9/17/2013phy 113 c fall 2013 -- lecture 71 phy 113 c general physics i 11 am – 12:15 pm mwf olin...
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![Page 1: 9/17/2013PHY 113 C Fall 2013 -- Lecture 71 PHY 113 C General Physics I 11 AM – 12:15 PM MWF Olin 101 Plan for Lecture 7: Chapter 7 -- The notion of work](https://reader034.vdocuments.net/reader034/viewer/2022051819/5516ac21550346a25b8b5894/html5/thumbnails/1.jpg)
PHY 113 C Fall 2013 -- Lecture 7 19/17/2013
PHY 113 C General Physics I11 AM – 12:15 PM MWF Olin 101
Plan for Lecture 7:
Chapter 7 -- The notion of work and energy1. Definition of work2. Examples of work3. Kinetic energy; Work-kinetic energy
theorem4. Potential energy and work; conservative
forces
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PHY 113 C Fall 2013 -- Lecture 7 29/17/2013
7.3,7.15,7.31,7.34
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PHY 113 C Fall 2013 -- Lecture 7 39/17/2013
Webassign questions for Assignment 6 -- #1
52. Consider a large truck carrying a heavy load, such as steel beams. … Assume that a 10,000-kg load sits on the flatbed of a 20,000-kg truck initially moving at vi=12 m/s. Assume that the load on the truck bed has a coefficient of static friction of mS=0.5. When the truck is braked at constant force, it comes to rest in a distance d. What is the minimum stopping distance d such that the load remains stationary relative to the truck bed throughout the breaking?
via
f
m
mgnfmaf SS if
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PHY 113 C Fall 2013 -- Lecture 7 49/17/2013
Webassign questions for Assignment 6 -- #1 -- continued
mgnfamf SS if
via m
f
gamgam SS or
iclicker exercise --Do we have enough information to calculate a?
A. YesB. No
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PHY 113 C Fall 2013 -- Lecture 7 59/17/2013
Webassign questions for Assignment 6 -- #1 -- continued
22
2
2 :algebra someAfter
2
1
:on acceleraticonstant For
ii
ii
i
vtvxtxa
attvxtx
atvtv
a
via m
f
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PHY 113 C Fall 2013 -- Lecture 7 69/17/2013
A block of mass 3 kg is pushed up against a wall by a force P that makes an angle of q=50o
with the horizontal. ms=0.25. Determine the possible values for the magnitude of P that allow the block to remain stationary.
0cos :forces Horizontal
0sin :forces Vertical s
NP
mgPN
f
mg
N
f
Webassign questions for Assignment 6 -- #4
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PHY 113 C Fall 2013 -- Lecture 7 79/17/2013
0cos :forces Horizontal
0sin :forces Vertical s
NP
mgPN
f
mg
N
f
N 48.57
N 72.31
50cos25.050sin
N 8.93
cossin :for Solving
0sincos
cos
s
s
ooP
mgPP
mgPP
PN
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PHY 113 C Fall 2013 -- Lecture 7 89/17/2013
Webassign questions for Assignment 6 -- #6
F
ra
aF
ˆ :case In this2
r
v
m
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PHY 113 C Fall 2013 -- Lecture 7 99/17/2013
Preparation for the introduction of work:
Digression on the definition of vector “dot” product
q
0 then ,90 if that Note
cos
BA
BAo
AB
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PHY 113 C Fall 2013 -- Lecture 7 109/17/2013
Digression: definition of vector “dot” product -- continued
q
(scalar) 5.37120cos)15)(5(
120 ,15 ,5 :Example
cos
o
o
BA
BA
BA
AB
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PHY 113 C Fall 2013 -- Lecture 7 119/17/2013
Digression: definition of vector “dot” product – component form
q
yyxx
yxyx
BABA
BBAA
BA
jiBjiA ˆˆ and ˆˆ Suppose
Note that the result of a vector dot product is a scalar.
103412
ˆ31̂ and ˆ4ˆ2 :Example
BA
jiBjiA
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PHY 113 C Fall 2013 -- Lecture 7 129/17/2013
rFr
rdW
f
i
fi
Definition of work: F
dr
ri rj
cal 0.239 J 1
JoulesmetersNewtons Work
: workof Units
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PHY 113 C Fall 2013 -- Lecture 7 139/17/2013
Units of work:
work = force · displacement = (N · m) = (joule)
• Only the component of force in the direction of the displacement contributes to work.
