Transcript Slide 1

Piri Reis University 2011-2012/ Physics -I
Physics for Scientists &
Engineers, with Modern
Physics, 4th edition
Giancoli
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Piri Reis University 2011-2012 Fall Semester
Physics -I
Chapter 7 and 8
Work and Energy
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Piri Reis University 2011-2012
Lecture VII and VIII
I. Work Done by a Constant Force
II. Kinetic Energy, and the Work-Energy Principle
III. Potential Energy
IV. Conservative and Nonconservative Forces
V. The Conservation of Mechanical Energy
VI. Problem Solving Using Conservation of Mechanical
Energy
VII. Other Forms of Energy; Energy Transformations and the
Law of Conservation of Energy
VIII. Energy Conservation with Dissipative Forces: Solving
Problems
IX. Power
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I. Work Done by a Constant Force
The work done by a constant force is defined as the distance moved
multiplied by the component of the force in the direction of displacement:
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I. Work Done by a Constant Force
W = Fd cosθ
Simple special case: F & d are parallel:
θ = 0, cosθ = 1

W = Fd
Example: d = 50 m, F = 30 N
W = (30N)(50m) = 1500 N m
F
θ
d
Work units: Newton - meter = Joule
1 N m = 1 Joule = 1 J
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I. Work Done by a Constant Force
In the SI system, the units of work are joules:
As long as this person does not lift or lower
the bag of groceries, he is doing no work on it.
The force he exerts has no component in the
direction of motion.
W = 0:
Could have d = 0  W = 0
Could have F  d
 θ = 90º, cosθ = 0
W=0
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I. Work Done by a Constant Force
Solving work problems:
1. Draw a free-body diagram.
2. Choose a coordinate system.
3. Apply Newton’s laws to determine any unknown forces.
4. Find the work done by a specific force.
5. To find the net work, either
find the net force and then find the work it does, or
find the work done by each force and add.
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Example
W = F|| d = Fd cosθ
m = 50 kg, Fp = 100 N, Ffr = 50 N, θ = 37º
Wnet = WG + WN +WP + Wfr = 1200 J
Wnet = (Fnet)x x = (FP cos θ – Ffr)x = 1200 J
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Example
ΣF =ma
FH = mg
WH = FH(d cosθ)
= mgh
FG = mg
WG = FG(d cos(180-θ))
= -mgh
Wnet = WH + WG = 0
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I. Work Done by a Constant Force
Work done by forces that oppose the direction of motion, such as friction,
will be negative.
Centripetal forces do no work, as
they are always perpendicular to the
direction of motion.
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II. Kinetic Energy, and the Work-Energy Principle
Energy was traditionally defined as the ability to do work. We now know
that not all forces are able to do work; however, we are dealing in these
chapters with mechanical energy, which does follow this definition.
Energy  The ability to do work
Kinetic Energy  The energy of motion
“Kinetic”  Greek word for motion
An object in motion has the ability to do work
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II. Kinetic Energy, and the Work-Energy Principle
Consider an object moving in straight line. Starts at speed v1. Due to presence
of a net force Fnet, accelerates (uniform) to speed v2, over distance d.
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II. Kinetic Energy, and the Work-Energy Principle
Newton’s 2nd Law:
Fnet= ma
(1)
1d motion, constant a 
(v2)2 = (v1)2 + 2ad

