1. Geo-synchronous Satellite (at the equator of Earth)
Height above the surface of Earth.
ℎ=𝑟− 𝑟 𝐸
Earth mass 𝑀 𝐸 =6𝑥 𝑘𝑔 and radius 𝑟 𝐸 =6380𝑘𝑚
𝑚 𝑠 𝑣 2 𝑟 =𝐺 𝑀 𝐸 𝑚 𝑠 𝑟 2  𝑣= 2𝜋𝑟 𝑇 and 𝑇=1 𝑑𝑎𝑦=86400 𝑠𝑒𝑐
𝑚 𝑠 (2𝜋𝑟) 2 𝑟 𝑇 2 =𝐺 𝑀 𝐸 𝑚 𝑠 𝑟 2 → 𝑟 3 = 𝐺 𝑀 𝐸 𝑇 2 4 𝜋 2 =7.54𝑥 𝑚 3
∴𝑟=4.23𝑥 10 7 𝑚
 ℎ=𝑟− 𝑟 𝐸 =36000 km≈6 𝑟 𝐸
b) What is the velocity?
𝑣= 𝐺 𝑀 𝐸 𝑟 =3070 𝑚 𝑠

2. Chapter 7: Work and Energy
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3. Goals for Chapter 7 Overview energy. Study work as defined in physics.
Relate work to kinetic energy.
Consider work done by a variable force.
Study potential energy.
Understand energy conservation.
Include time and the relationship of work to power.
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4. Work and Energy Work Energy Kinetic Energy Work-Energy Theorem
 Work done by a constant force
(scalar product)
 Work done by a varying force
(scalar product & integrals)
Kinetic Energy
Work-Energy Theorem
Work and Energy

5. Forms of Mechanical Energy
Work and Energy

6. An Overview of Energy Energy is conserved.
Kinetic energy describes motion and relates to the mass of the object and its speed squared.
Energy on earth originates from the sun.
Energy on earth is stored thermally and chemically.
Chemical energy is released by metabolism.
Energy is stored as potential energy in object height and mass and also through elastic deformation.
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7. CONSERVATION OF ENERGY
Work and Energy

8. What is "Work" as Defined in Physics?
Formally, work is the product of a constant force F through a parallel displacement s.
Work is the product of the component of the force in the direction of displacement and the magnitude s of the displacement.
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9. 𝑾= 𝑭 ∙ 𝒅 = 𝑭 ∙ 𝒅 cos ∝
[1 Joule]=[1 N][1 m]
Work = energy
𝑬=𝒎 𝒄 𝟐 (Einstein)
𝑬= 𝟏 𝟐 𝒎 𝒗 𝟐 (Kinetic energy)
Work – energy theorem
𝑾 𝒏𝒆𝒕 = 𝒌 𝟐 − 𝒌 𝟏 =∆𝒌
Force
𝟏 𝐉𝐨𝐮𝐥𝐞= 𝐤𝐠 𝐦 𝐬 𝟐 𝐦 𝟏𝟎 𝟑 𝐠 𝟏 𝐤𝐠 𝟏𝟎 𝟐 𝐜𝐦 𝟏 𝐦 𝟏𝟎 𝟐 𝐜𝐦 𝟏 𝐦 = 𝟏𝟎 𝟕 𝐠 𝐜𝐦 𝟐 𝐬 𝟐 = 𝟏𝟎 𝟕 erg
Power  [Watt]=[Joule/second]

10. Internal Energy Can be "Lost" as Heat
Atoms and molecules of a solid can be thought of as particles vibration randomly on spring like bonds. This vibration is an an example of internal energy.
Energy can be dissipated by heat (motion transferred at the molecular level). This is referred to as dissipation.
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11. Work by a Baseball Pitcher
A baseball pitcher is doing work on
the ball as he exerts the force over
a displacement.
v1 = 0
v2 = 44 m/s
Work and Energy

