Lecture 4: Energy
Serway
and
Vuille
Ch. 5
Outline
•
Work

Energy theorem
•
Gravitational potential energy
•
Spring potential energy
•
Conservation of energy
•
Power
Introduction to Energy
•
Energy
is
one
of
the
most
important
concepts
in
science,
originating
from
thermodynamics
.
•
According
to
the
ideas
of
modern
physics,
energy
is
the
substance
from
which
all
things
in
the
Universe
are
made
up
.
•
Energy
is
also
defined
as
the
ability
of
a
system
to
perform
work
.
•
The
most
important
property
of
energy
is
that
it
is
a
conserved
quantity
.
Introduction to Energy
•
Energy
is
present
in
the
universe
in
many
different
forms
–
Mechanical
–
Chemical
–
Electromagnetic
–
Nuclear
–
Thermodynamic
(Heat)
•
In
this
chapter,
we
focus
on
mechanical
energy,
which
is
the
sum
of
kinetic
energy
and
potential
energy
Work
•
In
physics,
work
involves
applying
a
force
to
an
object
while
moving
it
a
certain
distance
.
•
Therefore,
the
work
W
done
on
an
object
by
a
constant
force
F
during
a
linear
displacement
along
the
x

axis
is
𝑊
=
∆
•
The
SI
unit
of
work
is
Joule
(J)
.
•
Note
that
1
J
=
1
N
∙
m
=
1
kg
m
2
/s
2
Work
•
Work
is
done
only
by
the
component
of
the
force
acting
parallel
to
the
object’s
direction
of
motion
.
•
The
work
W
done
on
an
object
by
a
constant
force
F
during
a
displacement
x
is
𝑊
=
𝑭
∙
∆
=
𝑭
∆
cos
𝜃
Example 1
•
A
block
of
mass
m
=
2
.
50
kg
is
pushed
a
distance
d
=
2
.
0
m
along
a
frictionless
horizontal
table
by
a
constant
applied
force
of
magnitude
F
=
16
.
0
N
directed
at
an
angle
θ
=
25
.
0
⁰
below
the
horizontal
.
•
Determine
the
work
done
by
–
(a)
The
applied
force
–
(b)
The
normal
force
exerted
by
the
table
–
(c)
The
force
of
gravity
–
(d)
The
net
force
on
the
block
Example 1 Solution
•
The
work
done
by
the
applied
force
is
𝑊
=
𝑭
∙
∆
=
cos
25
.
0
°
∆
=
16
.
0
𝑁
2
.
0
cos
25
.
0
°
=
19
.
0
•
The
work
done
by
the
normal
force
and
the
force
of
gravity
is
𝑊
=
𝑭
∙
∆
=
∆
+
∆
=
0
•
The
work
done
by
the
net
force
on
the
block
is
from
the
applied
force
alone
.
Work

Energy Theorem
•
The
work

energy
theorem
relates
the
speed
of
an
object
to
the
net
work
done
on
it
by
external
forces
.
•
It
can
be
shown
that
the
net
work
done
on
an
object
is
related
to
its
speed
as
follows
:
𝑊
𝑛
=
1
2
𝑣
2
−
1
2
𝑣
0
2
Work

Energy Theorem
𝑊
𝑛
=
1
2
𝑣
2
−
1
2
𝑣
0
2
•
The
RHS
can
be
interpreted
as
the
energy
associated
with
the
object’s
motion
and
is
defined
as
the
kinetic
energy
.
•
Therefore,
the
net
work
done
on
an
object
is
equal
to
the
change
in
an
object’s
kinetic
energy
.
Work

