Transcript Lecture 9.1

Welcome back to Physics 215
Today’s agenda:
• Work
• Non-conservative forces
• Power
Physics 215 – Fall 2014
Lecture 09-1
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Current homework assignment
• HW7:
– Knight Textbook Ch.9: 54, 72
– Ch.10: 48, 68, 76
– Ch.11: 50, 64
– Due Wednesday, Oct. 22nd in recitation
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Work, Energy
• Newton’s Laws are vector equations
• Sometimes easier to relate speed of a
particle to how far it moves under a
force – a single equation can be used –
need to introduce concept of work
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What is work?
• Assume constant force in 1D
• Consider:
vF2 = vI2 + 2as
• Multiply by m/2 
(1/2)mvF2 - (1/2)mvI2 = mas
• But: F = ma
 (1/2)mvF2 - (1/2)mvI2 = Fs
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Work-Kinetic Energy theorem (1)
(1/2)mvF2 - (1/2)mvI2 = Fs
Points:
• W = Fs  defines work done on
particle
= force times displacement
• K = (1/2)mv2  defines kinetic energy
=1/2 mass times square of v
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Work-Kinetic Energy demo
• Cart, force probe, and motion detector
• Plot v2 vs. x – gradient 2F/m
• constant F (measure) -- pulling with string
• Weigh cart and masses in advance
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Conclusions from experiment
Although the motion of the two carts looks very different
(i.e., different amounts of time, accelerations, and final
speeds), there is a quantity that is the same for both at
the end of the motion. It is (1/2) mv2 and is called the
(final) kinetic energy of the carts.
Moreover, this quantity happens to have the same value
as F Ds, which is given the name work.
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Improved definition of work
• For forces, write F  FAB
• Thus W = Fs  WAB = FAB DsA is
work done on A by B as A undergoes
displacement DsA
• Work-kinetic energy theorem:
Wnet,A = SBWAB = DK
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Non-constant force …
Work done in small interval Dx
DW = F Dx
F
Total W done from A to B
S F Dx =
Area under curve!
F(x)
A Dx
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B
x
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The Work - Kinetic Energy Theorem
Wnet = DK = Kf - Ki
The net work done on an object is equal to
the change in kinetic energy of the object.
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Suppose a tennis ball and a bowling ball are
rolling toward you. The tennis ball is moving
much faster, but both have the same momentum
(mv), and you exert the same force to stop each.
Which of the following statements is correct?
1. It takes equal distances to stop each ball.
2. It takes equal time intervals to stop each ball.
3. Both of the above.
4. Neither of the above.
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Suppose a tennis ball and a bowling ball are
rolling toward you. Both have the same
momentum (mv), and you exert the same force to
stop each.
The distance taken for the bowling ball to stop is
1. less than.
2. equal to
3. greater than
the distance taken for the tennis ball to stop.
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Two carts of different mass are accelerated from rest
on a low-friction track by the same force for the same
time interval.
Cart B has greater mass than cart A (mB > mA). The final
speed of cart A is greater than that of cart B (vA > vB).
After the force has stopped acting on the carts, the
kinetic energy of cart B is
1.
2.
3.
4.
less than the kinetic energy of cart A (KB < KA).
equal to the kinetic energy of cart A (KB = KA).
greater than the kinetic energy of cart A (KB > KA).
“Can’t tell.”
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Revised definitions for
Work and Kinetic Energy
Work done on object 1 by object 2:
W(on 1 by 2) = F1,2 ·Dsof 1
Kinetic energy of an object:
K = mv [or:
1
2
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2
1
2
m(v·v)]
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Scalar (or “dot”) product of vectors
The scalar product is a way to combine two
vectors to obtain a number (or scalar). It is
indicated by a dot (•) between the two vectors.
(Note: component of A in direction n is just A•n)
A · B = AB cosq
= Ax Bx + Ay By + Az Bz
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Is the scalar (“dot”) product
of the two vectors
1.
2.
3.
4.
positive
negative
equal to zero
“Can’t tell.”
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Is the scalar (“dot”) product
of the two vectors
1.
2.
3.
4.
positive
negative
equal to zero
“Can’t tell.”
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Two identical blocks slide down two frictionless ramps.
Both blocks start from the same height, but block A is on
a steeper incline than block B.
The speed of block A at the bottom of its ramp is
1.
2.
3.
4.
less than the speed of block B.
equal to the speed of block B.
greater than the speed of block B.
“Can’t tell.”
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Solution
• Which forces do work on block?
• Which, if any, are constant?
• What is F•Ds for motion?
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Work done by gravity
Work W = -mg j•Ds
N
Ds
Therefore,
W = -mgDh
mg
j
N does no work!
i
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Work done on an object by gravity
W (on object by earth) = – m g Dh,
where Dh = hfinal – hinitial is the change in height.
Object moves
Change in
height Dh
Work done
on object
by earth
upward
positive
negative
downward
negative
positive
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Defining gravitational potential energy
W(on obj. by earth) = DK
0 = DK - W(on obj. by earth)
0 = DK + DUg
The change in gravitational potential energy of the objectearth system is just another name for the negative value
of the work done on an object by the earth.
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Curved ramp
Ds =
W = F•Ds =
Work done by gravity between 2
fixed pts does not depend on path
taken!
