy unit
Elec companies use:
Kilowatts
(103 W)
hours
(3600 s)
1 kilowatt-hour = 1kW-hr
= 103 W x 3.6x103 s = 3.6x106 Ws
J
HECO charges us about 15 cents /kW-hr
x

94. 7) Rate of doing work (Power) :-
1 kilowatt = 1000 watts
1 kW = 1000 W
1 kW = 1000 J s-1
Power is the rate of doing work.
If W is the work done in time t, then
work done W
Power = ---------------- or P = ---
time taken t
The unit of power is watt (W).
1 watt is the power of an agent which does work at the
rate of 1 joule per second.
1 watt = 1 joule / second or 1 W = 1 J s
-1

95. 8) Commercial unit of power :-
1 kW h = 1 kW x 1 h
= 1000 W x 1 h
= 1000 W x 3600 s
1 kW h
=
=
3600000 J
3.6 x 10
-6
J
The electrical energy used in homes and industries are
expressed kilowatt hour. The electrical energy used during
a month is expressed in ‘units’. Here 1 unit means 1
kilowatt hour.
The commercial unit of energy is kilowatt hour (kW h).
1 kilowatt hour is the energy used in one hour at the rate
of 1 kilowatt (or 1000 J s-1 ).