• Work is a scalar quantity.
• If the force is not constant, the integral form must be used.
• Work can be defined for a specific force or for a combination of forces
rFr
rdW
f
i
11 rFr
rdW
f
i
22 212121 )( WWdWf
i
rFFr
r
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PHY 113 C Fall 2013 -- Lecture 7 149/17/2013
iclicker question: A ball with a weight of 5 N follows the trajectory shown. What is the work done by gravity from the initial ri to final displacement rf?
(A) 0 J (B) 7.5 J (C) 12.5 J (D) 50 J
1m1m
2.5m
ri
rf
10 m
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PHY 113 C Fall 2013 -- Lecture 7 159/17/2013
mg
ri
rf
W=-mg(rf-ri)<0
mg
ri
rf
W=-mg(rf-ri)>0
Gravity does negative work:
Gravity does positive work:
0r 0r
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PHY 113 C Fall 2013 -- Lecture 7 169/17/2013
Work done by a variable force: rFr
rdW
f
i
fi
f
i
f
i
x
x
xfi dxFdW rFr
r
:case In this
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PHY 113 C Fall 2013 -- Lecture 7 179/17/2013
Example:
JmNmNdxFdWf
i
f
i
x
x
xfi 2525)4)(5( 21 rF
r
r
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PHY 113 C Fall 2013 -- Lecture 7 189/17/2013
Example – spring force: Fx = - kx
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PHY 113 C Fall 2013 -- Lecture 7 199/17/2013
x
F
Positive work
Negative work
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PHY 113 C Fall 2013 -- Lecture 7 209/17/2013
Detail:
22212
21 if
x
x
x
x
x
x
xfi
xxkkx
dxkxdxFdW
f
i
f
i
f
i
f
i
rFr
r
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PHY 113 C Fall 2013 -- Lecture 7 219/17/2013
More examples:
Suppose a rope lifts a weight of 1000N by 0.5m at a constant upward velocity of 4.9m/s. How much work is done by the rope?
(A) 500 J (B) 750 J (C) 4900 J (D) None of these
Suppose a rope lifts a weight of 1000N by 0.5m at a constant upward acceleration of 4.9m/s2. How much work is done by the rope?
(A) 500 J (B) 750 J (C) 4900 J (D) None of these
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PHY 113 C Fall 2013 -- Lecture 7 229/17/2013
FP
q
mg
n
fk
xi xf
Assume FP sin q <<mg
Work of gravity? 0
Work of FP? FP cos q (xf-xi)
Work of fk?-mkn (xf-xi)=-mk(mg- FP sin ) q (xf-xi)
Another example
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PHY 113 C Fall 2013 -- Lecture 7 239/17/2013
iclicker exercise:Why should we define work?
A. Because professor like to torture students.
B. Because it is always good to do workC. Because it will help us understand
motion. D. Because it will help us solve the energy
crisis.
Work-Kinetic energy theorem.
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PHY 113 C Fall 2013 -- Lecture 7 249/17/2013
Back to work:
F
dr
ri rj
rFr
rdW
f
i
fi
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PHY 113 C Fall 2013 -- Lecture 7 259/17/2013
Why is work a useful concept?
