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Chapter 8
Conservation of Energy
1
Three types of energy in a system
• Previously we only mentioned about K.E. and PE of a system.
• A system, apart from having KE and PE, could also has
another form of energy, known as internal energy, Eint (I.E.).
Mechanical Energy
system
=
+
+
Total energy of a system = K.E + PE + IE
Environment
2
Internal energy (IE)
• Unlike KE and PE, IE (represented by
the symbol Eint) is the energy
associated with the temperature of an
object.
• Example: if the book is dragged
through a distance of d, work is done
by the frictional force on the book,
and the result is the increase in the
temperature, and hence IE, of the
book.
• Here the work done by the frictional
force is the mechanism that is
responsible to transfer energy into
the book (defined as the system). The
result of such energy transfer into the
system is the increase in the
system’s IE.
• As long as there is a change in
temperature of the system, the
internal energy is changed.
3
Mechanisms to transfer energy across the boundary of
a system
•
•
•
1.
2.
3.
4.
5.
6.
The total energy within an system will change if there is
energy being transferred across the boundary.
The listed mechanism are the ways energy can be transferred
into or out of a system:
Ways (mechanisms) to transfer energy across the boundary
and change the total energy of the system:
work done by forces (we have seen this in previous chapters)
mechanical waves
heat
mass transfer
electrical transmission
electromagnetic radiation
4
Examples of mechanisms to
transfer energy across the
boundary of a system
• a) energy is transferred to the
book by work done by a force.
• b) energy leaves the radio from
the speaker by sound waves.
• c) energy transfer to the handle
by heat due to a temperature
difference.
5
Examples of mechanisms to
transfer energy across the
boundary of a system
• d) Energy enters the car
tank by mass transfer.
• e) energy enters the hair
dryer by electrical
transmission.
• f) energy leaves the light
bulb by electromagnetic
radiation.
6
The significance of interactions
•
•
•
•
•
•
•
One can think of the mechanisms as “interactions” that render energies to
be transferred across the boundary between the system and its
environment.
Can you imagine what will happen to a system where it does not interacts
with its environment?
Without the mechanisms to exchange energy across the system’s boundary,
a system becomes an isolated system. In such system the environment will
not be able to ‘influence’ the system.
Exchanging energies via interactions is a generic phenomena in any
system, be it macroscopic or microscopic. Interactions and exchange of
energies make our Universe a lively, diversified, varied and interacting
place.
Without interactions with its environment, a system will remain constant in
energy, and as a result, the changes in the states within the system become
less varied. In some cases this could mean the systems are like ‘dead
systems’ where the physical states never changes (e.g., “heat dead”).
Sometimes this isolation is desired (such as in the qubit system of a
quantum computer), but in other cases, interactions are needed so that the
energies a system can be varied or controlled.
Note that in nature there is no system that is truly isolated by definition,
except, maybe, the Universe itself (?).
7
Conservation of energy (CoE)
• Energy is one of those few conserved quantity
know in our Universe.
• This means energy cannot be created or
destroyed.
• The total energy in a system changes only as a
result energy transfer across the system
boundary via the mechanisms discussed before.
• The above statements about energy
conservation are based on empirical
observations by human from all authentic
experiments ever done in human history.
8
Mathematical statement of energy
conservation
•
•
•
•
•
•
•
•
•
•
SDEs= STt
SDEs refers to the sum of all contributions to the change in all forms of
energies, DEs, within a system.
STt refers to the total energy that are transferred across the systemenvironment boundary via mechanism t.
LHS, SDEs = DK + DU + DEint
RHS, SDTt = W + Q + TMW + TMT + TET + TER
For system which RHS is not all zero, the system is known to be a
nonisolated system; otherwise it is an isolated system.
Note: in the equation SDEs= STt, the LHS is the ‘effect’, the RHS is the
‘cause’.
Only if there is a cause will then be an effect. Or in other words, it’s the cause
that determine the effect, but not the other way round.