12. Consider Only Parallel F and S – Figure 7.9
Forces applied at angles must be resolved into components.
W is a scalar quantity that can be positive, zero, or negative.
If W > 0 (W < 0), energy is added to (taken from) the system.
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13. Applications of Force and Resultant Work – Figure 7.10
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14. Work Done by a Constant Force (I)
Work (W)
 How effective is the force in moving a body ?
 Both magnitude (F) and directions (q ) must be taken into account.
W [Joule] = ( F cos q ) d
Work and Energy

15. Clicker question
Q6.7
Two blocks are connected as shown. When released, the 6.00-kg block accelerates downward and the 8.00-kg block accelerates to the right. After each block has moved 2.00 cm, the force of gravity has done
A. work on both blocks, but more work on the 8.00-kg block than on the 6.00-kg block.
B. the same amount of work on both blocks.
C. work on both blocks, but less work on the 8.00-kg block than on the 6.00-kg block.
D. work on the 6.00-kg block only.
E. Not enough information is given to decide.
Answer: D

16. Clicker question Work Done by a Constant Force
Example: Work done on the bag by the person..
 Special case: W = 0 J
a) WP = FP d cos ( 90o )
b) Wg = m g d cos ( 90o )
A. total work>0
B.total work=0
C. total work<0
Work and Energy

17. Example 1A A 50.0-kg crate is pulled 40.0 m by a
constant force exerted (FP = 100 N and
q = 37.0o) by a person. A friction force Ff =
50.0 N is exerted to the crate. Determine
the work done by each force acting on the
crate.
Work and Energy

18. Example 1A (cont’d) F.B.D. WP = FP d cos ( 37o )
Wf = Ff d cos ( 180o )
Wg = m g d cos ( 90o )
WN = FN d cos ( 90o )
180o d
90o
Work and Energy

19. Example 1A (cont’d) WP = 3195 [J] Wf = -2000 [J] (< 0) Wg = 0 [J]
WN = 0 [J]
180o
Work and Energy

20. Example 1A (cont’d) The body’s speed increases. Wnet = SWi
= 1195 [J] (> 0)
The body’s speed increases.
Work and Energy

21. Work Done By Several Forces – Example 7.3
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22. Q6.4
Clicker question
A tractor driving at a constant speed pulls a sled loaded with firewood. There is friction between the sled and the road. The total work done on the sled after it has moved a distance d is
positive.
negative.
zero.
D. two of A, B, and C, depending on circumstances.
E. three of A, B, and C, depending on circumstances.
Answer: C

23. Work-Energy Theorem Wnet = Fnet d = ( m a ) d
= m [ (v2 2 – v1 2 ) / 2d ] d
= (1/2) m v2 2 – (1/2) m v1 2
= K2 – K1
Work and Energy

24. Work and Kinetic Energy
Unbalanced work causes kinematics.
Work-energy theorem:
The kinetic energy K of a particle with mass m moving with speed is During any displacement of the particle, the work done by the net external force on it is equal to its change in kinetic energy.
Although Ks are always positive, Wtotal may be positive, negative, or zero (energy added to, taken away, or left the same).
If Wtotal = 0, then the kinetic energy does not change and the speed of the particle remains constant.
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25. Kinetic energy
Chapter Summary 7.2
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26. Work and Energy Related – Example 7.4
Using work and energy to calculate speed.
Returning to the tractor pulling a sled problem of Example 7.3:
If you know the initial speed, and the total work done, you can determine the final speed after displacement s.
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27. Example 2 A car traveling 60.0 km/h to can brake to
a stop within a distance of 20.0 m. If the car
is going twice as fast, 120 km/h, what is its
stopping distance ?
(a)
(b)
Work and Energy

28. Example 2 (cont’d) (1) Wnet = F d(a) cos 180o
= - F d(a) = 0 – m v(a)2 / 2
 - F x (20.0 m) = - m (16.7 m/s)2 / 2
(2) Wnet = F d(b) cos 180o
= - F d(b) = 0 – m v(b)2 / 2
 - F x (? m) = - m (33.3 m/s)2 / 2
(3) F & m are common. Thus, ? = 80.0 m
Work and Energy