Work done:
Combine (1), (2), (3):
a = [(v2)2 - (v1)2]/(2d)
Wnet = Fnetd
(2)
(3)
 Wnet = mad = md [(v2)2 - (v1)2]/(2d)
OR
Wnet = (½)m(v2)2 – (½)m(v1)2
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II. Kinetic Energy, and the Work-Energy Principle
If we write the acceleration in terms of the velocity and the distance,
we find that the work done here is
We define the translational kinetic energy:
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II. Kinetic Energy, and the Work-Energy Principle
This means that the work done is equal to the change in the kinetic
energy:
• If the net work is positive, the kinetic energy increases.
• If the net work is negative, the kinetic energy decreases.
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Net work on an object = Change in KE.
Wnet = KE
 The Work-Energy Principle
Note: Wnet = work done by the net (total) force.
Wnet is a scalar.
Wnet can be positive or negative (because KE can be
both + & -)
Units are Joules for both work & KE.
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II. Kinetic Energy, and the Work-Energy Principle
Because work and kinetic energy can be equated, they must have the
same units: kinetic energy is measured in joules.
Work done on hammer:
Wh = KEh = -Fd
= 0 – (½)mh(vh)2
Work done on nail:
Wn = KEn = Fd
= (½)mn(vn)2 - 0
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III. Potential Energy
An object can have potential energy by virtue of its surroundings.
Familiar examples of potential energy:
• A wound-up spring
• A stretched elastic band
• An object at some height above the ground
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III. Potential Energy
In raising a mass m to a height h, the work
done by the external force is
We therefore define the gravitational potential
energy:
WG = -mg(y2 –y1)
= -(PE2 – PE1)
= -PE
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III. Potential Energy
Wext = PE
The Work done by an external force to move the
object of mass m from point 1 to point 2 (without
acceleration) is equal to the change in potential
energy between positions 1 and 2
WG = -PE
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III. Potential Energy
This potential energy can become kinetic energy if the object is dropped.
Potential energy is a property of a system as a whole, not just of the
object (because it depends on external forces).
If
, where do we measure y from?
It turns out not to matter, as long as we are consistent about where we
choose y = 0. Only changes in potential energy can be measured.
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III. Potential Energy
Potential energy can also be stored in a spring when it is compressed;
the figure below shows potential energy yielding kinetic energy.
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III. Potential Energy
The force required to compress or stretch a spring is:
FP = kx
(6-8)
Restoring force
“Spring equation or Hook’s law”
where k is called the spring constant,
and needs to be measured for each
spring.
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III. Potential Energy
The force increases as the spring is stretched or compressed further
(elastic) (not constant). We find that the potential energy of the
compressed or stretched spring, measured from its equilibrium position,
can be written:
1
1 2
W  F x  ( kx ) x  kx
2
2
(6-9)
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IV. Conservative and Nonconservative Forces
Conservative Force  The work done by that force depends only on initial
& final conditions & not on path taken between the initial & final positions
of the mass.

A PE CAN be defined for conservative forces
Non-Conservative Force  The work done by that force depends on the
path taken between the initial & final positions of the mass.
 A PE CAN’T be defined for non-conservative forces
The most common example of a non-conservative force is FRICTION
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IV. Conservative and Nonconservative Forces
If friction is present, the work done depends not only on the starting and
ending points, but also on the path taken. Friction is called a
nonconservative force.
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IV. Conservative and Nonconservative Forces
Potential energy can only be
defined for conservative forces.
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IV. Conservative and Nonconservative Forces
Therefore, we distinguish between the work done by conservative forces and
the work done by nonconservative forces.
We find that the work done by nonconservative forces is equal to the total
change in kinetic and potential energies:
Wnet = Wc + Wnc
Wnet = ΔKE = Wc + Wnc
Wnc = ΔKE - Wc = ΔKE – (– ΔPE)
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V. The Conservation of Mechanical Energy
If there are no nonconservative forces, the sum of the changes in the
kinetic energy and in the potential energy is zero – the kinetic and
potential energy changes are equal but opposite in sign.
KE + PE = 0
This allows us to define the total mechanical energy:
And its conservation:
(6-12b)
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 PE1 = mgh
KE1 = 0
The sum remains constant
 KE + PE = same
as at points 1 & 2
KE1 + PE1 = KE2 + PE2
0 + mgh = (½)mv2 + 0
v2 = 2gh
 PE2 = 0
KE2 = (½)mv2
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VI. Problem Solving Using Conservation of
Mechanical Energy
In the image on the left, the total mechanical
energy is:
The energy buckets (right) show how the
energy moves from all potential to all
kinetic.
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VI. Problem Solving Using Conservation of
Mechanical Energy
If there is no friction, the speed of a roller coaster will depend only on
its height compared to its starting point.
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VI. Problem Solving Using Conservation of
Mechanical Energy
For an elastic force, conservation of energy tells us:
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VII. Other Forms of Energy; Energy
Transformations and the Law of Conservation of
Energy
Some other forms of energy:
Electric energy, nuclear energy, thermal energy, chemical energy.
Work is done when energy is transferred from one object to another.
Accounting for all forms of energy, we find that the total energy neither
increases nor decreases. Energy as a whole is conserved.
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VIII. Energy Conservation with Dissipative Forces:
Solving Problems
If there is a nonconservative force such as friction, where do the kinetic
and potential energies go?
They become heat; the actual temperature rise of the materials involved
can be calculated.
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VIII. Energy Conservation with Dissipative Forces:
Solving Problems
We had, in general:
WNC = KE + PE
WNC = Work done by non-conservative forces
KE = Change in KE
PE = Change in PE (conservative forces)
Friction is a non-conservative force! So, if friction is present, we have
(WNC  Wfr)
Wfr = Work done by friction
In moving through a distance d, force of friction Ffr does work
Wfr = - Ffrd
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When friction is present, we have:
Wfr = -Ffrd = KE + PE = KE2 – KE1 + PE2 – PE1
– Also now, KE + PE  Constant!
– Instead, KE1 + PE1+ Wfr = KE2+ PE2
OR:
KE1 + PE1 - Ffrd = KE2+ PE2
• For gravitational PE:
(½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2 + Ffrd
• For elastic or spring PE:
(½)m(v1)2 + (½)k(x1)2 = (½)m(v2)2 + (½)k(x2)2 + Ffrd
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VIII. Energy Conservation with Dissipative Forces:
Solving Problems
Problem Solving:
1. Draw a picture.
2. Determine the system for which energy will be conserved.
3. Figure out what you are looking for, and decide on the initial and
final positions.
4. Choose a logical reference frame.
5. Apply conservation of energy.
6. Solve.
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IX. Power
Power is the rate at which work is done –
(6-17)
In the SI system, the units of power are watts:
British units: Horsepower (hp). 1hp = 746 W
The difference between walking and
running up these stairs is power – the
change in gravitational potential energy is
the same.
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IX. Power
Power is also needed for acceleration and for moving against the force of
gravity.
The average power can be written in terms of the force and the average
velocity:
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Summary of Chapter VII and VIII
• Work:
•Kinetic energy is energy of motion:
• Potential energy is energy associated with forces that depend on the position
or configuration of objects.
•
•The net work done on an object equals the change in its kinetic energy.
• If only conservative forces are acting, mechanical energy is conserved.
• Power is the rate at which work is done.
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Example 1 Pulling a Suitcase-on-Wheels
Find the work done if the force is 45.0-N, the angle is 50.0
degrees, and the displacement is 75.0 m.


