physics 111 lecture 07 - new jersey institute of technologyjanow/physics 111 spring...
TRANSCRIPT
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Physics 111 Lecture 07
Potential Energy & Energy Conservation
SJ 8th Ed.: Chap 7.6 –7.8, 8.1 –8.5
•Potential Energy
•Conservative Forces
•Determining Potential Energy Values
–Gravitational Potential Energy
–Elastic Potential Energy
•Conservation of Mechanical Energy
•Work Done by Kinetic Friction
(Non-Conservative Forces)
•Conservation of Energy (General)
–Isolated Systems
•Power
•Reading Energy Diagrams
–Finding the Force (Gradient)
–Turning Points
–Equilibrium Points
7.6
Po
ten
tial E
nerg
y o
f a S
ys
tem
7.7
Co
nserv
ati
ve a
nd
No
n-C
on
serv
ati
ve F
orc
es
7.8
Rela
tio
nsh
ip B
etw
een
Co
nserv
ati
ve
Fo
rces a
nd
Po
ten
tial E
nerg
y
7.9
En
erg
y D
iag
ram
s a
nd
Eq
uilib
riu
m
8.1
En
erg
y in
No
n-I
so
late
d S
yste
ms
8.2
En
erg
y f
or
Iso
late
d S
yste
ms
8.3
Kin
eti
c F
ricti
on
8.4
Ch
an
ges in
Mech
an
ical E
nerg
y f
or
No
n-c
on
serv
ati
ve f
orc
es
8.5
Po
wer
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Overview
Previou
sly:
•K
ine
tic
En
erg
y2
21m
vK
≡ ≡≡≡∑ ∑∑∑
≡ ≡≡≡s
ys
tem
a
for
jto
t
m
vK
2
21
rd
F d
Wr
or
= ===fo
rce
c
on
sta
nt
rF
Wr
or
∆ ∆∆∆= ===
ge
ne
ral
in
pa
th
rd
F
W∫ ∫∫∫
= ===r
or
forc
es
a
ll
by
d
on
e w
ork
n
et
K
W∆ ∆∆∆
= ===
po
sit
ion
s
pri
ng
in
sto
red
wo
rk
2 f
2 is
p
kx
21
kx
21
W− −−−
= ===
•G
ravit
ati
on
al
Wo
rk
•E
lasti
c,
sp
rin
g f
orc
e
•P
ath
in
dep
en
den
ce
•W
ork
–K
E T
he
ore
m
•W
ork
g
g
jm
gF
yF
W
r
vr
ov
− −−−= ===
∆ ∆∆∆= ===
Now
: M
ore p
owerf
ul v
iew u
sing
ene
rgy c
onse
rvation
•M
ech
an
ical
En
erg
y =
Kin
eti
c +
Po
ten
tial
)r(U
r
)r(U
K
Em
ec
h
r+ +++
≡ ≡≡≡
)
0
( W
E
so
me
tim
es
nc
me
ch
= ==== ===
∆ ∆∆∆
int
me
ch
tot
E E
E+ +++
≡ ≡≡≡
)
0 (
WE
so
me
tim
es
ex
tto
t= ===
= ===∆ ∆∆∆
•C
on
serv
ati
ve F
orc
es <� ���
Po
ten
tia
l E
nerg
y (
so
meti
mes F
ield
s)
•T
ota
l E
nerg
y =
Mech
an
ica
l +
Th
erm
al
an
d I
nte
rnal
•M
ech
an
ical
En
erg
y i
s c
on
sta
nt
wh
en
th
e w
ork
du
e t
o n
on
-co
nserv
ati
ve
fo
rces i
s z
ero
•T
ota
l E
nerg
y i
s c
on
sta
nt
if s
ys
tem
is i
so
late
d
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Potential Energy -Definition
Pict
ure:
•A p
art
icle in
a s
yst
em f
eels int
ern
al fo
rces
due
to
other
part
icles.
•The n
et
forc
e F
(x,y
,z)due
to
the r
est
of
the s
yst
em d
oes
wor
k o
n a p
art
icle
•while it
chang
es
position
.•
If
F(x
,y,z
)is “
cons
erv
ative
”th
e w
ork
is c
onve
rted t
o “p
otent
ial ene
rgy”
stor
ed
in t
he m
utua
l po
sition
s of
the p
art
icle a
nd t
he r
est
of
the s
yst
em.
•Po
tent
ial ene
rgy s
tora
ge is
reve
rsib
le:
ene
rgy s
tore
d w
hen
a p
art
icle m
oves
subje
ct t
o “c
onse
rvative
”fo
rce c
an
be r
eco
vere
d w
ithou
t loss
.
Pote
ntial ene
rgy i
s repr
ese
nted b
y a
sca
lar
func
tion
U(x
,y,z
) of
the
part
icle’s p
osition
r=
(x,y
,z).
