1. Topics in textbook Actual Work Definition of Work Potential Energy
Work on Particle
Potential Energy
Definition of Energy
Work-Energy Equation
Kinetics Energy
Definition of Energs

2. Three approaches for solving dynamics
Newton’s
2nd Law
Kinematics Eq:
path
1) Direct Method A
From 2nd Law
(kinetics Eq)
work (and potential energy)
of Force i along the path
2) Work and Energy
From 2nd Law
3) Impulse and Momentum
linear impulse of Force i
From 2nd Law

3. Work and Energy Usually convenient From 2nd Law change of
Newton’s
2nd Law
Work and Energy
Usually convenient
when F = F(s), and you want to find velocity at final state
(without finding acc. first).
From 2nd Law
change of
kinetic energy
kinetic
energy
at B
path B
Work of Force i along the path
kinetic
energy
at A A
Work of a force
during small displacement
Principle of work and Energy
kinetic energy

4. Work & Energy Work Energy “something” successful done by force.
The objective of force is to create the movement. a
Work
[Unit: Newton- Meter (J: joule) ]
“something” successful done by force.
- it depends on
and
( “indicator” of successiveness of force )
- “Indicator” has sign
Energy
- Capacity/Potential to do the work.
( the work is not yet done)

5. Work done on Particle Work by a force Work done over particle A
path P
(inactive force)
Work done over particle A
Since ,
the total work done on object is ……
sum of works
done by all forces
over the particle A

6. Note on work is positive when has the same direction.
path
is positive when
has the same direction.
Unit of work is N-m or Joule (J).
Active force is the force that does the work
Reactive force = constrain force that does not do the work

7. A (spring stretched length is 0.5m)
The 10-kg block rest on a smooth incline. If the spring is originally stretched 0.5 m, determine the total work done by all forces acting on the block when a horizontal force P = 400 N pushes the block up the plane s = 2 m. The block is not tipping. B
A (spring stretched length is 0.5m)
Pos B
Horizontal Force P: constant
Pos A
Weight W: constant.
“Active Force”
Spring Force Fs.: varying Force
Normal Force NB : constant
“Inactive Force”

8. The 10-kg block rest on a smooth incline
The 10-kg block rest on a smooth incline. If the spring is originally stretched 0.5 m, determine the total work done by all forces acting on the block when a horizontal force P = 400 N pushes the block up the plane s = 2 m.
P = 400 N constant along the path
scalar calculation (aware of +/- )
Spring = k(0.5+x) x=0 to 0.5
mg = (10)(9.81) constant along the path mg
Does not do the work
P =400N
Spring
Normal Force N
This force does no work since it is always perpendicular to the displacement.

10. Work done on Particle t Work done over particle A path sum of works
done by all forces
over the particle A P
Pos B
Pos A
Work done on particle P during path A->B,
is to increase kinetic energy of particle

11. Work and Energy Usually convenient From 2nd Law change of
Newton’s
2nd Law
Work and Energy
Usually convenient
when F = F(s), and you want to find velocity at final state
(without finding acc. first).
From 2nd Law
change of
kinetic energy
kinetic
energy
at B
path B
Work of Force i along the path
kinetic
energy
at A A
Work of a force
during small displacement
Principle of work and Energy
kinetic energy

12. Kinetics Energy Advantage
path A
T is the work done on a particle to accelerate it from rest to the velocity v
Principle of work and Energy
Unit of T is N-m or Joule (J)
Advantage
Scalar equation. (1 unknown)
Integral Equation (not instantaneous eq like 2nd Law)
No need to find acceleration first
Get change in velocity directly from active forces.
it can be applied to system of particles with
frictionless and non-deformable links

13. Motivation : Work-Energy Equation
Solve
for every instant t
(Newtonian Equation is instantaneous Equation)
We have to write FBD of
general position of time t
initial
final
New way: Use Work-Energy Equation

