PHYS 1441 – Section 002
Lecture #14
Monday, Oct. 25, 2010
Dr. Jaehoon Yu
•
•
•
•
•
•
Work – Kinetic Energy Theorem Revisited
Work and Energy Involving Kinetic Friction
Potential Energy
Gravitational Potential Energy
Elastic Potential Energy
Mechanical Energy Conservation
Today’s homework is homework #8, due 10pm, Tuesday, Nov. 2!!
Monday, Oct. 25, 2010
PHYS 1441-002, Fall 2010
Dr. Jaehoon Yu
1
• Quiz #3
Announcements
– Class average: 6.6/10
• Equivalent to 66/100
• Previous quizzes: 53/100 and 56/100
– Top score: 10
• 2nd non-comprehensive term exam
–
–
–
–
Date: Wednesday, Nov. 3
Time: 1 – 2:20pm in class
Covers: CH3.5 – CH6.7
There will be a review in class Monday, Nov. 1
• Bring your own problems to solve
• Mid-term grade discussion this Wednesday
– Dr. Yu’s office, CPB342
• A – F: 12:45 – 1:15
• F – N: 1:15 – 1:50
• N – Y: 1:50 – 2:20
• Colloquium this week
– On the subject of renewable energy
Monday, Oct. 25, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
2
Monday, Oct. 25, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
3
Reminder: Special Project
• Using the fact that g=9.80m/s2 on the Earth’s
surface, find the average density of the Earth.
– Use the following information only
11
2
2
• The gravitational constant G  6.67  10 N  m kg
3
• The radius of the Earth RE  6.37 10 km
• 20 point extra credit
• Due: This Wednesday, Oct. 27
• You must show your OWN, detailed work to
obtain any credit!!
Monday, Oct. 25, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
4
Work-Kinetic Energy Theorem
When a net external force by the jet engine does work on and
object, the kinetic energy of the object changes according to
W  KEf  KEo  1 mvf2  1 mvo2
2
2
Monday, Oct. 25, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
5
Ex. Deep Space 1
The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If
a 56.0-mN force acts on the probe parallel through a displacement of
2.42×109m, what is its final speed?
 Fcos  s 


v f  vo2  2
1
2
 F cos s
mvf2  12 mvo2
m

Solve for vf
275 m s2  2 5.60  10-2Ncos0o 2.42  109 m 474
v f  805m s
Monday, Oct. 25, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
6
Ex. Satellite Motion and Work By the Gravity
A satellite is moving about the earth in a
circular orbit and an elliptical orbit. For
these two orbits, determine whether the
kinetic energy of the satellite changes
during the motion.
For a circular orbit No change! Why not?
Gravitational force is the only external
force but it is perpendicular to the
displacement. So no work.
For an elliptical orbit Changes! Why?
Gravitational force is the only external
force but its angle with respect to the
displacement varies. So it performs work.
Monday, Oct. 25, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
7
Work and Energy Involving Kinetic Friction
• What do you think the work looks like if there is friction?
– Static friction does not matter! Why? It isn’t there when the object is moving.
– Then which friction matters? Kinetic Friction
Ffr
M
M
vi
vf
d
Friction force Ffr works on the object to slow down
The work on the object by the friction Ffr is
W fr  F fr d cos 180  Ffr d KE  Ffr d
The negative sign means that the work is done on the friction!!
The final kinetic energy of an object, taking into account its initial kinetic
energy, friction force and other source of work, is
KE f  KEi W Ffr d
t=0, KEi
Monday, Oct. 25, 2010
Friction,
Engine work
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
t=T, KEf
8
Ex. Downhill Skiing
A 58kg skier is coasting down a 25o
slope. A kinetic frictional force of
magnitude fk=70N opposes her motion.
At the top of the slope, the skier’s
speed is v0=3.6m/s. Ignoring air
resistance, determine the speed vf at
the point that is displaced 57m downhill.
What are the forces in this motion?
Gravitational force: Fg Normal force: FN
Kinetic frictional force: fk
What are the X and Y component of the net force in this motion?
Y component
F
From this we obtain
y
 Fgy  FN  mg cos 25  FN  0
FN  mg cos 25  58  9.8  cos 25  515N
What is the coefficient of kinetic friction?
Monday, Oct. 25, 2010
f k  k FN
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
70
fk
 0.14
k 

FN 515
9
Ex. Now with the X component
X component  Fx  Fgx  fk  mg sin 25  fk   58  9.8  sin 25  70   170N  ma
Total work by
W    Fx   s  mg sin 25o  f k  s   58  9.8  sin 25  70   57  9700J
this force
1 2
1 2
From work-kinetic
mv