W  F cos s  45.0 N  cos50.0 75.0 m 
 2170J
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Example 2
W  F cos0s  Fs
W  F cos180s  Fs
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Example 3 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?
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Example 3 Accelerating a Crate
The angle between the displacement
and the friction force is 0 degrees.


f s  ma  120kg 1.5 m s2  180N
W  180Ncos065 m  1.2 104 J
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Example 4 Deep Space 1
The mass of the space probe is 474-kg and its initial velocity
is 275 m/s. If the .056 N force acts on the probe through a
displacement of 2.42×109m, what is its final speed?
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Example 4 Deep Space 1
W  mv  mv
1
2
W
2
f
1
2
2
o
 F cos s
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Example 4 Deep Space 1
Fcos s 
1
2
mvf2  12 mvo2
.056Ncos0 2.42109 m  12 474kgvf2  12 474kg275m s2
v f  805m s
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Example 5

F

mg
sin
25
 fk
In this case the net force is 
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Example 6
W  F cos s
Wgravity  mgho  hf 
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Example 7
Wgravity  mgho  hf 
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Example 8 A Gymnast on a Trampoline
The 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?
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Example 8 A Gymnast on a Trampoline
W  12 mvf2  12 mvo2
mgho  h f    12 mvo2
Wgravity  mgho  hf 
vo   2 g ho  h f 


vo   2 9.80 m s 2 1.20 m  4.80 m   8.40 m s
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Example 9 A Daredevil Motorcyclist
A motorcyclist is trying to leap across the canyon by driving
horizontally off a cliff 38.0 m/s. Ignoring air resistance, find
the speed with which the cycle strikes the ground on the other
side.
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Example 9 A Daredevil Motorcyclist
Ef  Eo
mghf  mv  mgho  mv
1
2
2
f
1
2
2
o
ghf  12 v2f  gho  12 vo2
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Example 9 A Daredevil Motorcyclist
ghf  12 v2f  gho  12 vo2
v f  2 g ho  h f   vo2


v f  2 9.8 m s 35.0m  38.0 m s   46.2 m s
2
2
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Example 10 Fireworks
Assuming that the nonconservative force
generated by the burning propellant does
425 J of work, what is the final speed
of the rocket. Ignore air resistance.


Wnc  m ghf  12 m v2f 
m gh 
o
1
2
m vo2

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Example 10 Fireworks
Wnc  mghf  mgho  mv  mv
1
2
2
f
1
2
2
o
Wnc  mghf  ho  12 mv2f


425 J  0.20 kg 9.80 m s 2 29.0 m 
 12 0.20 kgv 2f
v f  61m s
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Example 11
Part a) constant speed
a = 0 ∑Fx = 0
F – FR – mg sinθ = 0
F = FR + mg sinθ
F = 3100 N
P = Fv
= 91 hp
Part b) constant acceleration
Now, θ = 0
∑Fx = ma
F – FR= ma
v = v0 + at
a = 0.93 m/s2
F = ma + FR = 2000 N
P = Fv
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= 82 hp
HOMEWORK
Giancoli, Chapter 7
3, 7, 8, 11, 14, 25, 32, 40, 45, 47
Giancoli, Chapter 8
7, 9, 10, 17, 20, 21, 23, 28, 36, 38
References
o “Physics For Scientists &Engineers with Modern Physics” Giancoli 4th edition,
Pearson International Edition
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