29. Satellite in a circular orbit
Does the Earth do work on the satellite?
Work and Energy

30. Clicker question
What can you say about the work the sun does on the earth during the earth's orbit? (Assume the orbit is perfectly circular.)
The sun does net negative work on the earth.
The sun does net positive work on the earth.
The work done on the earth is positive for half the orbit and negative for the other half, summing to zero.
The sun does no work on the earth at any point in the orbit.
Answer: d) The sun does no work on the earth at any point in the orbit.
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31. Forces on a hammerhead
Forces
Work and Energy

32. Forces on a hammerhead
𝑾=𝑭∙𝒅
𝑾=𝟐𝟎𝟎𝒌𝒈∙𝟗.𝟖 𝒎 𝒔 𝟐 =𝟏𝟗𝟔𝟎𝑵
Weight 𝒘− 𝑭 𝒇𝒓 =𝟏𝟗𝟔𝟎−𝟔𝟎=𝟏𝟗𝟎𝟎𝑵
𝑲 𝟐 − 𝑲 𝟏 = 𝟏 𝟐 𝒎 𝒗 𝟐 𝟐 −𝟎=𝑾= 𝒘− 𝑭 𝒇𝒓 ∙ 𝒅 𝟏𝟐 =𝟏𝟗𝟎𝟎∙𝟑=𝟓𝟕𝟎𝟎𝑵
 𝟏 𝟐 𝒎 𝒗 𝟐 𝟐 =𝟓𝟕𝟎𝟎𝑵 (Falling) and 𝒗 𝟐 = 𝟐∗𝟓𝟕𝟎𝟎 𝟐𝟎𝟎 =𝟕.𝟓𝟓 𝒎 𝒔
𝑲 𝟑 − 𝑲 𝟐 = 𝒘− 𝑭 𝒇𝒓 − 𝑭 𝑵 ∙ 𝒅 𝟐𝟑 (Hitting)
 𝑭 𝑵 =𝒘− 𝑭 𝒇𝒓 − 𝑲 𝟑 − 𝑲 𝟐 𝒅 𝟐𝟑 =𝟏𝟗𝟔𝟎−𝟔𝟎− 𝟎−𝟓𝟕𝟎𝟎 𝟎.𝟎𝟕𝟒
∴ 𝑭 𝑵 =𝟕𝟗𝟎𝟎𝟎𝑵
So, the normal force is 40 times larger than weight of the hammer.

33. Work Done By a Varying Force
In Section 7.2, we defined work done by a constant force.
Work by a changing force is sometimes considered.
On a graph of force as a function of position, the total work done by the force is represented by the area under the curve between the initial and final positions.
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34. Spring Force (Hooke’s Law)FS
Spring Force
(Restoring Force):
The spring exerts its force in the
direction opposite
the displacement. FP
x > 0
Natural Length
x < 0
FS(x) = - k x
Work and Energy

35. Work Done to Stretch a SpringFS FP
FS(x) = - k x
Natural Length
Area ( triangle)
= baseline * height/ 2 W
k x · x/2=A
Work and Energy

36. Work Done By a Varying Force
Hooke's law:
As seen, this is a prime example of a varying force. The work done by a stretching/compressing a spring is equal to the area of the shaded triangle, or
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37. Figure 7.20
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38. Figure 7.21
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39. Example 1A A person pulls on the spring, stretching it
3.0 cm, which requires a maximum force
of 75 N. How much work does the person
do ? If, instead, the
person compresses
the spring cm,
how much work
does the person do ?
Work and Energy

40. Example 1A (cont’d) (a) Find the spring constant k k = Fmax / xmax
= (75 N) / (0.030 m) = 2.5 x 103 N/m
(b) Then, the work done by the person is
WP = (1/2) k xmax2 = 1.1 J
Work and Energy