If there is not causes, i.e., there is not mechanisms / interactions to perform
energy transfer (the RHS), the total energy (the LHS) would not change.
However, the absence of the LHS does not mean the individual term in the
LHS, DK + DU + DEint, must remain constant.
Separately, DK, DU, DEint can still varies in time but the their sum must
remain zero, i.e., DK + DU + DEint = 0, if the RHS, SDTt = 0
9
Work-KE theorem is a special case
of conservation of energy
• In the special case where the RHS of the
conservation of energy equation, SDEs= STt,
where there the only mechanism present to
transfer energy across the system boundary is
supplied by W (work done by a force) only, and
further, the only form of energy change in the
system is the kinetic energy, DK, the
conservation of energy equation is reduced to
the work-KE theorem:
LHS: S(DEs) = DE = DK
RHS: = STt = W
 DK = W
10
Graphical representation of
conservation of energy
STt = W + Q + TMW
+ TMT + TET + TER
System’s initial total energy
SEs= K + U + Eint
STt
system’s total
energy changes to
System’s final total energy
SEs+ SDEs =
K + U + Eint
+
(DK + DU + DEint)
Environment
System
DE = (DK + DU + DEint)
= STt
11
Nonisolated system (NIS) and
isolated system (IS)
• A nonisolated system (NIS) is one for which
energy crosses the boundary of the system
during some time interval due to an interaction
with the environment.
• By mathematical definition, for NI systems, not
all the terms in the RHS of SDEs= STt, i.e. STt,
are zero.
• Systems in which all terms in STt are zero, are
known as isolated system (IS).
• By the above definition, the work-KE theorem,
which requires STt = W  0, applies only to NIS.
12
Some examples of nonisolated
systems
Environment
The total energy of these NISs changes as
energy crosses the boundary due to interaction Boundary
with the environment.
Fapp
(external
force)
NI System
Environment
Boundary
j
NI System
Boundary
Environment
ra
Book-Earth system
NI System
13
Isolated system
• If a system does not interact with its
environment, (i.e., energy does not
crosses the border of the system due to
the absence of interaction), it’s an isolated
system.
14
Graphical illustration of non-isolated
contained in an isolated system
Isolated system. This is
the environment to the
non-isolated system.
system boundaries
Non isolated
system
“The rest of the universe” is
the environment to the IS
No energy transfer across the
border of the isolated system
with “the rest of the universe”
15
Book in a NIS
•
•
Consider a book and threat it as a NIS
subjected to the external force (the gravitational
force) Fg. Let the book falls from the height yi to
y f.
Transfer of energy across the system’s border
is only contributed by the work done by the
gravity, W:
T  W  Q  T
t
MW
TMT TET TER  W ;
NIS
yi
Fg
Dr  y f  y i  yf j  yi j
t
W  Fg  Dr  mgj  ( yf  yi  j  mgyi  mgy j
•
The change in the total energy of the book
system only happens in the energy forms of its
KE, but not in PE (since the system is a block,
without any internal forces – hence there is no
change in the ‘configuration’ of the system) or
IE (assumed no temperature change in the
block)
 DE
s 1,2,3
•
•
•
•
s
 DK  DU  DEint  DK
Applying CoE SDEs= STt to the book system:
LHS, SDEs= DK
RHS, STt = W = mgyi - mgyf = - (mgyf - mgyi)
Equating both sides, DK + (mgyf - mgyi) = 0
yf
Fg
j
y=0
16
Book-Earth IS
•
•
•
•
•
•
•
Now let us consider the book-Earth system
and treat it as an IS.
In the book-Earth system, a change in KE will
yi
happen when the book falls.
In principle, the change in KE in the bookEarth system receives contribution from both
the book and the Earth:
Dr  y f  y i  yf j  yi j
DKb-E = DKE + DKbook
Since we assume the Earth to be still, DKE= 0,
DKb-E = DKbook= DK
where DKbook is the same DK as in the book
system.
