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Chapter 6: Conservation of Energy
Energy and Work: loose definitions
• Very common terms, beware: everyday life
meanings are not identical what these terms mean
in physics
Work
Kinetic Energy
• Loose definitions in physics:
• A moving body is said to have kinetic energy; the
faster the object moves, the more energy it has.
• Something capable of increasing kinetic energy of
an object is said to have potential energy stored.
• In order to increase object’s kinetic energy, one
needs to apply a force and do some work.
Work done by gravitational force
GPE
Work
Units
Work = Energy: Joule (J) 1 J = kg ⋅ m
s2
Force · Time = ?
Force · Distance = ?
F
Both are actually very much
relevant:
• F·t characterizes the
change in momentum (to
be discussed later)
• F·∆r characterizes the
change in energy, i.e.
2
∆rx
θ
Power: Watt (W)
J kg ⋅ m 2
1W= =
s
s3
1 horsepower = 746 Watt
W = F∆r cosθ
Energy = Power x time: 1 kWatt-hour = 3.6 MJ
W arning!
For a constant force F acting over a displacement ∆r
Do not confuse W ork, W att, W eight
Example continued:
It is only the force in the direction of the displacement that does work.
The work done by the normal force N is:
WN = 0
An FBD for the box at left:
The normal force is perpendicular to the displacement.
F
θ
∆rx
y
∆rx
N
θ
w
The work done by gravity (w) is:
x
Wg = 0
F
The force of gravity is perpendicular to the displacement.
The work done by the force F is:
WF = Fx ∆rx = (F cos θ )∆x
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In general, the work done by a force F is defined as
The net work done on the box is:
W = F∆r cos θ
Wnet = WF + WN + Wg
= (F cos θ )∆x + 0 + 0
where F is the magnitude of the force, ∆r is the magnitude of the object’s
displacement, and θ is the angle between F and ∆r (drawn tail-to-tail).
= (F cos θ )∆x
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What did you see?
Why do we tell you to memorize this?
a) Lot of mechanical problems can’t be solved
exactly but we still want to understand part
of the problem.
b) Because these concepts help understand
complex mechanical systems. You can
understand why by studying the history a
little
First, some questions about billiards
http://www.youtube.com/watch?v=p_aOWYYnIHo
http://comp.uark.edu/~jgeabana/mol_dyn/KinThI.html
There are many physical systems with many
particles where velocities change a lot. Gottfried
Wilhelm Leibnitz (Newton’s competitor) ~ 1680:
Sometimes even when velocities change,
something stays the same:
Total kinetic energy of all the particles didn’t
change in gas simulation:
N
1
2
K = ∑ mi vi
2
i =1
“Conservation laws”
Statement about a physical quantity
which doesn’t change with time.
Energy, momentum…
Kinetic Energy
K=
1 2
mv
2
is an object’s translational
kinetic energy.
This is the energy an object has because of its
state of motion.
It can be shown that, in general
Wnet = ∆K .
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Example
Accelerating free object (const. a!)
The extinction of the dinosaurs and the majority of species on Earth in the
Cretaceous Period (65 Myr ago) is thought to have been caused by an asteroid
striking the Earth near the Yucatan Peninsula. The resulting ejecta caused
widespread global climate change.
If the mass of the asteroid was 1016 kg (diameter in the range of 4-9 miles)
and had a speed of 30.0 km/sec, what was the asteroid’s kinetic energy?
1
d = v0t + at
r
d
r
F
W = Fd cos θ = Fd = ma ⋅ d
r
v0
r
v0
2
2
(
)(
1
1
K = mv 2 = 1016 kg 30 ×103 m/s
2
2
= 4.5 ×10 24 J
v = v0 + at
)
2
Kinetic energy:
K=
mv 2
2
 v − v0 a  v − v0  2 

at 2 
mv 2 mv02
W = ma ⋅ d = ma  v0 t +
+ 
−
 = ma  v0
  = ... =
2 
2
2
2
a
a





This is equivalent to ~109 Megatons of TNT.
= K − K0
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Decelerating free object
W = Fd cos θ = − Fd = − ma ⋅ d
1
d = v0t − at
2
Sliding an object along a table
W = Fnet d cos θ = Fnet , x d = ma x ⋅ d
r
F
r
v0
r
d
2
v = v0 − at

