PHYS 1441 – Section 001
Lecture # 9
Wednesday, June 11, 2008
Dr. Jaehoon Yu
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Wednesday, June 11,
2008
Scalar Product
Work-Kinetic Energy Theorem
Work and Energy Involving Kinetic Friction
Potential Energy
Gravitational Potential Energy
Conservative and Non-conservative Forces
Conservation of Mechanical Energy
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
1
Announcements
• Quiz results
– Class Average: 5.4/8
• Equivalent to 68/100
– Top score: 8/8
• Second term exam
– 8 – 10am, Tuesday, June 17, in SH103
– Covers CH4.1 – What we finish tomorrow, June 12
(CH7?)
– Practice tests have been posted on the class web page
– Dr. Satyanand will conduct a help session 8 – 10am,
Monday, June 16 in class
June 11,
PHYS 1441-001, Summer 2008
• Wednesday,
Reading
assignment:
CH6.9
2008
Dr. Jaehoon Yu
2
Special Project for Extra Credit
• Prove the following equivalence mathematically
ur ur
ur ur
A  B  Ax Bx  Ay By  A B cos 
ur


– where A  Ax i  Ay j
ur


– and B  Bx i  By j
– 20 point extra credit
– Due: Wednesday, June 18
Wednesday, June 11,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
3
Work Done by a Constant Force
A meaningful work in physics is done only when a sum of
forces exerted on an object made a motion to the object.
F
M
y

Free Body
Diagram
M
d
Which force did the work?
How much work did it do?

ur
Force F Why?
 
r
F
r
FN
ur ur
W   F  d  Fd cos
x
r
r
Fg  Mg
Unit?
N m
 J (for Joule)
Physically meaningful work is done only by the component
What does this mean? of the force along the movement of the object.
Wednesday, June 11,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
Work is an energy transfer!!
4
Scalar Product of Two Vectors
• Product of magnitude of the two vectors and the cosine of the
ur ur
u
r u
r
angle between them
A  B  A B cos 
• Operation is commutative
ur ur
ur ur ur ur
A  B  A B cos   B A cos  

ur ur
B A

ur ur ur
ur ur ur ur
• Operation follows the distribution A  B  C  A  B  A  C
law of multiplication
• Scalar products of Unit Vectors
 
 
 
 
 
 
i  i  j j  k  k  1 i  j  j k  k  i 
0
• How does scalar product look in terms of components?
ur



A  Ax i  Ay j  Az k
ur



B  Bx i  By j  Bz k
 
 
 
ur ur  





  
 
A  B   Ax i  Ay j  Az k    Bx i  By j  Bz k    Ax Bx i  i  Ay By j j  Az Bz k  k  cross terms

 
 

ur ur
A  B  Ax Bx  Ay By  Az Bz
Wednesday, June 11,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
=0
5
Example of Work by Scalar Product
A particle moving on the xy plane undergoes a displacement d=(2.0i+3.0j)m as a
constant force F=(5.0i+2.0j) N acts on the particle.
Y
d
F
X
a) Calculate the magnitude of the displacement
and that of the force.
ur
d  d x2  d y2  2.02  3.02  3.6m
ur
F  Fx2  Fy2  5.02  2.02  5.4 N
b) Calculate the work done by the force F.
ur ur
W  F d 


 
  
  2.0  5.0 i  i  3.0  2.0 j j  10  6  16( J )
 2.0 i  3.0 j    5.0 i  2.0 j 

 

Can you do this using the magnitudes and the angle between d and F?
ur ur ur ur
W  F  d  F d cos 
Wednesday, June 11,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
6
Ex. 2 Bench Pressing and The
Concept of Negative Work
A weight lifter is bench-pressing a barbell whose weight is
710N a distance of 0.65m above his chest. Then he lowers it
the same distance. The weight is raised and lowered at a
constant velocity. Determine the work in the two cases.
What is the angle between the force and the displacement?
Lifting
up
W   F cos0 s  Fs
 710  0.65  460  J 
Lowering
down
  s   Fs
W   F cos180
 710  0.65  460  J 
What does the negative work mean?
Wednesday, June 11,
2008
The gravitational force does the
work on the weight lifter!
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
7
Ex. 3 Accelerating Crate
A truck is accelerating at a rate of +1.50
m/s2. The mass of the crate is 120-kg and
it does not slip. The magnitude of the
displacement is 65 m. What is the total
work done on the crate by all of the forces
acting on it?
What are the forces acting in this motion?
Gravitational force on the crate,
weight, W or Fg
Normal force force on the crate, FN
Static frictional force on the crate, fs
Wednesday, June 11,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
8
Ex. 3 Continued…
Let’s figure out what the work done by
each force in this motion is.
Work done by the gravitational force on the crate, W or Fg
W


