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Chapter
6
Work
and
Energy
October
8
and
October
13,
2009
Work
and
Energy
  Energy
is
one
of
the
most
important
concepts
in
physics
  Alterna've
approach
to
analyze
some
mechanics
situa'ons
  Ini'ally
considered
the
energy
of
mo'on
(kine'c
energy)
  But
many
other
forms
(with
names)
  Many
applica'ons
beyond
mechanics
  Thermodynamics
(movement
of
heat)
  Quantum
mechanics...
  Chemistry
  Biology
  ……………….(every
science)
  Very
useful
tool
in
understanding
our
physical
world.
  You
will
learn
new
(some'mes
much
easier)
ways
to
solve
problems.
Energy
and
Newton’s
Laws
The
concept
of
energy
turns
out
to
be
even
more
general
than
Newton’s
Laws.
But
the
importance
of
mechanical
energy
in
classical
mechanics
is
a
consequence
of
Newton’s
laws.
Defini'on
of
Work:
Constant
Force
 
W = F ⋅ Δ r = FΔr cosθ = Fr Δr
Vector
“Dot Product”
Work is a scalar.
θ
Fr
Defini'on
of
Work….
Only
the
component
of
F
parallel
to
the
displacement
is
doing
work.
θ
F cos θ
Ques'on
You
li[
an
object
1
m
from
the
floor
to
a
shelf.
The
work
done
by
the
earth’s
gravita'onal
field
on
the
object
is:
A.  posi've
B.  zero
C.  nega've
Ques'on
You
li[
an
object
1
m
from
the
floor
to
a
shelf.
The
work
done
by
the
earth’s
gravita'onal
field
on
the
object
is:
A.  posi've
B.  zero
C.  nega've
correct
The object’s displacement is opposite in direction to
the force of gravity on the object. So the work done
by the gravitational force on the object is negative.
Ques'on
You
push
a
heavy
box
1m
across
a
rough
floor.
Which
statements
are
correct?
A.  You
have
done
posi've
work
on
the
box.
B.  You
have
done
nega've
work
on
the
box.
C.  The
earth’s
gravita'onal
force
has
done
nega've
work
on
the
box.
E.  You
have
done
nega've
work
on
the
outside
world.
Ques'on
You
push
a
heavy
box
1m
across
a
rough
floor.
Which
statements
are
correct?
A.  You
have
done
posi've
work
on
the
box.
correct
B.  You
have
done
nega've
work
on
the
box.
C.  The
earth’s
gravita'onal
force
has
done
nega've
work
on
the
box.
E.  You
have
done
nega've
work
on
the
outside
world.
Defini'on
of
Work
Total
work,
W,
of
a
force
F
ac'ng
through
a
displacement
Δx
=
dx
î
is:
 
