L7 Kinetic Energy and Work F1819 - UNIVERSE OF ALI OVGUN
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Physics 101 Assist. Prof. Dr. Ali ÖVGÜN EMU Physics Department www.aovgun.com Classroom Lecture 7 Kinetic Energy and Work
Transcript of L7 Kinetic Energy and Work F1819 - UNIVERSE OF ALI OVGUN
Physics 101
Assist. Prof. Dr. Ali ÖVGÜNEMU Physics Department
www.aovgun.com
Classroom Lecture 7 Kinetic Energy and Work
November 25, 2019
Why Energy?q Why do we need a concept of energy?q The energy approach to describing motion is
particularly useful when Newton’s Laws are difficult or impossible to use
q Energy is a scalar quantity. It does not have a direction associated with it
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What is Energy?q Energy is a property of the state of a system,
not a property of individual objects: we have to broaden our view.
q Some forms of energy:n Mechanical:
n Kinetic energy (associated with motion, within system)n Potential energy (associated with position, within system)
n Chemicaln Electromagneticn Nuclear
q Energy is conserved. It can be transferred from one object to another or change in form, but cannot be created or destroyed
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Kinetic Energyq Kinetic Energy is energy associated with the
state of motion of an objectq For an object moving with a speed of v
q SI unit: joule (J)1 joule = 1 J = 1 kg m2/s2
2
21mvKE =
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Why ?221mvKE =
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Work Wq Start with Work “W”
q Work provides a link between force and energyq Work done on an object is transferred to/from itq If W > 0, energy added: “transferred to the
object”q If W < 0, energy taken away: “transferred from
the object”
xFmvmv xD=- 20
2
21
21
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Definition of Work Wq The work, W, done by a constant force on an
object is defined as the product of the component of the force along the direction of displacement and the magnitude of the displacement
n F is the magnitude of the forcen Δ x is the magnitude of the
object’s displacementn q is the angle between
xFW Dº )cos( q
and DF x! !
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Work Unitq This gives no information about
n the time it took for the displacement to occurn the velocity or acceleration of the object
q Work is a scalar quantityq SI Unit
n Newton • meter = Joulen N • m = Jn J = kg • m2 / s2 = ( kg • m / s2 ) • m
xFW Dº )cos( q
xFmvmv D=- )cos(21
21 2
02 q
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Work: + or -?q Work can be positive, negative, or zero. The
sign of the work depends on the direction of the force relative to the displacement
n Work positive: W > 0 if 90°> q > 0°n Work negative: W < 0 if 180°> q > 90°n Work zero: W = 0 if q = 90°n Work maximum if q = 0°n Work minimum if q = 180°
xFW Dº )cos( q
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Example: When Work is Zeroq A man carries a bucket of water
horizontally at constant velocity.q The force does no work on the
bucketq Displacement is horizontalq Force is verticalq cos 90° = 0
xFW Dº )cos( q
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Example: Work Can Be Positive or Negative
q Work is positive when lifting the box
q Work would be negative if lowering the boxn The force would still be upward,
but the displacement would be downward
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Work Done by a Constant Forceq The work W done on a system by
an agent exerting a constant force on the system is the product of the magnitude F of the force, the magnitude Δr of the displacement of the point of application of the force, and cosθ, where θ is the angle between the force and displacement vectors:
qcosrFrFW D=D׺!!
F!
II
F!
IIIr!D
F!
Ir!D
F!
IVr!D
r!D
0=IW
qcosrFWIV D=rFWIII D=
rFWII D-=
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Work and Forceq An Eskimo pulls a sled as shown. The total mass
of the sled is 50.0 kg, and he exerts a force of 1.20 × 102 N on the sled by pulling on the rope. How much work does he do on the sled if θ = 30°and he pulls the sled 5.0 m ?
JmN
xFW
2
2
102.5)0.5)(30)(cos1020.1(
)cos(
´=
´=
D=!
q
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Work Done by Multiple Forcesq If more than one force acts on an object, then
the total work is equal to the algebraic sum of the work done by the individual forces
n Remember work is a scalar, sothis is the algebraic sum
=ånet by individual forcesW W
rFWWWW FNgnet D=++= )cos( q
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Work and Multiple Forcesq Suppose µk = 0.200, How much work done on
the sled by friction, and the net work if θ = 30°and he pulls the sled 5.0 m ?
q
q
sin0sin,
FmgNFmgNF ynet
-=
=+-=
JmNsmkg
xFmgxNxfxfW
kk
kkfric
2
2
2
103.4)0.5)(30sin102.1/8.90.50)(200.0(
)sin()180cos(
´-=
´-
×-=
D--=D-=
D-=D=
!
