Download Slide lecture for chapter 7

Chapter 7: Energy of a System
Chapter 7 Goals:
• To learn what constitutes a system of bodies
• To define the work done by a force when a body moves
• To generalize this definition to higher dimensions and/or
position-varying forces and/or curvy paths
• To define the kinetic energy and its changes, as related
to the net force
• To understand the circumstances in which a force
possesses a potential energy, and to express it
• To introduce the notion of mechanical energy
conservation
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What is a System?
• may be a single body, and all the forces that act on the body
• they are dubbed external forces
• may be a collection of bodies, and all the third-law pairs
that act on the collection, and all external forces that act on
the collection
• the part of the universe that is exterior to the boundary is
the environment
• external forces get in through the boundary
• other stuff cross the boundary: mass, heat, charge…
• a system is often a rigid (or not so rigid) extended body
• The universe = system  environment
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Four increasingly complex scenarios involving
the combination of force and motion
• 1d constant force, motion in 1d
• 1d varying force, motion in 1d
• vector constant force, vector displacement
• finally, a varying vector force with motion along a curvy
path
• sometimes, the path may be regarded as merely the
displacement: but only if the force is conservative!!
• a conservative force is one for which the work is pathindependent!!
• note that work is done in a process
• if the force is the net force, cool things can be said:
kinetic energy can be defined and utilized!!
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Scenario the first
• 1d constant force, motion in 1d
• Work in moving from xi to xf under the influence of F:
W [i  f ] : Fx  "force times distance"
Scenario the Second
• consider a force that varies with a body’s position F(x)
• if the body moves a small displacement dx, so small that
F(x) doesn’t change much, in the same (or opposite)
direction as the force, the small bit of work done by the
force is
dW := F(x) dx
• this is of course the (signed) area of the thin strip of
width dx and height F(x) on a graph of F versus x
• for a non-small displacement (from xi to xf), the work
done by the force is the integral of dW (area under F(x)):
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Work as an integral in 1d
xf

W [ xi  x f ] : F ( x) dx
xi
• the work is the signed
area unde r the graph of
F(x)
• could be positive,
negative or zero
• note that if F(x) is a
constant (first scenario)
then ‘force times
distance’ is the area of a
simple rectangle
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The definition of Kinetic Energy,
and the Work-Energy Theorem
xf
• if it’s the net force we get Wnet [i  f ] :
 Fnet ( x) dx
xi
• Since Fnet = ma = m dv/dt we have (watch closely!!)
xf
Wnet [i  f ] :

xi
xf
Fnet ( x) dx  m

xi
xf
xf
2 f
xi
xi
i
dv
dx
mv
dx  m
dv  m v dv 
dt
dt
2


1 2
• Define Kinetic Energy of a body K : mv
2
• Thus we have the Work-Energy Theorem: the work
done on a body by the net force that acts in some
process is equal to the change in kinetic energy of the
body during that process, or Wnet = K:= Kf - Ki
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Units of Work and Energy
• [W] = N-m = kg-m2/s2 = Joule (J)
• If a force of 1 N acts on a body, and the body moves in
the direction of the force a distance of 1 m, the force does
1 J of work (pretty small…)
• If a force of 1 dyne (= 1 g-cm/s2) acts, and the body
moves1 cm, the force does 1 erg of work (really small…)
• If a force of 1 lb… moves 1 ft… 1 foot-pound of work
• 1 calorie (cal) = 4.186 J [mechanical equivalent of heat]
• 1 British Thermal Unit (BTU) = amount of heat needed
to raise 1 lb of water 1 °F = 1055 J
• 1 kcal = 1 Cal = 1 food calorie
• 1 electron-Volt (ev) = 1.602x10-19 J
• 1 kilowatt-hr = another common energy unit…
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Scenario the Third
• consider a constant vector force that acts on a body, F
• the force does not vary with a body’s position r
• let the body move a vector displacement r:=rf –ri
• done by the force is W[if] := F •r = |F||r| cos q
where q is the angle between F and r when arranged
tail-to-tail
• thus the work is a scalar, no matter what
• the more the motion is ‘with’ the force, the bigger the
work done by the force
• if the motion is perpendicular to the force, NO work!
• if the motion is ‘against’ the force, then W < 0
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Reminder of scalar product facts
• dot product can be thought of as magnitude of one, times
the component of the other along the line of the first
• in cartesian coordinates, A•B = AxBx + AyBy + AzBz
• for any vector A , A•A = A2
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Here the forces are constant. Which ones do
positive work? Which negative work? Which
none?
W  F  r  ' force dotted into displacement'
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Scenario the Fourth
• consider a vector force that acts on a body, F(r)
• the force varies with body’s position r as it moves along
its path, so the answer may well depend on the path
• if the body moves a small vector displacement dr, so
small that F doesn’t change much, the small bit of work
done by the force is dW := F ∙dr = |F|| dr| cos q
• for a non-small displacement (from ri to rB), the work
done by the force is the integral of dW
r
f

