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Chapter 6
Work and Energy
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Contents of Chapter 6
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Kinetic Energy, and the Work-Energy Principle
Potential Energy
Conservative and Nonconservative Forces
Mechanical Energy and Its Conservation
Problem Solving Using Conservation of Mechanical
Energy
• Other Forms of Energy and Energy Transformations;
the Law of Conservation of Energy
• Energy Conservation with Dissipative Forces: Solving
Problems
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Work is force x distance but the force must move the
object in same direction that the force is being
applied.
6-3 Kinetic Energy and the
Work-Energy Principle
Energy was traditionally defined as the ability to do
work. We now know that not all forces are able to do
work; however, we are dealing in these chapters with
mechanical energy, which does follow this definition.
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6-3 Kinetic Energy and the
Work-Energy Principle
• If we write the acceleration in terms of the velocity
and the distance, we find that the work done here is
(6-2)
• We define the kinetic energy:
(6-3)
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6-3 Kinetic Energy and the
Work-Energy Principle
• This means that the work done is equal to the change
in the kinetic energy:
(6-4)
• If the net work is positive, the kinetic energy
increases.
• If the net work is negative, the kinetic energy
decreases.
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6-3 Kinetic Energy and the
Work-Energy Principle
Because work and kinetic energy can be equated, they
must have the same units: kinetic energy is measured in
joules.
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Example 6-4
A 145g baseball is thrown with a speed of 25m/s.
A) What is its KE?
Given:mass=145g=0.145kg; v=25m/s
Formula: KE=½mv2
Substitution: KE=½(0.145kg)(25m/s)2
Answer w/unit: 45.3J
B) How much work was done on the ball to make it reach this
speed , if it started from rest?
Given:mass=145g=0.145kg; final v=25m/s;initial vo=0m/s
Formula: Wnet= ½mv22 - ½mv12
Substitution: Wnet=½(0.145kg)(25m/s)2 - ½(0.145kg)(0m/s)2
Answer w/unit: 45.3J
Remember: Use W-E Theorem only when solving for net work!
Example 6-5
How much work is required to accelerate a 1000kg car
from 20m/s to 30m/s?
Given:mass=1000kg; final v=30m/s; initial vo=20m/s
Formula: Wnet= ½mv22 - ½mv12
Substitution:
Wnet=½(1000kg)(30m/s)2 - ½(1000kg)(20m/s)2
Answer w/unit: 2.50 x 105J
Example 6-6
An automobile traveling 60km/h can brake to a stop
within a distance of 20m. If the car is going twice as
fast, what is its stopping distance? The maximum
braking force is approximately independent of speed.
Since speed is
squared in the
equation and the
speed is doubled,
the stopping
distance will be 4x
as great.
20m x 4 = 80m
6-4 Potential Energy
An object can have potential energy by virtue of its
surroundings.
Familiar examples of potential energy:
• A wound-up spring
• A stretched elastic band
• An object at some height above the ground
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6-4 Potential Energy
In raising a mass m to a height h,
the work done by the external
force is
(6-5a)
We therefore define the
gravitational potential energy:
(6-6)
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6-4 Potential Energy
This potential energy can become kinetic energy if the
object is dropped.
Potential energy is a property of a system as a whole, not
just of the object (because it depends on external forces).
If PEG = mgy, where do we measure y from?
It turns out not to matter, as long as we are consistent
about where we choose y = 0. Only changes in potential
energy can be measured.
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Example 6-7
A 1000kg roller-coaster car moves from point A to point
B and then to point C.
a) What is its gravitational PE at B and C relative to
point A?
Given:mass=1000kg; yB=10m; yC=-15m; g=9.81m/s2
B
Formula: PE=mgy
Substitution:
A
PEB=1000kg (9.81m/s2) (10m)
PEC=1000kg (9.81m/s2) (-15m)
Answer w/unit:
PEB= 9.80 x 104J
PEC= -1.47 x 105J
C
Example 6-7
A 1000kg roller-coaster car moves from point A to point
B and then to point C.
b) What is the change in PE when it goes from B to C?