By definition, the potential energy of the bookEarth system is defined as U(y) = mgy.
Hence, mgyf – mgyi = DUg is the change in the
potential energy of the book-Earth system
when the book falls from yb to ya.
From the previous slide, we have the relation
DK + (mgyf - mgyi) = 0
which can be expressed as
DK + DUg = 0
IS
Fg
yf
Fg
j
y=0
17
Interpretation of the statement
DK + DUg = 0
• This is the equation that holds within the isolated bookEarth system, where external forces are absent.
• In such an isolated system, the sum of the changes in
KE and PE is zero.
• Since we define the mechanical energy of a system as
Emech = K + U
DK + DUg = 0  DEmech  DK + DUg = 0
• This is a mathematical statement of the Conservation of
Mechanical Energy (CoME) of an isolated system.
• In a physical process, the total mechanical energy within
an isolated system is conserved and not changing with
time.
Emech = Kf + Uf = Ki+ Ui
18
When Emech is not conserved
• Different ways to express CoME in an isolated
system:
Emech, i = Emech, f
Ki + Ui = Kf + Uf
DEmech = DK + DU = 0
• However, if there are nonconservative forces
acting within the IS system, Emech is not
conserved.
• In that case, DEmech 0, but we have to use
DEsystem = 0 instead (to be discussed later)
19
Example of application of CoME
• In the ball-Earth system,
conservation of mechanical
energy says
Emech = Kf + Uf = Ki+ Ui
• In terms of initial and final
mechanical energies, it reads
• ½mvi+mgyi=½mvf+mgyf
• The ball is released from rest,
at the height h.
• We wish to find the final velocity
of the ball when its
instantaneous height from the
ground is y (see figure)
20
•
•
•
•
•
•
Example of application of CoME
(cont.)
Write down the CoME for the ball-Earth
system:
Ei = Ki +Ui = 0 + mgh
Ef = Kf +Uf = ½ mvf2 + mgy
In writing down the equation, we have
assumed implicitly that the zero
potential energy configuration of the
ball-Earth system corresponds to y = 0.
Equating Ei to Ef :
mgh = ½ mvf2 + mgy
v f  2g ( h  y 
Note that the same answer could also
be obtained by solving the Newton’s
second law of the ball under the
gravitational pull. It should be clear that
applying the CoME method is far more
easier and more economical.
21
CoME in the block-spring system
• We consider the compressed
block-spring system as an
isolated system (frictionless,
and negligible spring mass).
(The hand nor the external
force from the hand on the
block is part of the system)
• When the block is
compressed by xmax from its
relaxed position (x = 0), and
the block is at rest, the total
initial mechanical energy of
the block-spring system is
Ei = Ui + Ki = ½ kxmax2
Active figure 8.4
22
CoME in the block-spring system
• When the compressed spring
is released, the spring
expands back to its relaxed
length, hence the PE of the
spring-block system now
becomes zero, i.e., Uf = 0. But
the block will gain in KE. The
final mechanical energy (at the
configuration x = 0) becomes
Ef = Kf
• By CoME, Ei = Ef,
½k(xmax)2 = ½mvf2
• If xmax, k, m are known, vf the
speed of the block when it
passes through the position x =
0 can be calculated.
23
CoME in the block-spring system
• CoME gives
½k(xmax)2 = ½mvf2
 -DU = DK
• In the block-spring IS, the
potential energy of the
spring-block system (which
is stored in the spring
when it is compressed to
x=xmax) is converted into
the KE of the block.
24
Example: The spring-loaded
popgun (i)
• The spring constant of
the popgun spring is not
known.
• When the spring is
compressed by 0.120 m
and shoot upward, it
sends its projectile of
mass 35 g to a maximum
height 20.0 m above.
Determine the spring
constant by ignoring
frictional effects.
25
Solution
• First choose the system to include
the gun + spring + bullet + Earth.
(note: the choice of system is not
unique – choose according to
convenience.)
