Ch07 ssm

Chapter 7
The Conservation of Energy
Conceptual Problems
1 • Two cylinders of unequal mass are connected by a massless cord that
passes over a frictionless peg (Figure 7-34). After the system is released from rest,
which of the following statements are true? (U is the gravitational potential
energy and K is the kinetic energy of the system.) (a) ΔU < 0 and ΔK > 0,
(b) ΔU = 0 and ΔK > 0, (c) ΔU < 0 and ΔK = 0, (d) ΔU = 0 and ΔK = 0,
(e) ΔU > 0 and ΔK < 0.
Determine the Concept Because the peg is frictionless, mechanical energy is
conserved as this system evolves from one state to another. The system moves and
so we know that ΔK > 0. Because ΔK + ΔU = constant, ΔU < 0. )(a is correct.
Estimation and Approximation
15 •• Assume that your maximum metabolic rate (the maximum rate at
which your body uses its chemical energy) is 1500 W (about 2.7 hp). Assuming a
40 percent efficiency for the conversion of chemical energy into mechanical
energy, estimate the following: (a) the shortest time you could run up four flights
of stairs if each flight is 3.5 m high, (b) the shortest time you could climb the
Empire State Building (102 stories high) using your Part (a) result. Comment on
the feasibility of you actually achieving Part (b) result.
Picture the Problem The rate at which you expend energy, that is do work, is
defined as power and is the ratio of the work done to the time required to do the
work.
(a) Relate the rate at which you can
expend energy to the work done in
running up the four flights of stairs:
t
W
P
Δ
Δ
=ε ⇒
P
W
t
ε
Δ
Δ =
where ε is the efficiency for the
conversion of chemical energy into
mechanical energy.
The work you do in climbing the
stairs increases your gravitational
potential energy:
mghW =Δ
Substitute for ΔW to obtain:
P
mgh
t
ε
=Δ (1)
119

Chapter 7120
Assuming that your mass is 70 kg,
substitute numerical values in
equation (1) and evaluate Δt:
( )( )( )
( )( )
s16
W150040.0
m3.54m/s81.9kg70
Δ
2
≈
×
=t
(b) Substituting numerical values in equation (1) yields:
( )( )( )
( )( )
min8.6s409
W150040.0
m3.5021m/s81.9kg70
Δ
2
≈=
×
=t
The time of about 6.8 min is clearly not reasonable. The fallacy is that you cannot
do work at the given rate of 1500 W for more than very short intervals of time.
17 •• The chemical energy released by burning a gallon of gasoline is
approximately 1.3 × 105
kJ. Estimate the total energy used by all of the cars in the
United States during the course of one year. What fraction does this represent of
the total energy use by the United States in one year (currently about
5 × 1020
J)?
Picture the Problem There are about 3 × 108
people in the United States. On the
assumption that the average family has 4 people in it and that they own two cars,
we have a total of 1.5 × 108
automobiles on the road (excluding those used for
industry). We’ll assume that each car uses about 15 gal of fuel per week.
Calculate, based on the assumptions identified above, the total annual
consumption of energy derived from gasoline:
( ) J/y105.1
gal
kJ
103.1
weeks
52
weekauto
gal
15auto101.5 1958
×=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⋅
×
y
Express this rate of energy use as a
fraction of the total annual energy
use by the United States:
%3
J/y105
J/y101.5
20
19
≈
×
×
Force, Potential Energy, and Equilibrium
25 •• The force Fx is associated with the potential-energy function U = C/x,
where C is a positive constant. (a) Find the force Fx as a function of x. (b) Is this
force directed toward the origin or away from it in the region x > 0? Repeat the
question for the region x < 0. (c) Does the potential energy U increase or decrease
as x increases in the region x > 0? (d) Answer Parts (b) and (c) where C is a
negative constant.

