Download Tuesday, Sept. 30, 2014

PHYS 1443 – Section 004
Lecture #11
Tuesday, Sept. 30, 2014
Dr. Jaehoon Yu
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Newton’s Law of Universal Gravitation
Work Done by a Constant Force
Scalar Product
Work Done by a Varying Force
Work-Kinetic Energy Theorem
Work under friction
Today’s homework is homework #6, due 10pm, Tuesday, Oct. 7!!
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
1
Special Project #4
• Two protons are separated by 1m.
– Compute the gravitational force (FG) between the two
protons (10 points)
– Compute the electric force (FE) between the two protons
(10 points)
– Compute the ratio of FG/FE (5 points) and explain what
this tells you (5 point)
• You must specify the formulae for each of the
forces and the values of necessary quantities, such
as mass, charge, constants, etc, in your report
• Due: Beginning of the class, Tuesday, Oct. 7
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
3
Newton’s Law of Universal Gravitation
People have been very curious about the stars in the sky, making
observations for a long time. The data people collected, however, have
not been explained until Newton has discovered the law of gravitation.
Every object in the universe attracts every other object with a force that
is directly proportional to the product of their masses and inversely
proportional to the square of the distance between them.
How would you write this
law mathematically?
G is the universal gravitational
constant, and its value is
m1 m2
Fg  2
r12
With G
G  6.673 10
11
m1m2
Fg  G
r122
Unit?
N  m2 / kg 2
This constant is not given by the theory but must be measured by experiments.
This form of forces is known as the inverse-square law, because the magnitude of the
force is inversely proportional to the square of the distances between the objects.
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
4
Free Fall Acceleration & Gravitational Force
The weight of an object with mass m
is mg. Using the force exerting on a
particle of mass m on the surface of
the Earth, one can obtain
What would the gravitational
acceleration be if the object is at
an altitude h above the surface of
the Earth?
mg
g
M Em
RE2
ME
G
RE2
G
M Em  G M Em
Fg  mg '  G
2
2


R

h
r
E
ME
g'  G
from the
RE  h2 Distance
center of the Earth
What do these tell us about the gravitational acceleration?
to the object at the
altitude h.
•The gravitational acceleration is independent of the mass of the object
•The gravitational acceleration decreases as the altitude increases
•If the distance from the surface of the Earth gets infinitely large, the weight of the
object approaches 0.
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
5
Ex. for Gravitational Force
The international space station is designed to operate at an altitude of 350km. Its designed
weight (measured on the surface of the Earth) is 4.22x106N. What is its weight in its orbit?
The total weight of the station on the surface of the Earth is
FGE  mg
ME
M Em
6
G
2  4.22  10 N
RE
Since the orbit is at 350km above the surface of the Earth,
the gravitational force at that altitude is
FO  mg ' = G
MEm
( RE + h )
2
=
RE2
( RE + h )
2
FGE
Therefore the weight in the orbit is
FO
=
R
( RE + h )
( 6.37 ´ 10 )
6 2
2
E
2
FGE =
Tuesday, Sept. 30, 2014
( 6.37 ´ 10
6
+ 3.50 ´ 10
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
)
5 2
´ 4.22 ´ 10 6 = 3.80 ´ 10 6 N
6
Example for Universal Gravitation
Using the fact that g=9.80m/s2 on the Earth’s surface, find the average density of the Earth.
Since the gravitational acceleration is
Fg
G
M Em
RE2
 mg
Solving for g
Solving for ME
Therefore the
density of the
Earth is
g
ME
M
 G 2  6.67 1011 E2
RE
RE
RE 2 g
ME 
G
2

ME

VE
RE g
3g
G


4GRE
4
3
RE

3  9.80
3
3


5
.
50

10
kg
/
m
4  6.67 1011  6.37 106
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
7
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
ur
F
ur
FN

d
x
Fg = Mg
Which force did the work?
How much work did it do?
Force
Why?
W
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.
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
Work is an energy transfer!!
8
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.
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
9
Scalar Product of Two Vectors
• Product of magnitude of the two vectors and the cosine of the
angle between them
• Operation is commutative
• Operation follows the distribution
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?
 
 
 


  Ax Bx i  i  Ay By j j  Az Bz k  k  cross terms


Ax Bx  Ay By  Az Bz
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
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.
a) Calculate the magnitude of the displacement
and that of the force.
Y
d
F
d x2  d y2 
X
2.02  3.02  3.6m
Fx2  Fy2  5.0  2.0  5.4 N
2
2
b) Calculate the work done by the force F.
W


 
  
  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?
W
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
11
Example of Work by a Constant Force
A man cleaning a floor pulls a vacuum cleaner with a force of magnitude F=50.0N at
an angle of 30.0o with East. Calculate the work done by the force on the vacuum
cleaner as the vacuum cleaner is displaced by 3.00m to East.
F
0o
M
d
W
M
W = 50.0 ´ 3.00 ´ cos30 = 130J
No
Does work depend on mass of the object being worked
on?
Why ? This is because the work done by the force bringing the
object to a displacement d is constant independent of the
mass of the object being worked on. The only difference
would be the acceleration and the final speed of each of the
objects after the completion of the work!!
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
12
Ex. 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
(
Ffr
)
Work done on the crate by Fp:
WG = FG × x = -mg cos -90 × x = 0J
WN = FN × x = mg cos90 × x = 100 × cos90 × 40 = 0J
Wp = F p × x = F p cos37 × x = 100 × cos37 × 40 = 3200J
Work done on the crate by Ffr:
W fr = F fr × x = F fr cos180 × x = 50 × cos180 × 40 = -2000J
Work done on the crate by FG
Work done on the crate byFN
( )
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
Tuesday, Sept. 30, 2014
Wnet =
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
13
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
  s   Fs
W   F cos180
( )
= -710 × 0.65 = -460 J
What does the negative work mean?
Tuesday, Sept. 30, 2014
The gravitational force does the
work on the weight lifter!
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
14
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
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
15
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
1.5m
s
ma

fs 


  180N
(
)
(
)
W  f s  s  éë 180N cos0 ùû 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!
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
16
Work Done by a 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
Tuesday, Sept. 30, 2014
0
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
17
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
M
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
W
 ma cos 0 s   ma  s
s
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
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
Work done by the net force causes
change in the object’s kinetic energy.
18
Work-Kinetic Energy 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  mv  mv
1
2
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
2
f
1
2
2
o
19
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
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
Tuesday, Sept. 30, 2014
W  K f  Ki  K
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
Nm=Joule
20
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
12  3.0cos0  36  J 
W
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
Tuesday, Sept. 30, 2014
vf 
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
2W
2  36

 3.5m / s
m
6.0
21
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   s  12 mvf2  12 mvo2


v f  vo2  2   F cos   s m
=
Solve for vf
( 275 m s)2 + 2 ( 5.60 ´ 10-2N) cos 0 ( 2.42 ´ 109 m )
v f  805 m s
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
22
474
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.
Tuesday, Sept. 30, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
23