Ug=0
• Then choose a reference
configuration where the gravitational
PE of the system is 0. This is chosen
as the configuration in which the
bullet is placed at position A in the
figure (a).
• Note that the system contains two
kinds of PE, Us and Ug.
H
Earth
26
Solution
• Apply CoME to the system for the
“before” and “after”
configurations, of which the bullet
is at A and at C respectively:
Ug=0
• Emech,A = Emech,C
 KA + UgA + UsA = KC + UgC + UsC
LHS: KA + UgA + UsA = UsA = ½kx2
RHS: KC + UgC + UsC = UgC = mgH
LHS = RHS
 k = 2mgH/x2 = …= 953 N/m
H
Earth
27
Example: The spring-loaded
popgun (ii)
• Determine the speed of the ball at the
instant it passes through the point B.
Solution:
Ug=0
• EA = EB
 KA + UgA + UsA = KB + UgB + UsB
LHS:
KA + UgA + UsA = UsA = ½kx2
RHS:
KB + UgB + UsB = KB + UgB = ½mvB2 + mgx
LHS = RHS
½kx2 = mgh + ½mvB2
vB = (kx2/m - 2mgh) ½ = …= 18.7 m/s
Earth
28
CoME in block-spring system
• Consider the block-spring as an IS (frictionless, spring mass ignored,
no arises in temperature). The block attached to the spring will
oscillates between x = -A and x = A about the equilibrium position x = 0,
if the block is displaced from x = 0 by an amplitude A.
• When the block oscillates, the PE stored in the spring and the KE of the
block are exchanged into each other due to the CoME.
• The mechanical energy of the system is conserved. At any position,
CoME of the system reads Emech = Us + Kblock
• (i) At the amplitude positions, i.e. positions where the spring’s
displacement is maximum, x = A: Emech is comprised of Us only as
Kblock = 0 at these ‘turning point’ positions.
• (ii) At the EB position, i.e. positions where the spring displacement is
zero (spring is relaxed), x = 0: Emech is comprised of Kblcok only as Us =
0 at these ‘turning point’ positions. All of the PE of the spring stored at x
= A is transformed into Kblcok at x = 0.
2A
EB
x = -A
x=0
Active figure 8.16
x=A
29
The CoME in the block-spring system is
similar to this one
• In figure 10-13,
• the PE role is played
by the gravitational
PE instead of the
spring PE in the
previous system.
30
Spring-glider system with external
forces
• A glider with mass m =
0.2 kg on an air track
(frictionless) is attached
to a spring with spring
constant k = 5.00 N/m.
• If the glider is exerted by
a constant force with
magnitude 0.610 N on the
i direction, what is the
velocity of the glider when
it passes through x = 0.08
m? (x is the displacement
measured from the EB
position)
Fapp = 0.61N
+i
31
Solution
• We choose the spring-glider to be a
NI system.
• Assume no change in Ein of the
system in the process.
• Apply CoE: SDE = SWi
SDE = SD(U+K+Ein)=DEmech= DU + DK
SWi = WF  0 (work done by external
force only).
• The work done by the external force
WF on the spring-glider system will
will change the PE and KE of the
system according to:
DEmech = DK + DU = WF
where WF = Fappd = +Fappd
• Here, Emech of the system is not
conserved.
Fapp = 0.61N
+i
32
Solution
• At the initial configuration,
Ei = Ki + Ui = 0 + 0 = 0
• At the final configuration (where x
= 0.008 m),
Ef = Kf + Uf = ½mvf2 + ½kx2
• DEmech = E2 - E1 = DK + DU =
(Kf - Ki) + (Uf - Ui) = ½mvf2+½kx2
• Work done by external force,
WF=Fappd
• Equating WF = DEmech
+Fappd =½mvf2 + ½kx2 
vf2  2Fappd/m - kx2/m
vf  …0.67 m/s
Fapp = 0.61N
+i
33
Effect of frictional forces
• We wish to find out how the presence of nonconservative force in an
IS would modify the energy make-up of it. We use kinetic frictional
(KF) force as illustration. In the following analysis, the systems
contain no PE.
