DET
Work W = Fd = _____
W
W
PE
stored up
due to
position
___________
KE
due to
motion
___________
of entire
object
Q
within
molecular
and atomic
bonds
_______
PE + KE + Q
Total energy of a system: ET =________________
Work W = Fd = DET = D(
=
)
DET
Work W = Fd = _____
W
W
PE
stored up
due to
position
___________
KE
due to
motion
___________
of entire
object
Q
within
molecular
and atomic
bonds
_______
PE + KE + Q
Total energy of a system: ET =________________
Work W = Fd = DET = D( PE + KE + Q
= DPE + DKE + DQ
)
DPE =
gravitational:
DPE = mgDh
elastic:
DPE = ½ kx2
Work W = Fd = DET
acceleration
motion of entire object:
DKE = ½ mv2
heat
DQ _________
internal energy
 D____________
 DKE of atoms/molec.
bonds
or DPE or ________
temp
 increase_________.
phase
or change__________
What kind of energy does the work change if you….
1/ …lift an object at a constant speed?
 constant v  no D_____
KE
gravitational
Q
 ____________________
DPE or D___
2/ …accelerate an object at a constant height?
 constant h  no D____
PE
KE or D____
Q
 D_____
3/ …drag an object across a surface at a constant speed
and height?
 constant v  no D______
KE
PE
 constant h  no D______
friction
 D___
Q only  energy from work against___________
phase
goes into raising __________
temp. or changing _________
Work done on a system  DET
However, if NO work is done on a system, then its
total energy ET cannot change—it stays the same.
In other words, get rid of the "external work,"
and the total energy ET will stay the same forever.
X
PE
stored up
due to
position
KE
due to
motion
of entire
object
X
Q
within
molecular
and atomic
bonds
This is what we call an isolated system.
Conservation
The Law of the _______________________
of Energy:
The total energy of an isolated system of bodies
remains _________________.
constant
after
before
=
ET
=
PE + KE + Q
=
ET'
PE' + KE' + Q'
total
The __________
energy of a system is neither
increased nor decreased in any process. Energy
created or _______________
cannot be __________
destroyed . It can only
transformed (changed) from one type to another.
be_______________
Q' . Then:
0  Q = ____
If there is no friction, DQ = ____
PE + KE
=
mechanical
_____________
energy before
PE' + KE '
mechanical
______________
energy afterward
Notice the similarity between:
…the Conservation of Energy:
PE
+ KE
=
PE' + KE'
…and the Conservation of Linear Momentum:
p1 +
p2
=
p1'
+
p2'
•Momentum conservation is used to determine what
happens to objects after they collide or explode apart.
•Energy conservation is also used to determine what
happens to objects at a later time.
•Both momentum and energy are useful ideas
because they allow us to ignore forces, which can
be very complicated.
Ex: A pendulum swings back and forth. It position at
two points is shown below. Ignore friction. What energy
does it have at each position?
Use:
ET =
ET'
0
At max. height, v = ____
0
so KE will = ______
When it reaches
maximum height:
PE = 20 J (given)
KE =_____
0J
ET = 20 J
Notice:
PEtop
=>
KEbottom
When it reaches
minimum height:
PE' = ? 0
KE' = ? 20 J
ET' = ? 20 J
Ex: A 0.25-kg box is released from rest and slides
down a frictionless incline. Find its speed as it
arrives at the bottom of the incline.
Use:
ET =
ET'
4.0 m
When the box is released at the top:
PE = mgh = (0.25 kg)(9.81m/s2)(4.0 m)
0
KE = 0
because vi = ____
ET = 9.81 J
= 9.81 J
Just before it hits bottom:
PE' = ? 0 (h = 0)
KE' = ? 9.81 J
ET ' = ? 9.81 J
PEtop => KEbottom
Now use the equation for KE to find v:
KE = (1/2) mv2
9.81 J = (0.5) (0.25 kg)v2
78.5 = v2
8.9 m/s = v
Ex: Drop a 2.0 kg rock off of a 4.0-m cliff.
Use g ≈10 m/s2 to simplify.
Do not ignore air resistance.
PE = mgh = (2.0 kg)(10m/s2)(4.0 m) = 80 J
0
KE =_____
ET = 80 J
4.0 m
Just before it hits:
PE = 0
Suppose KE = 75 J (PEtop ≠ KEbottom)
ET = 75 J + ????
How much energy is "missing?" 5 J
And where did it go?
air resistance  Q  heat
How will it affect v at bottom?
v will be less
Ex: The mass m is not really needed to find the
speed v when using the Conservation of Energy
with no friction:
PEtop => KEbottom
top
h
v=?
bottom
ET (top)
PE + KE
PE + 0
mgh
gh
=
=
=
=
=
v =
ET (bottom)
PE' + KE'
0 + KE'
(1/2)mv2
(1/2)v2
(2gh)1/2  indep. of m!!!
Ex: A spring with a spring constant of 220 N/m is
compressed a distance 0.035 m as shown below. A
mass of 0.027 kg is place against it on a frictionless
slide. When the mass is released: 1/ how fast will it
go as it leaves the spring, and 2/ how high up the slide
will it go before it stops and comes back down?
m
energy stored
in spring
PEs
(1/2)kx2
energy of mass
as it starts moving
energy of mass
at highest point

KE

=
(1/2)mv2
=
PEg
mgh
1/ How fast will the mass be going as it leaves the
spring?
KE
PEs
=
(1/2)mv2
(1/2)kx2 =
2
(1/2)(0.027
kg)v
2
=
(1/2)(220 Nm)(0.035 m)
0.135 J =
v =
(0.0135)v2
3.2 m/s
2/ How high up the slide will it go?
PEs
=
(1/2)kx2
=
0.135 J
=
PEg
mgh
(0.027 kg)(9.8 m/s2)h
0.135 J =
(0.265) h
h = 0.51 m
Ex. Mr. Butchko is fired out of a cannon at 3 different
angles with the same speed from a cliff.
1/ For which angle will he hit the ground with the
most speed?
2/ For which angle will he hit the ground in the least
time?
same PE
All 3 cases begin with the:
________
same
_________KE
1
same
_________E
T
2
3
When he reaches the bottom in all 3 cases,
he will have the:
same PE
________
same v
same
_________KE
same
_________E
T
3 he reaches the ground in the least time.
In case _____
Open your
Review Book packet
to pages: 88-89
Do problems
#67-80