•The P
E
belong
s to
the s
yst
em a
s a w
hole.
•U
(r)of
a p
art
icle d
epe
nds
explicitly
on t
hat
part
icles’
loca
tion
.
•U
(r)depe
nds
implicitly a
lso
on loc
ation
s of
oth
er
part
s of
the s
yst
em.
•
When
PE c
hang
es,
KE m
ay c
hang
e a
nd/o
r ot
her
non-
cons
erv
ative
exte
rnal
forc
es
may d
o wor
k.
)r(d
U
rd
)r(F
dW
rr
or
r− −−−
≡ ≡≡≡= ===
NET FORCE DUE TO
REST OF SYSTEM
WORK DONE ON
PARTICLE
POTENTIAL
ENERGY CHANGE
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Conservative and Non-Conservative Forces
Non
-co
nserv
ative
for
ces:
•ca
n be d
issipa
tive
(fr
iction
, dra
g):
they t
rans
form
mech
anica
l ene
rgy
to h
eat
and
caus
e t
he int
ern
al ene
rgy E
intof
syst
em t
o rise
irr
eve
rsib
ly.
•ca
n be c
onta
ct f
orce
s •ca
n no
t be r
epr
ese
nted b
y a
pote
ntial ene
rgy f
unct
ion
Con
serv
ative
for
ces
trans
form
kinetic
to p
otent
ial ene
rgy a
nd b
ack
aga
in r
eve
rsib
ly•
There
is
a p
otent
ial ene
rgy f
unct
ion
for
such
a f
orce
•Examples:
gra
vity
, elast
ic (sp
ring
), e
lect
rost
atic
PE a
nd p
otent
ial (“vo
ltage
”).
A
B
Path
1
Path
2
F
F
F
F
F
F
•T
he w
ork
do
ne o
n a
part
icle
mo
vin
g a
rou
nd
an
yclo
sed
path
is z
ero
.
e.g
. W
AB
,1 +
WB
A,2
= 0
Fo
r a c
on
serv
ati
ve f
orc
e•
An
y p
ath
is r
evers
ible
i.e.,
fo
r an
y p
ath
W
AB
,pa
thn
=
-W
BA
,pa
thn
•T
he w
ork
do
ne o
n a
part
icle
mo
vin
g b
etw
een
an
y
en
dp
oin
ts d
oes n
ot
dep
en
d o
n t
he p
ath
taken
.
i.e.,
WA
B,1
= W
AB
,2 =
….W
AB
,n
Fo
r a n
on
-co
nserv
ati
ve f
orc
e:
wo
rk d
on
e a
rou
nd
a c
losed
path
is >
0.
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Example: Gra
vita
tion
al Po
tent
ial Ene
rgy
•The s
yst
em is
the a
pple p
lus
the E
art
h
)r(d
U
rd
)r(F
dW
gg
g
rr
or
r− −−−
≡ ≡≡≡= ===
FORCE DUE TO
REST OF SYSTEM
WORK DONE ON PARTICLE BY
CONSERVATIVE FORCE
POTENTIAL
ENERGY CHANGE
y
)r(F
W
)r(
Ug
gg
ro
rr
r∆ ∆∆∆
− −−−= ===
∆ ∆∆∆− −−−
= ===∆ ∆∆∆
jy
y
jm
g
)r(F
forc
e
co
ns
tan
tg
∆ ∆∆∆± ±±±
= ===∆ ∆∆∆
− −−−= ===
rr
r
•PE
inc
rease
s when
Fgdoe
s ne
gative
wor
k•PE
decr
ease
s when
Fgdoe
s po
sitive
wor
k
( (((
U
g= ===
∆ ∆∆∆ri
sin
g
y
mg
∆ ∆∆∆+ +++
fallin
g
y
mg
∆ ∆∆∆− −−−
•The z
ero
point
of
PE h
ere
is
arb
itra
ry –
choo
se a
refe
renc
e leve
l
)y
mg
(y
)y(
U)
yU
U
y
)r(F
W
if
ig
fg
gext
ext
− −−−= ===
− −−−= ===
∆ ∆∆∆= ===
∆ ∆∆∆= ===
∆ ∆∆∆r
or
r
•is a
n exte
rnalfo
rce,
slightly g
reate
r th
an
mg
,so
boo
k mov
es
up (sp
eed r
emains
~ z
ero
)
F ex
t
r
F e
xt
r
Act
ion
of e
xte
rnal, n
on-co
nserv
ative
forc
e:
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Finding the Potential Energy
rd
F d
Wr
or
= ===fo
rce
c
on
sta
nt
rF
Wr
or
∆ ∆∆∆= ===
∆ ∆∆∆g
en
era
l
inp
ath
rd
F
W∫ ∫∫∫
= ===∆ ∆∆∆
ro
r
Reca
ll: The w
ork
don
e b
y n
on-co
nserv
ative
for
ces
can
depe
nd o
n th
e p
ath
:
can
be t
he n
et
forc
e,
a d
issipa
tive
for
ce,
or a
con
serv
ative
for
ceFr
is t
he w
ork d
one b
y
(w
hate
ver
might
be)
F rW
∆ ∆∆∆F r
For
a c
onse
rvative
for
ce t
here
can
be a
pot
ent
ial ene
rgy f
unct
ion:
∆ ∆∆∆W
cdepe
nds
only o
n th
e e
ndpo
ints
, no
t th
e p
ath
deta
ils
cW
U
∆ ∆∆∆− −−−
= ===∆ ∆∆∆