Fg cos  90o  s  0
Work done by Normal force force on the crate, FN
W


Fg cos  90o  s  0
Work done by the static frictional force on the crate, fs
2
120
kg
1.5
m
s
ma

fs 


  180N
4
W  f s  s  180N  cos 0   65 m   1.2 10 J
Which force did the work? Static frictional force on the crate, fs
How?
By holding on to the crate so that it moves with the truck!
Wednesday, June 11,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
9
Kinetic Energy and Work-Kinetic Energy Theorem
• Some problems are hard to solve using Newton’s second law
– If forces exerting on an object during the motion are complicated
– Relate the work done on the object by the net force to the change of the
speed of the object
M
SF
M
Suppose net force SF was exerted on an object for
displacement d to increase its speed from vi to vf.
The work on the object by the net force SF is
r r
W   F  d   ma cos 0  d  ma  d
d
v 2f  v02
Using the kinematic 2ad  v 2  v 2
ad 
2
f
0
equation of motion
Kinetic
1 2
1 2 1 2
1
2
2
KE  mv
Work W   ma  d  2 m  v f  v0   2 mv f  2 mv0
Energy
2
vi
vf
1
2
1
2

Work W  mv 2f  mvi2  KE f  KEi  KE
Wednesday, June 11,
2008

Work done by the net force causes
change in the object’s kinetic energy.
PHYS 1441-001, Summer 2008Work-Kinetic Energy
Dr. Jaehoon Yu
10
Theorem
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  KE f  KE o  12 mvf2  12 mvo2
Wednesday, June 11,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
11
Ex. 4 Deep Space 1
The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If
the 56.0-mN force acts on the probe parallel through a displacement of
2.42×109m, what is its final speed?
  F  cos   s  12 mvf2  12 mvo2


 5.60 10 N  cos 0  2.42 10 m    474 kg  v
-2
Wednesday, June 11,
2008
o
9
1
2
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
2
f

1
2
 474 kg  275 m s 
v f  805 m s
12
2
Ex. 6 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.
Wednesday, June 11,
2008
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
13
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  Ffr 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 other source of work, is
KE f  KEi W  F fr d
t=0, KEi
Wednesday, June 11,
2008
Friction,
PHYS 1441-001, Summer 2008Engine work
Dr. Jaehoon Yu
t=T, KEf
14
Ex. 5 Downhill Skiing
A 58kg skier is coasting down a 25o
slope. A kinetic frictional force of
magnitude fk=70N opposes her motion.
Neat 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?
Wednesday, June 11,
2008
f k  k FN
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
70
fk
 0.14
k 

FN 515
15
Ex. 5 Now with the X component
X component
Total work by
this force
From work-kinetic
energy theorem
 F  F  f  mg sin 25  f   58  9.8  sin 25  70   170N  ma
W    F   s  mg sin 25  f   s   58  9.8  sin 25  70   57  9700J
x
gx
k
k
x
k
W  KE f  KEi
1 2
1
mv f  W  KEi W  mv02
2
2
2
2
2W  mv02
2  9700  58   3.6 
2W

mv
2
0
Solving for vf v 
vf 
 19 m s

f
m
58
m
Fx 170

What is her acceleration?
 Fx  ma

a
 2.93 m s 2
m
58
Wednesday, June 11,
2008
KE f 
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
16
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
F
vi=0
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 gd  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
Wednesday, June 11,
2008
Solving the equation
for vf, we obtain
vf 
PHYS 1441-001, Summer 2008
Dr. Jaehoon Yu
2W
2 10

 1.8m / s
m
6.0
17