Conservation of Energy 121
Picture the Problem Fx is defined to be the negative of the derivative of the
potential-energy function with respect to x, that is dxdUFx −= . Consequently,
given U as a function of x, we can find Fx by differentiating U with respect to x.
(a) Evaluate :
dx
dU
Fx −= 2
x
C
x
C
dx
d
Fx =⎟
⎠
⎞
⎜
⎝
⎛
−=
(b) Because C > 0, if x > 0, Fx is positive and F
r
points away from the origin. If
x < 0, Fx is still positive and F
r
points toward the origin.
(c) Because U is inversely proportional to x and C > 0, U(x) decreases with
increasing x.
(d) When C < 0, if x > 0, Fx is negative and F
r
points toward the origin. If x < 0,
Fx is negative and F
r
points away from the origin.
Because U is inversely proportional to x and C < 0, U(x) becomes less negative as
x increases and U(x) increases with increasing x.
29 •• The potential energy of an object constrained to the x axis is given by
U(x) = 8x2
– x4
, where U is in joules and x is in meters. (a) Determine the force
associated with this potential energy function. (b) Assuming no other forces
act on the object, at what positions is this object in equilibrium? (c) Which of
these equilibrium positions are stable and which are unstable?
xF
Picture the Problem is defined to be the negative of the derivative of the
potential-energy function with respect to x, that is
xF
dxdUFx −= . Consequently,
given U as a function of x, we can find by differentiating U with respect to x.
To determine whether the object is in stable or unstable equilibrium at a given
point, we’ll evaluate
xF
22
dxUd at the point of interest.
(a) Evaluate the negative of the
derivative of U with respect to x:
( )
( )( )224164
8
3
42
−+=−=
−−=−=
xxxxx
xx
dx
d
dx
dU
Fx
(b) The object is in equilibrium
wherever :0net == xFF
( )( )
.m2and,0,m2arepoints
mequilibriuthe0224
−=
⇒=−+
x
xxx

Chapter 7122
(c) To decide whether the equilibrium
at a particular point is stable or
unstable, evaluate the 2nd derivative
of the potential energy function at the
point of interest:
( ) 23
2
2
1216416 xxx
dx
d
dx
Ud
−=−=
Evaluating 22
dxUd at x = −2 m, 0 and x = 2 m yields the following results:
x, m 22
dxUd Equilibrium
−2 −32 Unstable
0 16 Stable
2 −32 Unstable
Remarks: You could also decide whether the equilibrium positions are stable
or unstable by plotting F(x) and examining the curve at the equilibrium
positions.
33 •• A straight rod of negligible mass is mounted on a frictionless pivot, as
shown in Figure 7-38. Blocks have masses m1 and m2 are attached to the rod at
distances and . (a) Write an expression for the gravitational potential energy
of the blocks-Earth system as a function of the angle θ made by the rod and the
horizontal. (b) For what angle θ is this potential energy a minimum? Is the
statement ″systems tend to move toward a configuration of minimum potential
energy″ consistent with your result? (c) Show that if
1l 2l
1 1 2 2m m=l l , the potential
energy is the same for all values of θ. (When this holds, the system will balance at
any angle θ. This result is known as Archimedes’ law of the lever.)
Picture the Problem The gravitational potential energy of this system of two
objects is the sum of their individual potential energies and is dependent on an
arbitrary choice of where, or under what condition(s), the gravitational potential
energy is zero. The best choice is one that simplifies the mathematical details of the
expression for U. In this problem let’s choose U = 0 where θ = 0.
(a) Express U for the 2-object system
as the sum of their gravitational
potential energies; noting that because
the object whose mass is m2 is above
the position we have chosen for U = 0,
its potential energy is positive while
that of the object whose mass is m1 is
negative:
( )
( ) θ
θθ
θ
sin
sinsin
1122
1122
21
gmm
gmgm
UUU
ll
ll
−=
−=
+=