Change in potential
energy of system
Work done by conservative
force on ANY path
)r(
U)
r(U
rd
)r(F
U
i
f
r rc
i,f
f i
rr
ro
rr
v v
− −−−= ===
− −−−= ===
∆ ∆∆∆∫ ∫∫∫
Path
may b
e c
hos
en
to m
ake
the int
egr
ation
sim
ple
A c
onse
rvative
for
ce is
the g
radient
of
a p
otent
ial ene
rgy f
unct
ion
rd
F
dU
c
ro
r− −−−
= ===
dx
dU
F cx
− −−−= ===
similar
for
y &
z
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Example: Elast
ic P
otent
ial Ene
rgy d
ue t
o a S
pring
Block
is
att
ach
ed t
o a s
pring
)x(
dU
xd
)x(
F d
Ws
ss
rr
or
r− −−−
≡ ≡≡≡= ===
FORCE SPRING
EXERTS ON BLOCK
WORK DONE ON
BLOCK BY SPRING
CHANGE IN SPRING
POTENTIAL ENERGY
v
v
Fs
Fs
x
k )
x(F
rr
r− −−−
= ===
Upp
er:
•Block
sta
rts
at
x =
0 w
ith v
righ
tward
•Spr
ing
compr
ess
es,
Fsis left
ward
•Nega
tive
wor
k d
one o
n block
•Po
tent
ial ene
rgy o
f sy
stem inc
rease
s, K
E d
ecr
ease
s•Block
sto
ps a
t so
me p
oint
•Po
tent
ial Ene
rgy is
stor
ed r
eve
rsib
lyin d
efo
rmed s
hape
of
the s
pring
•Usu
ally c
hoo
se x
= 0
as
zero
of
U(x
)
Low
er:
•Spr
ing
forc
e s
till
left
ward
•Block
acq
uire
s vleft
ward
•Po
sitive
wor
k don
e o
n block
•Po
tent
ial ene
rgy o
f sy
stem d
ecr
ease
s•PE
gro
ws
aga
in a
s block
pass
es
x =
0
21
kx
)x(
U2
el
= ===r
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Potential Energy Function for the Block-Spring System
d
x)
x(F
)x(
U)
x(U
i)
(f,
U
f i
x xs
is
fs
s∫ ∫∫∫
− −−−= ===
− −−−≡ ≡≡≡
∆ ∆∆∆•1 d
imens
iona
l mot
ion
along
x
kx
(x
)F
s− −−−
= ===•Hoo
ke’s L
aw r
est
oring
forc
e
f i
f i
x x21
x xs
kx
d
x
x k
i)
(f,
U
2+ +++
= ===+ +++
= ===∆ ∆∆∆
∫ ∫∫∫•Sub
stitut
e:
2kx
)x(
U 21
s= ===
potential function
indefinite
integral
kx
kx
i)
(f,
U
i
21f
21s
22
− −−−= ===
∆ ∆∆∆
∆ ∆∆∆U
sis p
ositive if
|xf| >
|x
i| w
heth
er
spring
is
stre
tched o
r co
mpr
ess
ed
The p
otent
ial ene
rgy f
unct
ion
itse
lf is:
U
kx
)
(xU
021
fs
+ +++= ===
2
para
bola
positive
Usu
ally c
hoo
se:
refe
renc
e leve
l
0
U
0)
(xU
0s
= ===≡ ≡≡≡
= ===po
tent
ial well
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Application to Gravitation
Ove
r sm
all
region
s ne
ar
the E
art
h’s s
urfa
ce is
cons
tant
gr
)r(
U)
r(U
rd
)r(F
i)
(f,
U
ig
fg
r rg
g
f i
rr
ro
rr
v v
− −−−= ===
− −−−= ===
∆ ∆∆∆∫ ∫∫∫
jg
gr
r− −−−
≡ ≡≡≡
jm
gF g
rr
− −−−≡ ≡≡≡
Gra
vita
tion
is
cons
erv
ative
so
∆ ∆∆∆U d
epe
nds
only o
n end
points
, no
t pa
th
Replace
path
by s
tairca
se-like
segm
ent
s:•hor
izon
tal st
eps
that
are
wor
kless
–∆ ∆∆∆
U =
0•ve
rtical rise
rs f
or w
hich ∆ ∆∆∆
Ui=
mg
∆ ∆∆∆y
i
•ca
n ch
oose
refe
renc
e leve
l –
only
diffe
renc
es
of U
are
meaning
ful
In
example,
roller
coast
er
car
forc
ed t
o st
ay o
n th
e t
rack
desc
ribed b
y
)x(f
y
= ===
PE f
unct
ion
can
be w
ritt
en
in t
erm
s of
x:
f(x)
m
g
m
gy
U
(x)
= ==== ===
x
Ug(x
)
U’
= 0
U =
0
mg
yi
mg
yf
E
) y-
g(y
m
rd
j
mg
i)(f
,U
i
f
r rg
f i
= ===+ +++
= ===∆ ∆∆∆
∫ ∫∫∫v v
ro
r
But
betw
een
end
points
rf, r
ith
epo
tent
ial diffe
renc
e is
as
abov
edue
to
path
ind
epe
ndenc
e
WO
RK
DO
NE
BY
GR
AV
ITY
ON
PA
RT
ICL
E
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
A
B
CD
E
Where is the Potential Energy Largest?