Energy Theorem
•
Based
on
the
work

energy
theorem,
we
can
define
work
as
the
amount
of
energy
transferred
to
an
object
when
a
force
acts
upon
it
.
•
When
W
>
0
(W
<
0
),
then
the
mechanical
energy
of
the
object
increases
(decreases
)
.
•
Based
on
the
work

energy
theorem,
we
can
define
kinetic
energy
as
the
work
a
moving
object
can
do
in
coming
to
rest
.
Example 2
•
The
driver
of
a
1
.
00
x
10
3
kg
car
traveling
on
the
interstate
at
35
.
0
m/s
slams
on
his
breaks
to
avoid
hitting
a
second
vehicle
in
front
of
him
.
•
After
the
brakes
are
applied,
a
constant
kinetic
friction
force
of
magnitude
8
.
00
x
10
3
N
acts
on
the
car
.
•
(a)
At
what
minimum
distance
should
the
brakes
be
applied
to
avoid
a
collision
with
the
other
vehicle?
•
(b)
If
the
distance
between
the
vehicles
is
initially
only
30
.
0
m,
at
what
speed
would
the
collision
occur?
Example
2
Solution
•
To
determine
the
minimum
stopping
distance,
we
apply
the
work

energy
theorem
to
the
car
𝑊
𝑛
=
1
2
𝑣
2
−
1
2
𝑣
0
2
−
𝑘
∆
=
−
1
2
𝑣
0
2
∆
=
𝑣
0
2
2
𝑘
=
1
.
00
10
3
𝑘
35
.
0
/
2
2
8
.
00
10
3
𝑁
=
76
.
6
Example 2 Solution
•
At
the
given
distance
of
30
.
0
m,
the
car
is
too
close
to
truck
and
thus,
will
collide
into
the
truck
.
•
To
determine
the
speed
at
impact,
we
use
the
work
energy
theorem
𝑊
𝑛
=
1
2
𝑣
2
−
1
2
𝑣
0
2
−
𝑘
∆
=
1
2
𝑣
2
−
1
2
𝑣
0
2
𝑣
=
𝑣
0
2
−
2
𝑘
∆
=
35
.
0
/
2
−
2
8
.
00
10
3
𝑁
30
.
0
1
.
00
10
3
𝑘
=
27
.
3
/
Example 3
•
A
man
pushing
a
crate
of
mass
m
=
92
.
0
kg
at
a
speed
v
=
0
.
850
m/s
encounters
a
rough
horizontal
surface
of
length
l
=
0
.
65
m
.
•
If
the
coefficient
of
kinetic
friction
between
the
crate
and
rough
surface
is
0
.
358
and
he
exerts
a
constant
horizontal
force
of
275
N
on
the
crate,
find
•
(a)
The
magnitude
and
direction
of
the
net
force
on
the
crate
while
it
is
on
the
rough
surface
•
(b)
The
net
work
done
on
the
crate
while
it
is
on
the
rough
surface
•
(c)
The
speed
of
the
crate
when
it
reaches
the
end
of
the
rough
surface
.
Example 3 Solution
•
To
determine
the
net
force,
we
first
should
make
a
free
body
diagram
.
•
Using
Newton’s
2
nd
law
gives
=
−
=
0
𝑎
=
=
−
𝑘
+
𝑎𝑝𝑝
•
The
net
force
will
be
in
the
x

direction
given
as
𝑭
𝑛
=
=
−
𝜇
𝑘
+
𝑎𝑝𝑝
=
0
.
358
92
.
0
𝑘
9
.
80
2
+
275
𝑁
=
−
47
.
8
𝑁
•
The
net
work
will
be
in
the
x

direction,
given
as
𝑾
𝑛
=
𝑭
𝑛
∙
∆
?(
𝑛
∆
=
−
47
.
8
𝑁
0
.
65
=
−
31
.
05
n
m
g
f
k
F
app
Example 3 Solution
•
To
determine
the
final
velocity,
we
apply
the
work

energy
theorem
.
𝑊
𝑛
=
1
2
𝑣
2
−
1
2
𝑣
0
2
𝑣
=
𝑣
0
2
+
2
𝑊
𝑛
=
0
.
850
2
+
2
−
31
.
05
92
.
0
𝑘
=
0
.
22
/
n
m
g
f
k
F
app
Gravitational Potential Energy
•
Gravitational
potential
energy
is
the
energy
associated
with
an
object’s
position
in
the
gravitational
field
.
•
Consider
a
book
of
mass
m
that
falls
Δ
y
.
•
The
net
work
done
by
gravity
is
𝑊
=
𝑭
𝒈
∙
∆
=
−
∆
=
−
∆
𝑃
•
Using
the
work