W = ∑ Fi ⋅ Δx i
Fx
i
x2
= ∫ Fx dx
x1
Fi
x
Δxi
Work = Area under F(x) curve
Work
done
by
a
spring
force
Force
exerted
by
a
spring
extended
a
distance
x
is:
Fx
=
‐kx
(Hooke’s
law)
Force
done
by
spring
on
an
object
when
the
displacement
of
spring
is
changed
from
xi
to
xf
is:
Defini'on
of
Kine'c
Energy
Kine'c
energy
(K.E.)
of
a
par'cle
of
mass
m
moving
with
speed
v
is
defined
as
K.E.
=
½mv2
Kine'c
energy
is
a
useful
concept
because
of
the
Work/Kine'c
Energy
theorem,
which
relates
the
work
done
on
an
object
to
the
change
in
kine'c
energy.
First do some simple examples, then show general
relationship between work and kinetic energy.
Work/Kine'c
Energy
Theorem:
Special
case
of
constant
force
v1
v2
m
Work/Kine'c
Energy
Theorem:
Special
case
of
constant
force
v1
v2
m
Work/Kine'c
Energy
Theorem:
Special
case
of
constant
force
v1
v2
m
Problem:
Work
and
Energy
Two
blocks
have
masses
m1
and
m2,
where
m1
>
m2.
They
are
sliding
on
a
fric'onless
floor
and
have
the
same
kine'c
energy
when
they
encounter
a
long
rough
stretch
with
μ>0,
which
slows
them
down
to
a
stop.
Which
one
will
go
farther
before
stopping?
(A)
m1
(B)
m2 (C)
They
will
go
the
same
distance
m1
m2
Problem:
Work
and
Energy
Two
blocks
have
masses
m1
and
m2,
where
m1
>
m2.
They
are
sliding
on
a
fric'onless
floor
and
have
the
same
kine'c
energy
when
they
encounter
a
long
rough
stretch
with
μ>0,
which
slows
them
down
to
a
stop.
Which
one
will
go
farther
before
stopping?
(A)
m1
(B)
m2 (C)
They
will
go
the
same
distance
Solution method on
following slides
m1
m2
Problem:
Work
and
Energy
(Solu'on)
The
work‐kine'c
energy
theorem
says
that
for
any
object,
WNET
=
ΔK.
In
this
example,
the
only
force
that
does
work
is
fric'on
(since
both
N
and
mg
are
perpendicular
to
the
block’s
mo'on).
N
f
m
mg
Problem:
Work
and
Energy
(Solu'on)
The
net
work
done
to
stop
a
box
is
‐fD
=
‐μmgD.
The
work‐kine'c
energy
theorem
says
that
for
any
object,
WNET
=
ΔK,
so
WNET=Kf‐Ki=0‐Ki.
Since
the
boxes
start
out
with
the
same
kine'c
energy,
we
have
μm1gD1=μm2gD2
and
D1/D2=m2/m1.
Since
m1>m2,
we
must
have
D2>D1.
m1
D1
m2
D2
A
falling
object
What
is
the
speed
of
an
object
that
starts
at
rest
and
then
falls
a
ver'cal
distance
H?
v0 = 0
H
v
A
falling
object
What
is
the
speed
of
an
object
that
starts
at
rest
and
then
falls
a
ver'cal
distance
H?
Work
done
by
gravita'onal
force
WG
=
FΔr
=
mgH
v0 = 0
H
Work/Kine'c
Energy
Theorem:
WG = mgH = 12 mv2
⇒ v = 2gH
v
The
scalar
product
(or
dot
product)
ba
θ
θ
ab
Examples
of
dot
products
y
x
z
Suppose
Then
Proper'es
of
dot
products
ay
ax
More
proper'es
of
dot
products
Which
of
the
statements
below
is
correct?
A.  The
scalar
product
of
two
vectors
can
be
nega've.
B.  AcB=c(BA),
where
c
is
a
constant.
C.  The
scalar
product
can
be
non‐zero
even
if
two
of
the
three
components
of
the
two
vectors
are
equal
to
zero.
E.  All
of
the
above
statements
are
correct.
Which
of
the
statements
below
is
correct?
A.  The
scalar
product
of
two
vectors
can
be
nega've.
B.  AcB=c(BA),
where
c
is
a
constant.
C.  The
scalar
product
can
be
non‐zero
even
if
two
of
the
three
components
of
the
two
vectors
are
equal
to
zero.
E.  All
of
the
above
statements
are
correct.
Work/KE
Theorem
(Variable
Force,
3‐d)
Work done by a force exerted over a path is
 
W = ∫ F ⋅ d r

r2
r1




t
t
f dv d r
f dv 
dv 
= m ∫
⋅ d r = m ∫ ⋅ dt = m ∫ ⋅ v dt
r (t i ) dt
t i dt dt
t i dt
tf d ⎛1
2⎞
= m ∫ ⎜ ⋅ v ⎟dt
⎠
t i dt ⎝ 2
1 2 1 2
1 2 1 2
= mvf − mvi
ΔW = mvf − mvi
2
2
2
2

r (t f )
Net work done on object = change in kinetic energy of object
What
about
mul'ple
forces?
Comments
about
work
W
=
F
Δr
• 
The
'me
interval
over
which
the
force
acts
is
not
relevant
• No
work
is
done
if:
 