7-1
: In
th
e f
igu
re, th
e g
ravit
ati
on
al p
ote
nti
al en
erg
y o
f a p
art
icle
mo
vin
g
alo
ng
th
e t
rack v
ari
es. A
rran
ge p
oin
ts A
, B
, C
, D
, E
in
ord
er
of
decre
asin
g p
ote
nti
al en
erg
y.
A.
C,
D, E
, A
, B
B.
E,
D,
C,
B,
A
C.
A,
B,
C,
D, E
D.
E,
A,
C,
D,
B
E.
B,
D, C
, A
, E
7-2
: In
th
e f
igu
re t
he b
lock is s
ho
wn
at
its e
qu
ilib
riu
m p
oin
t. T
he e
lasti
c
po
ten
tial en
erg
y o
f th
e b
lock v
ari
es a
s it
str
etc
hes o
r co
mp
resses t
he
sp
rin
g. A
rran
ge p
oin
ts A
, B
, C
, D
, E
in
ord
er
of
incre
asin
g p
ote
nti
al
en
erg
y.
AC
DB
A.
C,
D,
A,
B
B.
D,
C, B
, A
C.
B,
C,
A,
D
D.
A,
C,
D,
B
E.
B,
D, C
, A
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Total Energy E
totof a system -Definition
•Kinetic
Ene
rgy:
Ass
ociate
d w
ith m
ovement
of
members
of
a s
yst
em
•Po
tent
ial Ene
rgy:
Dete
rmined b
y t
he c
onfigu
ration
of
the s
yst
em
•Int
ern
al Ene
rgy:
Relate
d t
o th
e t
empe
ratu
re o
f th
e s
yst
em
int
tot
tot
tot
E U
KE
+ ++++ +++
= ===
Etotcan change only if energy transfers into or out of the system
Non
-isolate
d s
yst
ems:
•Ene
rgy c
an
cros
s th
e s
yst
em b
ound
ary
•Tot
al ene
rgy o
f th
e s
yst
em c
hang
es
Iso
late
d s
yst
ems:
•Ene
rgy d
oes
not
cros
s th
e b
ound
ary
of
the s
yst
em
•Tot
al ene
rgy o
f th
e s
yst
em is
cons
tant
•The t
otal ene
rgy o
f th
e U
nive
rse is
cons
tant
–ene
rgy c
an
not
be c
reate
dor
dest
royed.
Ene
rgy t
rans
fers
mus
t balanc
e t
he a
ccou
nts
of s
yst
ems)
(E
sfe
rsE
nerg
yT
ran
tot∑ ∑∑∑
= ===∆ ∆∆∆
0
Eto
t= ===
∆ ∆∆∆
INCLUDES THERMAL &
CHEMICAL ENERGY
Types of energy transfers into or out of a system
Mech
anica
l W
ork:
Inc
lude o
nly W
ork
due
to
Non
-co
nserv
ative
for
ces
er
et
mt
mw
sfe
rsE
nerg
yT
ran
T
T
T
T
Q
W
)(
+ ++++ +++
+ ++++ +++
+ +++= ===
∑ ∑∑∑Heat
Mech
Wave
sM
ass
Tra
nsfe
rElect
rica
l Tra
nsfe
rEM
Radiation
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Non
-co
nserv
ative
for
ces
may b
e d
oing
mech
anica
l wor
k on
syst
em
•Syst
em m
ay n
ot b
e iso
late
d•
Wn
cinclud
es
wor
k don
e b
y c
onta
ct f
orce
s and
frict
ion
•Tre
at
these
as
exte
rnal to
the s
yst
em,
neglect
oth
er
trans
fers
•Oft
en
neglect
chang
es
in int
ern
al ene
rgy
Energy Conservation (for mechanics)
nc
tot
WE
≡ ≡≡≡∆ ∆∆∆
Change in total
energy
Work done by non-
conservative forces
For
a p
art
icle:
UK
Em
ec
h+ +++
≡ ≡≡≡
For
a s
yst
em:
∑ ∑∑∑∑ ∑∑∑
+ +++≡ ≡≡≡
forc
es
pa
rtic
les
me
ch
UK
E
Pote
ntial ene
rgy is
exch
ang
ed r
eve
rsib
ly w
ith k
inetic
ene
rgy.