Consider Newton’s second law:
Ftotal = m a Ftotal · dr= m a · dr
dtdtd
mdtdtd
dtd
mddtd
mdmdf
i
f
i
f
i
f
i
f
i
total vvrv
rv
rarFr
r
r
r
r
r
r
r
r
r
Wtotal = ½ m vf2 - ½ m vi
2
Kinetic energy (joules)
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PHY 113 C Fall 2013 -- Lecture 7 269/17/2013
Introduction of the notion of Kinetic energy
Some more details:
Consider Newton’s second law:
Ftotal = m a Ftotal · dr= m a · dr
dtdtd
mdtdtd
dtd
mddtd
mdmdf
i
f
i
f
i
f
i
f
i
t
t
t
ttotal v
vrvr
vrarF
r
r
r
r
r
r
Wtotal = ½ m vf2 - ½ m vi
2
Kinetic energy (joules)
22
21
21
21
if
f
i
t
tmvmvmddmdt
dtd
mf
i
f
i
vvvvv
v v
v
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PHY 113 C Fall 2013 -- Lecture 7 279/17/2013
Kinetic energy: K = ½ m v2
units: (kg) (m/s)2 = (kg m/s2) m
N m = joules
Work – kinetic energy relation:
Wtotal = Kf – Ki
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PHY 113 C Fall 2013 -- Lecture 7 289/17/2013
Kinetic Energy-Work theorem
22
2
1
2
1if
f
i
totalfi mvmvdW rF
iclicker exercise:Does this remind you of something you’ve seen recently?A. YesB. No
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PHY 113 C Fall 2013 -- Lecture 7 299/17/2013
Kinetic Energy-Work theorem
22
2
1
2
1if
f
i
totalfi mvmvdW rF
222
:constant is if :Note
ifif
ififtotal
f
i
totalfi
total
vv
mdW
rra
rrarrFrF
F
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PHY 113 C Fall 2013 -- Lecture 7 309/17/2013
Kinetic Energy-Work theorem
22
2
1
2
1if
f
i
totalfi mvmvdW rF
Example: A ball of mass 10 kg, initially at rest falls a height of 5m. What is its final velocity?
i
f
h
??fv
0iv
22
2
1
2
1iffi mvmvmghW
0
smmghv f /899.9)5)(8.9)(2(2
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PHY 113 C Fall 2013 -- Lecture 7 319/17/2013
ExampleA block, initially at rest at a height h, slides down a frictionless incline. What is its final velocity?
hh=0.5m
22
2
1
2
1iffi mvmvmghW
0
smmghv f /13.3)5.0)(8.9)(2(2
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PHY 113 C Fall 2013 -- Lecture 7 329/17/2013
ExampleA block of mass m slides on a horizontal surface with initial velocity vi, coming to rest in a distance d.
vi vf=0
d
1. Determine the work done during this process.2. Analyze the work in terms of the kinetic friction
force.
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PHY 113 C Fall 2013 -- Lecture 7 339/17/2013
Example -- continuedvi vf=0
d
222
2
1
2
1
2
1iiffi mvmvmvW
f
gd
vmgdmv
mgdfdfdxdW
ikki
k
x
x
fi
f
i
f
i
2or
2
1
22
rFr
r
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PHY 113 C Fall 2013 -- Lecture 7 349/17/2013
Example A mass m initially at rest and attached to a spring compressed a distance x=-|xi|, slides on a frictionless surface. What is the velocity of the mass when x=0 ?
k
smv
mxmNkkgm
xm
kv
mvmvkxW
f
i
if
ififi
/63.02.05.0
5
2.0 /5 5.0For
2
1
2
1
2
1 222
0
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PHY 113 C Fall 2013 -- Lecture 7 359/17/2013
Special case of “conservative” forces conservative non-dissipative
if
f
i
fi UUdW rrrF
F
write topossible isit , forces edissipativ-nonFor
iiff
ifif
f
i
fi
mgyUmgyU
mgymgyyymgdW
)( and )(
:Earth of surfacenear gravity of Example
rr
rF
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PHY 113 C Fall 2013 -- Lecture 7 369/17/2013
2
212
21
2212
21
)( and )(
:force spring of Example
iiff
if
x
x
f
i
fi
kxUkxU
kxkxdxxkdWf
i
rr
rF
k
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PHY 113 C Fall 2013 -- Lecture 7 379/17/2013
iclicker exercise:Why would you want to write the work as the difference between two “potential” energies?
A. Normal people wouldn’t.B. It shows a lack of imagination.C. It shows that the work depends only on the
initial and final displacements, not on the details of the path.
dx
dUF
dU
x
ref
that Note
:functionenergy potential Define
rFrr
r
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PHY 113 C Fall 2013 -- Lecture 7 389/17/2013
(constant) 2
1
2
1
:Or2
1
2
1
22
22
EUmvUmv
mvmvUUW
iiff
ififfi
rr
rr
Work-Kinetic Energy Theorem for conservative forces:
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PHY 113 C Fall 2013 -- Lecture 7 399/17/2013
Energy diagrams 22
2
1
2
1kxmvE
2max2
1221
2max2
1
max
,0(0)
:0 when :Note
,0
: when :Note
kxmvU
x
kxEv
xx
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PHY 113 C Fall 2013 -- Lecture 7 409/17/2013
Example: Model potential energy function U(x) representing the attraction of two atoms