F
=
0 or
 
Δr
=
0 or
 
F
is
perpendicular
to
Δr
More
comments
about
work
W
=
F
Δr
No
work
is
done
if
F
and
Δr
are
perpendicular.
No
work
is
done
by
tension
T
No
work
is
done
by
normal
force
N.
Ques'on
The
work
needed
to
maintain
uniform
circular
mo'on
in
a
horizontal
plane
is:
A.  zero
B. 
posi've
and
constant
C. 
nega've
and
constant
E. 
nega've
and
variable
Ques'on
The
work
needed
to
maintain
uniform
circular
mo'on
in
a
horizontal
plane
is:
A.  zero
B. 
posi've
and
constant
C. 
nega've
and
constant
E. 
nega've
and
variable
Force applied (towards the center of the circle) is
perpendicular to the direction of motion at all points
in the uniform circular motion.
Power
Defini'on
of
power:
Power
is
the
rate
at
which
a
force
does
work.
dW
Power
P =
dt
   
dW = F ⋅ d r = F ⋅ vdt
dW  
⇒P=
= F⋅v
dt
Unit of power: 1 watt = 1 joule/second
Ques'on
A
sports
car
accelerates
from
zero
to
30
mph
in
1.5
s.
How
long
does
it
take
for
it
to
accelerate
from
zero
to
60
mph,
assuming
the
power
of
the
engine
to
be
independent
of
velocity
and
neglec'ng
fric'on?
Ques'on
A
sports
car
accelerates
from
zero
to
30
mph
in
1.5
s.
How
long
does
it
take
for
it
to
accelerate
from
zero
to
60
mph,
assuming
the
power
of
the
engine
to
be
independent
of
velocity
and
neglec'ng
fric'on?
d 1 2
Power
is
constant,
so
,
a
constant.
(
2 mv ) = C
dt
Integra'ng
with
respect
to
'me,
and
no'ng
that
the
2
1
ini'al
velocity
is
zero,
one
gets
.
So
genng
2 mv = Ct
to
twice
the
speed
takes
4
'mes
as
long,
and
the
'me
to
reach
60
mph
is
4×1.5
=
6s.
Ques'on
A
cart
on
an
air
track
is
moving
at
0.5
m/s
when
the
air
is
suddenly
turned
off.
The
cart
comes
to
rest
a[er
traveling
1
m.
The
experiment
is
repeated,
but
now
the
cart
is
moving
at
1
m/s
when
the
air
is
turned
off.
How
far
does
the
cart
travel
before
coming
to
rest?
Ques'on
A
cart
on
an
air
track
is
moving
at
0.5
m/s
when
the
air
is
suddenly
turned
off.
The
cart
comes
to
rest
a[er
traveling
1
m.
The
experiment
is
repeated,
but
now
the
cart
is
moving
at
1
m/s
when
the
air
is
turned
off.
How
far
does
the
cart
travel
before
coming
to
rest?
Work
done
by
fric'on
force
over
a
distance
d
is
‐Fd.
If
ini'al
velocity
is
v,
then
work/kine'c
energy
theorem
says
that
the
stopping
distance
D
is
determined
by
FD=½mv2.
If
the
ini'al
velocity
is
doubled,
then
the
stopping
distance
goes
up
by
a
factor
of
4.
So
the
stopping
distance
for
the
larger
ini'al
velocity
is
4
m/s.
Power
P
is
required
to
li[
a
body
a
distance
d
at
constant
speed
v.
What
power
is
required
to
li[
the
body
a
distance
2d
at
constant
speed
3v?
A.  P
B.  2P
C.  3P
E.  3P/2
Power
P
is
required
to
li[
a
body
a
distance
d
at
constant
speed
v.
What
power
is
required
to
li[
the
body
a
distance
2d
at
constant
speed
3v?
A.  P
B.  2P
C.  3P
E.  3P/2
[Power = Fv; force needed does not change (since speed is
still constant).]
If
a
fighter
jet
doubles
its
speed,
by
what
factor
should
the
power
from
the
engine
change?
(Assume
that
the
drag
