presentation chapter6
TRANSCRIPT
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CHAPTER 6
WORK and KINETIC ENERGY
Work Done by a Constant Force
Work Done by a Variable Force
Work- KineticEnergy Theorem
Center of Mass Work
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The energy is one of the most important concept that helps us describe
many processes in the world around us.
Falling water releases stored gravitational potential energy turninginto a kinetic energy of motion. This mechanical energy can be used
to spin turbines and alternators doing work to generate electrical
energy.
Human beings transform the stored chemical energy of food intovarious forms necessary for the maintenance of the functions of the
various organ system, tissues and cells in the body.Burning gasoline in car engines converts chemical energy stored in
the atomic bonds of the constituent atoms of gasoline into heat that
then drives a piston of combustion engine.Stretching or compressing a spring stores elastic potential energy that
can be released as kinetic energy.
The process of vision begins with stored atomic energy released as
electromagnetic radiation (light) that is detected by exciting atoms in theeye.
In physics Work it is transfer energy by force.
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WORK DONE BY CONSTANT FORCE
In physics Work it is transfer energy by force.
The Work is scalar value and can be positive negative and zero.
The work done by constant force is equals to the component force in the
direction of the displacement times the magnitude of the displacement.
Or another definition Work done by constant force is scalar (or dot)product of force vector and displacement vector.
coslFlFW
Assuming risinx direction.
xFxFW x cos
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The SI unit of work is thejoule (J), which is equals
The product of Newton and meter.
1J = 1N*m
In the U.S. customary system, the unit of work is
foot-pound.
1ft*lb = 1.356J
Unit of work that use in nuclear and quantum physics.
is electron volt (eV).
1 eV = 1.62*10^-19J
For the multiple forces, that act on object we will have.
.....332211 xFxFxFWtotal
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When force Fnetapplied to object and object move under influence
of this force with acceleration afrom A to B along the x axis.
The physical value Kmv 22 is scalar represented energy associated
with motion and call Kinetic Energy and
Work Kinetic Energy Theorem Kmvmv
W if
22
22
Completed work is equal change in kinetic energy of object
222
122
2 AB
v
v
v
v
B
A
AB
net
B
A
netAB
mvmvmvdvvmdtv
dt
dvmW
dt
dxv
dt
dvmmaFdxFW
B
A
B
A
AB
dlFldFdW cos
ABAB KKW
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WORK IN SCALAR (or DOT) PRODUCT NOTATION
The workdWdone by forceFfor displacement dlis.
dlFldFdW cos
2
1
ldFW
Definition of Work
cosABBA
Definition for scalar or dot product of two vectors.zzyyxxzyxzyx BABABAkBjBiBkAjAiABA
dzFdyFdxFkdzjdyi dxkFjFi FldFdW zyxzyx
2222
21
22
21
22
21
22
21
22
2
1
2
1
2
1
2
1
2
1
2
1
vvmvvmvvmvvm
dvmvdvmvdvmvdzFdyFdxFW
zzyyxx
zzyyxxzyx
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kBjBiBkAjAiABA zyxzyx
02cos
10cos
jijiiiiizzyyxx BABABABA
Component of vector in specific direction
of unit vector:
xzyx AikAjAiAiA )(
Same rules is applied to sum of vectors
xxx
CBAiCiBACBA )(
dt
BdAB
dt
AdBA
dt
d
The rules for differentiating
Dot Product of Two Vectors
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Negative work means
A the kinetic energy of the object increases.
B the applied force is variable.
C the applied force is perpendicular to the displacement.
D the applied force is opposite to the displacement.
E nothing; there is no such thing as negative work.
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Two athletes numbers 1 and 2 run a race in beginning they have same
kinetic energy, but athlete #2 is running faster than the 1st. When athlete
#1 increase speed by 25 percent, he is running at the same speed as theathlete #2. If mass athlete #1 is 85 kg, what is mass athlete #2?
