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Ch6. Work and Energy
Work Done by a Constant Force
Force Fpoints in the same direction as the
resulting displacement s
W=Fs
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SI Unit of Work: newton.meter=joule(J)
System Force * Distance = Work
SI newton(N) meter(m) joule(J)
CGS dyne(dyn) centimeter(cm) erg
BE pound(lb) foot(ft) foot.pound(ft.lb)
sFW )cos(
Units of Measurement for Work
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Example 1. Pulling a Suitcase-
on-Wheels
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Find the work done by a 45.0 N force in
pulling the suitcase at an angle fora distance s=75.0 m 0.50
JmNsFW 2170)0.75(0.50cos)0.45()cos(
No work done due to Fsin
F and s in the same direction positive work
F and s in the opposite direction negative work
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Example 2. Bench-Pressing
The weight lifter is bench-pressing a barbell whoseweight is 710N. In part (b) of the figure, he raises
the barbell a distance of 0.65m above his chest, and
in part (c) he lowers it the same distance.
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The weight is raised and lowered at a constant
velocity. Determine the work done on thebarbell by the weight lifter during (a) the
lifting phase and (b) the lowering phase.
(a)
(b)
JmNsFW 460)65.0(0cos)710()cos(
JmNsFW 460)65.0(180cos)710()cos(
cos1800 = -1
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Example 3. Accelerating a Crate
A 120kg crate on the flatbed of a
truck that is moving with an
acceleration of a=+1.5m/s2along
the positive x axis. The crate does
not slip with respect to the truck,
as the truck undergoes a
displacement whose magnitude iss=65m. What is the total work
done on the crate by all of the
forces acting on it?
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Forces that act on the crate:
(1) the weight W=mg of the crate,
(2) the normal force FNexerted by the flatbed,
(3) the static frictional force fs.
Nsmkgmafs 180)/5.1)(120( 2
JmNsfW s4102.1)65(0cos)180()cos(
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Check your understanding 1A suitcase is hanging straight down from your hand as you
ride an escalator. Your hand exerts a force on the suitcase
and this force does work.Which one of the following
statements is correct?
(a) The work is negative when you ride up the escalator and
positive when you ride down the escalator.
(b) The work is positive when you ride up the escalator and
negative when you ride down the escalator.
(c) The work is positive irrespective of whether you ride up or
down the escalator.
(d) The work is negative irrespective of whether you ride up
or down the escalator.
Answer: (b)
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The Work-Energy Theorem and
Kinetic Energy
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Concepts At a Glance:
In physics, when a net force performs work on anobject, there is always a result from the effort.
The result is a change in the Kinetic energy.
What is Kinetic Energy?
Energy associated with motion.
KE=(1/2)mv2
The relationship that relates work to the change
in kinetic energy is known as the Work-energy
theorem.
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Definition of Kinetic Energy:
The kinetic energy KE of an object with mass m
and speed vis given by
2
2
1mvKE
SI Unit of Kinetic Energy: joule(J)
asvvf 2
2
0
2 +a
vv
s
f
2
2
0
2
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massF)(Work done by net ext. force
2
0
2
2
1
2
1)( mvmvsF f Work done by
net ext. forceFinal KE Initial KE
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The Work-Energy Theorem
When a net external force does work W on
an object, the kinetic energy of the object
changes from its initial value of KE0to a final
value of KEf, the difference between the twovalues being equal to the work:
20
20
21
21 mvmvKEKEW ff
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Example 4. Deep Space 1
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The space probe Deep Space 1 was launched October 24,
1998. Its mass was 474kg. The goal of the mission was totest a new kind of engine called an ion propulsion drive,
which generates only a weak thrust, but can do so for
long periods of time using only small amounts of fuel. The
mission has been spectacularly successful. Consider theprobe traveling at an initial speed of v0=275m/s. No forces
act on it except the 56.0-mN thrust of its engine. This
external force F is directed parallel to the displacement s
of magnitude 2.42*109m. Determine the final speed of the
probe, assuming that the mass remains nearly constant.