force
on
the
plane
is
propor'onal
to
the
square
of
the
plane’s
speed.)
A.  by
half
B.  unchanged
C.  doubled
E.  8
'mes
If
a
fighter
jet
doubles
its
speed,
by
what
factor
should
the
power
from
the
engine
change?
(Assume
that
the
drag
force
on
the
plane
is
propor'onal
to
the
square
of
the
plane’s
speed.)
A.  by
half
B.  unchanged
C.  doubled
E.  8
?mes
Magnitude of power is Fv. When the velocity v is
doubled, the drag force goes up by a factor of 4, and
Fv goes up by a factor of 8
Center‐Of‐Mass
Work
For systems of particles that are not all moving at the same
velocity, there is a work-kinetic energy relation for the center of
mass.
So
where
Net work done on
collection of objects
is the translational
kinetic energy
=
change in translational
kinetic energy of system
Problem
(6‐62)
The
magnitude
of
the
single
force
ac'ng
on
a
par'cle
of
mass
m
is
given
by
F
=
bx2,
where
b
is
a
constant.
The
par'cle
starts
from
rest
at
x=0.
A[er
it
travels
a
distance
L,
determine
its
(a)
kine'c
energy
and
(b)
speed.
Problem
(6‐62)
The
magnitude
of
the
single
force
ac'ng
on
a
par'cle
of
mass
m
is
given
by
F
=
bx2,
where
b
is
a
constant.
The
par'cle
starts
from
rest
at
x=0.
A[er
it
travels
a
distance
L,
determine
its
(a)
kine'c
energy
and
(b)
speed.
work done by force:
  L 2
b 3
W = ∫ F ⋅ dx = ∫ bx dx = L
0
0
3
L
So (a) kinetic energy is K=bL3/3
(b) Since ½mv2=bL3/3,
2bL3
v=
3m
Ques'on
A
1kg
block
slides
4
m
down
a
fric'onless
plane
inclined
at
30
degrees
to
the
horizontal.
What
is
the
speed
of
the
block
as
it
leaves
the
inclined
plane?
30°
Ques'on
A
1kg
block
slides
4
m
down
a
fric'onless
plane
inclined
at
30
degrees
to
the
horizontal.
What
is
the
speed
of
the
block
as
it
leaves
the
inclined
plane?
30°
No friction, so only force that does work is the gravitational
force. Work done by gravity is W = mgL sinθ.
By Work/Kinetic Energy Theorem, we have
mgL sinθ = ½mv2, so v = (2gL sinθ)1/2 = 6.3 m/s
A
person
pushes
against
a
spring,
the
opposite
end
of
which
is
arached
to
a
fixed
wall.
The
spring
compresses.
Is
the
work
done
by
the
spring
on
the
person
posi've,
nega've
or
zero?
A. 
posi've
B. 
nega've
C. 
zero
A
person
pushes
against
a
spring,
the
opposite
end
of
which
is
arached
to
a
fixed
wall.
The
spring
compresses.
Is
the
work
done
by
the
spring
on
the
person
posi've,
nega've
or
zero?
A. 
posi've
B. 
nega?ve
C. 
zero
A
light
rope
runs
through
two
fric'onless
pulleys
of
negligible
mass.
A
mass,
m,
is
hung
from
one
of
the
pulleys
and
a
force,
F,
is
applied
to
one
end
of
the
rope
so
that
the
mass
moves
at
a
constant
speed.
What
is
the
force
needed
to
move
the
mass
at
a
constant
speed?
A. 
0
B. 
½
mg
C. 
mg
E. 
4mg
A
light
rope
runs
through
two
fric'onless
pulleys
of
negligible
mass.
A
mass,
m,
is
hung
from
one
of
the
pulleys
and
a
force,
F,
is
applied
to
one
end
of
the
rope
so
that
the
mass
moves
at
a
constant
speed.
What
is
the
force
needed
to
move
the
mass
at
a
constant
speed?
A. 
0
B. 
½
mg
C. 
mg
E. 
4mg