Transcript Lect11&12

Chapter 6
Work and Energy
Work (dot product)
Work by gravity, by spring
Kinetic energy, power
Work-kinetic energy theorem
C.M. system
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Definition of Work: Constant Force
Ingredients: Force (F), displacement (Δr)
Work, W, of a constant force F
acting through a displacement Δr :
W  F  r  Fr cos   Fr r
Vector
“Dot Product”
Work is a scalar.
Units are Newton meter: Joule = N x m
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Work and Force Direction
Work is done in lifting the box (you do +work; gravity -)
No work is done on the bucket when held still,
or to move horizontally
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More on “dot product” (or scalar product)
a
Definition:
ba
.
a b = ab cos 
= a[b cos  ] = aba

b
a
= b[a cos  ] = bab

Some properties:
.b = b .a
q(a . b) = (qb) . a = b .(qa)
a .(b + c) = (a . b) + (a . c)
ab
b
a
(q is a scalar)
(c is a vector)
The dot product of perpendicular vectors is 0 !!
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Examples of dot products
y
.i=j.j=k.k=1
i.j=j.k=k.i=0
i
j
i
k
x
z
Suppose
a=1i+2j+3k
b=4i -5j+6k
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Then
. b = 1x4 + 2x(-5) + 3x6 = 12
a . a = 1x1 +
2x2 + 3x3 = 14
b . b = 4x4 + (-5)x(-5) + 6x6 = 77
a
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More properties of dot products

Components:
a = ax i + ay j + az k = (ax , ay , az) = (a . i, a . j, a . k)