(b) Differentiate U with respect to θ
and set this derivative equal to zero
to identify extreme values:
( ) 0cos1122 =−= θ
θ
gmm
d
dU
ll
from which we can conclude that
cosθ = 0 and θ = cos−1
0.
To be physically meaningful,
.22 πθπ ≤≤− Hence:
2πθ ±=
Express the 2nd
derivative of U with
respect to θ and evaluate this
derivative at :2πθ ±=
( ) θ
θ
sin11222
2
gmm
d
Ud
ll −−=
If we assume, in the expression for U
that we derived in (a), that
m2l2 – m1l1 > 0, then U(θ) is a sine
function and, in the interval of
interest,
22 πθπ ≤≤− ,
takes on its minimum value when
θ = −π/2:
and0
2
2
2
>
−π
θd
Ud
2atminimumais πθ −=U
and0
2
2
2
<
π
θd
Ud
2atmaximumais πθ =U
(c) If m2l2 = m1l1, then: 02211 =− ll mm
and
.oftindependen0 θ=U
Remarks: An alternative approach to establishing that U is a maximum at
θ = π/2 is to plot its graph and note that, in the interval of interest, U is
concave downward with its maximum value at θ = π/2. Similarly, it can be
shown that U is a minimum at θ = −π/2 (Part (b)).
The Conservation of Mechanical Energy
41 • A 16-kg child on a 6.0-m-long playground swing moves with a speed
of 3.4 m/s when the swing seat passes through its lowest point. What is the angle
that the swing makes with the vertical when the swing is at its highest point?
Assume that the effects due to air resistance are negligible, and assume that the
child is not pumping the swing.

Chapter 7124
Picture the Problem Let the system
consist of Earth and the child. Then
Wext = 0. Choose Ug = 0 at the child’s
lowest point as shown in the diagram to
the right. Then the child’s initial energy
is entirely kinetic and its energy when
it is at its highest point is entirely
gravitational potential. We can
determine h from conservation of
mechanical energy and then use
trigonometry to determineθ.
θ
0g =U
L hL −
h
iv
r
Using the diagram, relate θ to h and
L:
⎟
⎠
⎞
⎜
⎝
⎛
−=⎟
⎠
⎞
⎜
⎝
⎛ −
= −−
L
h
L
hL
1coscos 11
θ (1)
Apply conservation of mechanical
energy to the system to obtain:
0ΔΔext =+= UKW
or, because Kf = Ug,i = 0,
0fg,i =+− UK
Substituting for Ki and Ug,f yields:
02
i2
1
=+− mghmv ⇒
g
v
h
2
2
i
=
Substitute for h in equation (1) to
obtain: ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−= −
gL
v
2
1cos
2
i1
θ
Substitute numerical values and
evaluate θ :
( )
( )( )
°=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−= −
26
m6.0m/s9.812
m/s3.4
1cos 2
2
1
θ
45 •• A ball at the end of a string moves in a vertical circle with constant
mechanical energy E. What is the difference between the tension at the bottom of
the circle and the tension at the top?

Picture the Problem The diagram
represents the ball traveling in a
circular path with constant energy. Ug
has been chosen to be zero at the
lowest point on the circle and the
superimposed free-body diagrams show
the forces acting on the ball at the top
(T) and bottom (B) of the circular path.
We’ll apply Newton’s second law to
the ball at the top and bottom of its path
to obtain a relationship between TT and
TB and conservation of mechanical
energy to relate the speeds of the ball at
these two locations.
R
m
m
m
0g =U
gm
r
gm
r
TT
r
v
r
BT
r
Apply to the ball at
the bottom of the circle and solve for
T
∑ = radialradial maF
B:
R
v
mmgT
2
B
B =−
and
R
v
mmgT
2
B
B += (1)
Apply to the ball at
the top of the circle and solve for T
∑ = radialradial maF
T: R
v
mmgT
2
T
T =+
and
R
v
mmgT
2
T
T +−= (2)
Subtract equation (2) from equation
(1) to obtain:
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
+−−
+=−
R
v
mmg
R
v
mmgTT
2
T
2
B
TB
mg
R
v
m
R
v
m 2
2
T
2
B
+−= (3)
Using conservation of mechanical
energy, relate the energy of the ball
at the bottom of its path to its
mechanical energy at the top of the
circle:
( )Rmgmvmv 22
T2
12
B2
1
+=
or
mg
R
v
m
R
v
m 4
2
T
2
B
=−
Substituting in equation (3) yields: mgTT 6TB =−