W

mv0
W  KEi 
KE f 
f
energy theorem W  KE f  KEi
2
2

Solving for vf
2W  mv02
2
vf 
m
What is her acceleration?
Monday, Oct. 25, 2010

vf 
F
x
 ma
2  9700  58   3.6 
 19 m s
58
2W  mv

m
Fx 170


 2.93 m s 2
a
m
58
2
0
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
2
10
Example of Work Under Friction
A 6.0kg block initially at rest is pulled to East along a horizontal surface with coefficient
of kinetic friction k=0.15 by a constant horizontal force of 12N. Find the speed of the
block after it has moved 3.0m.
Fk M
vi=0
F
Work done by the force F is
r r
WF  F d cos  12  3.0cos 0  36  J 
M
vf
d=3.0m
Work done by friction Fk is
Thus the total work is
r r
r r
Wk  Fk  d  Fk d cos 
ur
k mg d cos
 0.15  6.0  9.8  3.0 cos180  26 J 
W  WF  Wk  36  26  10( J )
Using work-kinetic energy theorem and the fact that initial speed is 0, we obtain
1 2
W  WF  Wk  mv f
2
Monday, Oct. 25, 2010
Solving the equation
for vf, we obtain
vf 
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
2W
2 10

 1.8m / s
m
6.0
11
Potential Energy
Energy associated with a system of objects  Stored energy which has the
potential or the possibility to work or to convert to kinetic energy
What does this mean?
In order to describe potential energy, U,
a system must be defined.
The concept of potential energy can only be used under the
special class of forces called the conservative force which
results in the principle of conservation of mechanical energy.
EM  KEi  PEi KE f  PE f
What are other forms of energies in the universe?
Mechanical Energy
Chemical Energy
Biological Energy
Electromagnetic Energy
Nuclear Energy
Thermal Energy
These different types of energies are stored in the universe in many different forms!!!
If one takes into account ALL forms of energy, the total energy in the entire
Monday, Oct.is
25,conserved.
2010
PHYStransforms
1441-002, Fall 2010
Dr. Jaehoon
universe
It just
from
one form to another.
Yu
12
Gravitational Potential Energy
This potential energy is given to an object by the gravitational field in the
system of Earth by virtue of the object’s height from an arbitrary zero level
When an object is falling, the gravitational force, Mg, performs the work
on the object, increasing the object’s kinetic energy. So the potential energy
of an object at a height y,the potential to do work, is expressed as
m
mg
hi
r r
r r
r r
PE  Fg  y  Fg y cos   Fg y  mgh
m
hf
The work done on the object by
the gravitational force as the
brick drops from yi to yf is:
What does
this mean?
Monday, Oct. 25, 2010
PE  mgh
Wg  PEi  PE f
 mghi  mghf  PE
(since PE  PE  PE )
f
i
Work by the gravitational force as the brick drops from yi to yf
is the negative change of the system’s potential energy
 Potential energy was spent in order for the
gravitational
force
to Jaehoon
increase the brick’s kinetic energy.
PHYS 1441-002, Fall
2010 Dr.
13
Yu
Ex. A Gymnast on a Trampoline
The gymnast leaves the trampoline at an initial height of 1.20 m and reaches
a maximum height of 4.80 m before falling back down. What was the initial
speed of the gymnast?
Monday, Oct. 25, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
14
Ex. Continued
From the work-kinetic energy theorem
W
1
2
mv  mv
2
f
2
o
1
2
Work done by the gravitational force
Wgravity  mg  ho  h f
Since at the maximum height, the
final speed is 0. Using work-KE
theorem, we obtain
mg  ho  h f    mv
2
o
1
2
vo  2 g  ho  h f 
 vo  2  9.80 m s2  1.20 m  4.80 m   8.40 m s
Monday, Oct. 25, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
15

Example for Potential Energy
A bowler drops bowling ball of mass 7kg on his toe. Choosing the floor level as y=0, estimate
the total work done on the ball by the gravitational force as the ball falls on the toe.
Let’s assume the top of the toe is 0.03m from the floor and the hand
was 0.5m above the floor.
Ui  mgyi  7  9.8  0.5  34.3J U f  mgy f  7  9.8  0.03  2.06J
Wg  U   U f  U i  32.24 J  30 J
M
b) Perform the same calculation using the top of the bowler’s head as the origin.
What has to change?
First we must re-compute the positions of the ball in his hand and on his toe.
Assuming the bowler’s height is 1.8m, the ball’s original position is –1.3m, and the toe is at –1.77m.
Ui  mgyi  7  9.8   1.3  89.2J U f  mgy f  7  9.8  1.77  121.4J
Wg  U   U f  U i   32.2 J  30 J
Monday, Oct. 25, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
16