Given: PEB= 9.80 x 104J; PEC=-1.47 x 105J
Formula: ΔPE=PEf – Pei=PEC - PEB
B
Substitution:
ΔPE=-1.47 x 105J – 9.80 x 104J
A
Answer w/unit:
-2.45 x 105J
C
Example 6-7
A 1000kg roller-coaster car moves from point A to point B and then to
point C.
c) Repeat parts a) and b), but take the reference point to be at point 3.
a) Given:mass=1000kg; yB=25m; yC=0m; g=9.81m/s2
Formula: PE=mgy
Substitution:
PEB=1000kg (9.81m/s2) (25m)
B
PEC=1000kg (9.81m/s2) (0m)
Answer w/unit:
PEB= 2.45 x 105J; PEC= 0J
A
b) Given: PEB= 2.45 x 105J; PEC=0J
Formula: ΔPE=PEf – Pei=PEC - PEB
Substitution:
ΔPE=0J – 2.45 x 105J
Answer w/unit:
C
-2.45 x 105J
6-4 Potential Energy
Potential energy can also be stored in a spring when it is
compressed; the figure below shows potential energy
yielding kinetic energy.
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6-4 Potential Energy
The force required to
compress or stretch a spring
is:
(6-8)
where k is called the spring
constant, and needs to be
measured for each spring.
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6-4 Potential Energy
The force increases as the spring is stretched or
compressed further. We find that the potential energy of
the compressed or stretched spring, measured from its
equilibrium position, can be written:
(6-9)
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Elastic Potential Energy Problew
(complete in notebook)
A spring has a a spring constant k of 88.0 N/m. How
much must this spring be compressed to store 45.0J of
potential energy?
Given:spring constant (k)=88.0 N/m; PEel=45.0J
Formula: PEel=½kx2
Substitution: 45.0J=½(88.0 N/m)(x)2
Answer w/unit: 1.01m
6-5 Conservative and Nonconservative Forces
If friction is present, the work done depends not only on
the starting and ending points, but also on the path taken.
Friction is called a nonconservative force.
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6-5 Conservative and Nonconservative Forces
Potential energy can
only be defined for
conservative forces.
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6-5 Conservative and Nonconservative Forces
Therefore, we distinguish between the work done by
conservative forces and the work done by
nonconservative forces.
We find that the work done by nonconservative forces is
equal to the total change in kinetic and potential
energies:
(6-10)
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6-6 Mechanical Energy and Its Conservation
If there are no nonconservative forces, the sum of the
changes in the kinetic energy and in the potential energy
is zero—the kinetic and potential energy changes are
equal but opposite in sign.
This allows us to define the total mechanical energy:
E = KE + PE
And its conservation:
(6-12b)
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6-7 Problem Solving Using Conservation of
Mechanical Energy
In the image on the left, the total
mechanical energy is:
E = KE + PE = ½ mv2 + mgy
The energy buckets on the right
of the figure show how the
energy moves from all potential
to all kinetic.
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Example 6-8
If the original height of the stone in the figure is h=3.0m,
calculate the stone’s speed when it has fallen to 1.0m
above the ground.
Given: y1=3.0m; y2=1.0m; v1=0m/s
Strategy: Apply conservation of ME and with only gravity
acting on the stone, we choose the ground as a
reference level where h=0.
Formula: ½mv21 + mgy1= ½mv22 + mgy2
(m’s cancel out (same mass) and since v1=0, the first part
of the equation cancels)
Substitution: 0 + gy1= ½v22 + gy2
(9.81m/s2)3.0m = ½v22 + 9.81m/s2(1.0m)
Answer w/unit: 6.26m/s
6-7 Problem Solving Using Conservation of
Mechanical Energy
If there is no friction, the speed of a roller coaster will
depend only on its height compared to its starting height.