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)1042.2(0cos)100.56()cos( 93
mNsFW
=1.36*108J
28
0 )/275)(474(
2
1)1036.1( smkgJKEWKEf ++
=1.54*108J
mKEv ff )(2 kg
J474
)1054.1(2 8
=806 m/s
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Example 5. Downhill Skiing
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A 58 kg skier is coasting down a 250slope. A kinetic
frictional force of magnitude fk=70N opposes hermotion. Near the top of the slope, the skiers speed
is v0=3.6m/s. Ignoring air resistance, determine the
speed vfat a point that is displaced 57m downhill.
NsmkgfmgF k 7025sin)/80.9)(58(25sin 2
=+170 N
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KEf=W+KE0=9700J+(1/2)(58kg)(3.6m/s)2=10100J
m
KEv
f
f
)(2 sm
kg
J/19
58
)10100(2
JmNsFW 9700)57(0cos)170()cos(
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Check your understanding 2
A rocket is at rest on the launch pad. When the
rocket is launched, its kinetic energy increases.
Is the following statement true or false?
The amount by which the kinetic energy
increases is equal to the work done by the force
generated by the rockets engine.
Answer: False
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Conceptual Example 6. Work and
Kinetic EnergyA satellite moving abut the earth in a
circular orbit and in an elliptical
orbit. The only external force that acts
on the satellite is the gravitational
force. For these two orbits, determine
whether the kinetic energy of the
satellite changes during the motion.
KE changes in the elliptical orbit,
but not in the circular orbit.
In circular motion, F S alwaysno work done.
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Gravitational Potential Energy
Work Done by the Force of Gravity
Wgravity=(mg cos00)(h0-hf)=mg(h0-hf)
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Example 7. A Gymnast on a
Trampoline
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A gymnast springs vertically upward from a
trampoline. The gymnast leaves the trampoline at a
height of 1.20m and reaches a maximum height of
4.80m before falling back down. All heights are
measured with respect to the ground. Ignoring air
resistance, determine the initial speed v0with whichthe gymnast leaves the trampoline.
)80.420.1(/80.9(2)(2 200 mmsmhhgv f
=8.40 m/s
Wgravity=mg (h0-hf)
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Gravitational Potential Energy
Wgravity= mgh0 - mghf
Initial
gravitationa
l potential
energy PE0
Final
gravitational
potential
energy PEf
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Definition of Gravitational Potential Energy
The gravitational potential energy PE is the energy
that an object of mass m has by virtue of its
position relative to the surface of the earth. That
position is measured by the height h of the objectrelative to an arbitrary zero level:
PE=mgh
SI Unit of gravitational potential energy: joule (J)
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Conservative Versus Non-
conservative Forces
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Definition of a Conservative Force
Version 1: A force is conservative when the work it
does on a moving object is independent of the path
between the objects initial and final positions.
Version 2: A force is conservative when it does no
net work on an object moving around a closed path,
starting and finishing at the same point.
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20
20
21
21)( mvmvWhhmg fncf +
Work done by external forces =
KE
i.e,2
0
2
2
1
2
1
mvmvWW fncc +
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PEKEWnc +
Wnc = (KEf - KE0) + (PEf - PE0)Change in
kinetic energy
Change in gravitational
potential energy
Net work
done by non-
conservative
forces
)()2
1
2
1
( 02
0
2
mghmghmvmvW ffnc +
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The Conservation of Mechanical
Energy
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SupposeWnc=0J, so Ef = E0
mvf2+mghf= mv0
2+mgh0
The principle of Conservation of Mechanical Energy
The total mechanical energy (E=KE+PE) of an
object remains constant as the object moves,
provided that the net work done by external non-conservative forces is zero, Wnc=0J
2
1
2
1
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Example 8. A Daredevil
Motorcyclist
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A motorcyclist is trying to leap across the canyon by
driving horizontally off the cliff at a speed of 38.0m/s. Ignoring air resistance, find the speed with
which the cycle strikes the ground on the other side.