Chapter 7126
55 •• A pendulum consists of a string of length L and a bob of mass m. The
bob is rotated until the string is horizontal. The bob is then projected downward
with the minimum initial speed needed to enable the bob to make a full revolution
in the vertical plane. (a) What is the maximum kinetic energy of the bob?
(b) What is the tension in the string when the kinetic energy is maximum?
Picture the Problem Let the system consist of Earth and the pendulum bob. Then
Wext = 0. Choose Ug = 0 at the bottom of the circle and let points 1, 2 and 3
represent the bob’s initial point, lowest point and highest point, respectively. The
bob will gain speed and kinetic energy until it reaches point 2 and slow down
until it reaches point 3; so it has its maximum kinetic energy when it is at point 2.
We can use Newton’s second law at points 2 and 3 in conjunction with
conservation of mechanical energy to find the maximum kinetic energy of the bob
and the tension in the string when the bob has its maximum kinetic energy.
m
m 0g =U
gm
r
gm
r
v
r
T
r
3
1
2
2
L
m
1
(a) Apply ∑ to the
bob at the top of the circle and solve
for :
= radialradial maF
2
3v
L
v
mmg
2
3
= ⇒ gLv =2
3
energy to the system to express the
relationship between K2, K3 and U3:
02323 =−+− UUKK
or, because U2 = 0,
0323 =+− UKK
Solving for K2 yields: 33max2 UKKK +==
Substituting for K3 and U3 yields: ( )LmgmvK 22
32
1
max +=
Substitute for and simplify to
obtain:
2
3v ( ) mgLmgLgLmK 2
5
2
1
max 2 =+=

(b) Apply to the
bob at the bottom of the circle
and solve for T
cradial maF =∑
2:
L
v
mmgTF
2
2
2net =−=
and
L
v
mmgT
2
2
2 += (1)
Use conservation of mechanical energy
to relate the energies of the bob at
points 2 and 3 and solve for K2:
0where0 22323 ==−+− UUUKK
and
( )LmgmvUKK 22
32
1
332 +=+=
Substitute for and K2
3v 2 to obtain: ( ) ( )LmggLmmv 22
12
22
1
+= ⇒ gLv 52
2 =
Substitute for in equation (1) and
simplify to obtain:
2
2v
mg
L
gL
mmgT 6
5
2 =+=
59 ••• A pendulum is suspended from the ceiling and attached to a spring
fixed to the floor directly below the pendulum support (Figure 7-48). The mass of
the pendulum bob is m, the length of the pendulum is L, and the force constant is
k. The unstressed length of the spring is L/2 and the distance between the floor
and ceiling is 1.5L. The pendulum is pulled aside so that it makes an angle θ with
the vertical and is then released from rest. Obtain an expression for the speed of
the pendulum bob as the bob passes through a point directly below the pendulum
support.
Picture the Problem Choose Ug = 0 at
point 2, the lowest point of the bob’s
trajectory and let the system consist of
Earth, the ceiling, the spring, and the
pendulum bob. Given this choice, there
are no external forces doing work to
change the energy of the system. The
bob’s initial energy is partially
gravitational potential and partially
potential energy stored in the stretched
spring. As the bob swings down to
point 2 this energy is transformed into
kinetic energy. By equating these
energies, we can derive an expression
for the speed of the bob at point 2.
L
θ
θ2
2/L
cosL θ
1
2x
2/L

Chapter 7128
energy to the system as the
pendulum bob swings from point 1
to point 2:
0ΔΔΔ sgext =++= UUKW
or, because K1 = Ug,2 = Us,2 = 0,
0s,1g,12 =−− UUK
Substituting for K2, Ug,1, and Us,2
yields:
( ) 0cos1 2
2
12
22
1
=−−− kxmgLmv θ (1)
Apply the Pythagorean theorem to the lower triangle in the diagram to obtain:
( ) ( )[ ]
[ ]
( )θ
θθθ
θθ
cos3
coscos3sin
cossin
4
132
2
4
922
2
2
3222
2
1
−=
+−+=
−+=+
L
L
LLx
Take the square root of both sides of
the equation to obtain:
( )θcos34
13
2
1
−=+ LLx
Solving for x yields: ( )[ ]2
1
4
13
cos3 −−= θLx
Substitute for x in equation (1) to obtain:
( )[ ] ( )θθ cos1cos3
2
2
1
4
132
2
12
22
1
−+−−= mgLkLmv
Solving for v2 yields:
( ) ( )2
2
1
4
13
2 cos3cos12 −−+−= θθ
m
k
L
g
Lv
Total Energy and Non-conservative Forces
65 •• The 2.0-kg block in Figure 7-49 slides down a frictionless curved
ramp, starting from rest at a height of 3.0 m. The block then slides 9.0 m on a
rough horizontal surface before coming to rest. (a) What is the speed of the block
at the bottom of the ramp? (b) What is the energy dissipated by friction?
(c) What is the coefficient of kinetic friction between the block and the horizontal
surface?
Picture the Problem Let the system include the block, the ramp and horizontal
surface, and Earth. Given this choice, there are no external forces acting that will
change the energy of the system. Because the curved ramp is frictionless,
mechanical energy is conserved as the block slides down it. We can calculate its