• Consider a block in contact with an Earth surface moving initially
with velocity vi. We first take the block as an NIS.
34
Microscopics of KF forces
• Kinetic frictional forces act
between the contact points at the
block-Earth interface. KF,
microscopically, is not a single
constant force but a ‘dynamic’
concept. They continuously break
and form at various contact points
in the interface.
• When the interface moves along
the horizontal direction, the forces
connecting the welder points are
induced. These forces are
generically not in the horizontal or
vertical directions. The sum of
these contact forces on the block
along the directions opposite to the
direction of motion of the block
manifests themselves as kinetic
friction, fk:
fc,1
fc,2
fc,3
fc,5
fc,4
35
KF on the block system (NIS)
• Assume the only external forces acting on the block (NIS) are kinetic
frictional forces.
• The block has initial KE Ki and is moving through a total path length d
along +i. The KF force fk will tend to oppose the motion of the block
since fk on the block is in the –i direction.
• By considering the block as an NIS, the CoE reads:
fk  d   f k  d  DENIS  D ( Eint ,NIS  U  K   DEint ,NIS  DK
DK   f k  d  DEint ,NIS
• Note that the block has no PE since there is no ‘relative configuration’
within the system boundary.
Eint,NIS
E’int,NIS
36
KF on the block-Earth system (IS)
• Now consider the block + Earth as an IS. CoE reads
DEIS  D ( Eint  U  K   DEint,IS  DK  0
 DK  DEint ,IS   ( DEint,NIS  DEint,ENV 
• Note that the block-Earth has no PE as there are no
conservative forces acting within the system.
Eint,ENV
Eint,NIS
37
Work done by friction increases the
internal energy of the environment
•
DK for the block-Earth system is the same as for the
block system. Equating DK in both systems:
DK   ( DEint,ENV  DEint,IS    f k  d  DEint,IS
 DEint,ENV  f k  d  0
• Interpretation: After the frictional force is displaced
through a total distance d, the internal energy of the
‘environment’ (everything but the block) increases
(i.e., the temperature of the Earth and air increase).
This increase in IE is due to the conversion of KE via
the frictional force as an energy transfer mechanism.
38
The most generic case:
Conservation of energy in the
presence of frictional forces
• Generically, in the presence of frictional forces in a NIS
system containing PE, the CoE reads
W  f
i
k
 d  DE  D ( Eint ,NIS  U  K   DEint ,NIS  DEmech
i
where DEmech  D (U  K   DU  DK
• Wi are the energy transfer across the system boundary via
the mechanism i.
• In the special case where there is no change in the Ein,NIS
of the NIS, DEin,NIS = 0, and Wi  0 ,
i
DEmech   f k  d
39
Example: When a crate slides
down a ramp
• A 3.00-kg crate slides down a ramp.
The ramp is 1.00 m in length and
inclined at an angle of 30.0° as
shown in the figure. the crate starts
from rest at the top, experiences a
constant friction force of magnitude
5.00 N, and continues to move a
short distance on the horizontal
floor after it leaves the ramp.
• Use the energy method to
determine the speed of the crate at
the bottom of the ramp.
40
Solution
• Choose an appropriate system: CrateEarth + ramp as a NIS; Gravitational
force is internal force. Frictional force is
considered ‘external force’ to the NIS.
• Assume no change in IE of the NIS,
then, CoE reads:
DEmech= -fk d
• By definition,
DEmech  DK + DU = (Kf – Ki) + (Uf - Ui)
• Uf = 0; Ki = 0; Ui = mgh; Kf = ½mvf2;
• DEmech = ½mvf2 – mgh = -fk d
• vf2 = 2(gh - fk d/m)
• vf = … = 2.54 m/s
NIS
Initial state
final state
41
Solution
• How far does the crate slide
on the horizontal floor if it
continues to experience a
friction force of magnitude
5.00 N?
• Solution
• Apply CoE, but now DU’ = 0.