(1/2)mvf2
+mghf= (1/2) mv02
+mgh0
)(2 02
0 ff hhgvv +
)0.350.70)(/80.9(2)/9.38( 22 mmsmsmvf +
=46.2 m/s
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Check your understanding 3
Some of the following situations are consistent with the principleof conservation of mechanical energy, and some are not.
Which ones are consistent with the principle?
(a) An object moves uphill with an increasing speed.
(b) An object moves uphill with a decreasing speed.
(c) An object moves uphill with a constant speed.
(d) An object moves downhill with an increasing speed.
(e) An object moves downhill with a decreasing speed.
(f) An object moves downhill with a constant speed.
(b) And (d)
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Conceptual Example 9. The
Favorite Swimming HoleA rope is tied to a tree limb
and used by a swimmer to
swing into the water below.The person starts from rest
with the rope held in the
horizontal position, swings
downward, and then lets goof the rope.
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Three forces act on him: his weight, the tension in the rope, and
the force due to air resistance. His initial height h0and final
height hf are known. Considering the nature of these forces,
conservative versus non-conservative, can we use the principle
of conservation of mechanical energy to find his speed vfat the
point where he lets go of the rope?
If Wnc= 0 (work due to nonconservative forces)
conservation of energy
Tension and air resistance --- non conservative
So no work done by T.
T is always r to the circular path.
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Work done by air resistance is nonzero.
So ideally no.
But if ignore air resistance,
ff mghmvmghmv ++ 2
0
2
02
1
2
1
)(2 02
0 ff hhgvv +
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Example 10. The Steel Dragon
The tallest and fastest roller coaster in the world is now
the Steel Dragon in Mie, Japan. The ride includes a
vertical drop of 93.5m. The coaster has a speed of
3.0m/s at the top of the drop. Neglect friction and findthe speed of the riders at the bottom.
(1/2)mvf2+mghf= (1/2) mv0
2+mgh0
Ef E0
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)(2 02
0 ff hhgvv +
)5.93)(/80.9(2)/0.3( 2 msmsmvf +
= 42.9m/s (about 96 mi/h)
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Example 11. The Steel Dragon,
RevisitedIn example 10, we ignored non-conservative forces,
such as friction. In reality, however, such forces are
present when the roller coaster descends. The actualspeed of the riders at the bottom is 41.0 m/s, which is
less than that determined in example 10. Assuming
again that the coaster has a speed of 3.0 m/s at the top,
find the work done by non-conservative forces on a55.0kg rider during the descent from a height h0to a
height hf, where h0hf=93.5m
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)
2
1()
2
1( 0
2
0
2mghmvmghmvW ffnc ++
)()(2
10
2
0
2
ffnc hhmgvvmW
EfE0
)5.93)(/80.9)(0.55(])/3()/41)[(0.55(
2
1 222 msmkgsmsmkgWnc
J4400
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Example 12. FireworksA 0.20kg rocket in a fireworks
display is launched from rest and
follows an erratic flight path to
reach the point P. Point P is 29m
above the starting point. In theprocess, 425J of work is done on
the rocket by the non-conservative
force generated by the burning
propellant. Ignoring air resistanceand the mass lost due to the
burning propellant, find the speed
vfof the rocket at the point P.
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)2
1()
2
1( 0
2
0
2mghmvmghmvW ffnc ++
m
hhmgmvW
vfnc
f
)](2
1[2
0
2
0 +
kg
msmkgsmkgJ
vf20.0
)]29)(/80.9)(20.0()/0)(20.0(2
1425[2
22 +
=61m/s
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Power
The idea of power incorporates both the concepts ofwork and time, for poweris work done per unit time.