speed at the bottom of the ramp by using conservation of energy. The potential
energy of the block at the top of the ramp or, equivalently, its kinetic energy at the
bottom of the ramp is converted into thermal energy during its slide along the
horizontal surface.
(a) Let the numeral 1 designate the
initial position of the block and the
numeral 2 its position at the foot of
the ramp. Choose Ug = 0 at point 2
and use conservation of energy to
relate the block’s potential energy at
the top of the ramp to its kinetic
energy at the bottom:
thermmechext EEW Δ+Δ=
or, because Wext = Ki = Uf = ΔEtherm = 0,
00 2
22
1
=Δ−= hmgmv ⇒ hgv Δ= 22
evaluate v2:
( )( )
m/s7.7
m/s67.7m3.0m/s9.812 2
2
=
==v
(b) The energy dissipated by
friction is responsible for
changing the thermal energy of
the system:
0
ΔΔΔΔΔ thermf
=
++=++ UKEUKW
Because ΔK = 0 for the slide: ( ) 112f UUUUW =−−=−= Δ
Substituting for yields:1U hmgW Δf =
evaluate :1U
( )( )( )
J59
J9.58m3.0m/s9.81kg2.0 2
f
=
==W
(c) The energy dissipated by
friction is given by:
xmgsfE Δ=Δ=Δ ktherm μ
Solving for kμ yields:
xmg
E
Δ
Δ
= therm
kμ
evaluate kμ : ( )( )( )
33.0
m9.0m/s9.81kg2.0
J58.9
2k
=
=μ

Chapter 7130
69 •• The initial speed of a 2.4-kg box traveling up a plane inclined 37° to
the horizontal is 3.8 m/s. The coefficient of kinetic friction between the box and
the plane is 0.30. (a) How far along the incline does the box travel before coming
to a stop? (b) What is its speed when it has traveled half the distance found in Part
(a)?
Picture the Problem The box will slow down and stop due to the dissipation of
thermal energy. Let the system be Earth, the box, and the inclined plane and apply
the work-energy theorem with friction. With this choice of the system, there are
no external forces doing work to change the energy of the system. The pictorial
representation shows the forces acting on the box when it is moving up the
incline.
gF
r
nF
r
θ
θ
x
y
kf
r
0g =U
0
0
0
=
x
L
x
=
1
h
m
m
(a) Apply the work-energy theorem
with friction to the system: therm
thermmechext
EUK
EEW
Δ+Δ+Δ=
Δ+Δ=
Substitute for ΔK, ΔU, and
to obtain:
thermEΔ LFmghmvmv nk
2
02
12
12
1
0 μ++−= (1)
Referring to the free-body diagram,
relate the normal force to the weight
of the box and the angle of the
incline:
θcosn mgF =
Relate h to the distance L along
the incline:
θsinLh =
Substitute in equation (1) to obtain:
0sincos 2
02
12
12
1
k =+−+ θθμ mgLmvmvmgL (2)

