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PHYS 1441 – Section 001
Lecture #12
Tuesday, June 24, 2014
Dr. Jaehoon Yu
•
•
•
•
•
Work done by a constant force
Multiplication of Vectors
Work-Kinetic Energy Theorem
Work and Energy Involving Kinetic
Friction
Potential Energy
Today’s homework is homework #7, due 11pm, Friday, June 27!!
Tuesday, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
1
Announcements
• Term exam #2
–
–
–
–
In class this Wednesday, June 25
Non-comprehensive exam
Covers CH 4.7 to what we finish today, Tuesday, June 24
Bring your calculator but DO NOT input formula into it!
• Your phones or portable computers are NOT allowed as a replacement!
– You can prepare a one 8.5x11.5 sheet (front and back) of
handwritten formulae and values of constants for the exam  no
solutions, derivations or definitions!
• No additional formulae or values of constants will be provided!
• Quiz 3 results
– Class average: 27.4/45
• Equivalent to: 61/100
• Previous results: 69.3/100 and 59.6/100
– Top score: 45/45
Tuesday, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
2
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
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, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
Work is an energy transfer!!
3
Let’s think about the meaning of work!
• A person is holding a grocery bag and
walking at a constant velocity.
• Are his hands doing any work ON the bag?
– No
– Why not?
– Because the force hands exert on the bag, Fp,
is perpendicular to the displacement!!
– This means that hands are not adding any
energy to the bag.
• So what does this mean?
– In order for a force to have performed any
meaningful work, the energy of the object the
force exerts on must change due to that force!!
• 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, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
4
Work done by a constant force
s
W  F ×s
= F cosq s
(
Tuesday, June 24, 2014
)
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
cos 0  1
cos 90  0
cos180  1
5
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, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
=0
6
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, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
7
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.
W
(
)
= 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?
Tuesday, June 24, 2014
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?8
PHYS 1441-001, Summer 2014
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
(
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 by FN
( )
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, June 24, 2014
Wnet =
PHYS 1441-001, Summer 2014
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 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? The gravitational force does the
work on the weight lifter!
Tuesday, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
10
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, June 24, 2014
Work done by the net force causes
change in the object’s kinetic energy.
PHYS 1441-001, Summer 2014Work-Kinetic Energy
Dr. Jaehoon Yu
11
Theorem
Work-Kinetic Energy Theorem
When a net external force by the jet engine does work on an
object, the kinetic energy of the object changes according to
W  KEf - KE o = mv - mv
1
2
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2
f
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
1
2
2
o
12
Ex. Deep Space 1
The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If
a 56.0-mN force acts on the probe parallel through a displacement of
2.42×109m, what is its final speed?
(
é
ë
)
2
2
1
1
ù
F
cos
q
s
=
mvf - 2 mvo
å
2
û
v f = vo2 + 2
(å F cosq ) s
Tuesday, June 24, 2014
m
=
Solve for vf
( 275 m s)2 + 2 ( 5.60 ´10-2N ) cos0 ( 2.42 ´10 9 m )
v f = 805 m s
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
13
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, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
14
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 = -Ffr d KE  -Ffr d
The negative sign means that the work is done on the friction!!
The final kinetic energy of an object, including its initial kinetic energy,
work by the friction force and all other sources of work, is
KE f  KEi W -Ffr d
t=0, KEi
Tuesday, June 24, 2014
Friction,
PHYS 1441-001, Summer 2014Engine work
Dr. Jaehoon Yu
t=T, KEf
15
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
vi=0
F
Work done by the force F is
M
d=3.0m
Work done by friction Fk is
Thus the total work is
12  3.0cos 0  36  J 
WF =
vf
= 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
Tuesday, June 24, 2014
Solving the equation
for vf, we obtain
vf 
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
2W
2 10

 1.8m / s
m
6.0
16
Ex. Downhill Skiing
A 58kg skier is coasting down a 25o
slope. A kinetic frictional force of
magnitude fk=70N opposes her motion.
At 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?
Tuesday, June 24, 2014
f k  k FN
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
70
fk
 0.14
k 

FN 515
17
Ex. Now with the X component
X component  Fx  Fgx  f k  mg sin 25  f k   58  9.8  sin 25  70   170N  ma
Total work by
W    Fx   s  mg sin 25 - f k × s =  58  9.8  sin 25  70   57  9700J
this force
1 2
1 2
From work-kinetic
mv