41. Clicker question
The two cylinders in the figure have masses mA > mB. The larger cylinder is attached to a spring that is stretched a distance s from its equilibrium length. A latch keeps the spring stretched and the system is stationary. What is the system's total mechanical energy?
a) mAghA + mBghB – 1/2khA2 b) mAghA – mBghB + 1/2khA2 c) mAghA + mBghB – 1/2ks2 d) mAghA + mBghB + 1/2ks2
Answer: d) mAghA + mBghB + 1/2ks2
s=compression distance of spring
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42. Example 2 A 1.50-kg block is pushed against a spring
(k = 250 N/m), compressing it m, and
released. What will be the speed of the
block when it separates from the spring at
x = 0? Assume mk =
0.300.
FS = - k x
(i) F.B.D. first !
(ii) x < 0
Work and Energy

43. Example 2 (cont’d) (a) The work done by the spring is
(b) Wf = - mk FN (x2 – x1) = ( )
(c) Wnet = WS + Wf = x 0.200
(d) Work-Energy Theorem: Wnet = K2 – K1
 4.12 = (1/2) m v2 – 0
 v = 2.34 m/s
WS =1/2 k x2 = ½ (250N/m) (0.2 m2 )= J
Work and Energy

44. Potential Energy on an Air Track with Mass and Spring
Refer to Example 7.8.
Knowing the initial state of our system and thus the total mechanical energy, we use this to find the final state at any position.
Using conservation of total mechanical energy:
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45. Conversion and Conservation – Figures 7.27 and 7.28
As kinetic and potential energy are interconverted, dynamics of the system may be solved.
𝐊 𝐢 = 𝟏 𝟐 𝐦 𝐯 𝟐 =𝟔𝟒 𝟎𝟎𝟎𝐉 𝐔 𝐢 =𝐦𝐠𝐡=𝟓𝟖 𝟖𝟎𝟎 𝐉 𝐔 𝐢 + 𝐊 𝐢 = 𝐔 𝐟 + 𝐊 𝐟
J J = 0 + 𝟏 𝟐 𝐤 −𝟑 𝐦 𝟐 𝐤=𝟐.𝟕𝟑𝐱 𝟏𝟎 𝟐 N/m
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46. Clicker question
K increases as it slides down the ramp and remains constant along the horizontal surface.
K remains constant as it slides down the ramp and decreases along the horizontal surface.
K increases as it slides down the ramp and deceases along the horizontal surface
How about the potential energy with the same questions??
Figure 7.28
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47. Q6.6
Clicker question
A block of mass m is placed at rest on a frictionless incline of angle . It is released and slides a distance s along the incline before reaching the floor. The floor exerts a kinetic friction force f on the block. After reaching the floor the block slides a distance x (different from s) and stops. The net work done on the block during its entire motion is m
floor s x
Answer: B
A. positive B. zero C. negative.
D. either positive or negative, depending on the value of f.
E. either positive, negative, or zero, depending on the value of f.

48. Potential Energy and Energy Conservation
Conservative/Nonconservative Forces
 Work along a path
(Path integral)
 Work around any closed path
Potential Energy
Mechanical Energy Conservation
Energy Conservation

49. Gravitational potential energy
Figure 7.23
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50. Q7.1
Clicker question
A piece of fruit falls straight down. As it falls, the
A. gravitational force does positive work on it and the gravitational potential energy increases.
B. gravitational force does positive work on it and the gravitational potential energy decreases.
C. gravitational force does negative work on it and the gravitational potential energy increases.
D. gravitational force does negative work on it and the gravitational potential energy decreases.
E. answer depends on whether or not air resistance is present.
Answer: B

51. Potential Energy
In cases of conservative forces (gravity or elastic forces), there can be "stored" energy due to the spatial arrangement of a system, or potential energy.
Gravitational potential energy (Ugrav), near the surface of the earth can be written:
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52. Potential Energy
The change in the potential energy due to conservative forces is related to the work done by the net force:
If only conservative forces act, then by the work-energy theorem we can define the total mechanical energy:
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53. Work Done by the Gravitational Force (III)
Wg < 0 if y2 > y1
Wg > 0 if y2 < y1
The work done by the gravitational
force depends only on the initial and
final positions..
Energy Conservation