Solving equation (2) for L yields:
( )θθμ sincos2 k
2
0
+
=
g
v
L
Substitute numerical values and evaluate L:
( )
( )( )[ ]
m87.0m8747.0
sin37cos370.30m/s9.812
m/s3.8
2
2
==
°+°
=L
(b) Let L
v
2
1 represent the box’s speed when it is halfway up the incline.
Then equation (2) becomes:
( ) ( ) 0sincos 2
12
02
12
2
1
2
1
k
2
1 =+−+ θθμ LmgmvmvLmg L
Solving for L
v
2
1 yields : ( )θμθ cossin k
2
0
2
1 +−= gLvv L
Substitute numerical values and evaluate L
v
2
1 :
( ) ( )( ) ( )[ ]] m/s7.2cos370.30sin37m0.8747m/s9.81m/s3.8 22
f =°+°−=v
Mass and Energy
75 • You are designing the fuel requirements for a small fusion electric-
generating plant. Assume 33% conversion to electric energy. For the deuterium–
tritium (D–T) fusion reaction in Example 7-18, calculate the number of reactions
per second that are necessary to generate 1.00 kW of electric power.
Picture the Problem The number of reactions per second is given by the ratio of
the power generated to the energy released per reaction. The number of reactions
that must take place to produce a given amount of energy is the ratio of the energy
per second (power) to the energy released per second.
In Example 7-18 it is shown that the energy per reaction is 17.59 MeV. Convert
this energy to joules:
( )( ) J1028.18J/eV101.602MeV17.59MeV59.17 1319 −−
×=×=

Chapter 7132
Assuming 33% conversion to electric energy, the number of reactions per second
is:
( )( ) sreactions/101.1
J/reaction1028.1833.0
J/s1000 15
13
×≈
× −
General Problems
85 • You are in charge of ″solar-energizing″ your grandfather’s farm. At the
farm’s location, an average of 1.0 kW/m2
reaches the surface during the daylight
hours on a clear day. If this could be converted at 25% efficiency to electric
energy, how large a collection area would you need to run a 4.0-hp irrigation
water pump during the daylight hours?
Picture the Problem The solar constant is the average energy per unit area and
per unit time reaching the upper atmosphere. This physical quantity can be
thought of as the power per unit area and is known as intensity.
Letting represent the intensity
of the solar radiation at the surface of
Earth, express as a function of
power and the area on which this
energy is incident:
surfaceI
surfaceI
A
P
I =surfaceε ⇒
surfaceI
P
A
ε
=
where ε is the efficiency of conversion
to electric energy.
evaluate A:
( )( )
2
2
m12
kW/m0.10.25
hp
W746
hp0.4
=
×
=A
93 •• In a volcanic eruption, a 2-kg piece of porous volcanic rock is thrown
straight upward with an initial speed of 40 m/s. It travels upward a distance of
50 m before it begins to fall back to Earth. (a) What is the initial kinetic energy of
the rock? (b) What is the increase in thermal energy due to air resistance during
ascent? (c) If the increase in thermal energy due to air resistance on the way
down is 70% of that on the way up, what is the speed of the rock when it returns
to its initial position?

Picture the Problem Let the system consist of Earth, the rock and the air. Given
this choice, there are no external forces to do work on the system and Wext = 0.
Choose Ug = 0 to be where the rock begins its upward motion. The initial kinetic
energy of the rock is partially transformed into potential energy and partially
dissipated by air resistance as the rock ascends. During its descent, its potential
energy is partially transformed into kinetic energy and partially dissipated by air
resistance.
(a) The initial kinetic energy of the
rock is given by:
2
i2
1
i mvK =
evaluate Ki:
( )( ) kJ1.6m/s40kg2.0
2
2
1
i ==K
(b) Apply the work-energy theorem
with friction to relate the energies of
the system as the rock ascends:
0therm =Δ+Δ+Δ EUK
or, because Kf = 0,
0thermi =Δ+Δ+− EUK
Solving for yields:thermEΔ UKE Δ−=Δ itherm
Substitute numerical values and evaluate thermEΔ :
( )( )( ) kJ0.6kJ0.619m50m/s9.81kg2.0kJ6.1Δ 2
therm ==−=E
(c) Apply the work-energy theorem
with friction to relate the energies of
the system as the rock descends:
0Δ70.0ΔΔ therm =++ EUK
Because Ki = Uf = 0: 0Δ70.0 thermif =+− EUK
Substitute for the energies to obtain: 0Δ70.0 therm
2
f2
1
=+− Emghmv
Solve for to obtain:fv
m
E
ghv therm
f
Δ40.1
2 −=
Substitute numerical values and evaluate :fv
( )( ) ( ) m/s23
kg2.0
kJ0.6191.40
m50m/s9.812 2
f =−=v