if n
o ot
her
forc
es
are
act
ing
tot
tot
UK
∆ ∆∆∆− −−−
= ===∆ ∆∆∆
The p
otent
ial ene
rgy inc
ludes
only
cons
erv
ative
for
ces
(PE’s e
xist)
act
ing
on s
yst
em
Typica
lly a
llco
nserv
ative
for
ces
are
inc
luded in
the m
ech
anica
l ene
rgy
e.g
., U
gU
sU
ele
cU
ma
g…
Mechanical Energy -Definition
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Principle of Mechanical Energy Conservation
E
m
ech
0= ===
∆ ∆∆∆
Tot
al mech
anica
l ene
rgy is
cons
tant
, but
ind
ividua
l pa
rticles
can
exch
ang
e k
inetic
and
pot
ent
ial ene
rgy
AD
VA
NT
AG
E O
F
EN
ER
GY
A
PP
RO
AC
H:
•N
eed
to
lo
ok o
nly
at
fin
al
an
d i
nit
ial
sta
tes
•D
eta
ils o
f fo
rce
s m
ay o
fte
n b
e i
gn
ore
d
U
KU
K f
fi
i+ +++
= ===+ +++
For
a s
ingle p
art
icle:
tot
tot
UK
∆ ∆∆∆− −−−
= ===∆ ∆∆∆
U
KU
K f
tot,
f,to
ti,
tot
ito
t,+ +++
= ===+ +++
For
a s
yst
em o
f pa
rticles:
or
or
E co
nsta
nt
mech
= ===
Holds
for
isolate
d s
yst
ems
i.e.,
0
Wn
c
= ===and
when
Ein
tis c
onst
ant
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Example: Con
serv
ation
of
Emech
in a
Pend
ulum
Osc
illato
r
θ θθθT
mg∆ ∆∆∆
sco
ns
tan
t
U
K
E
gm
ec
h= ===
+ +++= ===
Isolated system
Conservative forces only
due to gravitation
No dissipation
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Example: B
all in
free F
all
A b
all o
f mass
mis d
ropp
ed f
rom r
est
at
a h
eight
h abov
e t
he g
roun
d.
What
is t
he b
all’s s
peed
when
it is
at
height
yabov
e t
he g
roun
d.
Neglect
air r
esist
anc
e.
The s
yst
em is
the b
all +
the E
art
hThe s
yst
em is
isolate
dThe o
nly f
orce
act
ing
(gra
vity
) is c
onse
rvative
App
ly M
ech
anica
l Ene
rgy C
onse
rvation
E
m
ech
0= ===
∆ ∆∆∆)
fin
al
(E
)in
it(
E m
ec
hm
ec
h= ===
gf
fg
ii
UK
UK
+ +++= ===
+ +++
mg
y
mv
mg
h
mv
f
21i
21+ +++
= ===+ +++
22
]y
[h
g2
v
vi
f− −−−
+ +++= ===
22
Not
e:
0
v
i= ===
]y
[h
g2
v f
− −−−= ===
Sup
pose
ball is
thro
wn
upward
ins
tead w
ith s
peed
vi
App
roach
is
the s
ame,
resu
lt is
]y
[h
g2
v
v
2 if
− −−−+ +++
= ===
Ene
rgy c
onta
ins
velocity
squ
are
d,
resu
lts
do
not
depe
nd o
n direct
ion
of v
i
SAME AS KINEMATICS
FORMULAS (IN THISCASE)
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Example: Spe
ed a
t Bot
tom o
f a W
ate
r Slide
The c
hild (mass
m) st
art
s at
rest
at
the t
op -
a
vert
ical dista
nce h
abov
e t
he g
roun
d. The s
lide is
friction
less
. Find s
peed a
t bot
tom.