14. SP3/3 The block A is released from rest, and pulls the 200-kg log up
SP3/3 The block A is released from rest, and pulls the 200-kg log up. Determine the velocity of the block A when it hits the ground
Static? The pulley C is moving T T 2T 2T
200(g) T N
125(g)
constant at any time t A

16. Three approaches for solving dynamics
Newton’s
2nd Law
Kinematics:
Direct Method
From 2nd Law
work (and potential energy)
of Force i along the path
Work and Energy
From 2nd Law
Impulse and Momentum
linear impulse of Force i
From 2nd Law

17. Advantage of Work-Energy Eq.
Principle of work and Energy
Advantage of Work-Energy Eq.
No need to find acceleration first
Work-Energy Eq.
Get change in velocity directly from active forces.
can be applied to system of particle using frictionless and non-deformable links
path A
Usually convenient
when F = F(s)

19. How to calculate Work In general In xyz-coord In nt-coord In rq-coord
path
In general
displacement x force component in the direction of displacement
In xyz-coord
scalar
(be careful
Of +/-)
displacement in the direction of force
x force component
In nt-coord
In rq-coord

20. M3/107) Calculate the work done on 10-kg object with the constant Force ( F= 8N ) during the curve path AB.
x-y y x mg
F=8N (const) N
Does not do the work
Ans
If F is not constant, how to calculate it?

21. If F is not constant y x
More general case or

22. M3/107) Calculate the work done by F during the curve path AB.
n-t
Fn doest not effect works! s
engine thrust a

23. M3/107) Calculate the work done by F during the curve path AB.
r-q
r-q reference point
central force

24. M3/107) Calculate the work done by F during the curve path AB.
n-t co-ord s N
engine thrust
central force
r-q reference point

25. M3/121) The 0.2-kg slider moves freely along the fixed curved rod from A to B in the vertical plane under the action of the constant 5-N tension in the cord. If the slider is released from rest at A, calculate its velocity v as it reaches B. mg F
Work-Energy Eq.
general
position N
Does not
do the work
r-q coordinate
reference point

26. Hibbeler Ex. 14.1 Given: x1 = 0.5 m P = 400 N s = 2 m
Find: Total work done by all forces
Plan: 1) Draw a FBD
2) Calculate for work done by each forces (Horizontal P, Spring force Fs, Weight W, and Normal force N). Then sum all works together.

27. Hibbeler Ex (cont.)

28. Hibbeler Ex (cont.)

29. Hibbeler Ex (cont.)

30. Work and Curvilinear Motion

32. Advantage of Work-Energy Eq.
No need to find acceleration first
Get change in velocity directly from active forces.
Work-Energy Eq. is a scalar equation.
it can be applied to system of particles using frictionless and non-deformable links

33. Work on frictionless connected particles
Only the external forces are needed to calculate the total work on a system of particles.
(If frictions exist, the sum of action and reaction of the friction may not be zero.)
internal force R and –R will have the same displacement.
So, the sum of these works are zero.

34. initial state
The system starts from rest at Configuration 1. Find the velocity of A at configuration where d = 0.5 m . F is 20 N (constant)
Final state
N causes
no work!
System selection is not so good
(you have to calculate Tension T for its work)

35. M3/131) The ball is released from position A with a velocity of 3m/s and swings in a vertical plane. At the bottom position, the cord strikes the fixed bar at B, and the ball continues to swing in the dashed arc. Calcuate the velocity v of the ball as it passes position C.
Work-Energy Eq. T
does
no work
system mg
r-q coordinate
reference point
system

37. Power Power is defined as time rate of work
path
Power is defined as time rate of work A
(scalar quantity)
For a machine, power tells how much work it can do in a period of time.
(small machine can deliver lots of energy given enough time)
Unit of power: Watt (W) = J/s = N-m/s

38. Mechanical Efficiency
If energy applied to the machine occurs during the same time interval at which it is removed.
Since machine consists of moving parts which may have frictions, so extra energy or power is needed to overcome the frictions.