Work done by gravitational force in moving some
object along any path is independent of the path
depending only on the change in vertical height
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Hot wheels demo
One hot wheels car, car A, rolls down the incline and
travels straight ahead, while the other car, B, goes
through a loop at the bottom of the incline. When the cars
reach the end of their respective tracks, the relative
speeds will be:
1. vA > vB
2. vA < vB
3. vA = vB
4. Can’t tell
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A person carries a book horizontally at constant
speed. The work done on the book by the
person’s hand is
1.
2.
3.
4.
positive
negative
equal to zero
“Can’t tell.”
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A person lifts a book at constant speed. Since
the force exerted on the book by the person’s
hand is in the same direction as the displacement
of the book, the work (W1) done on the book by
the person’s hand is positive.
The work done on the book by the earth is:
1.
2.
3.
4.
negative and equal in absolute value to W1
negative and less in absolute value than W1
positive and equal in absolute value to W1
positive and less in absolute value than W1
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A person lifts a book at constant speed. The
work (W1) done on the book by the person’s
hand is positive.
Work done on the book by the earth:
Net work done on the book:
Change in kinetic energy of the book:
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Work-Kinetic Energy theorem
(final statement)
W1net = Fnet •Ds1 = Kf - Ki
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A person lifts two books (each of mass m) at
constant speed. The work done on the upper
book by the lower book is positive. Its magnitude
is W = m g Ds.
Is there work done on the lower book
by the upper book?
1.
2.
3.
4.
Yes, positive work.
Yes, negative work.
No, zero work.
No, since the lower book does work on the upper book, this is
not a meaningful question.
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A person lifts two books at constant
speed. The work done on the upper
book by the lower book is positive.
Work done on lower book by upper
book:
Net work done on the lower book:
Change in kinetic energy of the lower
book:
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Conservative forces
• If the work done by some force (e.g.
gravity) does not depend on path the
force is called conservative.
• Then gravitational potential energy Ug
only depends on (vertical) position of
object Ug = Ug(h)
• Elastic forces also conservative –
elastic potential energy U = (1/2)kx2...
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Nonconservative forces
• friction, air resistance,...
• Potential energies can only be defined
for conservative forces
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Nonconservative forces
• Can do work, but cannot be represented by a
potential energy function
• Total mechanical energy can now change due to
work done by nonconservative force
• For example, frictional force leads to decrease of
total mechanical energy -- energy converted to
heat, or internal energy
• Total energy = total mech. energy + internal
energy -- is conserved
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Conservation of (mechanical)
energy
• If we are dealing with a potential energy
corresponding to a conservative force
0 = DK+DU
• Or
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K + U = constant
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Conservation of total energy
The total energy of an object or system is said
to be conserved if the sum of all energies
(including those outside of mechanics that have
not yet been discussed) never changes.
This is believed always to be true.
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Demo
Vertical spring and mass with damping
• Damping from air resistance
• Amplitude of oscillations decays
• Oscillations still about x = -xeq
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Many forces
• For a particle which is subject to several
(conservative) forces F1, F2 …
E = (1/2)mv2 + U1 + U2 +… is constant
• Principle called
Conservation of total mechanical energy
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Summary
• Total (mechanical) energy of an isolated
system is constant in time.
• Must be no non-conservative forces
• Must sum over all conservative forces
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A compressed spring fires a ping pong ball vertically
upward. If the spring is compressed by 1 cm initially
the ball reaches a height of 2 m above the spring.
What height would the ball reach if the spring were
compressed by just 0.5 cm? (neglect air resistance)
1. 2 m
2. 1 m
3. 0.5 m
4. we do not have sufficient information to calculate the
new height
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A 5.00-kg package slides 1.50 m down a long ramp that is inclined at 12.0 below the
horizontal. The coefficient of kinetic friction between the package and the ramp is mk =
0.310. Calculate: (a) the work done on the package by friction; (b) the work done on
the package by gravity; (c) the work done on the package by the normal force; (d) the
total work done on the package. (e) If the package has a speed of 2.20 m/s at the top of
the ramp, what is its speed after sliding 1.50 m down the ramp?
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Summary
• Work is defined as dot product of force
with displacement vector
• Each individual force on an object gives
rise to work done
• The kinetic energy only changes if net
work is done on the object, which
requires a net force
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Power
Power = Rate at which work is done
W
Average power =
Dt
W
Inst. power = lim
Dt ®0 Dt
Units of power:
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A sports car accelerates from zero to 30 mph in
1.5 s. How long does it take to accelerate from
zero to 60 mph, assuming the power (=DW / Dt)
of the engine to be constant?
(Neglect losses due to friction and air drag.)
1.
2.
3.
4.
2.25 s
3.0 s
4.5 s
6.0 s
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Power in terms of force and velocity
W F × Ds
Power = =
Dt
Dt
Ds
=F×
= F ×v
Dt
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A locomotive accelerates a train from rest to a
final speed of 40 mph by delivering constant
power. If we assume that there are no losses
due to air drag or friction, the acceleration of the
train (while it is speeding up) is
1.
2.
3.
decreasing
constant
increasing
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A ball is whirled around a horizontal circle at
constant speed.
If air drag forces can be neglected, the power
expended by the hand is:
1.
2.
3.
4.
positive
negative
zero
“Can’t tell.”
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Reading assignment
• Extended objects, center of mass
• Begin Chapter 12 in textbook
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