Chapter 7134
95 •• A block of mass m is suspended from a wall bracket by a spring and is
free to move vertically (Figure 7-54). The +y direction is downward and the
origin is at the position of the block when the spring is unstressed. (a) Show that
the potential energy as a function of position may be expressed as
mgykyU −= 2
2
1
, (b) Using a spreadsheet program or graphing calculator, make
a graph of U as a function of y with k = 2 N/m and mg = 1 N. (c) Explain how
this graph shows that there is a position of stable equilibrium for a positive value
of y. Using the Part (a) expression for U, determine (symbolically) the value of y
when the block is at its equilibrium position. (d) From the expression for U, find
the net force acting on m at any position y. (e) The block is released from rest
with the spring unstressed; if there is no friction, what is the maximum value of y
that will be reached by the mass? Indicate ymax on your graph/spreadsheet.
Picture the Problem Let the system include Earth, the block and the spring.
Given the potential energy function as a function of y, we can find the net force
acting on the given system using dydUF /−= . The maximum extension of the
spring; that is, the lowest position of the mass on its end, can be found by
applying the work-energy theorem. The equilibrium position of the system can be
found by applying the work-energy theorem with friction … as can the amount of
thermal energy produced as the system oscillates to its equilibrium position. In
Part (c), setting dU/dy equal to zero and solving the resulting equation for y will
yield the value of y when the block is in its equilibrium position
(a) The potential energy of the
oscillator is the sum of the
gravitational potential energy of the
block and the energy stored in the
stretched spring:
sg UUU +=
Letting the zero of gravitational
potential energy be at the oscillator’s
equilibrium position yields:
mgykyU −= 2
2
1
where y is the distance the spring is
stretched.

(b) A graph of U as a function of y follows. Because k and m are not specified, k
has been set equal to 2 and mg to 1.
-0.4
-0.2
0.0
0.2
0.4
0.6
0.8
1.0
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
y (m)
U(J)
(c) The fact that U is a minimum near y = 0.5 m tells us that this is a position of
stable equilibrium.
Differentiate U with respect to y to
obtain:
( ) mgkymgyky
dy
d
dy
dU
−=−= 2
2
1
Setting this expression equal to
zero for extrema yields:
0=− mgky ⇒
k
mg
y =
(d) Evaluate the negative of the
derivative of U with respect to y:
( )
mgky
mgyky
dy
d
dy
dU
F
+−=
−−=−= 2
2
1
(e) Apply conservation of energy
to the movement of the mass
from y = 0 to :maxyy =
0therm =Δ+Δ+Δ EUK
Because ΔK = 0 (the object starts
from rest and is momentarily at rest
at ) and (no friction), it
follows that:
maxyy =
( ) ( ) 00Δ max =−= UyUU
Because U(0) = 0: ( ) 0max =yU ⇒ 0max
2
max2
1
=− mgyky

Chapter 7136
Solve for to obtain:maxy
k
mg
y
2
max =
On the graph, ymax is at (1.0, 0.0).
99 ••• To measure the combined force of friction (rolling friction plus air
drag) on a moving car, an automotive engineering team you are on turns off the
engine and allows the car to coast down hills of known steepness. The team
collects the following data: (1) On a 2.87° hill, the car can coast at a steady
20 m/s. (2) On a 5.74° hill, the steady coasting speed is 30 m/s. The total mass of
the car is 1000 kg. (a) What is the magnitude of the combined force of friction at
20 m/s (F20) and at 30 m/s (F30)? (b) How much power must the engine deliver to
drive the car on a level road at steady speeds of 20 m/s (P20) and 30 m/s (P30)?
(c) The maximum power the engine can deliver is 40 kW. What is the angle of
the steepest incline up which the car can maintain a steady 20 m/s? (d) Assume
that the engine delivers the same total useful work from each liter of gas, no
matter what the speed. At 20 m/s on a level road, the car gets 12.7 km/L. How
many kilometers per liter does it get if it goes 30 m/s instead?
Picture the Problem We can use Newton’s second law to determine the force of
friction as a function of the angle of the hill for a given constant speed. The power
output of the engine is given by vF
rr
⋅= fP .
FBD for (a):
gF
r
nF
r
θ
x
y
fF
r
FBD for (c):
gF
r
nF
r
θ
x
y
F
r
fF
r
(a) Apply to the car:∑ = xx maF 0sin =− Fmg θ ⇒ θsinmgF =
Evaluate F for the two speeds: ( )( ) ( )
N491
2.87sinm/s9.81kg1000 2
20
=
°=F
and
( )( ) ( )
N981
5.74sinm/s9.81kg1000 2
30
=
°=F