W

mv0
KE

W  KEi 
f
f
energy theorem W  KE f  KEi
2
2
(
Solving for vf
)
2W  mv02
2
vf 
m
What is her acceleration?
Tuesday, June 24, 2014
vf 
F
x
 ma
2  9700  58   3.6 
 19 m s
58
2W  mv

m
Fx 170


a
 2.93 m s 2
m
58
2
0
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
2
18
Potential Energy
Energy associated with a system of objects  Stored energy which has the
potential or the possibility to work or to convert to kinetic energy
What does this mean?
In order to describe potential energy, U,
a system must be defined.
The concept of potential energy can only be used under the
special class of forces called the conservative force which
results in the principle of conservation of mechanical energy.
EM  KEi  PEi  KE f  PE f
What are other forms of energies in the universe?
Mechanical Energy
Chemical Energy
Biological Energy
Electromagnetic Energy
Nuclear Energy
Thermal Energy
These different types of energies are stored in the universe in many different forms!!!
If one takes into account ALL forms of energy, the total energy in the entire
Tuesday, June
2014
1441-001, Summer
universe
is24,conserved.
It justPHYS
transforms
from2014
one form to another.
Dr. Jaehoon Yu
19
Gravitational Potential Energy
This potential energy is given to an object by the gravitational field in the
system of Earth by virtue of the object’s height from an arbitrary zero level
When an object is falling, the gravitational force, Mg, performs the work
on the object, increasing the object’s kinetic energy. So the potential energy
of an object at a height y, the potential to do work, is expressed as
m
mg
hi
PE 
m
 mgh
The work done on the object by
the gravitational force as the
brick drops from yi to yf is:
hf
What does
this mean?
Tuesday, June 24, 2014
PE  mgh
Wg  PEi  PE f
 mghi  mgh f  PE
(since
DPE = PE f - PEi)
Work by the gravitational force as the brick drops from yi to yf
is the negative change of the system’s potential energy
 Potential energy was spent in order for the
gravitational
force
to2014
increase the brick’s kinetic energy.
PHYS 1441-001,
Summer
20
Dr. Jaehoon Yu
Ex. A Gymnast on a Trampoline
A gymnast leaves the trampoline at an initial height of 1.20 m and reaches a
maximum height of 4.80 m before falling back down. What was the initial
speed of the gymnast?
Tuesday, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
21
Ex. Continued
From the work-kinetic energy theorem
W
1
2
mv  mv
2
f
2
o
1
2
Work done by the gravitational force
Wgravity  mg  ho  h f
Since at the maximum height, the
final speed is 0. Using work-KE
theorem, we obtain
mg  ho  h f    mv
2
o
1
2
vo  2 g  ho  h f 
 vo  2  9.80m s2  1.20 m  4.80 m   8.40m s
Tuesday, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
22

Example for Potential Energy
A bowler drops bowling ball of mass 7kg on his toe. Choosing the floor level as y=0, estimate
the total work done on the ball by the gravitational force as the ball falls on the toe.
Let’s assume the top of the toe is 0.03m from the floor and the hand
was 0.5m above the floor.
U i  mgyi  7  9.8  0.5  34.3J U f  mgy f  7  9.8  0.03  2.06J
Wg  U   U f  U i   32.24J  30J
M
b) Perform the same calculation using the top of the bowler’s head as the origin.
What has to change?
First we must re-compute the positions of the ball in his hand and on his toe.
Assuming the bowler’s height is 1.8m, the ball’s original position is –1.3m, and the toe is at –1.77m.
U i  mgyi  7  9.8   1.3  89.2J U f  mgy f  7  9.8   1.77   121.4J
Wg  U   U f  U i   32.2J  30J
Tuesday, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
23
Elastic Potential Energy
Potential energy given to an object by a spring or an object with
elasticity in the system that consists of an object and the spring.
The force spring exerts on an object when it is
distorted from its equilibrium by a distance x is
The work performed on the
object by the spring is
Ws = ò
xf
xi
Fs  kx
Hooke’s Law
x
1 2ù f
é
( -kx )dx = ê- kx ú = - 1 kx 2f + 1 kxi2 = 1 kxi2 - 1 kx 2f
2
2
2
2
ë 2 û xi
The potential energy of this system is
1 2
U s º kx
2
The work done on the object by the
spring depends only on the initial and
final position of the distorted spring.
The gravitational potential energy, Ug
Where else did you see this trend?
What do you see from
the above equations?
So what does this tell you about the elastic force?
Tuesday, June 24, 2014
A conservative force!!!
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
24
Conservative and Non-conservative Forces
The work done on an object by the gravitational force does not
depend on the object’s path in the absence of a retardation force.
N
When directly falls, the work done on the object by the gravitation force is
l
h
mg