Definition of Average Power
Average power P is the average rate at which work Wis done, and it is obtained by dividing W by the time t
required to perform the work.
tW
TimeWorkP
SI Unit of Power: joule/s=watt (W)
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Check your understanding 4Engine A has a greater power rating than engine B. Which one
of the following statements correctly describes the abilities
of these engines to do work?
(a) Engine A and B can do the same amount of work in the
same amount of time.
(b) In the same amount of time, engine B can do more work
than engine A.
(c) Engine A and B can do the same amount of work, but A cando it more quickly.
Answer: (c)
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Units of Measurement for Power
System Work / Time = Power
SI joule (J) / second (s) = watt (W)
CGS erg / second (s) = erg persecond(erg/s)
BE foot.pound / second (s) = foot.pound
per second
(ft.lb) (ft.lb/s)
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Time
energyinchangeP
wattsondpoundfoothorsepower 7.745sec/5501
t
Fs
t
W
vFP
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smssmsmatvvf /0.23)00.5)(/60.4()/0( 2
0 ++
)0.78(1082.5)/5.11)(5060( 4 hpWsmNvFP
ff vvvv 2
1)(2
10 +
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Other Forms of Energy and the
Conservation of EnergyExamples: electrical energy, heat, chemical energy,
nuclear energy.
In general, energy of all types can be converted
from one form to another.
The principle of conservation of energy
Energy can neither be created nor destroyed, but can
only be converted from one form to another.
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Work Done by a Variable ForceThe work done by a variable force in moving an
object is equal to the area under the graph of Fcos
versus s.
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++ 2211 )cos()cos( sFsFW
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Example 14. Work and the
Compound BowFind the work that the archer must do in drawing backthe string of the compound bow from 0 to 0.500 m
Jsquare
J
squaresW 5.60)250.0)(242(
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Example 15. Skateboarding and
Work
The skateboarder is coasting down a ramp, and there
are three forces acting on her: her weight W(magnitude = 675 N), a frictional force f ( magnitude =
125 N). That opposes her motion, and a normal force
FN (magnitude =612 N). Determine the net work done
by the three forces when she coasts for a distance of9.2 m.
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Force F Angle S W= (Fcos sW 675N 65.00 9.2m W=(675N)(cos 65.00)(9.2m)=+2620J
f 125N 180.00 9.2m W=(125N)(cos 180.00)(9.2m)=-1150J
FN 612N 90.00 9.2m W=(612N)(cos 90.00)(9.2m)=0J
The net work done by the three forces is
+2626J+(-1150J)+0J=+1470J
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Concepts & Calculations Example 16.
Conservation of Mechanical Energy and
the Work-Energy Theorem
hA
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A 0.41kg block sliding from A to B along a frictionless
surface. When the block reaches B, it continues to
slide along the horizontal surface BC. The blockslows down, coming to rest at C. The kinetic energy
of the block at A is 37J, and the height of A and B
are 12m and 7m above the ground.
(a) What is the kinetic energy of the block when itreaches B?
(b) How much work does the kinetic frictional force do
during the BC segment of the trip?
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A---B motion: What are the forces?
Weight---------------
What is the non conservative work (Wnc) ?
Wnc=0 FN displacement
Conservation of energy valid for A---B?