• DE’mech  DK’ = K’f – K’i = 0 ½mvf2
DE’mech= -fk d’
• -fk d’ = - ½mvf2
• d’ = mvf2 / (2fk ) = … = 1.94 m
Initial state
final state
fk
Ki’ =½mvf2
Kf’ = 0
42
Example: Change of ME in the
presence of KF: Block-spring collision
• A block with a mass 0.80
kg collides with the spring
(with negligible mass) at
a speed of vA=1.2 m/s
from the left. Given that
the spring constant is k =
50 N/m, with a coefficient
of friction between the
block and the surface mk=
0.5.
• What is the maximum
compression in the spring?
43
Solution
•
•
•
Define the initial position of the spring at x
= xi = 0; Final position at xf = xmax
The change in the total mechanical energy
between the initial and final configuration
is DEmech = DK + DU
DEmech = (Kf – Ki) + (Uf – Ui)
= ( 0 - ½mvi2) + (½k xmax2)
= ½kxmax2 - ½mvf2
•
According to conservation of energy, the
frictional force (as an external force) will
cause the total mechanical energy of the
spring-block system (NIS) to change by an
amount -ƒk xmax
•
Equate DEmech = -ƒk xmax
½kxmax2 - ½mvf2 = -ƒk xmax
½kxmax2 + ƒk xmax - ½mvf2 = 0
xmax = … = 0.092 m and -0.25 m
Which answer to choose?
•
44
Connected block
• The system is chosen to be an
NIS consisting of the two
blocks + spring + the Earth +
the string.
• KF force, as an external force,
acts on block when it moves
along the horizontal surface.
• The system starts from the
rest and m2 falls through a
vertical distance of h before
the blocks come to rest.
• Calculate the coefficient of KF.
NIS
45
Solution
•
Here gravitational and the spring’s elastic PE
are involved.
•
The initial total PE of the (NI) system is
Ui = Ui (blok 2) + Ui (blok 1) = 0 + 0 = 0.
The final total PE of the (NI) system is
Uf = Uf (block 2) + Uf (block 1)
= -m2gh + ½kh2
•
•
•
•
•
•
•
•
The change in mechanical energy fo the
system is
DEmech = DK + DU = (Kf – Ki) + (Uf – Ui)
= (0) + (-m2gh + ½kh2 )
 -m2gh + ½kh2
During the process the KF fore is being
dragged through a distance of h. Hence the
work done by the KF on the NIS is
-ƒk h = -mkNh = -mkm1gh
Equate DEmech = -ƒk h
-m2gh + ½kh2  -mkm1gh
mk= m2g - ½kh/m1g
This is a useful way to measure KF
coefficient.
Reference level, Ui = 0
46
Example
• A mass on a horizontal frictionless surface is connected
to a spring, as illustrated in Figure 8a. The mass is
displaced a distance A from its equilibrium position and
released. The speed of the mass at the moment it
crosses the equilibrium position is measured. The same
spring and mass are now hung in the vertical position as
illustrated in Figure 8b, and the mass now sits at a new
equilibrium position, a distance d from the spring’s
equilibrium position, as shown. If the mass is now
displaced a distance A from its new equilibrium position
and released, what is the speed of the mass when it
crosses its new equilibrium position compared to that in
the horizontal case?
47
Figures 8a, 8b
Figure 8a
+j
-j
xi   ( d  A j
48
Solution
• To answer the question, apply work-KE
theorem to each case to find out the KE of
the block at their respective EB position in
the spring systems.
