PHYS 1443 – Section 002
Lecture #12
Monday, Oct. 13, 2008
Dr. Jaehoon Yu
•
•
•
•
Work done by a Constant Force
Work done by a Varying Force
Work and Kinetic Energy Theorem
Potential Energy and the Conservative Force
–
–
•
Gravitational Potential Energy
Elastic Potential Energy
Conservation of Energy
Today’s homework is homework #7, due 9pm, Monday, Oct. 22!!
Monday, Oct. 13, 2008
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
1
Announcements
• Reading assignments
– CH. 8 – 5, 8 – 6 and 8 – 7
• No colloquium this week
• 2nd term exam on Wednesday, Oct. 22
– Covers from Ch. 1 to what we cover up to this Wednesday, Oct.
15
– Time: 1 – 2:20pm in class
– Location: SH103
– Jason will conduct a summary session Monday, Oct. 20
• Please do NOT miss the exam
• Will have a mid-term grade discussion
– as soon as I am done with grading the exam and putting
together your overall grades
Monday, Oct. 13, 2008
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
2
Reminder: Special Project
• Derive the formula for the gravitational
acceleration (gin) at the radius Rin   RE  from
the center, inside of the Earth. (10 points)
• Compute the fractional magnitude of the
gravitational acceleration 1km and 500km
inside the surface of the Earth with respect to
that on the surface. (6 points, 3 points each)
• Due at the beginning of the class Wednesday,
Oct. 15
Monday, Oct. 13, 2008
PHYS 1443-002, Fall 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
q
M
FN
Free Body
Diagram
q
d
Which force did the work? Force F
r
Why?
 
Monday, Oct. 13, 2008
x
FG  M g
ur ur
How much work did it do? W   F  d  Fd cosq
What does this mean?
F
Unit?
N m
 J (for Joule)
Physical work is done only by the component of
the force along the movement of the object.
PHYS 1443-002, Fall 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
u
r u
r
u
r u
r
angle between them
A B cos q
A B 
• Operation is commutative
ur ur
ur ur ur ur
A  B  A B cos q  B A cos q 

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
Monday, Oct. 13, 2008
PHYS 1443-002, Fall 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 q
Monday, Oct. 13, 2008
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
6
Work Done by Varying Force
• If the force depends on the position of the object in motion,
→ one must consider the work in small segments of the displacement
where the force can be considered constant
W  Fx  x
– Then add all the work-segments throughout the entire motion (xi xf)
xf
W   Fx  x
xf
lim  Fx  x 
In the limit where x0
x 0
xi
xi

xf
xi
Fx dx  W
– If more than one force is acting, the net work done by the net force is
W (net ) 
   F  dx
xf
ix
xi
One of the position dependent forces is the force by the spring Fs kx
The work done by the spring force is
Hooke’s Law
1 2
Fs dx   x   kx  dx  kx max
W
 xmax
2
max
0
Monday, Oct. 13, 2008
0
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
7
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
Suppose net force SF was exerted on an object for
displacement d to increase its speed from vi to vf.
M
The work on the object by the net force SF is
r r
W   F  d   ma  d cos0   ma  d
d
1
v f  vi
v f  vi t Acceleration a 
Displacement d 
t
2
  v f  vi  1
1 2
1 2 1 2 Kinetic


W

mv

mv


m
v

v
t

mv


ma
d

KE

 

f
i
f
i
Work

Energy
2
2
2
  t  2
vi
vf




Haven’t we seen this equation somewhere?
1 2 1 2
mv f  mvi  KE f  KEi  KE
W

Work
2
2
Monday, Oct. 13, 2008
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
Work done by the net force causes
change of the object’s kinetic energy.
8
Work-Kinetic Energy Theorem
Example for Work-KE Theorem
A 6.0kg block initially at rest is pulled to East along a horizontal, frictionless surface by a
constant horizontal force of 12N. Find the speed of the block after it has moved 3.0m.
M
F
M
vi=0
vf
Work done by the force F is
ur ur ur ur
W  F  d  F d cosq  12  3.0cos0  36  J 
d
1 2 1 2
From the work-kinetic energy theorem, we know W  mv f  mvi
2
2
1 2
Since initial speed is 0, the above equation becomes W  mv f
2
Solving the equation for vf, we obtain
Monday, Oct. 13, 2008
vf 
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
2W
2  36

 3.5m / s
m
6.0
9
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
Monday, Oct. 13, 2008
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
Friction,
Engine work
t=T, KEf
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 mk=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 cosq  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 cosq 
ur
mk mg d cos q
 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. 13, 2008
Solving the equation
for vf, we obtain
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
vf 
2W
2 10

 1.8m / s
m
6.0
11
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 written as the product of
magnitudes of the net force vector, the magnitude of the
displacement vector and the angle between them.
W


ur ur
F i gd 

ur
Fi

ur
d cos q
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, Oct. 13, 2008
W  K f  Ki  K
PHYS 1443-002, Fall 2008
Dr. Jaehoon Yu
Nm=Joule
12