22221
21121 vmvm
Using the definition of kinetic energy2
2
112
v
vmm
Express the condition on speed that enables to run athlete #1 at the
same speed as the athlete #2: v2 = 1.25v1
Substitute forv2 and simplify to obtain:2
1
2
1
112
25.1
1
25.1
m
v
vmm
kgkg 5425.1
185
2
2
m
Conceptual problem:
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Work Done by Variable Force
Straight Line Motion
Work Done by Variable Force StraightLine Motion
2
1
x
x
xdxFW
This is area under curveFversusx
i
ixix
xFlimWi
0
xFW x
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A 3.0-kg object moving along thex axis has a velocity of +2.0 m/s as
it passes through the origin. It is subjected to a single force,
that varies with position. How much work is done by
the force as the object moves fromx = 0.0 m tox = 4.0 m? What isthe speed of the object when it is atx = 4.0 m?
xkNFx sin3
mk 3
The force and displacement are parallel, the work done is the area
under the curve.
JkxkN
dxxkNdxFW x 3.4cos3
sin3
4
0
4
0
4
0
According to the work-kinetic energy theorem:
if KKW if KWK
smvm
Wv
mvWmv
if
if
/.622
22
2
22
What work is done by the force if object move 6m?
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Work Done by Spring
Hookes Law
kxFx
The force varies withx.
Work by spring force is.
22)(
21
22
2
1
2
1
xxkdxkxdxFW
x
x
x
x x
22
22
21
fikxkx
AAW
We can use geometry to calculate area
under the curve
22
22fi
kxkxW
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One end of a light spring (force constant k) is attached to the ceiling,
the other end is attached to an object of mass m. The spring initially
is vertical and unstressed. You then ease the object down to an
equilibrium position a distance h below its initial position. Next, yourepeat this experiment, but instead of easing the object down, you
release it, with the result that it falls a distanceHbelow the initial
position before momentarily stopping. Find h and H.
Applyyy amF
to the object in its equilibrium position:0gs FF
khF s ,g mgF 0mgkh
k
mgh
Apply the work-kinetic energy theorem to the spring-object
system to obtain: KW extnet, or, because the object begins and ends at rest, 0extnet, W
0springbygravityby WW hH 2021 2
kHmgH k
mg
H 2
1
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A block with mass m is pulled up a slope with constant velocity.
Calculate the amount of work done by the force after the block has
moved to a height h.
sin
0sin
mgF
mgFFFmaFgx
This force acts over a distance d. The value of d is fixed by the angle
and the height h: sindh
The work done by the force on the block is given by
mghh
mgdFWF
sin
sin
Notice that work by force on block das not
depend from angle and length of incline
and only function of height h.
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The work done on the block by the gravitational force is given by
mghmgddFW GG sin
The work done on the crate by the normal force N is zerosince N is perpendicular to d.
We conclude that the total work done on the block is given by
00 mghmghWWtotal
gF
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A 6.0-kg block slides 1.5 m down a frictionless incline that makes an
angle of 60 with the horizontal. (a) Find the work done by each force
when the block slides 1.5 m (measured along the incline). (b) What is
the total work done on the block? (c) What is the speed of the block afterit has slid 1.5 m, if it starts from rest? (d) What is its speed after 1.5 m, if
it starts with an initial speed of 2.0 m/s?
The work done by gF
cosggg sFsFWF J7mgsWgF 630cos
The work done bynF cosnnn sFsFWF
90 0WF n
(c) The total work done on the block is the sum of the work done by
JWWWng FFtot
76
(a)
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(c) Apply the work-kinetic energy theorem to the block to obtain:
ifgKKKW
F or, because vi= 0
fg KWF 2f21cos mvmgs
smgsv /5cos2f (d) IfKi 0, equation (1) becomes:
(1)
2
i
2
fif2
1
2
1g
mvmvKKWF
Solving forvf and simplifying yields:
smvgsv /4.5cos22
if
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You have two icy slops.
Assume that the surfaces of objects 1 and 2 of same mass are frictionless.
Which object will have greater final speed 1 or 2?
h Fnmg
1
ghv
mghmvmvmv
W
mghdymgW
mgdyjdyidxjmgdlgmdW
dlFldFdW
WWW
f
fif
total
hg
g
nnn
gntotal
2
222
0)2/cos(
222
0
2
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True or false:
(a) The scalar product cannot have units.
(b) If the scalar product of two nonzero vectors is zero, then they areparallel.
(c) If the scalar product of two nonzero vectors is equal to the
product of their magnitudes, then the two vectors are parallel.
(d) As an object slides up an incline, the sign of the scalar product ofthe force of gravity on it and its displacement is negative.
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(a) False. Work is the scalar product of force and displacement.