at 2 
mv 2 mv02
W = −ma ⋅ d = −ma  v0t −
−
= K − K0
 = ... =
2 
2
2

W = ... =
Work can be negative,
if you act with a force against displacement
r
d
r
Fg = mg
mv 2 mv02
−
2
2
Work done by gravitational force
Wg = Fg d cosθ = −mg∆y
∆r
FBD for rising
ball:
r
d
∆y
r
Fg = mg
x
w
r
F
r
v0
Work can be zero, if the force is normal to displacement
Example: A ball is tossed straight up. What is the work done by the force of
gravity on the ball as it rises?
y
a t2
d = vx 0t − x
2
vx = vx 0 − a x t
r
FN
Wg = w∆y cos 180°
= − mg∆y
Work done by gravity
• depends on the change in height only
• independent of the path!
∆y
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Gravitational Potential Energy
Part 1
Work done by gravitational force
alone (e.g. projectile in free fall)
r
v
Wg = −mg ( y − y0 )
Wg =
mv 2 mv02
−
(work-KE “theorem”)
2
2
mv 2 mv02
−
2
2
mv 2
mv02
+ mgy =
+ mgy0 = const
2
2
r
v0
y
Objects have potential energy because of their
location (or configuration).
r
Fg = mg
− mg ( y − y0 ) =
There are potential energies associated with
different (but not all!) forces. Such a force is called
a conservative force.
y0
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The change in gravitational potential energy (only near the surface of the
Earth) is
Wg = −mg∆y
Wcons = − ∆U
In general
Sneak preview:
total energy = “potential energy” + KE is conserved
Example: What is the change in gravitational potential energy of the box if it is
placed on the table? The table is 1.0 m tall and the mass of the box is 1.0 kg.
∆U g = mg∆y
where ∆y is the change in the object’s vertical
position with respect to some reference point that
you are free to choose.
First: Choose the reference level at the floor. U = 0 here.
∆U g = mg∆y = mg ( y f − yi )
(
)
= (1.0 kg ) 9.8 m/s 2 (1.0 m − 0 m ) = +9.8 J
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Energy Conservation laws
Example continued:
Mechanical energy
is
Now take the reference level (U = 0) to be on top of the table so that yi =
−1.0 m and yf = 0.0 m.
Whenever nonconservative forces do no work,
the mechanical energy of a system is
conserved. That is Ei = Ef or ∆K = −∆U.
∆U g = mg∆y = mg ( y f − yi )
(
E = K +U
)
= (1 kg ) 9.8 m/s 2 (0.0m − (− 1.0 m )) = +9.8 J
The results do not depend on the location
of U = 0.
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Example (text problem 6.28): A cart starts from position 4 with v = 15.0 m/s to the
left. Find the speed of the cart at positions 1, 2, and 3. Ignore friction.
Example continued:
E4 = E2
Or use
E3=E2
U 4 + K4 = U 2 + K2
1
1
mgy4 + mv42 = mgy2 + mv22
2
2
v2 = v42 + 2 g ( y4 − y2 ) = 18.0 m/s
E 4 = E3
E4 = E1
U 4 + K 4 = U 3 + K3
U 4 + K 4 = U1 + K1
1
1
mgy4 + mv42 = mgy3 + mv32
2
2
1
1
mgy4 + mv42 = mgy1 + mv12
2
2
v3 = v42 + 2 g ( y4 − y3 ) = 20.5 m/s
Or use
E3=E1
E2=E1
v1 = v42 + 2 g ( y4 − y1 ) = 24.8 m/s
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Example (text problem 6.84): A roller coaster car is about to roll down a track.
Ignore friction and air resistance.
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Example continued:
(b) What is the force exerted on the car by the track at the top of the loop?
m = 988 kg
40 m
FBD for the car:
y
20 m
Apply Newton’s Second Law:
y=0
∑F
(a) At what speed does the car reach the top of the loop?
y
Ei = E f
= − N − w = − mar = −m
x
U i + Ki = U f + K f
N
1
mgyi + 0 = mgy f + mv 2f
2
v f = 2 g ( yi − y f ) = 19.8 m/s
N +w=m
w
N =m
v2
r
v2
− mg = 2.9 × 104 N
r
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Example continued:
v2
r
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Example continued:
(c) From what minimum height above the bottom of the track can the car be
released so that it does not lose contact with the track at the top of the loop?
What is vmin?
v = vmin when N = 0. This means that
v2
r
v2
w = mg = m min
r
vmin = gr
N +w=m
Using conservation of mechanical energy:
Ei = E f
U i + Ki = U f + K f
1 2
mgyi + 0 = mgy f + mvmin
2
Solve for the starting height
v2
yi = y f + min
2g
The initial height must be
yi = y f +
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2
vmin
gr
= 2r +
= 2.5r = 25.0 m
2g
2g
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