K. E at B ? K. E at A
K. E at B < K. E at A
Normal force--------
conservative
non conservative
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(a) KEB + mghB = KEA + mghA
KEB = KEA + mg(hA-hB)
=37J+(0.41kg)(9.80m/s2)(12m-7m)
=57J
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Problem 2
REASONING AND SOLUTION Each locomotive does work
W= T(cos s= (5.00103N) cos 20.0 (2.00 103m) = 9.40106JThe net work is then
WT= 2W= 2T(cos s = 1.88 107J
T
T
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Problem 16
2820 m/s
apogee8450 m/s
perigee
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From the work-energy theorem,
1 1 12 2 2 2f 0 f 02 2 2
W mv mv m v v
1 2 2
2W (7420 kg) (2820 m/s) (8450 m/s) =
a)
b) J111035.2
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Problem 20
To find the work, employ the work-energy theorem,
W= KEf KE0
F = 24N
8.0 M
v0= 0 vf= 2.0 m/8
16 kg
FN
P
mg
fkfk= kFN FN-mg=0
FN=mg
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SOLUTION According to the work-energy theorem, we have
W = Wpull+ Wf= KEf KE0
Using Equation 6.1 [W= (Fcos ) s] to express each work
contribution, writing the kinetic energy as , and
noting that the initial kinetic energy is zero (the sled starts
from rest), we obtain
212mv
) ) 212kpull f
cos0 cos180P s f s mv
W W
+
Work done by the net force (pulling force P and fk)
W= Wpulling
+ Wf
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Solving for the coefficient of kinetic friction gives
) )
) ) ) )
) ) )
221 12 2
k 2
cos0 16 kg 2.0 m/s 24 N 8.0 m0.13
cos180 16 kg 9.80 m/s 8.0 m
mv P s
mg s
The angle between the force and the displacement is 0 for
the pulling force (it points in the same direction as the
displacement) and 180 for the frictional force (it points
opposite to the displacement). Equation 4.8 indicates that the
magnitude of the frictional force is fk = kFN, and we know
that the magnitude of the normal force is FN= mg. With these
substitutions the work-energy theorem becomes
) ) 212kpull f
cos0 cos180P s mg s mv
W W
+
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Problem 26
REASONING The work done by the weight of
the basketball is given by Equation 6.1
as , where F= mgis the
magnitude of the weight, is the angle
between the weight and the displacement, and
sis the magnitude of the displacement. The
drawing shows that the weight and
displacement are parallel, so that = 0.Thepotential energy of the basketball is given byEquation 6.5 as PE = mgh, where his the
height of the ball above the ground.
)cosW F ss
mg
6.1 m
1.5 m
=0.6g
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SOLUTION
a. The work done by the weight of the basketball is
)cosW F s = mg (cos 0)(h0hf)= (0.60 kg)(9.80 m/s2)(6.1 m 1.5 m) = 27 J
b. The potential energy of the ball, relative to the ground,
when it is released is
PE0= mgh0= (0.60 kg)(9.80 m/s2)(6.1 m) = 36 J
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d. The change in the balls gravitational potential energy is
PE = PEfPE0= 8.8 J36 J = 27 J
We see that the change in the gravitational potential
energy is equal to27 J = , where Wis the work
done by the weight of the ball (see part a).W
c. The potential energy of the ball, relative to the ground,
when it is caught is
PEf= mghf= (0.60 kg)(9.80 m/s2)(1.5 m) = 8.8 J
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Problem 34
Total mechanical energy
conserved?
2 21 1
f f 0 02 2
mv mgh mv mgh+ +
) )
)
2 2
f 0 2
14.0 m/s 13.0 m/s1.4 m
2 9.80 m/sh h
Yes (WconservativeFN displacement)
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Problem 41
REASONING Friction and
air resistance are being
ignored. The normal force
from the slide is
perpendicular to the motion,
so it does no work. Thus, no
net work is done by non-
conservative forces, and the
principle of conservation ofmechanical energy applies.
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SOLUTION Applying the principle of conservation of
mechanical energy to the swimmer at the top and the bottom
of the slide, we have
1
2mvf
2 +mghf
Ef
1
2mv0
2 +mgh0
E0
If we let hbe the height of the bottom of the slide above the
water, , and . Since the swimmer starts from
rest, m/s, and the above expression becomes
12vf2 +ghgH
v0 0hf h h0 H
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The vertical displacement of the swimmer after
leaving the slide is, from Equation 3.5b (with down
being negative),
Therefore, h= 1.23 m. Using these values of and h
in the above expression forH, we find
Hh+ vf2
2g 1.23 m + (10.0 m/s)
2
2(9.80 m/s2
) 6.33 m
vf
) ) mssmtay y 23.1500.0/80.92
1
2
1 222
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REASONING AND SOLUTION At the bottom of the
circular path of the swing, the centripetal force isprovided by the tension in the rope less the weight of the
swing and rider. That is,
C
2
F
mvT mg
r
Solving for the mass yields
2Tm
vg
r
+
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The energy of the swing is conserved if friction is ignored.