49
The horizontal spring
For the horizontal spring released from maximum displacement, xi
= A to reach xf = 0 at the EB position, CoME reads
DEmech  DK  DU  0
DK  Kf  Ki  2 mvf2  0  2 mvf2
1
1
DU  2 kxf2  2 kxi2  0  2 kA2   2 kA2
1
1
1
1
DK  DU  2 mvf2  2 kA2
1
 vf  A
1
k
m
Figure 8a
50
The vertical spring
• At equilibrium, the vertical
spring obeys the relation
mg = kd
• I.e., the gravitational force
on is balanced by the spring
force on the mass.
+j
-j
xi   ( d  A j
51
The vertical spring (cont.)
• Initially, the vertical spring is
displaced to the initial position xi =
-(A+d)j, and then released. Once
released, the mass will ‘shoot’
upwards. We wish to calculate the
speed when the mass passes through
the position xf = -dj (the EB position
for the vertical spring).
• There are two forces acting on the
block: gravity and spring force. Each
force does work on the block when it
moves from x=xi=-(d+A)j to x=xf=-dj:
xf
Wg 
F
g
xi
-j
xi   ( d  A j
d
xf
 dx 
+j
 mgdx  
mgdx 
 ( d  A
xi
mg  x ( d  A  mg  d  ( d  A   mgA
d
2

Ws  DUs  k ( x  x   k ( d  A  d 2 


1
2
2
i
2
f
1
2
52
The vertical spring (cont.)
W  W
1
1
1
2
2
2



W

k
(
A

d
)

kd

mgA

kA
 kdA  mgA
s
g
2
2
2

 12 kA2 ( kd  mg during EB
DK  K f  Ki  12 mv '2f  0  12 mv '2f
+j
Work-KE theorem:
W  DK 
1
2
kA  mv '
2
1
2
2
f
-j
xi   ( d  A j
k
m
v ' f in the verticle spring is the same
 v'f  A
as that of the horizontal spring when
the mass passes through their respective
EB position.
53
Example
An experiment is performed on a level road with a
car initially traveling at a speed v. The car suddenly
slams on the brakes, skids to a stop, and the
distance of the skid is measured to be 16 m. The
experiment is repeated on the same road with the
same tires at the same initial speed v, but with a car
that is half the mass of the original car. What is the
distance of the skid marks?
54
Solution
• The coefficient of kinetic friction, initial and final speed
are all the same in both cases except the mass of the
car.
• The change in kinetic energy is also the same in both
cases. Use CoE:  f k Dx  DK
2
 f k Dx   mk nDx   mk mg Dx  DK   12 mv 2 , so that Dx 
v
2 mk g
.
• Since the distance of the skid is independent of the mass
of the car, the skid distances will be the same, Dx = 16 m
55
Power
• In the previous discussion we were discussing about the
mechanisms of how energy can be transferred across
the boundary of a system.
• Energy transfer can occur across the boundary of a
system at various rate. The rate at which energy is being
transferred is known as the power.
• In the special case in which the only energy transfer
mechanism is via work done by a force, WF = Fdr, the
mean (average) power of energy transfer via the work
done by the force within an time interval Dt is defined as
WF
P
Dt
56
Instantaneous power
• In the limit Dt 0, the mean power becomes
the instantaneous power:
P
lim
Dt 0
WF dWF

Dt
dt
• For the special case where the force is
constant, and is displaced in the same
direction as the force, i.e., F // r, the
instantaneous power can be expressed as
P
dW d ( F  r )
dr

 F
 F  v  Fv
dt
dt
dt
where v is the velocity of the force being
displaced.
57
Generalisation of the definition of
power
• The definition of power of the work done by a
force can be generalised to other energy
transfer mechanisms.
• Generically, the power due to a given
mechanism is expressed as
dE
P
dt
• where dE/dt is the rate at which energy crosses
the system-environment boundary due to that
particular mechanism.
58
The unit for power
• The SI unit for power is watt.
• 1 watt = 1 joule / second = 1 kg.m2/s2
• It is also common to use the non SI unit
“horsepower ” (hp) : 1 hp = 746 W.