39. A car has a mass of 2 Mg and an engine efficiency of e = 0. 65
A car has a mass of 2 Mg and an engine efficiency of e = The car uniformly accelerates at 5 m/s2, starting from rest. During that constant acceleration, the wind outside creates a drag resistance on the car of FD = 1.2v2 N, where v is the velocity in m/s. Find the engine output input when t=4 s. a x
Constant
acceleration:

40. A 50-N load (B) is lifted up by the motor from rest until the distance is 10 m. The motor M has an efficiency of 0.76 and exerts a constant force of 30 N. Find the power supplied to the motor at that instant. Neglect the mass of the pulleys and cable.
v=?
F = 30 N
(const)
Energy
Approach
2F= 2(30)
50 N B
50N
s = 10 m
(start from rest)

42. summation of all forces
Work and Energy
kinetic
energy
at B
path B
kinetic
energy
at A A
summation of all forces
We found that ….
It is much easier to solve dynamic problem, if we think the work done by spring and gravity force in the form of Potential Energy
Work from
Gravity Force
Work from spring
Work from all other forces
(not spring & gravitation)
Gravitational Potential Energy
Elastic
Potential Energy
kinetic energy

43. Work of Gravity Force
Only depends on
position at
final state (2)
Only depends on
position at initial state (2)
> 0
any path 1
energy level (higher)
W=mg
Work done by W , only depends on the initial state position and final state position only, i.e. , it does not depends on actual path 2
energy level (lower) h
Think in Term of “Potential Energy”
(for convenience)
Fixed reference line
point function
Potential Energy
- Energy from gravity field
Work done by
Gravity Force:
from position 1
to position 2
Energy “Emission”:
from position 1 to position 2
Work = “Energy in Transfer”
Work by couple is
- positive if M has the same sense as dq
- negative if M has the inverse sense as dq
Work done over Object 43

44. when change in g is significant
Define as negative of work done from the position to
the potential energy at r is
from

45. position at initial state (1)
(direction may be opposite)
Work of Spring Force
Only depends on
position at
final state (2)
Only depends on
position at initial state (1) 1 L
natural length
(unstretched length)
Work done by Spring , depends only on the initial state and final state only, i.e. ,
it does not depends on actual path 2
any path
x : distance , stretched or compressed from natural length
Think in Term of “Energy”
(for convenience)
point function
Work done by Spring Force:
from position 1 to position 2
Energy Emission:
from position 1 to position 2
Work done by Gravity Force:
from position 1 to position 2
Energy Emission:
from position 1 to position 2

47. Work-Energy Equation FBD ** FBD N N Work-Energy Equation
(Use Energy Concept)
FBD
Not Recommended Method in this course N N
(Conservative Force)
Think of Energy
Work-Energy Equation
(2nd Form)
Work-Energy Equation
(1st Form)
Virtual work by non-conservative forces.
Energy Concept
final
location
initial
location

48. Work-Energy Equation
(1st Form)
FBD
Work-Energy Equation
(2nd Form) N
(Conservative Force)
Think of Energy
The virtual work done by external active forces (other than gravitational and spring forces accounted for in the potential energy terms) on an ideal mechanical system in equilibrium, equals the corresponding change in the total elastic and gravitational potential energy of the system for any and all virtual displacements consistent with the constraints.

49. M3/173) The 0.6-kg slider is released from rest at A and slides down under the influence of its own weight and of the spring of k = 120 N/m. Determine the speed of the slider and the normal force at point B. The unstrecthed length of the spring is 200 mm.
gravitational potential datum
At position B
gravitational
Potential datum

51. Hibbeler Ex (cont.)

52. H14/92) The 1-kg collar has a speed of 1. 5 m/s at A
H14/92) The 1-kg collar has a speed of 1.5 m/s at A. the attached spring has an unstretched length of 0.6 m and a stiffness of k = 150 N/m. When it reaches point B, determine its velocity, the rate of decrease in its speed, and the normal force of the collar to the rod. mg N F mg
F=kx N

53. M3/172) The cars of an amusement-park ride have a speed v=90 km/h at the lowest part of the track. Determine their speed v at the highest part of the track. Neglect energy loss due to friction.
CFO system
gravitational potential datum

