Download A marble moves along the x-axis. Its potential energy function U(x) is

A marble moves along the x-axis. Its
potential energy function U(x) is shown
in the figure.!
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CURVEBALL on the Definition of Conservative
Forces!
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The reverse of this statement is NOT TRUE!!
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Moral: Only if the work done around every possible closed
path is zero could we say that a force is conservative
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(impossible to demonstrate!).!
An example which serves as a warning (i.e. an example where
the converse is false)!
Suppose a variable force acting on an object moving in the
plane is given by:!
= xy xˆ + 2xyˆ
The work done by F in moving it around the
closed loop (0,0)->(3,0)->(3,2)->(0,0) is!
!
But this force is in fact non-conservative: move it
around the unit square: (0,0)->(1,0)->(1,1)->(0,1)->(0,0)!
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Power under Constant Force !
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!!
!!
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power =
Work Done F"s
=
= Fv
"t
"t
(v is mean velocity)
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power =
Work Done F"s
=
= Fv
"t
"t
(v is mean velocity)
Implications:!
(a) Power increases when the velocity of the object
it acts on increases.!
!
(b) Suppose your car maxes out at 550 Horsepower in
your new BMW, by this relationship, the faster you
travel (average velocity), the smaller the force it
generate ---" smaller acceleration -" it is hard to
increase velocity further (i.e., hard to accelerate
more on a high way when you are already fast!) !
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Instantaneous Power!
#-%!.:(.:#+,'-&)#
!!
!!
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"#
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!!L%&#F,'A#9,.&#0?#*%&#+,'-&#(.#*%()#>&'(,9#()M#
"W = F"x cos#
!!
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#!/,6.*#,+#F,'A#9,.&#9(<(9&9#0?#*%&#;/&#(*#*!A&)#*,#9,#(*
#),DDD#
!
P=
"W F"x cos#
=
"t
"t
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K.#*%&#1(/(*#*%!*#*%&#;/&#-%!.:&#)%'(.A)#*,#O&',@#F&#%!<&%
#x
dx
= F cos "
= Fv cos "
#t
dt
dW ! !
Instantaneous Power P =
= F •v
dt
P = F cos"
!
Example: P#!'/?#*!.A#%!)#!#>,F&'#,+#QRRR@RRR#$D##K+#*%&#/!))#
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!,+#*%&#*!.A#F%&.#(*#()#*'!<&11(.:#!*#7S"#TU#A/V%,6'###7Q"##WQ#A/V%3%
Solution: (a) 4*!'*#F(*%#!#>,F&'#&56!;,.#F%(-%#()#'&1!*&9#*,#+,'-&%
P = Fv cos0
36 km "1000 m/sec
= 10F
1 hour " 3600 sec/hour
F = 200,000 N
2000000 = F "
N,F#6)&#N&F*,.#KK#*,#),1<&#+,'#!--&1&'!;,.%
!
F = ma !
200,000 = 100,000 " a
a = 2 m/s2
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!
(b) G!''?#,6*#)(/(1!'#-!1-61!;,.@#+,'#<XWQ#A/V%@##F&#:&*#!XS#/V)
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!
!
!
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We should be thankful that friction keeps
us from getting that speeding ticket! !
Why is Power a limiting factor???!
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Kinetic Energy =
!!
1 2
mv
2
Q
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G,.)(9&'#*%&#F,'A#9,.&#0?#*%&#+,'-&#
!!
!
Work Done W =
W =m
!
(v + v 0 )
(v " v 0 )
2
1 2 1
1
2
mv " mv 0 = m(v + v 0 )(v " v 0 )
2
2
2
W = mv "v
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What does the following equation tell us?!
W = mv "v
Suppose mass of a remote-control toy car is 1 kg, moving
with an average speed of 10 m/s on the road, to make a
2 m/s change in speed, what is the required work by the
engine? What happens when the average speed is 20 m/s?
Case 1!
!# 2 = 20J
W = mv "v = 1#10
Case 2!
W = mv "v = 1# 20 # 2 = 40J
A lot more work needs to spent to bring the same speed
change
when the toy car moves with faster average speed.!
!
Power is work over time, so need more power as well!!
!
Energy Conversion: Energy cannot be created or destroyed
in a closed system (careful: mechanical energy of an object
can be lost due to frictional heating), but can be converted
from one form to another.!
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The fastest steam engine in the world is the
Mallard which achieved a speed of 126mph (202
km/h) on 3rd July 1938 over a 5000 m stretch
of track!
!!
!!
The current electric trains operating on the same line
today have a maximum speed of 225 km/h! !
Based on the following info, what was the power
of the engine and the amount of coal burnt
during the speed trial?!
!!
!!
!!
!!
Coal contains 30MJ/kg chemical
energy!
The burning and heat extraction
efficiencies were 80% and 50%
respectively!
The six coaches pulled gave a
frictional force of 10000 N!
-]%
!
Power of the engine:!
1000
= 561 kW
3600
!!To calculate the coal burned we need to know
the energy required!
!! Work = Power x time !
5000
W = 561,000 "
" 3600 = 50 MJ
202,000
Now the conversion efficiency of the heat energy into
mechanical work is 50% !
50MJ
Heat
Energy
Needed
=
= 100 MJ
!
50%
The conversion efficiency of coal’s chemical energy into
heat is 80% !
100MJ
Chem Energy Needed =
= 125 MJ
80%
! So the total number of kg of coal needed:!
125MJ
Chem Energy Needed =
= 4.2 kg
--%
30MJ/kg
!
Power = Force " velocity = 10,000 " 202 "
!
!
Forces and Potential Energy in 2D & 3D
Partial derivatives
(Math interlude)
(and the differential criterion for determining whether a 2D/3D force is conservative)
Suppose we have a function U of two variables x and y. Then we
can ask “how fast” does U change in the x-direction and how fast
does U change in the y-direction. So there are two natural rates of
change, or two partial derivatives we can take:
In practice, we compute partial derivatives exactly as you would an ordinary
derivative except that you treat every variable as a constant except the one you are
differentiating with respect to.
"U
= 2# y
"y
1)
Examples
2)
3)
!
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