W [i  f ] : F(r )  dr
r
i
• this is known as a line integral
• it has to ‘follow’ the path
• how can this be thought of as
an area under a graph?
• it is, in 3d, the sum of 3 areas!
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Visualizing the work line integral
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Re-expressing this complex idea
• write the force, and the arbitrary small
diplacement, using components/unit vectors
F  Fx ˆi  Fy ˆj  Fz kˆ and dr  dx ˆi  dy ˆj  dz kˆ
• using the fact that the unit vectors are perpendicular,
only three of the nine possible terms are non-zero:
r
W [i  f ] :
f
r
f
r
f
r
f
r
f
 dW   F(r)  dr   Fx dx   Fy dy   Fz dz
r
i
r
i
r
i
r
i
r
i
• in a sense, there is ‘x-work’ and ‘y-work’ and ‘z-work’
• see calculus III for the details of line integrals
• again, the result may well be path-dependent… but in
special cases it is not: conservative forces
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
The (~same ) definition of Kinetic Energy,
and the Work-Energy Theorem survives!!
r
f
• If the force is the net force we get
• Since Fnet = ma = m dv/dt we have
Wnet
rB
rf
rf
ri
ri
rA

Wnet : Fnet (r )  dr
r
i
dv
dr
 m  dr  m dv 
 m v  dv
dt
dt



r f



 m v x dvx  2 more terms, for y and z 
 ri

m 2
2

v xf  v 2yf  v zf2  v xi
 v 2yi  v zi2
2
m
m
m 2

v f  v f  v i  v i   K where K  v
2
2
2
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


 


Is it possible to ‘unintegrate’ the work, to obtain
the force?
• YES it is possible but only if the force is conservative
• in this situation, we DEFINE the work done by the
force as the negative of the change in a different type of
energy: the potential energy (of the force)
• so the kinetic energy is owned by the body’s mass and
speed, whereas the potential energy is owned by
‘potential energy field’, by virtue of the body’s position
• alternative definition: given a force F(x), and a second
force that ‘you’ exert that exactly opposes it, so Fyou = –
F(x), the change in the potential energy of F(x) is equal
to the work you do in overcoming (barely) F(x), as the
motion proceeds (backward, of course)
• there’s a bit of a pun here as you’ll see..
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How is this expressed mathematically?
• Assume that F(x) is conservative so that the potential
energy exists; call it U(x)
x
x
• ha ha ha
f
f
U : U ( x f )  U ( xi ) :   F ( x) dx   Fyou ( x) dx
x
i
x
i
• we see that since U is basically the integral of F, F must
be the derivative of U (some details need to be clearer)!
dU ( x)
F ( x) : 
where x is a position coordinate ( y, z...)
dx
• in three dimensions the vector force is expressed as the
gradient of the scalar potential energy function
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What are some potential energies in nature?
• example: the force of gravity Fg = − mg
• the 1d coordinate is y, and +y is ↑
yf
yf
yi
yi
U : U ( y f )  U ( yi ) :   (mg ) dy   mg dy  mg ( y f  yi )
• we assume initial y to be 0, and take U(yi) = 0 too
• we assume final y to be just y, so we get U(y)
•  U := Ug(y) = mgy
• so we conclude that the potential energy of an object a
height y above the ‘ground’ is mass x g x height
• check: what is (minus) this function’s space derivative?
Fg ( y)  
dU g ( y)
dy
d (mgy)
dy

