Transcript Wednesday, October 20, 2010

PHYS 1441 – Section 002
Lecture #13
Wednesday, Oct. 20, 2010
Dr. Jaehoon Yu
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•
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•
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Motion in Resistive Force
Work done by a constant force
Scalar Product of the Vector
Work with friction
Work-Kinetic Energy Theorem
Potential Energy
Wednesday, Oct. 20, 2010
PHYS 1441-002, Fall 2010
Dr. Jaehoon Yu
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Announcements
• 2nd non-comprehensive term exam
– Date: Wednesday, Nov. 3
– Time: 1 – 2:20pm in class
– Covers: CH3.5 – what we finish Monday, Nov. 1
• Physics faculty research expo today
Wednesday, Oct. 20, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Wednesday, Oct. 20, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
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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
• The gravitational constant G  6.67  1011 N  m2 kg 2
• The radius of the Earth R  6.37 103 km
E
• 20 point extra credit
• Due: Wednesday, Oct. 27
• You must show your OWN, detailed work to
obtain any credit!!
Wednesday, Oct. 20, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Motion in Resistive Forces
Medium can exert resistive forces on an object moving through it due
to viscosity or other types frictional properties of the medium.
Some examples?
Air resistance, viscous force of liquid, etc
These forces are exerted on moving objects in opposite direction of
the movement.
These forces are proportional to such factors as speed. They almost
always increase with increasing speed.
Two different cases of proportionality:
1. Forces linearly proportional to speed:
Slowly moving or very small objects
2. Forces proportional to square of speed:
Large objects w/ reasonable speed
Wednesday, Oct. 20, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Work Done by a Constant Force
A meaningful work in physics is done only when the net
forces exerted on an object changes the energy of the object.
F
M
y

Free Body
Diagram
M
d
Which force did the work?
How much work did it do?
ur
Force F
 
ur
F
ur
FN

x
Fg  M g
Why?
ur ur
W   F  d  Fd cos 
What kind? Scalar
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, Oct. 20, 2010
Work
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
is an energy transfer!!
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Let’s think about the meaning of work!
• A person is holding a grocery bag and
walking at a constant velocity.
• Is he doing any work ON the bag?
– No
– Why not?
– Because the force he exerts on the bag, Fp, is
perpendicular to the displacement!!
– This means that he is not adding any energy
to the bag.
• So what does this mean?
– In order for a force to perform any meaningful
work, the energy of the object the force exerts
on must change!!
• What happened to the person?
– He spends his energy just to keep the bag up
but did not perform any work on the bag.
Wednesday, Oct. 20, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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Work done by a constant force
s
r r
W  F s
 F cos s

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
cos 0  1
cos 90  0
cos180  1
PHYS 1441-002, Fall 2010 Dr. Jaehoon
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Scalar Product of Two Vectors
• Product of magnitude of the two vectors and the cosine of the
ur ur
ur ur
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, Oct. 20, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
=0
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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
2
2
d  d x2  d y2  2.0  3.0  3.6m
ur
2
2
F  Fx2  Fy2  5.0  2.0  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, Oct. 20, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
10
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 d

 45.0  cos50o  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?
Wednesday, Oct. 20, 2010
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?11
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
Ex. 6.1 Work done on a crate
A person pulls a 50kg crate 40m along a horizontal floor by a constant force Fp=100N, which
acts at a 37o angle as shown in the figure. The floor is rough and exerts a friction force
Ffr=50N. Determine (a) the work done by each force and (b) the net work done on the crate.
What are the forces exerting on the crate?
Fp
Ffr
FG=-mg
FN=+mg
Which force performs the work on the crate?
Fp
Work done on the crate by FG
Work done on the crate byFN
Work done on the crate by Fp:
Work done on the crate by Ffr:
Ffr
 
r r
r
WG  FG  x  mg cos 90o  x  0J
r r
o r
mg
cos
90
 x  100  cos90o  40  0J
WN  FN  x 
ur r ur
o
o r
W p  F p  x  F p cos37  x  100  cos37  40  3200J
ur r ur
r
W fr  F fr  x  F fr cos180o  x  50  cos180o  40  2000J
So the net work on the crate Wnet WN WG W p W fr  0  0  3200  2000  1200 J 
This is the same as
Wednesday, Oct. 20, 2010
Wnet  
 
r r r r r r r r
r r
F  x  FN  x FG  x Fp  x  F fr  x
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu

12
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 cos0  s 
Fs

 710  0.65  460 J
W   F cos180
  s   Fs

 710  0.65  460 J
What does the negative work mean?
Wednesday, Oct. 20, 2010
The gravitational force does the
work on the weight lifter!
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
13
Ex. Accelerating a 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
Wednesday, Oct. 20, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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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




W  f s  s   180N cos0  65 m  1.2  104 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, Oct. 20, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Yu
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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
ΣF
vi
M
vf
s
Using the kinematic
equation of motion
Suppose net force ΣF 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 ΣF is
r r
W   F  s   ma cos 0 s   ma  s
v 2f  v02
 
2as  v  v
2
f
2
0
as 
2
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
1
1
Work done by the net force causes
Work W  mv 2f  mvi2  KE f  KEi  KE
2
2
change in the object’s kinetic energy.
Wednesday, Oct. 20, 2010
PHYS 1441-002, Fall 2010 Dr. Jaehoon
Work-Kinetic Energy
Yu
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Theorem