54. Work Done by the Gravitational Force (IV)
Wg(ABCA)
= Wg(AB) +
Wg(BC) +
Wg(CA)
= mg(y1 – y2) +
0 +
mg(y2- y1)
= 0 C B dl A
Energy Conservation

55. Work Done by the Gravitational Force (V)
Wg = 0 for a closed path
The gravitational force is a conservative force.
Energy Conservation

56. Conservative Forces II – Figure 7.35
The work done by a conservative force is independent of the path taken.
When the starting and ending points are the same, the total work is zero.
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57. Work Done by Ff (I) W(friction)= - μmg L L depends on the path. LB
Path B
Path A LA
Energy Conservation

58. Work Done by Ff (II) The work done by the friction force
depends on the path length.
The friction force:
(a) is a non-conservative force;
(b) decreases mechanical energy of the system.
Wf = 0 (any closed path)
Energy Conservation

59. Conservative and Nonconservative Forces
In the previous section, we discussed that if we had only conservative forces acting, then we could used conservation of total mechanical energy.
If we have nonconservative forces which do work, we have to add this to the total energy:
Wother is the work done by nonconservative forces (e.g. friction).
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60. Chapter Summary 7.5
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61. A Solved Baseball Problem – Example 7.7
When the ball, with initial speed i is thrown straight upward, it slows down on the way up as the kinetic energy is converted to potential energy (mgy>0).
At the top, the kinetic energy is zero and potential energy is maximum.
On the way back down, the potential energy is converted back to kinetic energy, and the ball speeds up.
Conservation of total mechanical energy
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62. Q7.5
Clicker question
The two ramps shown are both frictionless. The heights y1 and y2 are the same for each ramp. A block of mass m is released from rest at the left-hand end of each ramp. Which block arrives at the right-hand end with the greater speed?
Answer: C
A. The block on the curved track arrives with greater speed.
B. The block on the straight track arrives with greater speed.
C. Both blocks arrive at the right-hand end with the same speed.
D. The answer depends on the shape of the curved track.

63. Example 7.14
Loading a crate (with friction it does not go all the way up)
K 1 = 1 2 m v 2 = (5) 2 =100 J; K 2 =0; U 1 =0; U 2 =mgh= =62.7 J
W other =− f s =−f d=−f(1.6)
Here; f is the unknown friction force
K 1 + U 1 + W other = K 2 + U 2
100 J+0 −f 1.6 = J  f=23 N
b) W other = W f =(−2) =-74.6 J
K 1 =100 J; U 1 = U 3 =0; K 3 = 1 2 m v 3 2 = v 3 2
100 J+0−74.6J= v
 v 3 =2.5 m s
The return velocity is smaller than the launch velocity due to friction

64. Figure 7.31
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65. Clicker question
On which slide will the speed on the bottom be greatest ? A B C
The speed will be the same
Figure 7.40
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66. Example 1 A 1000-kg roller-coaster car moves from
point A, to point B and then to point C.
What is its gravitational potential energy
at B and C
relative to
point A?
Energy Conservation

67. Wg(AC) = Ug(yA) – Ug(yC)
Wg(ABC) = Wg(AB) + Wg(BC)
= mg(yA- yB) + mg(yB - yC)
= mg(yA - yC) y B A dl B C A
Energy Conservation