• Example: a National aircon of 1 hp is sold
at the prize RM 950; 1.5 h.p. at RM 1400
(during sale)
• Power can also be used in expressing the
amount of energy or work done:
• 1 kWh = (1000 W)(3600 s) = 3.6 x106 J
59
Example: Power supply by a
motor driving an elevator
• An elevator has a mass of 1600
kg and is taking a number of
passengers with a total mass of
200 kg. A constant frictional
force of 4000 N constantly resists
the motion of the elevator when it
move upwards.
• What is the power that must be
supplied by the motor driving the
elevator so that the elevator can
move upwards at a constant
speed of 3.00 m/s? Assume no
temperature change in the Earthelevator system.
d
s
dy
60
•
•
•
•
•
•
Solution
First, ignoring the frictional force.
As the elevator moves upwards at the
said speed, the motor must supply
energy to the lift + Earth system at a rate
equal to the rate of change of the total
energy of the elevator-Earth system.
At constant speed v, the elevator
experiences a change in the PE at the
rate
dU/dt = d(mgy)/dt = mg(dy/dt) = mgv > 0
Since DE=DEmech = DU only,
 dEmech/dt = dU/dt = mgv > 0
This means that the elevator-Earth
system experiences a constant increase
in mechanical energy at the rate mgv.
Hence, ignoring the friction, the motor
must supply energy at a rate
P = mgv
to the elevator-Earth system to cause it
to change in mechanical energy at the
calculated rate.
d
s
dy
61
Solution
•
•
•
•
•
•
Now, add in the effect of the frictional force.
In the presence of the frictional force, the
motor needs to supply additional power to
compensate for the work done by the
frictional force, in addition to the change in
mechanical energy of the elevator-Earth
system:
We consider the power by the motor P ’
and the power by the frictional force
d/dt(fks) = -d/dt(fks) are external to the
elevator-Earth system.
Both powers sit on the LHS of the CoE
equation (SWi – f d = DE, and their sum
equals to the rate of change of the total
energy of the elevator-Earth system:
P ’ + d/dt(fks) = dE/dt = dEmech/dt
s is the displacement of the frictional force.
It points in the upward direction.
Be reminded that the work done by the
frictional force on the NIS is always
negative, Wf = fks = -fks < 0
d
s
dy
62
Solution
• The motor must now supply a power P ’
obeying the CoE:
P ’ + d/dt(fks) = dEmech/dt
•  P ’ = dU/dt – d/dt(fks)
= mgv – d/dt(fks)
= mgv – fkds/dt
= mgv – fkv = mgv + fkv
where we have used the relation ds/dt
= v.
d
s
dy
• Hence the power of the motor, in the
presence of frictional force, is
P ’ = mgv + fkv = v(mg + fk)
=(3m/s)[(1800 kg)(9.8m/s2)+4000 N] =
64920 W
63
Further question
• What is the power of the
motor if it is to accelerate
the elevator upwards with
an acceleration ay = + 1.00
m/s2 at the instance the
speed of the elevator is v?
d
s
dy
• Note: we expect that the
answer must be a larger
value that that calculated
with zero acceleration.
64
Solution
• In this case the rate of the
change of the mechanical
energy of the elevatorEarth system has to be
modified to include the
increase in the KE, in
addition to the PE:
d
s
dy
dE ' dE 'mech d
d 1
dy
 m d (v  v

 ( K  U    m ( v  v   mgy  
 mg
dt
dt
dt
dt  2
dt
dt
 2
m  dv d v 
 v 

 v   mv  a y  mgv  mva y  mgv
2  dt dt 
65
Solution
• In this case the power
that must be delivered
by the motor P’’ can
be obtained from the
CoE:
d
s
dy
dE '
P '' v  fk 
 mgv  mva y
dt
P '' vf k  mgv  mva y
P ''  v ( f k  mg   mva y
(
 64920 W  (1800kg ( 3m/s  1m/s 2
 70320 W

66