!
qµµ
JJJ
WWWWW gNfricFnet
0.9000103.4102.5 22
=++´-´=
+++=
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Kinetic Energyq Kinetic energy associated with the motion of
an object
q Scalar quantity with the same unit as workq Work is related to kinetic energy
2
21mvKE =
xFmvmv netD=- 20
2
21
21
net f iW KE KE KE= - = D
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Work-Kinetic Energy Theoremq When work is done by a net force on an object
and the only change in the object is its speed, the work done is equal to the change in the object’s kinetic energyn Speed will increase if work is positiven Speed will decrease if work is negative
net f iW KE KE KE= - = D
20
2
21
21 mvmvWnet -=
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Work and Kinetic Energyq The driver of a 1.00�103 kg car traveling on the interstate at
35.0 m/s slam on his brakes to avoid hitting a second vehicle in front of him, which had come to rest because of congestion ahead. After the breaks are applied, a constant friction force of 8.00�103 N acts on the car. Ignore air resistance. (a) At what minimum distance should the brakes be applied to avoid a collision with the other vehicle? (b) If the distance between the vehicles is initially only 30.0 m, at what speed would the collisions occur?
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Work and Kinetic Energyq (a) We knowq Find the minimum necessary stopping distance
Nfkgmvsmv k33
0 1000.8,1000.1,0,/0.35 ´=´===
233 )/0.35)(1000.1(21)1000.8( smkgxN ´-=D´-
22
21
21
iffricNgfricnet mvmvWWWWW -==++=
202
10 mvxfk -=D-
mx 6.76=D
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Work and Kinetic Energyq (b) We knowq Find the speed at impact.q Write down the work-energy theorem:
Nfkgmvsmv k33
0 1000.8,1000.1,0,/0.35 ´=´===
Nfkgmsmvmx k33
0 1000.8,1000.1,/0.35,0.30 ´=´===D
xfm
vv kf D-=22
02
22
21
21
ifkfricnet mvmvxfWW -=D-==
2233
22 /745)30)(1000.8)(1000.12()/35( smmN
kgsmvf =´
´-=
smvf /3.27=
Work Done By a SpringqSpring force
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xF kx= -
Spring at EquilibriumqF = 0
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Spring Compressed
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Fig. 7.9, p. 173
0lim
ff
ii
xx
x xxx xF x F dx
D ®D =å ò
max
0 212
f f
i i
x x
xx x
x
W F dx kxdx
kxdx kx-
= = -
= - =
ò ò
ò2 21 1
2 2f
i
x
i fxW kxdx kx kx= - = -ò
Work done byspring on block
Measuring Spring Constantq Start with spring at its
natural equilibrium length.q Hang a mass on spring and
let it hang to distance d (stationary)
q From
so can get spring constant. November 25, 2019
0xF kx mgmgkd
= - =
=
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Scalar (Dot) Product of 2 Vectors
q The scalar product of two vectors is written as n It is also called the dot
productq
n q is the angle between A and B
q Applied to work, this means
×A B! !
A B cos q× ºA B! !
cosW F r q= D = ×DF r! !
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Dot Productq The dot product says
something about how parallel two vectors are.
q The dot product (scalar product) of two vectors can be thought of as the projection of one onto the direction of the other.
q ComponentsA!
xAAiA
ABBA
==×
=×
q
q
cosˆcos
!
!!B!
zzyyxx BABABABA ++=×!!
qBA )cos( q
)cos( qBA
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Projection of a Vector: Dot Product
q The dot product says something about how parallel two vectors are.
q The dot product (scalar product) of two vectors can be thought of as the projection of one onto the direction of the other.
q ComponentsA!
B!
zzyyxx BABABABA ++=×!!
p/2
Projection is zero
xAAiA
ABBA
==×
=×
q
q
cosˆcos
!
!!
1ˆˆ ;1ˆˆ ;1ˆˆ0ˆˆ ;0ˆˆ ;0ˆˆ
=×=×=×
=×=×=×
kkjjii
kjkiji
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Derivationq How do we show that ?q Start with
q Then
q But
q So
zzyyxx BABABABA ++=×!!
kBjBiBB
kAjAiAA
zyx
zyx
ˆˆˆ
ˆˆˆ
++=
++=!
!
)ˆˆˆ(ˆ)ˆˆˆ(ˆ)ˆˆˆ(ˆ
)ˆˆˆ()ˆˆˆ(
kBjBiBkAkBjBiBjAkBjBiBiA
kBjBiBkAjAiABA
zyxzzyxyzyxx
zyxzyx
++×+++×+++×=
++×++=×!!
1ˆˆ ;1ˆˆ ;1ˆˆ0ˆˆ ;0ˆˆ ;0ˆˆ
=×=×=×
=×=×=×
kkjjii
kjkiji
zzyyxx
zzyyxx
BABABAkBkAjBjAiBiABA
++=
×+×+×=× ˆˆˆˆˆˆ!!
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Scalar Productq The vectorsq Determine the scalar product
q Find the angle θ between these two vectors
ˆ2ˆ and ˆ3ˆ2 jiBjiA +-=+=!!
?=×BA!!
!
""
3.60654cos
654
5134cos
52)1( 1332
1
22222222
==
==×
=
=+-=+==+=+=
-q
qABBA
BBBAAA yxyx
46-223(-1)2 =+=×+×=+=× yyxx BABABA!!
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