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Example 6-9
Assuming the height of the hill in the figure is
40m, nad the rollercoaster car starts from
rest at the top, calculate
a) The speed of the rollercoaster car at the
bottom of the hill
Given: y1=40m; y2=0m; v1=0m/s
Formula: ½mv21 + mgy1= ½mv22 + mgy2
(m’s cancel out (same mass) and since v1 and y2=0,
those parts of the equation cancels)
Substitution: 0 + gy1= ½v22 + 0
(9.81m/s2)40m = ½v22
Answer w/unit: 28.0m/s
b) at what height it will have half this speed.
Given: y1=40m; v1=0m/s; v2=14.0m/s
Formula: ½mv21 + mgy1= ½mv22 + mgy2
(m’s cancel out (same mass) and since v1, that part of the equation cancels)
Substitution: 0 + gy1= ½v22 + gy2 ;
(9.81m/s2)40m = ½(14.0)2 + (9.81m/s2)y2
Answer w/unit: 30.0m
Example 6-10
Two water slides at a pool are shaped
differently but start at the same height. Two
riders, Paul and Kathleen, start from rest at
the same time on different slides. Ignore
friction and assume both slides have the
same path length.
a) Which rider is traveling faster at the bottom?
Since both riders are traveling the same length,
they both will be traveling at the same
speed at the bottom.
b) Which rider makes it to the bottom first?
Kathleen will make it to the bottom first. She is
at a lower elevation than Paul throughout
the whole slide.
Kathleen
Example 6-11
Estimate the KE and the speed required for a 70kg pole
vaulter to just pass over a bar 5.0m high. Assume the
vaulter’s center of mass is initially 0.90m off the ground
and reaches its maximum height at the level of the bar
itself.
Strategy: Find speed then KE.
Find Speed
Given: y1=0.90m; y2=5m; v2=0m/s(max height)
Formula: (9.81m/s2)0.90m + mgy1= ½mv22 + mgy2
(m’s cancel out (same mass) and since v2, that part of the
equation cancels)
Substitution: ½v21 + gy1= 0 + gy2
½v21 + (9.81m/s2)0.90m = (9.81m/s2)5m
Answer w/unit: 8.96m/s
Find KE
Given: mass=70kg; speed=8.96m/s
Formula: KE= ½mv2
Substitution: KE = ½(70kg)(8.96m/s)2
Answer w/unit: 2810J
6-7 Problem Solving Using Conservation of
Mechanical Energy
For an elastic force, conservation of energy tells us:
(6-14)
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Example 6-12
A dart of mass 0.100kg is pressed against the spring of
a toy dart gun. The spring (k=250 N/m) is
compressed 6.0cm and released. If the dart
detaches from the spring when the latter reaches
its normal length (x=0), what speed does the dart
acquire?
Given: m=0.100kg; k=250 N/m;
x=6.0cm=0.06m; v1=0m/s; x2=0m
Formula:
½mv21 + ½kx12 = ½mv22 + ½kx22
(Since v1 and x2=0, those parts of the
equation cancels)
Substitution: 0 + ½kx12 = ½mv22 + 0
½(250 N/m)(0.06m)2 = ½(0.100kg)v22
Answer w/unit: 3.00m/s
Example 6-13
For this problem, y is used instead of x in the elastic equation for
consistency and image.