(b) False. Because ,cosABBA where is the angle betweenif the scalar product of the vectors is zero, then must be 90
(c) True. If the scalar product of the vectors is equal to the product
of their magnitudes, then must be 0 and the vectors are parallel.
(d) True. Because the angle between the gravitational force and the
displacement of the object is greater than 90, its cosine is negative
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A particle undergoes a displacement . During
this displacement a constant force acts on
the particle. Find (a) the work done by the force, and (b) thecomponent of the force in the direction of the displacement.
jNiNF 43
jmiml 52
Nll
W
l
lFFlFlF
FFlFlF
lll
JmNmNW
lFlFlFW
yx
ll
llFlF
yx
yyxx
6.2
coscos
145423
22
22
Solution:
Problem
a)
b)
lF
C t l bl
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Find the unit vector that is in the same direction as the
kjiA 0.10.10.2
Find the component of the vector kjiA
0.1
0.1
0.2
in the direction of the vector jiB 0.40.3
By definition, the unit vector that is in the same direction as and:A
kji
kjiuA
41.041.082.00.10.10.2
0.10.10.2
222
BBuAA
ofdirection
4.05
4
5
30.10.10.2
ofdirection
jikjiAB
Conceptual problem;
A
Au
A
jiji
B
BuB
5
4
5
3
0.40.3
0.40.3
22
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POWER
In physics, the time rate at which a force does work is called
the Power P.
Or another definition, the Power Pis work performed per unit timeby force.
dtldvwheredtvFldFdW
dt
dWP
where
vFdt
dWP
The SI unit of power is one joule per second (J/s) and called a watt (w).
Power define by formula
The modern pulsing lasers can output more than 1.0 GW of power. A
typical large modern electric generation plant typically produces 1.0 GW
of electrical power. Does this mean the laser outputs a huge amount of
energy? No, the power of laser may only last for a short time interval.
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You lift a package vertically upward a distanceL in time t. You
then lift a second package that has twice the mass of the first
package vertically upward the same distance while providing the
same power as required for the first package. How much timedoes lifting the second package take?
L
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Express the power you exert in lifting the package one meter
in tseconds:
t
W
t
WP
1
1
11
Express the power you develop in lifting a package of twice the mass
one meter in tseconds:
2
1
2
2
2
2
t
W
t
WP
Because you exert the same power in lifting both packages:
2
11
2 t
W
t
W tt 2 2
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A object of mass m moves from rest at t = 0 under the influence
of a single constant force . Find the power delivered by
the force at any time t.
F
Express the rate at which this force does work
The velocity of the object, in terms of its acceleration
tav
:
Using Newtons 2nd law, substitute for a
tm
Ft
m
FFt
m
FFP
2
vFP
Substitute forvin equation (1) and simplify to obtain:
(1)
tm
Fv
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A 7.5-kg box is being lifted by a light rope that is threaded through a
single, light, frictionless pulley that is attached to the ceiling. (a) If
the box is being lifted at a constant speedof 2.0 m/s, what is the
power delivered by the person pulling on the rope? (b) If the box is
lifted, at constant acceleration, from rest on the floor to a height of
1.5 m above the floor in 0.42 s, what average power is delivered by
the person pulling on the rope?
(a) The power exerted by the person pulling on the rope is given by:cosTvvTP T vand are in the same direction, TvP
ymaFT gSecond Newtowns law ay= 0,because 0mgT
kWmgvP 15.0
(b) The average power exerted by the person pulling the rope is given by:
t
yF
t
WP
av 2
21
0 tatvy yy
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or because the box starts from rest, 221 tay
y
2
2
t
yay
2
2
t
ymmaF
y
t
yF
t
W
P
av
kWt
ym
t
yt
ym
P 46.0
2
2
3
22
av
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Let show that the power delivered by the net force acting on
a particle equals the rate at which the kinetic energy of the
particle is changing.
dtdKvFP
mv
dt
dv
dt
dmP
vadt
vd
vvdt
vd
vvdt
d
vdt
d
vamvFP
2)(
2
22
2
2
Rate at which kinetic energy change.
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CENTER OF MASS WORK
In Chapter 5 we learn that for system of objects net force define
Center Mass WorkTranslation Kinetic Energy Relation
transK Translation Kinetic Energy
transcmnetcm
transcmcmcmcmnet
cmi
netinet
KldFW
dt
dKM v
dt
dvaMvF
MaFF
2
1
2
2