1. W W SYSTEM DET Work W = Fd = _____ Fd Fd KE due to ___________
of entire
object Q
within
molecular
and atomic
_______ PE
stored up
due to
___________
motion
bonds
position
mechanical energy
SYSTEM
PE + KE + Q
Total energy of a system: ET =________________
Work W = Fd = DET = D( ) =

3. W W SYSTEM DET Work W = Fd = _____ Fd Fd KE due to ___________
of entire
object Q
within
molecular
and atomic
_______ PE
stored up
due to
___________
motion
bonds
position
mechanical energy
SYSTEM
PE + KE + Q
Total energy of a system: ET =________________
Work W = Fd = DET = D( ) =
PE + KE + Q
DPE + DKE + DQ

4. motion of entire object: DKE = Work W = Fd = DET ½ mv2
gravitational:
DPE =
mgDh
DPE =
elastic:
DPE =
Dposition
½ kx2
acceleration
motion of entire object:
DKE =
Work W = Fd = DET
½ mv2
friction
heat
DQ _________
 D____________ energy
 DKE of atoms/molec.
or DPE or ________
 increase_________.
or change__________
internal
bonds
temp
phase

5. What kind of energy does the work change if you….
1/ …lift an object at a constant speed?
 constant v  no D_____
 ____________________ DPE or D___ KE
gravitational Q
2/ …accelerate an object at a constant height?
 constant h  no D____
 D_____ or D____ PE KE Q
3/ …drag an object across a surface at a constant speed
and height?
 constant v  no D______
 constant h  no D______
 D___ only  energy from work against___________
goes into raising __________ or changing _________ KE PE
friction Q
phase
temp.

6. X X 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. PE
stored up
due to
position KE
motion
of entire
object Q
within
molecular
and atomic
bonds
mechanical energy Fd X Fd X
This is what we call an isolated system.

7. The Law of the _______________________ of Energy:
The total energy of an isolated system of bodies
remains _________________.
Conservation
constant
before =
ET =
PE + KE + Q =
after
ET'
PE' + KE' + Q'
total
The __________ energy of a system is neither increased nor decreased in any process. Energy cannot be __________ or _______________ . It can only be_______________ (changed) from one type to another.
created
destroyed
transformed
If there is no friction, DQ = ____  Q = ____ . Then: = Q'
PE + KE
PE' + KE '
_____________
energy before
mechanical
mechanical
______________
energy afterward

8. Notice the similarity between:
…the Conservation of Energy:
PE KE = PE' KE'
…and the Conservation of Linear Momentum:
p 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.

9. Restatements of the…
Law of the Conservation of Energy:
"The total energy is neither __________________ nor
__________________________ in any process."
"Energy can be __________________ from one form to
another and _______________ from one body to another,
but the __________ amount remains ________________."
"Energy is neither _____________ nor _______________."
increased
decreased
transformed
transferred
total
the same
destroyed
created

10. 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'
At max. height, v = ____
so KE will = ______
When it reaches
maximum height:
PE = 20 J (given)
KE =_____
ET =
When it reaches
minimum height:
PE' = ?
KE' = ?
ET' = ?
0 J
20 J
20 J
PEtop =>
KEbottom
Notice:

11. 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'
0.25 kg
4.0 m
When the box is released at the top:
PE = mgh =
KE = because vi = ____
ET =
(0.25 kg)(9.81m/s2)(4.0 m)
= 9.81 J
9.81 J

12. PEtop => KEbottom Just before it hits bottom: PE' = ? 0 (h = 0)
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

13. 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 =
KE =_____
ET =
mgh =
(2.0 kg)(10m/s2)(4.0 m) = 80 J
80 J
4.0 m
Just before it hits:
PE =
Suppose KE = 75 J (PEtop ≠ KEbottom)
ET = 75 J + ????
How much energy is "missing?"
And where did it go?
How will it affect v at bottom?
5 J
air resistance  Q  heat
v will be less

14. PEtop => KEbottom 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) ET (bottom)
PE + KE PE' + KE'
PE KE'
mgh (1/2)mv2
gh (1/2)v2
v (2gh)1/2  indep. of m!!! =

15. Ex: A spring with a spring constant of 220 N/m is
compressed a distance m as shown below. A
mass of 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?
slide m
energy stored energy of mass energy of mass
in spring as it starts moving at highest point
 
= =
PEs KE
PEg
(1/2)mv2
(1/2)kx2
mgh

16. 1/ How fast will the mass be going as it leaves the
spring? = KE
PEs
(1/2)kx2
(1/2)mv2
(1/2)(0.027 kg)v2
(1/2)(220 N/m)(0.035 m)2
0.135 J
(0.0135)v2
v m/s
2/ How high up the slide will it go? =
PEg
PEs
(1/2)kx2
mgh
0.135 J
(0.027 kg)(9.8 m/s2)h
0.135 J
(0.265) h
h m

17. Ex. Mr. Siudy 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?
All 3 cases begin with the:
________ PE
_________KE
_________ET
same
same 1
same 2 3
When he reaches the bottom in all 3 cases,
he will have the:
________ PE
_________KE
_________ET
same
same
same v
same 3
In case _____ he reaches the ground in the least time.