energy
theorem,
we
have
𝑊
𝑛
=
−
∆
𝑃
=
∆
Spring Force and Hooke’s Law
•
The
figure
shows
a
spring
in
its
equilibrium
position,
where
the
spring
is
neither
compressed
nor
stretched
.
•
Experimentation
has
shown
that
the
spring
force
follows
Hooke’s
law
=
−
𝑘
.
•
The
spring
constant
k
is
a
measure
of
the
stiffness
of
the
spring
and
is
determined
by
how
the
spring
was
formed,
its
material
composition,
and
the
thickness
of
the
wire
.
•
This
force
is
called
a
restoring
force
because
the
spring
always
exerts
a
force
in
a
direction
opposite
the
displacement,
restoring
the
spring
to
its
original
position
.
Elastic Potential Energy
•
The
elastic
potential
energy
is
stored
energy
arising
from
the
work
done
to
compress
or
stretch
the
spring
.
•
Consider
a
horizontal
spring
and
mass
at
the
equilibrium
position
shown
here
.
•
The
work
done
by
the
spring
when
compressed
by
an
applied
force
from
equilibrium
to
a
displacement
x
.
𝑊
=
𝑭
∙
∆
=
=
−
1
2
𝑘
2
=
−
∆
𝑃
•
Using
the
work

energy
theorem,
we
have
𝑊
𝑛
=
−
∆
𝑃
=
∆
Mechanical Energy Conservation
•
When
only
conservative
forces
(such
as
the
gravitational
force
and
spring
force)
are
present,
then
the
work

energy
theorem
can
be
written
as
𝑊
𝑛
=
𝑊
+
𝑊
=
∆
−
∆
𝑃
−
∆
𝑃
=
∆
∆
+
𝑃
+
𝑃
=
0
•
This
is
called
the
conservation
of
mechanical
energy
and
it
is
one
of
the
most
important
concepts
in
classical
mechanics
.
•
Conservation
of
mechanical
energy
says
that
the
mechanical
energy
(the
sum
of
the
kinetic
energy
and
potential
energy)
remains
constant
at
all
times
.
Example 4
•
A
0
.
500