A ball of mass m=2.60kg, starting from rest, falls a
vertical distance h=55.0cm before striking a
vertical coiled spring, which it compresses an
amount Y=15.0cm. Determine the spring
constant of the spring. Assume the spring has
negligible mass. Measure all distances from the
point where the ball first touches the
uncompressed spring. (y=o at this point)
Given: m=2.60kg; y1=55.0cm=0.55m;
y3=-15cm=-0.15m; v1=0m/s; y2=0m; v3=0m/s
Formula:
½mv21+mgy1+½ky22 = ½mv23+mgy3+½ky32
(Since v1, y2, and v3=0, those parts of the
equation cancels)
Substitution: 0+mgy1+0=0+mgy3+½ky32
2.60kg(9.81m/s2)0.55m=(2.60kg)(9.81m/s2)(-0.15m)+½k-0.15m
Answer w/unit: 1590 N/m
Example 6-14
Your looking for total height fell (d). Y1=15m+x and x2=the length stretched.
Dave jumps off a bridge with a bungee cord tied
around his ankle. He falls for 15m before the
bungee cord begins to stretch. Dave’s mass is
75kg and we assume the cord obeys Hooke’s
law, F=-kx, with k=50N/m. If we neglect air
resistance, estimate how far below the bridge
Dave will fall before coming to a stop. Ignore
the mass of the cord.
Given: m=75kg; y1=15m+x; k=50N/m;
v1=0m/s; y2=0m; v2=0m/s
Formula:
½mv21+mgy1+½ky22 = ½mv23+mgy3+½ky32
You should notice the need
(Since v1, y2, and v2=0, those parts of the
for a Quadratic equation.
equation cancels)
Out of your two answers,
Substitution: 0+mgy1+0=0+0+½kx22
only 40.3 makes sense.
75kg(9.8m/s2)0.15m+x=½50x22=11025m+735x=25x2
0=25x2-735x-11025; Answers from Quad. Eq: 40.3 and -10.9
Total distance=15m + x=15m + 40.3m=
Answer w/unit: 55.3m
6-8 Other Forms of Energy
and Energy Transformations;
the Law of Conservation of Energy
Some other forms of energy:
• Electric energy, nuclear energy, thermal energy,
chemical energy.
Work is done when energy is transferred from one object
to another.
Accounting for all forms of energy, we find that the total
energy neither increases nor decreases. Energy as a
whole is conserved.
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6-9 Energy Conservation with Dissipative
Processes; Solving Problems
If there is a nonconservative force such as friction,
where do the kinetic and potential energies go?
They become heat; the actual temperature rise of the
materials involved can be calculated.
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6-9 Energy Conservation with Dissipative
Processes; Solving Problems
Problem solving:
1. Draw a picture.
2. Determine the system for which energy will be
conserved.
3. Figure out what you are looking for, and decide on
the initial and final positions.
4. Choose a logical reference frame.
5. Apply conservation of energy.
6. Solve.
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Example 6-15
The rollercoaster car in example 6-9 is found to
reach a vertical height of only 25m on the
second hill before coming to a stop. It traveled a
total distance of 400m. Estimate the average
friction
force on the
whose
is energy
1000kg.will be conserved. The system is
1. Determine
thecar,
system
formass
which
the rollercoaster and the Earth. The forces acting are gravity and friction.
PE is from gravity and friction will be determined by work formula (Ffrd).
2.
Figure out what you are looking for, and decide on the initial and final
positions. Point 1 is when the car begins and point 2 is when it stops at
25m.
3.
Choose a logical reference frame. At lowest point, y=0.
4.
Apply conservation of energy. Since ME isn’t conserved because of
friction, you will use the conservation equation that includes it.
5.
Solve.
a)
given: h1=40m; h2=25m; d=400m; m=1000kg;v1=0m/s;v2=0m/s
b)
formula: ½mv12+mgh1=½mv22+mgh2+Ffrd
c)
Substitution: 0+1000kg(9.81m/s2)40m=0+1000kg(9.81m/s2)25m+Ffr400m
d)
Answer w/unit: 368N
Summary of Chapter 6
• Kinetic energy is energy of motion: KE = ½ mv2
• Potential energy is energy associated with forces that
depend on the position or configuration of objects.
• The net work done on an object equals the change in
its kinetic energy.
• If only conservative forces are acting, mechanical
energy is conserved.
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