```Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Chapter 6: Conservation of Energy
•Work by a Constant Force
•Kinetic Energy
•Potential Energy
•Work by a Variable Force
•Springs and Hooke’s Law
•Conservation of Energy
•Power
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§6.1 The Law of Conservation of
Energy
The total energy of the Universe is unchanged by any
physical process.
The three kinds of energy are: kinetic energy, potential
energy, and rest energy. Energy may be converted from
one form to another or transferred between bodies.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§6.2 Work by a Constant Force
Work is an energy transfer by the application of a force. For
work to be done there must be a nonzero displacement.
The unit of work and energy is the joule (J). 1 J = 1 Nm =
1 kg m2/s2.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
It is only the force in the direction of the displacement that
does work.
An FBD for the box at left:
F

rx
y
rx
N

w
x
F
The work done by the force F is:
WF  Fx rx  F cos  x
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The work done by the force N is:
WN  0
The normal force is perpendicular to the displacement.
The work done by gravity (w) is:
Wg  0
The force of gravity is perpendicular to the displacement.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The net work done on the box is:
Wnet  WF  WN  Wg
 F cos  x  0  0
 F cos  x
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
In general, the work done by a force F is defined as
W  Fr cos
where F is the magnitude of the force, r is the magnitude
of the object’s displacement, and  is the angle between F
and r (drawn tail-to-tail).
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: A ball is tossed straight up. What is the work done
by the force of gravity on the ball as it rises?
y
r
FBD for
rising ball:
x
w
Wg  wy cos 180
 mgy
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: A box of mass m is towed up a frictionless incline at
constant speed. The applied force F is parallel to the incline.
What is the net work done on the box?
y
F
N
F
x


w
Apply Newton’s
2nd Law:
F
F
x
 F  w sin   0
y
 N  w cos   0
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
The magnitude of F is:
F  mg sin 
If the box travels along the ramp a distance of x the
work by the force F is
WF  Fx cos 0  mgx sin 
The work by gravity is
Wg  wx cos  90  mgx sin 
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
The work by the normal force is:
WN  Nx cos 90  0
The net work done on the box is:
Wnet  WF  Wg  WN
 mgx sin   mgx sin   0
0
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: What is the net work done on the box in the
previous example if the box is not pulled at constant speed?
F
x
 F  w sin   ma
 F  ma  w sin 
Proceeding as before:
Wnet  WF  Wg  WN
 ma  mgx sin    mgx sin   0
 ma x  Fnet x
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§6.3 Kinetic Energy
1 2
K  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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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: 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
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 2 1 16
K  mv  10 kg 30 103 m/s
2
2
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 4.5 10 J

2
This is equivalent to ~109 Megatons of TNT.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§6.4 Gravitational Potential Energy
Part 1
Objects have potential energy because of their location (or
configuration).
There are potential energies associated with different (but
not all!) forces. Such a force is called a conservative
force.
In general Wcons  U
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
The change in gravitational potential energy (only near the
surface of the Earth) is
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.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
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.
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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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
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.
U g  mgy  mg y f  yi 

 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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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Mechanical energy is
E  K U
Whenever nonconservative forces do no work, the
mechanical energy of a system is conserved. That is Ei=Ef
or K= -U.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 6.27): 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.
E4  E3
U 4  K 4  U 3  K3
1 2
1 2
mgy4  mv4  mgy3  mv3
2
2
v3  v42  2 g  y4  y3   20.5 m/s
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
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
E4  E1
U 4  K 4  U1  K1
Or use
E3=E1
1 2
1 2
mgy4  mv4  mgy1  mv1
2
2
E2=E1
v1  v42  2 g  y4  y1   24.8 m/s
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 6.82): A roller coaster car is about to
roll down a track. Ignore friction and air resistance.
m=988 kg
40 m
20 m
y=0
(a) At what speed does the car reach the top of the loop?
Ei  E f
U i  Ki  U f  K f
1
2
mgyi  0  mgy f  mv f
2
v f  2 g  yi  y f   19.8 m/s
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
(b) What is the force exerted on the car by the track at the top
of the loop?