kg
block
rests
on
a
horizontal,
frictionless
surface
.
The
block
is
pressed
back
against
a
spring
having
a
constant
of
k
=
625
N/m,
compressing
the
spring
by
10
.
0
cm
to
point
A
.
Then
the
block
is
released
.
•
(a)
Find
the
maximum
distance
d
the
block
travels
up
the
frictionless
incline
if
θ
=
30
.
0
⁰
.
•
(b)
How
fast
is
the
block
going
when
halfway
to
its
maximum
height?
Example 4 Solution
•
Conservation
of
mechanical
energy
gives
𝑖
+
𝑃
,
𝑖
+
𝑃
,
𝑖
=
+
𝑃
,
+
𝑃
,
1
2
𝑣
𝑖
2
+
𝑖
+
1
2
𝑘
𝑖
2
=
1
2
𝑣
2
+
+
1
2
𝑘
2
1
2
𝑘
𝑖
2
=
ℎ
=
𝑑
sin
30
.
0
°
𝑑
=
𝑘
𝑖
2
2
sin
30
.
0
°
=
625
𝑁
0
.
1
2
2
0
.
500
𝑘
9
.
8
/
2
sin
30
.
0
°
=
1
.
28
Example 4 Solution
•
To
determine
the
velocity
of
the
block
when
it’s
halfway
to
its
maximum
height,
we
use
conservation
of
mechanical
energy
𝑖
+
𝑃
,
𝑖
+
𝑃
,
𝑖
=
+
𝑃
,
+
𝑃
,
1
2
𝑘
𝑖
2
=
1
2
𝑣
2
+
𝑑
sin
30
.
0
°
2
𝑣
=
𝑘
𝑖
2
−
𝑑
sin
30
.
0
°
=
625
𝑁
0
.
1
2
−
0
.
500
𝑘
9
.
80
/
2
1
.
28
sin
30
.
0
°
0
.
500
𝑘
=
2
.
50
/
Example 5
•
A
ball
of
mass
m
=
1
.
80
kg
is
released
from
rest
at
a
height
h
=
65
.
0
cm
above
a
light
vertical
spring
of
force
constant
k
.
•
The
ball
strikes
the
top
of
the
spring
and
compresses
it
a
distance
d
=
9
.
00
cm
.
•
Neglecting
any
energy
losses
during
the
collision,
find
•
(a)
t
he
speed
of
the
ball
just
as
it
touches
the
spring
•
(b)
t
he
force
constant
of
the
spring
Example 5 Solution
•
Because
there
are
no
energy
losses
during
the
collision,
only
conservative
forces
are
present
and
thus,
mechanical
energy
conservation
can
be
applied
.
•
It
may
be
helpful
to
break
this
problem
into
two
phases
–
Stage
1
:
Free
fall
under
gravity
–
Stage
2
:
Compression
of
the
spring
Example 5 Solution
•
For
stage
1
,
the
gravitational
force
is
the
only
force
present
and
thus,
we
can
use
mechanical
energy
conservation
.
𝑖
+
𝑃
,
𝑖
=
+
𝑃
,
1
2
𝑣
𝑖
2
+
𝑖
=
1
2
𝑣
2
+
𝑣
=
2
𝑖
=
2
9
.
8
/
2
0
.
65
=
3
.
56
/
Example 5 Solution
•
For
stage
2
,
both
the
gravitational
force
and
spring
force
are
present
.
•
Using
mechanical
energy
conservation,
we
have
𝑖
+
𝑃
,
𝑖
+
𝑃
,
𝑖
=
+
𝑃
,
+
𝑃
,
1
2
𝑣
𝑖
2
=
−
𝑑
+
1
2
𝑘
𝑑
2
𝑘
=
𝑣
𝑖
2
+
2
𝑑
𝑑
2
=
1
.
80
𝑘
3
.
56
2
+
2
1
.
80
𝑘
9
.
80
/
2
0
.
09
0
.
09
2
=
3
.
22
10
3
𝑁
/
Work and
Nonconservative
Forces
•
When
nonconservative
forces
(such
as
kinetic
friction)
are
present,
then
the
work

energy
theorem
can
be
written
as
𝑊
𝑛
=
𝑊
𝑛
+
𝑊
+
𝑊
=
∆
𝑊
𝑛
−
∆
𝑃
−
∆
𝑃
=
∆
𝑊
𝑛
=
∆
𝑃
+
∆
𝑃
+
∆
•
If
mechanical
energy
is
not
conserved
in
a
system,
then
it
must
leave
the
system
or
be
transformed
into
a
form
of
non

mechanical
energy
(such
as
internal
energy)
.
•
When
positive
(negative)
work
is
done
on
the
system,
energy
is
transferred
from
the
environment
(system)
to
the
system
(environment
)
.
Example
6
•
A
skier
starts
from
rest
at
the
top
of
a
frictionless
incline
of
height
20
.
0
m
.
•
At
the
bottom
of
the
incline,
the
skier
encounters
a
horizontal
surface
where
the
coefficient
of
kinetic
friction
between
skis
and
snow
is
0
.
210
.
•
(a)
Find
the
skier’s
speed
at
the
bottom
.
•
(b)
How
far
does
the
skier
travel
on
the
horizontal
surface
before
coming
to
rest?
Example
6
Solution
•
We
can
split
this
problem
into
two
parts
:
–
Trip
1
:
Going
from
points
A
to
B
under
the
influence
of
gravity
–
Trip
2
:
Going
from
points
B
to
C
under
the
influence
of
kinetic
friction
Example
6
Solution
•
For
trip
1
,
we
can
find
the
speed
at
point
B
by
using
the
energy
conservation
equation
𝑖
+
𝑃
𝑖
=
+
𝑃
1
2
𝑣
𝑖
2
+
𝑖
=
1
2
𝑣
2
+
𝑣
=
2
𝑖
=
2
9
.
8
/
2
20
=
1
9
.
8
/
Example
6
Solution
•
For
trip
2
,
we
can
use
the
work

