Transcript Chapter 6: Energy

Chapter 6
Conservation of Energy
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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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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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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Work - Example
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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Work - Example
The work done by the normal 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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Work - Example
The net work done on the box is:
Wnet  WF  WN  Wg
 F cos  x  0  0
 F cos  x
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Work Done
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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Work - Example
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
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Wg  wy cos 180
 mgy
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Inclined Plane-V Constant
A box of mass m is towed up a frictionless incline at constant speed.
The applied force F is parallel to the incline.
y
Question: What is the net work done on the box?
N
F
F
x
An FBD for
the box:


Apply Newton’s 2nd Law:
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F
F
w
x
 F  w sin   0
y
 N  w cos   0
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Inclined Plane-V Constant
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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Inclined Plane-V Constant
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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Inclined Plane-Acceleration
Example: What is the net work done on the box in the
previous example if the box is not pulled at constant speed?
F
 F  w sin   ma
x
F  ma  w sin 
Proceeding as before:
New Term
Wnet  WF  Wg  WN
 ma  mg sin  x  mgx sin   0
 ma x  Fnet x
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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
Net Work = Change in K
Wnet  K .
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Kinetic Energy
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 widespread global climate change.
If the mass of the asteroid was 1016 kg (diameter in the range of 49 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
 4.5 10 24 J

2
This is equivalent to ~109 Megatons of TNT.
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Gravitational Potential Energy
Part 1- Close to Earth’s Surface
Potential energy is an energy of position.
There are potential energies associated with different
forces. Forces that have a potential energy associated
with them are called conservative forces.
In general Wcons  U
Not all forces are conservative, i.e. Friction.
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Gravitational Potential Energy
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.
You are free to choose to location of this where ever it is
convenient.
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GPE Problem
The table is 1.0 m tall and the mass of the box is 1.0 kg.
Ques: What is the change in gravitational potential energy of
the box if it is placed on the table?
U=0
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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GPE Problem
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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Total Mechanical Energy
Mechanical energy is
E  K U
The total mechanical energy of a system is conserved
whenever nonconservative forces do no work.
That is
Ei = Ef
or K = U.
Then if K increases U decreases and vice versa
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Mechanical Energy Problem
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
1
mgy4  mv42  mgy3  mv32
2
2
v3  v42  2 g  y4  y3   20.5 m/s
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Mechanical Energy Problem
E4  E2
U 4  K4  U 2  K2
Or use
E3=E2
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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Roller Coaster Problem
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
Ei  E f
(a) At what speed does the
car reach the top of the loop?
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U i  Ki  U f  K f
1
mgyi  0  mgy f  mv 2f
2
v f  2 g  yi  y f   19.8 m/s
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Roller Coaster Problem
Example continued:
(b) What is the force exerted on the car by the track at the top
of the loop?
Apply Newton’s Second Law:
FBD for the car:
y
x
N
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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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Roller Coaster Problem
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
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2
vmin
yi  y f 
2g
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Roller Coaster Problem
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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Nonconservative Forces
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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Gravitational Potential Energy
Part 2 - Away from Earth’s Surface
The general expression for gravitational potential energy is:
GM1M 2
U r   
r
where U r     0
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Gravitational Potential Energy
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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Planetary Motion
A planet of mass m has an elliptical orbit around the Sun. The
elliptical nature of the orbit means that the distance between the
planet and Sun varies as the planet follows its orbital path. Take
the planet to move counterclockwise from its initial location.
QUES: How does the speed of a planet vary as it orbits the Sun
once?
The mechanical energy of the
planet-sun system is:
B
r
A
1 2 GmM sun
E  mv 
 constant
2
r
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Planetary Motion
1 2 GmM sun
E  mv 
 constant
2
r
B
r
A
At point “B” the planet is the farthest from the Sun. At point “A” the
planet is at its closest approach to the sun.
Starting from point “B” (where the planet moves the slowest), as
the planet moves in its orbit r begins to decrease. As it decreases
the planet moves faster.
At point “A” the planet reaches its fastest speed. As the planet
moves past point A in its orbit, r begins to increase and the planet
moves slower.
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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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Example: What is the work done by the variable force shown
below?
Fx (N)
F3
The net work is then
W1+W2+W3.
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
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W3  F3 x3  x2 
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Spring Force
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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Spring Force
(a) A force of 5.0 N applied to the end of a spring cause the
spring to stretch 3.5 cm from its relaxed length.
Ques: 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 
x1  
Solving for x2: x2 
3.5 cm  4.9 cm.
F1
 5.0 N 
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Spring Force
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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Spring Force
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?
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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Elastic Potential Energy
The work done in stretching or compressing a
spring transfers energy to the spring.
Below is the equation of the spring potential energy.
1 2
U s  kx
2
The spring is considered the system
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Elastic Potential Energy
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.
Ques: 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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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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Power - Car Example
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. (Neglect friction and air resistance.)
Ques: What is the engine’s average power output?
E U  K
Pav 

t
t
1 2
mv
K 2 f


 9.3 105 watts
t
t
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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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