Slope
ang
le θ θθθ
is n
ot k
nown,
so
cann
ot s
olve
using
Newto
n’s
seco
nd L
aw a
lone
mg
N
θ θθθ
su
rface
TO
P
BO
TT
OM
App
ly C
onse
rvation
of
Mech
anica
l Ene
rgy
•Syst
em (Eart
h +
child +
slide) is iso
late
d•Nor
mal fo
rce N
(no
n-co
nserv
ative
) doe
s zero
wor
k•Net
forc
e (gr
avity
) is c
onse
rvative
E
m
ec
h0
= ===∆ ∆∆∆
)b
ot
(E
)to
p(
E m
ec
hm
ec
h= ===
)b
ot
(U
(b
ot)
K )
top
(U
)to
p(
K g
g+ +++
= ===+ +++
bo
tb
ot
top
top
mg
y
m
v
g
ym
mv
+ +++
= ===+ +++
2
212
21
y
SOLVE
]y
[y
g
v
vb
ot
top
top
bo
t− −−−
+ +++= ===
22
2
Same as kinematics formula
USE:
h
y
0 y
0v
top
bo
tto
p= ===
= ==== ===
2
gh
v b
ot
= ===
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Example: Bun
gee J
umpe
rA B
unge
e jum
per
(mass
m =
61 k
g) st
eps
off
the c
liff
a d
ista
nce s
abov
e t
he g
roun
d.
He f
ree-fa
lls
a
dista
nce
Lbefo
re t
he B
unge
e (exact
ly like a
spr
ing
with c
onst
ant
k) begins
to
stre
tch.
How
close
to
the
grou
nd d
own
he t
urn
aro
und?
Doe
s he h
it?
•Two
cons
erv
ative
for
ces
act
, and
no
others
•Gra
vity
Ug(y
) =
mg
y•Spr
ing
Us(x
) =
½kx
2
•The s
yst
em (ju
mpe
r + E
art
h +
Bun
gee) is iso
late
d
k =
160 N
/m
s
m =
61 k
g
s =
45 m
L =
25
m
Low point
Ef
Start point
Ei
•Q
uadra
tic
equ
ation
for
d:
02
22
Lkm
g
d
kmg
d= ===
− −−−− −−−
•Choo
se p
ositive r
oot:
m
.d
91
17
= ===
•Find low
point
h:
m
2.1
dL
s
h= ===
− −−−− −−−
= ===Close
! h
ope jum
per
is n
ot t
oo t
all
Note: Did not need detailed analysis of forces, acceleration, velocity
E
EE
i
fm
ec
h0
= ===− −−−
= ===∆ ∆∆∆
0
UU
Kg
s= ===
∆ ∆∆∆+ +++
∆ ∆∆∆+ +++
∆ ∆∆∆
•Spe
ed =
0 a
t ju
mpe
r’s
start
and
low
point
0
K
= ===∆ ∆∆∆
∴ ∴∴∴
d]
[L m
g
Ug
+ +++− −−−
= ===∆ ∆∆∆
d =
maxim
um B
unge
e s
tretc
h2
kd
U
21
s= ===
∆ ∆∆∆
Mech
anica
l Ene
rgy is
cons
erv
ed, meaning
….
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Friction in Energy Conservation
Kinetic
friction
for
ces
dissipa
te e
nerg
y b
y o
ppos
ing
the m
otion
•Reduc
e t
he m
ech
anica
l ene
rgy,
conv
ert
ing
it t
o heat
•W
ork
don
e b
y f
rict
ion
may b
e int
ern
al (inc
rease
s E
int) o
r exte
rnal
f
rd
f
W
f
le
ss
- w
ork
iss
kf
d
on
e
Wo
rk"
"fo
rce
.
fric
tio
n
kin
eti
ck
ro
rv
∫ ∫∫∫≡ ≡≡≡
≡ ≡≡≡
EXAM
PLE
vi
•Block
sliding
righ
tward
with f
rict
ion
and
exte
rnal fo
rce F
•Nor
mal fo
rce a
nd w
eight
do
no w
ork
as
they a
re
perp
end
icular
to t
he d
isplace
ment
along
x-axis
•No
chang
es
in g
ravita
tion
al po
tent
ial ene
rgy
•For
ces
are
con
stant
as
block
mov
es
by ∆ ∆∆∆
x
A) Con
sider
syst
em t
o be b
lock
alone
. Frict
ion
and
con
tact
for
ce F
are
bot
h e
xte
rnal
xF
xf
WE
kn
cm
ec
h∆ ∆∆∆
+ +++∆ ∆∆∆
− −−−= ===
= ===∆ ∆∆∆
xf
W
kf
∆ ∆∆∆+ +++
≡ ≡≡≡x
F
Wo
the
r∆ ∆∆∆
+ +++≡ ≡≡≡
B) Con
sider
syst
em t
o be b
lock
+ s
urfa
ce.
Frict
ion
is int
ern
al and
raises
inte
rnal ene
rgy.