 mg
 mg
dy
dy
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A New Force That’s Conservative: The Spring Force
There does NOT need to be a block, yet!!!
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The Wonders of the Spring Force
• it is much like tension or compression, depending
whether the spring is stretched longer than
equilibrium position, or squeezed shorter
• there is a very elegant expression for the force
exerted BY the spring: Hooke’s Law F = – k (x – x0)
• here, k is the spring constant
• x is the location of the movable end of the spring
• x0 is a constant too: the location of the movable
end when the spring is neither stretched nor
squeezed
• often, one takes x0 to be zero, and we will take the
potential energy of this force to be zero for x = x0
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A graph of Hooke’s law F(x)
Stiff spring
F exerted BY spring
Limp spring
x0
• Slope of the graph is –k
• [k] = N/m
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x
What kind of acceleration occurs to a mass I?
• At this instant, the spring is squeezed (x < 0) so its
force is to the right (F > 0)
• Therefore, acceleration of mass is to the right
• If it is moving to the right, it will be speeding up
and fly right through equilibrium, since a = 0 there
• If it is moving to the left, it will be slowing down
and eventually stop, since a is growing and opposed
to the motion {show Active Figure AF_1502}
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What kind of acceleration occurs to a mass II?
• The only horizontal force on the mass is the spring
(ignore up/down forces)
• Fnet = – kx = ma  a = – kx/m
• So the acceleration is in the opposite direction to
the position, and proportional to it!!
d 2x k
 x  0 a differential equation for x(t )
2
m
dt
{show Active Figure AF_1501}
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What would the potential energy be?
xf
xf
xi
xi

U s : U s ( x f )  U s ( xi ) :   (kx) dx  k  x dx  12 k x 2f  xi2

• so this is the change in the potential energy if the
spring end moves from xi to xf
• has nothing to do with the mass except for where it is
• we adjust the thinking now, and integrate from the
reference position x0 (which we have taken to be zero:
the equilibrium spring position) to the present position x:




U s : U s ( x)  U s ( x0 )  12 k x 2  x02  12 k x 2  0 2  12 kx2
 U s ( x)  12 kx2 where we take x0  0 and U s ( x0 )  0
 U s ( x)  12 k ( x  x0 )2 if x0  0 and so U s ( x0 )  0
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Recovering the Force
dU s ( x)
Fs ( x)  
dx
d 1

k ( x  x0 ) 2  k ( x  x0 )
dx 2
The Physics Resides in V(x), if it exists
• potential energies that depend only on position are
conservative
• the classic non-conservative force is friction
• it depends on the state of motion, not merely on
location
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Potential energy example: Spring
• Checking the integral as an area
• For simplicity, take x0 = 0
• The end of the spring
moves from xmax < 0 to 0
• The force of the spring
is positive, and the area
under the curve is too
• Area = ½ (base)(height)
= ½ (–xmax)((– k xmax)
1 2
Area  kxmax
2
1 2
So U   kxmax
2
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Potential energy functions U(x)
dU ( x)
• What information is encoded in U(x)? F ( x)  
dx
• for x > 0, slope of U is
positive, so F is to left
• for x < 0, slope of U is
negative, so F is to right
• A restoring force
• for x = 0, slope of U is zero,
so F is zero: stable equilibrium
point
• turning points at ± xmax,
where E = U K = 0
• unstable equilibrium point:
object tends to ‘fly away’
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Energetics of Simple Harmonic Motion
 x(t )  A sin t    [ x0  0]
1 2 A2
 U  kx 
sin 2 t   
2
2
sin and sin2
 v(t )  A cost   
2
1 2 m 2 A2
kA
 K  mv 
cos 2 t    
cos 2 t   
2
2
2


2
kA2
kA
E  K U 
sin 2 t     cos 2 t    
constant!!
2
2
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