68. Problem97. Riding a loop-the-loop starting at rest from a point A, find the minimum height h, so that the car will not fall of the track at the top of the circular part of the loop, which has a radius of 20m.
At the top of the loop at point B
𝐹 𝑟𝑎𝑑 =𝑚𝑔=𝑚 𝑣 2 𝑅 →𝑣= 𝑔𝑅 = 𝑣 𝐵  minimum velocity needed to stay on the track.
Calculate the height ℎ 𝐴 necessary to produce minimum speed 𝑣 𝐵
𝑈 𝐴 + 𝐾 𝐴 = 𝑈 𝐵 + 𝐾 𝐵
𝑚𝑔ℎ 𝐴 +0= 𝑚𝑔ℎ 𝐵 𝑚 𝑣 𝐵 2
𝑔ℎ 𝐴 =𝑔2𝑅 ( 𝑔𝑅 ) 2
ℎ 𝐴 = 5 2 𝑅
If h < 2.5R the car is moving too slow and falls off the track
If h > 2.5R than at point B more downward force than gravity is needed and this is provided by the normal force that the track exerts on the car. B

69. Conservation of Energy
(Sections 7.6 and 7.7) When only conservative forces act on an object, the total mechanical energy (kinetic plus potential) is constant; that is,Ki + Ui = Kf + Uf, where U may include both gravitational and elastic potential energies.
If some of the forces are non conservative, we label
their work as Wother. The change in total energy (kinetic plus potential)of an object, during any motion, is equal to the work Wother done by the non conservative forces: Ki + Ui + Wother = Kf + Uf (Equation 7.17).
Non conservative forces include friction forces, which usually act to decrease the total mechanical energy of a system.
Work and Energy

70. Climbing the Sears tower
Work and Energy

71. Power Power = rate of energy production and consumption = [ 𝐉 𝐬 ]=watt
𝟏 𝒉𝒑≅𝟕𝟒𝟔 𝑾𝒂𝒕𝒕
What do we pay to the electricity company for 100 watt bulb? (11 cents / 1 kwatt.h)
𝟏 𝒌𝒘𝒂𝒕𝒕∙𝒉= 𝟏𝟎 𝟑 𝒘𝒂𝒕𝒕∙𝒉 𝟑.𝟔𝒙 𝟏𝟎 𝟑 𝒔 𝟏𝒉 =𝟑.𝟔𝒙 𝟏𝟎 𝟔 𝑱∙ 𝟏𝒌𝒄𝒂𝒍 𝟒.𝟏𝟖𝒙𝟏𝟎 𝟑 𝑱 =𝟖𝟔𝟎𝒌𝒄𝒂𝒍
Is electricity cheap? 1 kWatt-hour costs 11 cents. The exercise bike shows 100watt. How long must you pedal to produce energy of 1 kWatt-hour?
𝟏𝟎𝟎 𝒘𝒂𝒕𝒕∙𝟏𝟎 𝒉𝒐𝒖𝒓=𝟎.𝟏 𝒌𝒘𝒂𝒕𝒕∙𝟏𝟎 𝒉𝒐𝒖𝒓=𝟏 𝒌𝒘𝒂𝒕𝒕∙𝒉  𝟏𝟏 𝒄𝒆𝒏𝒕𝒔
So; need to pedal for 10 hours.
A power climb: P=? and m=50kg
𝒎𝒈∙𝒉=𝟓𝟎 𝟗.𝟖 𝟒𝟒𝟑=𝟐.𝟏𝟕𝒙 𝟏𝟎 𝟓 𝑱 and 𝐭=𝟏𝟓 𝒎𝒊𝒏 𝟔𝟎 𝒔 𝟏 𝒎𝒊𝒏 =𝟗𝟎𝟎 𝒔
∴𝑷= 𝟐.𝟏𝟕𝒙 𝟏𝟎 𝟓 𝟗𝟎𝟎 =𝟐𝟒𝟏 𝑾𝒂𝒕𝒕

72. The Burj Khalifa is the largest man made structure in the world
and was designed by Adrian
Smith class of 1966
thebatt.com Febuary 25th

73. Clicker question
How many pillars of steel and concrete must be in the foundation? How long and wide it should be, so this building can be constructed on desert sand? What are the friction forces, which must be present, so it does not sink in the dessert sand?

74. Work and Energy