FBD for the car:
y
Apply Newton’s Second Law:
x
N
w
v2
 Fy   N  w  mar  m r
v2
N wm
r
v2
N  m  mg  2.9  10 4 N
r
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
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?
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
2
vmin
yi  y f 
2g
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
What is vmin?
v=vmin when N=0. This means that
v2
N wm
r
2
vmin
w  mg  m
r
vmin  gr
The initial height must be
2
vmin
gr
yi  y f 
 2r 
 2.5r  25.0 m
2g
2g
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
What do you do when there are nonconservative forces?
For example, if friction is present
E  E f  Ei  Wfric
The work done
by friction.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§6.5 Gravitational Potential Energy
Part 2
The general expression for gravitational potential energy is:
GM1M 2
U r   
r
where U r     0
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: What is the gravitational potential energy of a
body of mass m on the surface of the Earth?
GM e m
GM1M 2
U r  Re   

r
Re
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§6.6 Work by a Variable Force
Work can be calculated by finding the area underneath a plot
of the applied force in the direction of the displacement
versus the displacement.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: What is the work done by the variable force shown
below?
Fx (N)
F3
F2
F1
x1
x2
x3
x (m)
The work done by F1 is
W1  F1 x1  0
The work done by F2 is
W2  F2 x2  x1 
The work done by F3 is
W3  F3 x3  x2 
The net work is then W1+W2+W3.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
By hanging masses on a spring we find that stretch 
applied force. This is Hooke’s law.
For an ideal spring: Fx = -kx
Fx is the magnitude of the force exerted by the free end of
the spring, x is the measured stretch of the spring, and k is
the spring constant (a constant of proportionality; its units
are N/m).
A larger value of k implies a stiffer spring.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 6.48): (a) If forces of 5.0 N applied to
each end of a spring cause the spring to stretch 3.5 cm from
its relaxed length, how far does a force of 7.0 N cause the
same spring to stretch?
For springs Fx. This allows us to write
F1 x1
 .
F2 x2
F2
 7.0 N 
Solving for x2: x2 
x1  
3.5 cm  4.9 cm.
F1
 5.0 N 
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example continued:
(b) What is the spring constant of this spring?
F1 5.0 N
k

 1.43 N/cm.
x1 3.5 cm
Or
F2 7.0 N
k

 1.43 N/cm.
x2 4.9 cm
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 6.46): An ideal spring has k = 20.0
N/m. What is the amount of work done (by an external
agent) to stretch the spring 0.40 m from its relaxed length?
Fx (N)
kx1
x1=0.4 m
x (m)
W  Area under curve
1
1
1
2
 kx1 x1   kx12  20.0 N/m 0.4 m   1.6 J
2
2
2
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§6.7 Elastic potential energy
The work done in stretching/compressing a spring transfers
energy to the spring.
1 2
U s  kx
2
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example: A box of mass 0.25 kg slides along a horizontal,
frictionless surface with a speed of 3.0 m/s. The box
encounters a spring with k = 200 N/m. How far is the spring
compressed when the box is brought to rest?
Ei  E f
U i  Ki  U f  K f
1 2 1 2
0  mv  kx  0
2
2
 m
v  0.11 m
x  

k


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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
§6.8 Power
Power is the rate of energy transfer.
Average Power
E
Pav 
t
Instantaneous Power
P  Fv cos
The unit of power is the watt. 1 watt = 1 J/s = 1 W.
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Example (text problem 6.73): A race car with a mass of
500.0 kg completes a quarter-mile (402 m) race in a time of
4.2 s starting from rest. The car’s final speed is 125 m/s.
What is the engine’s average power output? Neglect friction
and air resistance.
E U  K
Pav 

t
t
1 2
mv
K 2 f


 9.3 105 watts
t
t
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Fisica Generale - Alan Giambattista, Betty McCarty Richardson
Summary
•Conservation of Energy
•Calculation of Work Done by a Constant or Variable
Force
•Kinetic Energy
•Potential Energy (gravitational, elastic)
•Power
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