The initial energy,E0, when the swing is released is
completely potential energy and isE0= mgh0,
Conservation of energy
PEini+ KEini= PEf+ KEf
h0= r(1cos 60.0) =12r
r=2h0
mgh0+ 0 = 0 + (1/2)mvf2
0= 2v gh gr f
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The expression for the mass now becomes
)
2
2
8.00 10 N40.8 kg
2 2 9.80 m/s
Tm
g
gr
ghT
gr
vTm
+
+
02 2
g
T
gg
T
2
+
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Problem 48
m=0.75kg
18.0 m/s
No air friction.
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a. Since there is no air friction, the only force that acts on the
projectile is the conservative gravitational force (its weight). The
initial and final speeds of the ball are known, so the
conservation of mechanical energy can be used to find the
maximum height that the projectile attains.
The conservation of mechanical energy, as expressed by
Equation 6.9b, states that
2 21 1f f 0 02 2
0f
mv mgh mv mgh
E E
+ +
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The mass mcan be eliminated algebraically from this
equation since it appears as a factor in every term. Solving
for the final height hf
gives
)2 21 0 f2f 0
v vh h
g
+
Setting h0= 0 m and vf= 0 m/s, the final height, in the
absence of air resistance, is
) )
)2 22 2
o ff 2
18.0 m/s 0 m/s16.5 m
2 2 9.80 m/ s
v vh
g
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The work-energy theorem is
) )2 21 1nc f 0 f 02 2W mv mv mgh mgh +
b. When air resistance, a non-conservative force, is present, it
does negative work on the projectile and slows it down.
Consequently, the projectile does not rise as high as when thereis no air resistance. The work-energy theorem, in the form of
Equation 6.6, may be used to find the work done by air friction.
Then, using the definition of work, Equation 6.1, the average
force due to air resistance can be found.
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where Wncis the non-conservative work done by air
resistance. According to Equation 6.1, the work can be
written as , where is the average
force of air resistance. As the projectile moves upward, the
force of air resistance is directed downward, so the angle
between the two vectors is = 180and cos =1. Themagnitude sof the displacement is the difference betweenthe final and initial heights, s= hfh0= 11.8 m. With these
substitutions, the work-energy theorem becomes
sFW Rnc )180cos(
) )2 21
R f o f 02F s m v v mg h h +
RF
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Solving for givesRF
) )
) ) ) ) ) ) )
2 21f o f 02
R
2 2 212
0.750 kg 0 m/s 18.0 m/s 0.750 kg 9.80 m/s 11.8 m2.9 N
11.8 m
m v v mg h hF
s
+
+
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Problem 57
REASONING AND SOLUTION One is the amount of
work or energy generated when one kilowatt of power
is supplied for a time of one hour. From
Equation 6.10a, we know that . Using the fact
that and that 1h = 3600 s, we have
W P t1 kW = 1.0 103J/s
1.0 kWh = (1.0 103J/s)(1 h) = (1.0103 J/s)(3600 s) = 3.6106J
bl
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Problem 60
REASONING AND SOLUTION
a. The power developed by the engine is
P= Fv= (2.00 102N)(20.0 m/s) = 4.00 103W
mg
37
friction
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The power developed by the engine is then
P= Fv= (Fa+ mgsin 37.0)v
P= [2.00 102N + (2.50 102kg)(9.80 m/s2)sin 37.0](20.0 m/s)
3.35 10 4 W=
b. The force required of the engine in order to
maintain a constant speed up the slope is
F= Fa+ mg sin 37.0