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Work, Energy & Power
Physics 11
Work
 In physics, work is
defined as the dot
product of force
and displacement
 The result is
measured in Joules
(J) and a Joule is a
Newton metre
 Work is a scalar
 
W  F  d
W  Fd cos 
Zero Work Conditions
W  Fd cos
 Work will equal zero if:
 Force applied is equal to zero
 Displacement is equal to zero
 The angle between the force and the
displacement is equal to ninety degrees
Comprehension Check
1. How much work is done if you push on a
wall with 3500N but the wall does not
move?
2. How much work is done carrying a book
down the hall at constant velocity?
3. If you pull a crate with a force of 550N at
an angle of 35° to the horizontal and it
moves 25m horizontally, how much work
was done?
Work Done By Changing Forces
Force vs Displacement
180
160
140
120
Force (N)
 How could you
determine the
work done in
the graph?
100
80
60
40
20
0
0
0.2
0.4
0.6
Displacement (m)
0.8
1
1.2
Work Done By Changing Forces
 How could you
determine the
work done in
the graph?
AREA UNDER
CURVE!
Positive and Negative Work
 Positive work
occurs when the
angle between the
force and
displacement is 0°90°
 Negative work
occurs when the
angle between the
force and
displacement is
90°-180°
F
d
F
d
Kinetic Energy
Physics 11
Work on an moving object
A 2kg object is moving at 10 m/s when a
force is applied to it accelerating it to
20m/s over a distance of 5m. What is the
work done on the object?
Work on a Moving Object
10m/s
20m/s
5m
v 2f  vi2  2ad

v
a
2
f
v
2
i
W  Fd

2d

F  ma

v
F m
2
f
 vi2
2d



 v 2f  vi2 
W  d m

2d 

v 2f  vi2
W m
2
1 2 1 2
W  mv f  mvi
2
2

If you Insist on Numbers…
1 2 1 2
W  mv f  mvi
2
2
1
1
2
2
W  2(20)  2(10)
2
2
W  400 J  100 J  300 J
Work-Energy Theorem
 These terms have a special name:
Kinetic Energy
 Work is the change in Kinetic Energy
 Energy is defined as the potential to do
work
1 2 1 2
W  mv f  mvi
2
2
W  KE f  KEi
W  KE
Work done by Friction
 A car is travelling at 100kph when the
driver sees a moose on the road
ahead. The driver slams on the
brakes, bringing the car to a stop just
before hitting the moose. If the car's
mass is 1200kg, how long does it
take to stop?   0.72
Work done by Friction
KE  W f
1 2
mvi  F f d
2
1 2
mvi   FN d  mg d
2
mvi2
 d
2  mg
vi2
d 
2 g
Numbers again…
 vi2
(27.8) 2
d 
 54.6m

2g 2(0.72)(9.81)
Potential Energy
Physics 11
Potential Energy and Work
What is the work done on a 12kg
object to raise it from the ground to
a height of 1.5m?
Potential Energy and Work
What is the work done on a 12kg
object to raise it from the ground to
a height of 1.5m?
W
W
W
W
 
 F  d
 
 mgd
 (12)(9.81)(1.5)
 176 J
PE  W
PE  mgh
Potential Energy
 The work done on the system can be
recovered, so it must be stored. This stored
energy is Potential Energy, Ep or PE.
 Energy is defined as the potential to do
work.
 KE: a moving object can apply a force through a
distance
 PE: An object in the air, when dropped, can
apply a force through a distance (dropping
textbook)
Forms of Energy: Brainstorm
Kinetic
 Moving things
Physics 11
Chemistry
Physics 12
Potential
 Gravitational
 Elastic (springs)
 Chemical (stored in molecular bonds)
 Thermal (heat)
 Electrical (separated charges)
 Nuclear (E=mc2)
Spring Energy
Physics 11
Hooke’s Law
Fs  kx
Δ Fs: Applied force
Δx : displacement of the spring from the
equilibrium position
k: the spring constant
Where is the Energy?
 If a vertical spring is stretched 12cm
by the weight of a 0.10kg. How much
energy is in the spring?
E p  Wg  mgx  (0.10kg)(9.81 sm2 )(0.12m)  1.2 J
 Derive an expression for the energy
stored in a spring.
Analogy with Gravity
Fspring  kx
Fgravity  mg
F
F
d
d
E p  W  Fg d
E p  W  Fspringx
E p  mgd
E p  12 (kxmax ) xmax
E p  mgh
E p  12 k xmax 
2
What we know about Energy so Far
Spring
Kinetic
KE  mv
1
2
Work
W  E
2
PE  kx
1
2
Gravitational
Potential
PE  mgh
2
Work Energy Theorem,
Power,
and Efficiency
Physics 11
Conservation of Energy
 The total energy of a closed system is
constant.
 The total energy in the system is the
sum of the kinetic and potential
energy of it’s various parts.
EKi  EPi  EKf  EPf
or
KEi  PEi  KE f  PE f
Examples
KEi  PEi  KE f  PE f
 Book Drop
0  mghi  12 mv 2f  0
h
v
 Collision into a spring KEi  PEi  KE f  PE f
v
x
1
2
mvi2  0  0  12 kx 2f
 Car coasting up a hill KEi  PEi  KE f  PE f
v
v
h
1
2
mvi2  0  12 mv 2f  mgh f
Energy is a “State Function”
 Each type/form of energy we have seen so
far is a ‘state function’
 A state function only cares about the
current state of the closed system
 How the system got into that state does not
matter  path-independent
m
h
All Paths result in the same final velocity
v
Consider the following situation
m
h
r
k
a) What is the velocity at the top of the
loop?
b) What is the maximum compression
of the spring?
Consider the following situation
m
h
r
k
a) What is the velocity at the top of the
loop?
EKi  EGi  EKf  EGf
0  mghi  12 mv2f  0
Consider the following situation
m
r
k
a) b
b) What is the maximum compression
of the spring?
KEi  PEi  KE f  PE f
0  mghi  0  0  0  12 kx 2f
Rate of Change in Energy
m
h
All Paths result in the same final velocity
v
 So what is different about each of these
paths?
 The time it takes to go down the longer ramp
will be greater due to a lower acceleration
 Power:
W E
P

t t
J 
 s   W 
Efficiency
 Not all systems are closed.
 Energy lost through friction
 Efficiency of energy conversion in
open systems.
Eout
Efficiency 
100%
Ein
Efficiency 
Ef
Ei
100%
Efficiency
 Not all systems are closed.
 Energy lost through friction
 Consider the following:
 A rocket has 3.50x103 J of chemical potential
energy. The stored chemical energy is
transformed into gravitational potential energy
when it is launched. What is the efficiency of the
rocket’s transformation of energy if the 0.5kg
rocket travels 1.00x102 m.