Wg  Fg incline  l  mg sin   l
When sliding down the hill
of length l, the work is
How about if we lengthen the incline by a
factor of 2, keeping the height the same??
Wg  mgh
 mg  l sin    mgh
Still the same
amount of work
Wg  mgh
So the work done by the gravitational force on an object is independent of the path of
the object’s movements. It only depends on the difference of the object’s initial and
final position in the direction of the force.
Forces like gravitational and
elastic forces are called the
conservative force
Tuesday, June 24, 2014
1.
2.
If the work performed by the force does not depend on the path.
If the work performed on a closed path is 0.
Total mechanical energy is conserved!!
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
EM  KEi  PEi  KE f  PE f
25
Conservation of Mechanical Energy
Total mechanical energy is the sum of kinetic and potential energies
m
mg
h
Let’s consider a brick of
mass m at the height h
from the ground
The brick gains speed
h1
What does
this mean?
What is the brick’s potential energy?
PE = mgh
What happens to the energy as
the brick falls to the ground?
m
E  KE+PE
DPE = PE f - PEi = -Fs
By how much?
v  gt
1 2 1 22
So what?
The brick’s kinetic energy increased K  mv  mg t
2
2
And? The lost potential energy is converted to kinetic energy!!
The total mechanical energy of a system remains
constant in any isolated systems of objects that
interacts only through conservative forces:
KE +
Principle of mechanical energy conservation i
Tuesday, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
Ei  E f
å PE = KE + å PE
i
f
26
f
Example
A ball of mass m at rest is dropped from the height h above the ground. a) Neglecting the air
resistance, determine the speed of the ball when it is at the height y above the ground.
m
PE
KE
mgh
0
mg
h
m
mvi2/2
Ki + PEi = K f + PE f 0  mgh  1 mv2  mgy
2
1 2
mv  mg  h  y 
2
v  2 g h  y 
Using the
principle of
mechanical
energy
conservation
mgy mv2/2 mvf2/2
b) Determine the speed of the ball at y if it had initial speed vi at the
time of the release at the original height h.
y
0
Again using the
principle of mechanical
energy conservation
but with non-zero initial
kinetic energy!!!
This result look very similar to a kinematic
expression, doesn’t it? Which one is it?
Tuesday, June 24, 2014
Ki + PEi = K f + PE f
1 2
1 2
mvi  mgh  mv f  mgy
2
2


1
m v 2f  vi2  mg  h  y 
2
2

v

v
f
i  2 g h  y 
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
27
Power
• Rate at which the work is done or the energy is transferred
– What is the difference for the same car with two different
engines (4 cylinder and 8 cylinder) climbing the same hill?
–  The time… 8 cylinder car climbs up the hill faster!
Is the total amount of work done by the engines different? NO
Then what is different? The rate at which the same amount of work
performed is higher for 8 cylinders than 4.
Average power P  W = Fs  F s  F v
t
t
Dt
Unit? J / s  Watts
Scalar
quantity
1HP  746Watts
What do power companies sell? 1kWH  1000Watts  3600s  3.6 106 J
Energy
Tuesday, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
28
Energy Loss in Automobile
Automobile uses only 13% of its fuel to propel the vehicle.
67% in the engine:
Why?
•
•
•
Incomplete burning
Heat
Sound
16% in friction in mechanical parts
4% in operating other crucial parts
such as oil and fuel pumps, etc
13% used for balancing energy loss related to moving the vehicle, like air
resistance and road friction to tire, etc
Two frictional forces involved in moving vehicles
Coefficient of Rolling Friction;  =0.016
Air Drag
mcar  1450kg Weight  mg  14200 N
 n   mg  227 N
1
1
f a  D  Av 2   0.5 1.293  2v 2  0.647v 2
2
2
Total power to keep speed v=26.8m/s=60mi/h
Power to overcome each component of resistance
Tuesday, June 24, 2014
Total Resistance
ft  f r  f a
P  ft v   691N   26.8  18.5kW
Pr  f r v  227   26.8  6.08kW
Pa 2014
 f av
PHYS 1441-001, Summer
Dr. Jaehoon Yu
 464.7 26.8  12.5kW29
Human Metabolic Rates
Tuesday, June 24, 2014
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
30
Ex. The Power to Accelerate a Car
A 1.10x103kg car, starting from rest, accelerates for 5.00s. The
magnitude of the acceleration is a=4.60m/s2. Determine the average
power generated by the net force that accelerates the vehicle.
What is the force that
accelerates the car?
Since the acceleration
is constant, we obtain
From the kinematic
formula
Thus, the average
speed is
And, the
average
power is



3
2
F  ma  1.10 10  4.60 m s  5060 N
v 
v0  v f

0  vf

vf
2
2
2
v f  v0  at  0   4.60 m s 2    5.00s   23.0 m s
vf
2

23.0
 11.5 m s
2
4
5060
N

11.5
m
s

5.82

10
W




P  Fv =
Tuesday, June 24, 2014
 78.0hp
PHYS 1441-001, Summer 2014
Dr. Jaehoon Yu
31