PHYS 1443 – Section 001
Lecture #12
Monday, March 21, 2011
Dr. Jaehoon Yu
•
•
Work and Energy Involving Kinetic Friction
Potential Energy and the Conservative Force
–
–
•
Gravitational Potential Energy
Elastic Potential Energy
Conservation of Energy
Today’s homework is homework #7, due 10pm, Tuesday, Mar. 29!!
Announcements
• Mid-term grade discussion on Wednesday, Mar. 23
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–
–
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Class this Wednesday is replaced by the mid-term grade discussion
Last names start with A - F: 1 - 1:25pm
Last names start with G - M: 1:25 - 1:50pm
Last names start with N - R: 1:50 - 2:15pm
Last names start with S - Z: 2:15 - 2:40pm
• Change of the 2nd non-comprehensive term exam date
– From Wednesday, Mar. 30 to Wednesday, Apr. 6
– Final comprehensive exam: 11am, Monday, May 9
– Mark your calendars
• Quiz next Wednesday, Mar. 30
– Beginning of the class
– Covers up to what we finish next Monday, Mar. 28
• Colloquium Wednesday
Monday, March 21, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
2
Reminder: Special Project
• Two protons are separated by 1m.
– Compute the gravitational force (FG) between the two
protons (3 points)
– Compute the electric force (FE) between the two
protons (3 points)
– Compute the ratio of FG/FE (3 points) and explain
what this tells you (1 point)
• Due: Bring to the mid-term grade discussion,
Wednesday, Mar. 23
Monday, March 21, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
4
Work and Kinetic Energy
A meaningful work in physics is done only when the sum of
the forces exerted on an object made a motion to the object.
What does this mean?
However much tired your arms feel, if you were
just holding an object without moving it you
have not done any physical work to the object.
Mathematically, the work is defined as the
scalar product of the net force vector, and
the displacement vector!
W


ur ur
F i gd 


ur ur
F i d cos
Kinetic Energy is the energy associated with the motion and capacity to perform work.
Work causes change of energy after the completion Work-Kinetic energy theorem
1 2
K  mv
2
Monday, March 21, 2011
W  K f  Ki  K
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
Nm=Joule
5
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
 
r r
W fr  Ffr gd  F fr d cos 180   F fr d KE  F fr 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 all other sources of work, is
KE f  KEi W  F fr d
t=0, KEi
Monday, March 21, 2011
Friction,
PHYS 1443-001, Spring 2011 Engine work
Dr. Jaehoon Yu
t=T, KEf
6
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 mk=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
mk 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, March 21, 2011
Solving the equation
for vf, we obtain
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
vf 
2W
2 10

 1.8m / s
m
6.0
7
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, March 21, 2011
f k  mk FN
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
70
fk
 0.14
mk 

FN 515
8
Ex. Now with the X component
X component  Fx  Fgx  f k  mg sin 25  f k   58  9.8  sin 25  70   170N  ma
Total work by
Fx   s  mg sin25o  fk  s   58  9.8  sin 25  70   57  9700J
W



this force
1 2
1 2
From work-kinetic
mv

W

mv0
KE

W  KEi 
f
f
energy theorem W  KE f  KEi
2
2

Solving for vf

2W  mv02
2
vf 
m
What is her acceleration?
Monday, March 21, 2011
vf 
F
x
 ma
2  9700  58   3.6 
 19 m s
58
2W  mv

m
Fx 170


a
 2.93 m s 2
m
58
2
0
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
2
9
Potential Energy & Conservation of Mechanical 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
Electromagnetic Energy
Biological Energy
Nuclear 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, March
2011
1443-001, Spring
universe
is 21,
conserved.
It justPHYS
transforms
from2011
one form to another.
Dr. Jaehoon Yu
10
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 height h, 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
The work done on the object by
the gravitational force as the
brick drops from hi to hf is:
hf
What does
this mean?
Monday, March 21, 2011
PE  mgh
Wg  PEi  PE f
 mghi  mgh f  PE
(since PE  PE
f
 PEi )
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 PHYS
increase
theSpring
brick’s
1443-001,
2011 kinetic energy.
11
Dr. Jaehoon 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, March 21, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
12
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, March 21, 2011
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
13

Conservative Forces and Potential Energy
The work done on an object by a conservative force is
Wc 
equal to the decrease in the potential energy of the system
What does this
statement tell you?

xf
xi
Fx dx  U
The work done by a conservative force is equal to the negative
change of the potential energy associated with that force.
Only the changes in potential energy of a system is physically meaningful!!
We can rewrite the above equation
in terms of the potential energy U
xf
U  U f  U i   x Fx dx
i
So the potential energy associated
with a conservative force at any
given position becomes
U f  x     Fx dx  U i Potential energy
What can you tell from the
potential energy function above?
Since Ui is a constant, it only shifts the resulting
Uf(x) by a constant amount. One can always
change the initial potential so that Ui can be 0.
Monday, March 21, 2011
xf
xi
PHYS 1443-001, Spring 2011
Dr. Jaehoon Yu
function
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