Derivatives:
Apply to velocity
So if v is constant
(like for uniform circular motion):
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Which of the statements below is correct?
A. The scalar product of two vectors can be negative.
B. AcB=c(BA), where c is a constant.
C. The scalar product can be non-zero even if two of
the three components of the two vectors are equal
to zero.
E. All of the above statements are correct.
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What about multiple forces?
Suppose FNET = F1 + F2 and the
displacement is Δr.
The work done by each force is:
W1
= F . Δr
W2
1
= F . Δr
F1
FNET
Δr
2
WTOT = W1 + W2
= F . Δr + F . Δr
= (F + F ) . Δr
. Δr It’s the total force that matters!!
=F
1
1
WTOT
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F2
2
2
TOT
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Question 1
You are towing a car up a hill with constant velocity.
The work done on the car by the normal force is:
1. positive
2. negative
3. zero
correct
FN
T
Normal force is perpendicular to displacement
cos  = 0
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W
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Question 2
You are towing a car up a hill with constant velocity.
The work done on the car by the gravitational force is:
1. positive
2. negative
correct
3. zero
FN
T
There is a non-zero component of gravitational
force pointing opposite the direction of motion.
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Question 3
You are towing a car up a hill with constant velocity.
The work done on the car by the tension force is:
correct
1. positive
2. negative
3. zero
FN
T
T is pointing in the direction of motion therefore, work done by this force is positive.
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Question 4
You are towing a car up a hill with constant velocity.
The total work done on the car by all forces is:
1. positive
2. negative
3. zero
correct
FN
T
Constant velocity implies that there is no net force
acting on the car, so there is no work being done overall
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Lifting an object:
A 3000–kg truck is to be loaded onto a ship by a crane that exerts an upward
force of 31 kN on the truck. This force, which is strong enough to overcome the
gravitational force and keep the truck moving upward, is applied over a distance
of 2.0 m. Find
(a) the work done on the truck by the crane,
(b) the work done on the truck by gravity,
(c) the net work done on the truck.
(a) the work done by the crane:
Wc = Fc d = 31 kN x 2 m = 62 kJ
(b) the work done by gravity:
Wg = Fg d = (-31 kN) x 2 m = -62 kJ
(c) the net work done on the truck:
W = Wc + Wg = 0
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Work by a variable force:
Total work, W, of a force F acting
through a displacement Δx = dx î is an integration:
W   Fi  x i
Fx
i
Fi
x
xi
Work = Area under F(x) curve
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Work done by a spring force
Force exerted by a spring extended
a distance x is:
Fx = -kx
(Hooke’s law)
Force done by spring on an object when the
displacement of spring is changed from xi to xf is:
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A person pushes against a spring, the
opposite end of which is attached to a fixed
wall. The spring compresses. Is the work
done by the spring on the person positive,
negative or zero?
A. positive
B. negative
C. zero
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Energy
Energy is that quality of a substance or object
which “causes something to happen”;
or “capability of exerting forces”;
or “ability to do work”…
The vagueness of the definition is due to the fact
that energy can result in many effects.
Electrical Energy
Chemical Energy
Mechanical Energy
Nuclear Energy
It is convertible into other forms without loss
(i.e it is conserved)
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Kinetic Energy
The “energy of motion”.
Work done on the object increases its energy,
-- by how much? (i.e. how to calculate the value?)
W=Fd
F = ma
a=
2
2d
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d
2
V - V0
2
V - V0
2
1
K.E. = mV
2
0
2
W = (ma) d
W=m
2
V = V 2+ 2 a d
2d
W=
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1
mV
2
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-
1 m
2
2
V0
Definition of Kinetic Energy
Kinetic energy (K.E.) of a particle of mass m moving
with speed v is defined as
K.E. = ½mv2
Kinetic energy is a useful concept because of the
Work/Kinetic Energy theorem, which relates the work
done on an object to the change in kinetic energy.
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A falling object
What is the speed of an object that starts at rest and
then falls a vertical distance H?
Work done by gravitational force
WG = FΔr = mgH
v0 = 0
r
mg
j
H
Work/Kinetic Energy Theorem:
WG  mgH  12 mv 2
 v  2gH
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v
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A skier of mass 50 kg is moving at speed 10 m/s at point P1 down
a ski slope with negligible friction. What is the skier’s kinetic
energy when she is at point P2, 20 m below P1?
A. 2500 J
B. 9800 J
C. 12300 J
D.13100 J
E. 15000 J
P1
H
P2
Work-kinetic energy theorem:
KEf – KEi = Wg = mgH
KEf = ½×50×102 + 50×9.8×20 = 2500 J +
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Energy and Newton’s Laws
The importance of mechanical energy in classical
mechanics is a consequence of Newton’s laws.
The concept of energy turns out to be even more
general than Newton’s Laws.
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Problem: Work and Energy
Two blocks have masses m1 and m2, where m1 > m2. They are
sliding on a frictionless floor and have the same kinetic energy
when they encounter a long rough stretch with μ>0, which slows
them down to a stop.
Which one will go farther before stopping?
(A) m1
(B) m2
(C) They will go the same distance
Solution next page 
m1
m2
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Problem: Work and Energy (Solution)
The net work done to stop a box is -fD = -μmgD.
The work-kinetic energy theorem says that for any object,
WNET = ΔK so WNET=Kf-Ki=0-Ki.
Since the boxes start out with the same kinetic energy, we have
μm1gD1=μm2gD2 and D1/D2=m2/m1.
Since m1>m2, we must have D2>D1.
m1
m2
D1
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D2
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Problem:
The magnitude of the single force acting on a particle of
mass m is given by F = bx2, where b is a constant. The
particle starts from rest at x=0. After it travels a distance L,
determine its (a) kinetic energy and (b) speed.
work done by force:
b 3
W   F  dx   bx dx  L
0
0
3
L
L
2
So (a) kinetic energy is K=bL3/3
½mv2=bL3/3,
(b) Since
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2bL3
v
3m
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Power:
P: Work done by unit time.
Units: Watt
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Question
A sports car accelerates from zero to 30 mph in 1.5 s.
How long does it take for it to accelerate from zero to
60 mph, assuming the power of the engine to be
independent of velocity and neglecting friction?
Power is constant, so
d 1 2
, a constant.

2 mv   C
dt
Integrating with respect to time, and noting that the
initial velocity is zero, one gets 12 mv 2  Ct . So getting
to twice the
speed takes 4 times as long, and the time
to reach 60 mph is 4×1.5 = 6s.
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If a fighter jet doubles its speed, by what factor should the
power from the engine change? (Assume that the drag
force on the plane is proportional to the square of the
plane’s speed.)
A. by half
B. unchanged
C. doubled
E. 8 times
Magnitude of power is Fv. When the velocity v is
doubled, the drag force goes up by a factor of 4, and
Fv goes up by a factor of 8
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Recall: Center of Mass
Definition of the center of mass:
m r

m
i i
Rcm
i
i
 acm
m a

.
m
i
i
i
Because
i i
i
Fi  mi ai ,
ma


i i

 Fi   mi ai    mi   i   M TOT aCM
i
i
i
  mi 
i
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Center-Of-Mass Work
For systems of particles that are not all moving at the same
velocity, there is a work-kinetic energy relation for the center of
mass.
So
where
K trans
1 2
 mvCM
2
Net work done on collection
of objects
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is the translational kinetic energy
=
change in translational kinetic
energy of system
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