(b) The power an engine must
deliver on a level road in order to
overcome friction loss is given by:
vFP f=
Evaluate this expression for
v = 20 m/s and 30 m/s:
( )( ) kW9.8m/s20N49120 ==P
and
( )( ) kW29m/s30N98130 ==P
(c) Apply to the car:∑ = xx maF ∑ =−−= 0sin fFmgFFx θ
Solving for F gives: fsin FmgF += θ
Relate F to the power output of the
engine and the speed of the car: v
P
F =
Equate these expressions for F to
obtain:
fsin Fmg
v
P
+= θ
Solving for θ yields:
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
= −
mg
F
v
P
f
1
sinθ
evaluate θ for :20f FF =
( )( )
°=
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
−
= −
8.8
m/s9.81kg1000
N491
m/s20
kW40
sin 2
1
θ
(d) Express the equivalence of the
work done by the engine in driving
the car at the two speeds:
( ) ( )30302020engine sFsFW Δ=Δ=
Let ΔV represent the volume of fuel
consumed by the engine driving the
car on a level road and divide both
sides of the work equation by ΔV to
obtain:
( ) ( )
V
s
F
V
s
F
Δ
Δ
=
Δ
Δ 30
30
20
20

Chapter 7138
Solve for
( )
V
s
Δ
Δ 30
:
( ) ( )
V
s
F
F
V
s
Δ
Δ
=
Δ
Δ 20
30
2030
evaluate
( )
V
s
Δ
Δ 30
:
( ) ( )
km/L6.36
km/L12.7
N981
N49130
=
=
Δ
Δ
V
s
105 •• A high school teacher once suggested measuring the magnitude of
free-fall acceleration by the following method: Hang a mass on a very fine thread
(length L) to make a pendulum with the mass a height H above the floor when at
its lowest point P. Pull the pendulum back so that the thread makes an angle θ0
with the vertical. Just above point P, place a razor blade that is positioned to cut
through the thread as the mass swings through point P. Once the thread is cut, the
mass is projected horizontally, and hits the floor a horizontal distance D from
point P. The idea was that the measurement of D as a function of θ0 should
somehow determine g. Apart from some obvious experimental difficulties, the
experiment had one fatal flaw: D does not depend on g! Show that this is true,
and that D depends only on the angle θ0.
Picture the Problem The pictorial
representation shows the bob swinging
through an angle θ0 before the thread is
cut and the ball is launched
horizontally. Let its speed at position 1
be v1. We can use conservation of
mechanical energy to relate v1 to the
change in the potential energy of the
bob as it swings through the angle θ0.
We can find its flight time Δt from a
constant-acceleration equation and then
express D as the product of v1 and Δt.
x
y
L
H
D
L(1 − cos )θ
θ0
0
U = 0g
0
P
0
1
Relate the distance D traveled
horizontally by the bob to its launch
speed v1 and time of flight Δt:
tvD Δ= 1 (1)
Use conservation of mechanical
energy to relate its energies at
positions 0 and 1:
00101 =−+− UUKK
or, because U1 = K0 = 0,
001 =−UK

Substitute for K1 and U0 to obtain: ( ) 0cos1 0
2
12
1
=−− θmgLmv
Solving for v1 yields: ( )01 cos12 θ−= gLv
In the absence of air resistance, the
horizontal and vertical motions of
the bob are independent of each
other and we can use a constant-
acceleration equation to express the
time of flight (the time to fall a
distance H):
( )2
2
1
0 tatvy yy Δ+Δ=Δ
or, because Δy = −H, ay = −g, and
v0y = 0,
( )2
2
1
tgH Δ−=− ⇒
g
H
t
2
=Δ
Substitute in equation (1) and
simplify to obtain: ( )
( )0
0
cos12
2
cos12
θ
θ
−=
−=
HL
g
H
gLD
which shows that, while D depends on
θ0, it is independent of g.

Ch07 ssm

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