PHYS 1441 – Section 002
Lecture #13
Monay, Mar. 23, 2009
Dr. Jaehoon Yu
•
•
•
•
Work done by a constant force
Work-Kinetic Energy Theorem
Work with friction
Potential Energy
Today’s homework is homework #7, due 9pm, Tuesday, Mar. 31!!
Monday, Mar. 23, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
1
Announcements
• Reading Assignment
– CH6.2
• Mid-term exam
– Comprehensive exam
• Covers CH1.1 – CH6.3 + Appendix A
–
–
–
–
Date: This Wednesday, Mar. 25
Time: 1 – 2:20pm
In class – SH103
Please do not miss the exam!!!
Monday, Mar. 23, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
2
Special Project Reminder
• 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
• the gravitational constant is G  6.67 1011 N  m2 kg 2
• The radius of the Earth is RE  6.37  103 km
• 20 point extra credit
• Due: Monday, Mar. 30
• You must show your OWN, detailed work to
obtain any credit!!
Monday, Mar. 23, 2009
PHYS 1441-002, Spring 2009 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
Free Body
Diagram
M
d
Which force did the work?
How much work did it do?
ur
Force F Why?
 
ur
F
ur
FN
q
x
Fg  M g
ur ur
W   F  d  Fd cosq
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.
Monday, Mar. 23, 2009
PHYS 1441-002, Spring 2009 Dr. Work
Jaehoon Yu
is an energy transfer!!
4
Work Done by a Constant Force
W  Fs
1 N  m  1 joule  J 
Monday, Mar. 23, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
5
Work done by a constant force
W   F cos q  s
cos 0  1
cos 90  0
cos180  1
Monday, Mar. 23, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
6
Ex. Pulling A Suitcase-on-Wheel
Find the work done by a 45.0N force in pulling the suitcase in the
figure at an angle 50.0o for a distance s=75.0m.


ur ur
W  F d 


ur
ur
 F cos q d
  45.0  cos50   75.0  2170J
Does work depend on mass of the object being worked on?
Why don’t I see the mass
term in the work at all then?
Monday, Mar. 23, 2009
Yes
It is reflected in the force. If an object has smaller
mass, it would take less force to move it at the same
acceleration than a heavier object. So it would take
less work. Which makes perfect sense, doesn’t it?7
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
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 q
• 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, Mar. 23, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
=0
8
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, Mar. 23, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
9
Ex. 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?
W   F cosq0  s  Fs
 710  0.65  460  J 
q  s   Fs
W   F cos180
 710  0.65  460  J 
What does the negative work mean?
Monday, Mar. 23, 2009
The gravitational force does the
work on the weight lifter!
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
10
Ex. Accelerating Crate
The 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
Monday, Mar. 23, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
11
Ex. Continued…
Let’s figure 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


FN 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!
Monday, Mar. 23, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
12
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  s   ma  s
d
v 2f  v02
Using the kinematic 2as  v 2  v 2
as 
2
f
0
equation of motion
Kinetic
1 2
1 2 1 2
1
2
2
KE  mv
Work W   ma  s  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
Monday, Mar. 23, 2009

Work done by the net force causes
change in the object’s kinetic energy.
PHYS 1441-002, Spring 2009 Dr.
Work-Kinetic Energy
Jaehoon Yu
13
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
Monday, Mar. 23, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
14
Ex. 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 q  s  12 mvf2  12 mvo2


v f  vo2  2   F cos q  s m
Solve for vf
5.60 10 Ncos 0 2.42 10 m  474 kgv
-2
Monday, Mar. 23, 2009

9
1
2
2
f
 12 474 kg 275 m s
v f  805 m s
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
2
15
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, Mar. 23, 2009
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
16
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, Mar. 23, 2009
Friction,
PHYS 1441-002, Spring 2009 Dr.Engine work
Jaehoon Yu
t=T, KEf
17
Ex. 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?
Monday, Mar. 23, 2009
f k  k FN
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
70
fk
 0.14
k 

FN 515
18
Ex. 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
Monday, Mar. 23, 2009
KE f 
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
19
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 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
k 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, Mar. 23, 2009
Solving the equation
for vf, we obtain
vf 
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
2W
2 10

 1.8m / s
m
6.0
20