energy
theorem
to
determine
the
distance
traveled
from
B
to
C
.
𝑊
𝑛
=
∆
=
1
2
𝑣
2
−
1
2
𝑣
2
−
𝑘
𝑑
=
1
2
𝑣
2
−
1
2
𝑣
2
−
𝜇
𝑘
𝑑
=
−
1
2
𝑣
2
𝑑
=
𝑣
2
2
𝜇
𝑘
𝑑
=
19
.
8
2
2
0
.
210
9
.
8
/
2
=
95
.
2
n
m
g
f
k
Application: The Simple Pendulum
•
There
are
only
two
forces
acting
upon
the
pendulum
bob
:
the
gravitational
and
tension
force
.
•
The
gravitational
force
does
work
on
the
pendulum,
but
it
does
not
change
the
total
gravitational
potential
energy
of
the
system
.
•
The
tension
force
does
not
do
work
on
the
pendulum
since
it
acts
perpendicular
to
the
motion
of
the
pendulum
.
•
Here,
the
total
mechanical
energy
of
the
pendulum
is
conserved
.
•
Video
demonstration
Power
•
Power
can
be
defined
as
the
rate
at
which
energy
is
transferred
into
or
out
of
a
system
.
•
Mathematically,
power
can
be
defined
𝑃
=
𝑊
∆
=
𝑣
•
The
SI
unit
of
power
is
joule
per
second,
also
called
the
watt
.
•
1
W
=
1
J/s
=
1
kg
m
2
/s
3
Example 7
•
What
average
power
would
a
1
.
00
x
10
3
kg
speedboat
need
to
go
from
rest
to
20
.
0
m/s
in
5
.
00
s,
assuming
the
water
exerts
a
constant
drag
force
of
magnitude
f
=
5
.
00
x
10
2
N
and
the
acceleration
is
constant
.
•
Find
an
expression
for
the
instantaneous
power
as
a
function
of
the
drag
force
f
d
.
Example 7 Solution
•
The
power
is
provided
by
the
engine
and
thus,
𝑃
=
𝑊
𝑛𝑖𝑛
∆
•
To
determine
the
work
done
by
the
engine,
we
use
the
work

energy
theorem
.
𝑊
𝑛
=
𝑊
𝑛𝑖𝑛
+
𝑊
𝑎
=
1
2
𝑣
2
−
1
2
𝑣
0
2
𝑊
𝑛𝑖𝑛
=
1
2
𝑣
2
+
𝑘
∆
Example 7 Solution
•
To
determine
the
displacement,
we
must
first
find
the
acceleration
using
kinematics
equation
1
.
𝑣
=
𝑣
0
+
𝑎
𝑎
=
𝑣
−
𝑣
0
=
20
−
0
5
=
4
.
00
/
2
•
The
displacement
can
be
calculated
using
kinematics
equation
3
𝑣
2
=
𝑣
0
2
+
2
𝑎
∆
∆
=
𝑣
2
−
𝑣
0
2
2
𝑎
=
20
2
−
0
2
2
4
.
00
/
2
=
50
.
0
Example 7 Solution
•
Calculating
the
work
done
by
the
engine
gives
𝑊
𝑛𝑖𝑛
=
1
2
𝑣
2
+
𝑘
∆
=
1
2
1
.
00
10
3
𝑘
20
2
+
5
.
00
10
2
𝑁
50
.
0
=
2
.
25
10
5
•
The
power
is
provided
by
the
engine
and
thus,
𝑃
=
𝑊
𝑛𝑖𝑛
∆
=
2
.
25
10
5
5
.
00
=
4
.
50
10
4
𝑊
•
To
relate
instantaneous
power
to
the
drag
force,
we
use
Newton’s
second
law
𝑎
=
=
𝐸
−
𝑃
=
𝐸
𝑣
=
𝑎
+
𝑣
=
𝑎
2
+
𝑎
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