For
ce F
is
exte
rnal
xf
E
xF
EE
kin
tin
tm
ec
h∆ ∆∆∆
= ===∆ ∆∆∆
∆ ∆∆∆+ +++
= ===∆ ∆∆∆
+ +++∆ ∆∆∆
xF
W
oth
er
∆ ∆∆∆+ +++
≡ ≡≡≡
Gene
ralize
for
A) or
B):
W
W
WE
o
ther
fn
cm
ech
∑ ∑∑∑+ +++
− −−−= ===
= ===∆ ∆∆∆
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Example: B
lock
Pulled o
n a R
ough
Sur
face
vi
Find t
he s
peed o
f th
e b
lock
aft
er
it h
as
mov
ed a
hor
izon
tal dista
nce o
f3.0
m,
start
ing
from
rest
. Ass
ume:
m =
6.0
kg
F =
12
Nµ µµµ
k=
0.1
5 W
W
WE
o
ther
fn
cm
ech
∑ ∑∑∑+ +++
= ==== ===
∆ ∆∆∆
App
ly m
ech
anica
l ene
rgy c
onse
rvation
for
ano
n-isolate
d s
yst
em w
ith f
rict
ion
22
i21
f21
if
gm
ech
mv
mv
K K
U
K
E
− −−−= ===
− −−−= ===
∆ ∆∆∆+ +++
∆ ∆∆∆= ===
∆ ∆∆∆
x
fW
kf
∆ ∆∆∆− −−−
= ===x
F
W
oth
er
∆ ∆∆∆+ +++
= ===m
g
N
mg
- N
F
y= ===
⇒ ⇒⇒⇒= ===
= ===∑ ∑∑∑
0
Find f
k
89
06
15
0.
x.
x.
gm
N
f
kk
k= ===
µ µµµ= ===
µ µµµ= ===
Gra
vita
tion
al po
tent
ial
ene
rgy is
cons
tant
0 U
g
= ===∆ ∆∆∆
N 8.2
f k
= ===
x
F
xf
mv
mv
k
i21
f21
∆ ∆∆∆+ +++
∆ ∆∆∆− −−−
= ===2
2
]f
F [
xm2
v
v k
if
− −−−∆ ∆∆∆
+ +++= ===
2
Sub
stitut
e n
umerica
l va
lues:
] .
1
2
.0[
6.02
v
f8
28
30
− −−−+ +++
= ===
m
/s
1.8
.78
v
f≈ ≈≈≈
= ===1
Fn
et,
ext
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Example: R
oller
Coa
ster
with F
rict
ion y
0
yf
v0
The r
oller
coast
er
car
start
s with s
peed v
0when
its
height
is y
0abov
e t
he g
roun
d.
It
reach
es
a
maxim
um h
eight
yfbefo
re s
liding
back
. There
is
friction
. F
ind t
he inc
rease
in
therm
al ene
rgy o
f th
e s
yst
em.
Values:
m = 100 kg
V0= 20 m/s
y0= 25 m
yf= 30 m
•Syst
em =
Eart
h +
tra
ck +
car
•Frict
ion
is int
ern
al -
incr
ease
s E
int
•Can
not
use W
f=
fs∆ ∆∆∆
x(o
r an
inte
gral) t
o eva
luate
wor
k d
one b
y f
rict
ion
as
friction
forc
e v
aries
in a
n un
know
n way.
Syst
em is
isolate
d,
so…
0
W
E
E
E
nc
int
me
ch
tot
= ==== ===
∆ ∆∆∆+ +++
∆ ∆∆∆= ===
∆ ∆∆∆
ff
21
21in
tm
gy
mv
mg
y
m
v
E
− −−−− −−−
+ +++= ===
∆ ∆∆∆2
0
2 0
0
v
f= ===
Eva
luate
:30)
5100x9.8
x(2
x
E
21in
t− −−−
+ +++= ===
∆ ∆∆∆2
20
10
0
Jo
ule
s
15,1
00
E
int
= ===∆ ∆∆∆
How
high w
ould
the c
ar
go if
friction
were
abse
nt?
f21
mg
y'
mg
y
m
v
− −−−
+ +++= ===
0
2 00
m 45.4
v
y
y'
2g1
f= ===
+ +++= ===
2 00
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Example: Con
nect
ed B
lock
s in M
otion
with F
rict
ion
Cor
d is
un-st
retc
hable a
nd m
ass
less
. The s
yst
em is
initially a
t re
st w
ith t
he s
pring
neither
stre
tched n
or
compr
ess
ed.
The h
ang
ing
mass
falls
a d
ista
nce h
and
comes
to r
est
. F
ind t
he s
liding
friction
coe
fficient
µ µµµk
.
