1. Chapter 6 Work and Energy
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2. Contents of Chapter 6 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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3. Work is force x distance but the force must move the object in same direction that the force is being applied.

4. 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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5. 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
We define the kinetic energy:
(6-2)
(6-3)
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6. 6-3 Kinetic Energy and the Work-Energy Principle
This means that the work done is equal to the change in the kinetic energy:
If the net work is positive, the kinetic energy increases.
If the net work is negative, the kinetic energy decreases.
(6-4)
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7. 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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8. Example 6-4 A 145g baseball is thrown with a speed of 25m/s.
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
Remember: Use W-E Theorem only when solving for net work!

9. 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

10. 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

11. 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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12. 6-4 Potential Energy
In raising a mass m to a height h, the work done by the external force is We therefore define the gravitational potential energy:
(6-5a)
(6-6)
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13. 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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14. Example 6-7
A 1000kg roller-coaster car moves from point A to point B and then to point C.
What is its gravitational PE at B and C relative to point A?
Given:mass=1000kg; yB=10m; yC=-15m; g=9.81m/s2
Formula: PE=mgy
Substitution:
PEB=1000kg (9.81m/s2) (10m)
PEC=1000kg (9.81m/s2) (-15m)
Answer w/unit:
PEB= 9.80 x 104J
PEC= x 105J B A C

15. 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
Substitution:
ΔPE=-1.47 x 105J – 9.80 x 104J
Answer w/unit:
-2.45 x 105J B A C

16. 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)
PEC=1000kg (9.81m/s2) (0m)
Answer w/unit:
PEB= 2.45 x 105J; PEC= 0J
b) Given: PEB= 2.45 x 105J; PEC=0J
Formula: ΔPE=PEf – Pei=PEC - PEB
ΔPE=0J – 2.45 x 105J
-2.45 x 105J B A C

17. 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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18. 6-4 Potential Energy
The force required to compress or stretch a spring is:
where k is called the spring constant, and needs to be measured for each spring.
(6-8)
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19. 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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20. 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

21. 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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22. 6-5 Conservative and Nonconservative Forces
Potential energy can only be defined for conservative forces.
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23. 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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24. 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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25. 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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26. 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 = ½v m/s2(1.0m)
Answer w/unit: 6.26m/s

27. 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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28. 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
The speed of the rollercoaster car at the bottom of the hill
at what height it will have half this speed.
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
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

29. 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

30. 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

31. 6-7 Problem Solving Using Conservation of Mechanical Energy
For an elastic force, conservation of energy tells us:
(6-14)
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32. 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

33. Example 6-13 Given: m=2.60kg; y1=55.0cm=0.55m;
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

34. 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
(Since v1, y2, and v2=0, those parts of the
equation cancels)
Substitution: 0+mgy1+0=0+0+½kx22
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 m=
Answer w/unit: 55.3m
You should notice the need for a Quadratic equation. Out of your two answers, only 40.3 makes sense.

35. 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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36. 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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37. 6-9 Energy Conservation with Dissipative Processes; Solving Problems
Problem solving:
Draw a picture.
Determine the system for which energy will be conserved.
Figure out what you are looking for, and decide on the initial and final positions.
Choose a logical reference frame.
Apply conservation of energy.
Solve.
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38. 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 car, whose mass is 1000kg.
Determine the system for which energy will be conserved. The system is 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).
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.
Choose a logical reference frame. At lowest point, y=0.
Apply conservation of energy. Since ME isn’t conserved because of friction, you will use the conservation equation that includes it.
Solve.
given: h1=40m; h2=25m; d=400m; m=1000kg;v1=0m/s;v2=0m/s
formula: ½mv12+mgh1=½mv22+mgh2+Ffrd
Substitution: kg(9.81m/s2)40m=0+1000kg(9.81m/s2)25m+Ffr400m
Answer w/unit: 368N

39. 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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