N
m1g
int
sg
int
me
ch
tot
EU
UK
EE
E
∆ ∆∆∆+ +++
∆ ∆∆∆+ +++
∆ ∆∆∆+ +++
∆ ∆∆∆= ===
∆ ∆∆∆+ +++
∆ ∆∆∆= ===
= ===∆ ∆∆∆
0
0= ===
∆ ∆∆∆K
Sub
stitut
e:
2k
h
U21
s= ===
∆ ∆∆∆g
hm
U
2g
− −−−= ===
∆ ∆∆∆
gh
m
h
f
W
E
kk
fin
t1
µ µµµ= ===
= ===+ +++
= ===∆ ∆∆∆
gh
mkh
gh
m
k2
1
2
210
µ µµµ+ +++
+ +++− −−−
= ===
gm
kh
gm
2k
1
21− −−−
= ===µ µµµ
Variation
: Sup
pose
µ µµµkis k
nown
but
want
to
find
hSolve
qua
dra
tic
abov
e.
Roo
ts a
re:
]m
m [
k2g
h
or
0
h k
21
µ µµµ− −−−
= ==== ===
What does the
system do next?
Define
the s
yst
em t
o be b
oth b
lock
s + E
art
h +
spr
ing
+ t
able +
cor
dSyst
em is
isolate
d:
tota
l ene
rgy is
cons
tant
, but
frict
ion
conv
ert
spo
tent
ial and
kinetic
ene
rgy t
o heat
(int
ern
al ene
rgy).
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Power
Power
is t
he r
ate
of
ene
rgy t
rans
fer
(using
mech
anica
l wor
k fo
r no
w).
Ave
rage
Pow
er:
tW
P avg
∆ ∆∆∆∆ ∆∆∆≡ ≡≡≡
Insta
nta
neo
us
P
ow
er
tW
Lim
tdW
d
P
t∆ ∆∆∆∆ ∆∆∆
= ===≡ ≡≡≡
→ →→→∆ ∆∆∆
0
Po
wer
is s
cala
r[P
] =
Jo
ule
s/s
= W
att
s t
P
W
a
vg
∆ ∆∆∆= ===
∆ ∆∆∆
Wo
rk d
on
e i
n t
ime ∆ ∆∆∆
t
td
P
W
d = ===
Infi
nit
esim
al w
ork
is d
on
e in
tim
e d
t
Wo
rk d
on
e b
y f
orc
e F
an
d d
isp
lacem
en
t d
r r
d
F
W
d r
or
= ===
d
trd
F
dtW
d
P
r
or
= ==== ===
∴ ∴∴∴
v
F
P
ro
r= ===
Units
and
Con
vers
ions
ft
.lb
/sec
0.7
38
Jo
ule
/s 1
W
att
= ===
≡ ≡≡≡1
ft
.lb
/sec
550
W
att
s746
hp
1
ho
rsep
ow
er
= ==== ===
≡ ≡≡≡1
J.
10
x
.6
sec
3600
x
w
att
s1000
ho
ur
-kilo
watt
63
1= ===
≡ ≡≡≡
td
P
W
∫ ∫∫∫
= ===∆ ∆∆∆
P n
ot
necessari
lyco
nsta
nt
Co
pyri
gh
t R
. J
an
ow
–S
pri
ng
2012
Example: Po
wer
don
e b
y f
orce
s act
ing
on a
box
Two
cons
tant
for
ces
act
as
show
n on
a b
ox
sliding
acr
oss
a f
rict
ionless
sur
face
F1
= 2
.0 N
, F
2=
4.0
N,
v =
3.0
m/s
ec
Find t
he p
ower
trans
ferr
ed b
y e
ach
for
ce
and
the n
et
power.
Is
Pn
etch
ang
ing?
v
F
P
ro
r= ===
App
ly ins
tant
ane
ous
power
form
ula
Watt
s6.0
)co
s(1
80
3.0
2.0
v
F
P o
11
− −−−= ===
× ×××× ×××
= ==== ===
ro
r
Watt
s6.0
)co
s(6
03.0
4.0
v
F
P o
22
+ +++= ===
× ×××× ×××
= ==== ===
ro
r
Watt
s0.0
6.0
6.0
P
P
P 2
1n
et
= ===+ +++
− −−−= ===
+ +++= ===
F1is d
rawing
ene
rgy
F2is s
upplying
ene
rgy
Net
power
trans
fer
= 0
, so
KE is
cons
tant
as
is v
Since
v is
cons
tant
, th
e n
et
power
is a
lso
cons
tant
Now
let
the m
agn
itud
e o
f F
2=
6.0
N.
W
hat
chang
es?
Watt
s9.0
)co
s(6
03.0
6.0
v
F
P o
22
+ +++= ===
× ×××× ×××
= ==== ===
ro
ru
nc
ha
ng
ed
1 -
W
att
s6.0
P − −−−
= ===
Watt
s3.0
9.0
6.0
P
ne
t= ===
+ +++− −−−
= ===Net
power
trans
fer
is p
ositive,
soKE is
incr
easing
as
is v
Since
v is
chang
ing,
the n
et
power
is a
lso
chang
ing