55. Hibbeler Ex (cont.)
Datum
an = v2/,  = 20 m

57. Integral Equation (not instantaneous equation like 2nd Law)
Advantage
Integral Equation (not instantaneous equation like 2nd Law)
Scalar equation. (easy to handle with1 unknown)
Get change in velocity directly.
(No need to find acceleration first)
Handle with only active forces.
it can be applied to system of particles with
frictionless and non-deformable links
We will see this later, when applying at system of particles

58. Work on frictionless connected particlesB C O
Only the external forces are needed to calculate the total work on a system of particles.
(If frictions exist, the sum of action and reaction of the friction may not be zero.)
internal force R and –R will have the same displacement.
So, the sum of these works are zero.

59. Work on frictionless connected particles
Only the external forces are needed to calculate the total work on a system of particles.
(If frictions exist, the sum of action and reaction of the friction may not be zero.)
internal force R and –R will have the same displacement.
So, the sum of these works are zero.

60. initial state
Final state
The system starts from rest at Configuration 1. Find the velocity of A at configuration where d = 0.5 m . F is 20 N (constant) T T T T F F
Object A
Object B
system
T is internal force F
(excluding from Work Calculation)
We have no interest in T, thus object separation
(separating object A and B) is not good in this problem.

61. (Active Force Diagram )
Internal forces (action/reaction pair) are cancelled with each other.
Work on Connected Rigid Bodies
AFD
(Active Force Diagram )
Active Force:
Forces that do work. A B C O
Constrain Force (reactive force) causes no work because the movement is not possible.
friction-less system
AFD

62. Work on Connected Rigid Bodies
Internal forces (action/reaction pair) are cancelled with each other.
Work on Connected Rigid Bodies
AFD
(Active Force Diagram )
Active Force:
Forces that do work. A B C O
Constrain Force (reactive force) causes no work because the movement is not possible.
friction-less system
AFD

63. (Active Force Diagram )
Internal forces (action/reaction pair) are cancelled with each other.
Work on Connected Rigid Bodies
AFD
(Active Force Diagram )
Active Force:
Forces that do work. A B C O
Constrain Force (reactive force) causes no work because the movement is not possible.
friction-less system
AFD

65. M3/158) If the system is released from rest, determine the speeds of both masses after B have move 1 m. Neglect friction and the masses of pulleys.
System: block A + block B + cord+ 2 Pulleys up
Position A: at rest
(assume)
Position B: block B moves down as 1 meter
datum
unsolvable

66. M3/158) If the system is released from rest, determine the speeds of both masses after B have move 1 m. Neglect friction and the masses of pulleys.
System: block A + block B + cord+ 2 Pulleys up
Position A: at rest
(assume)
Position B: block B moves down as 1 meter
datum

67. H14/16) Block A rest on a surface which has friction
H14/16) Block A rest on a surface which has friction. Determine the distance d cylinder B must move down so that A has a speed of
starting from rest.
20 N
System: block A + block B + cord+ 2 Pulleys
50 N
Position A: at rest
Position B: block B moves down as d meter

68. 3/168) The system is released from rest with q=180, where the uncompressed spring of stiffness k= 900 N/m is just touch the underside of 4-kg collar. Determine the angle q corresponding to the maximum spring compression.
O2-1
O2-2 O1 r L r
datum
System: O1+O2+O3+4 rods
Position A: at rest with q=180
Position B: maximum compression

70. Power Power is defined as time rate of work
path
Power is defined as time rate of work A
(scalar quantity)
For a machine, power tells how much work it can do in a period of time.
(small machine can deliver lots of energy given enough time)
Unit of power: Watt (W) = J/s = N-m/s

71. Mechanical Efficiency
If energy applied to the machine occurs during the same time interval at which it is removed.
Since machine consists of moving parts which may have frictions, so extra energy or power is needed to overcome the frictions.

72. A 50-N load (B) is lifted up by the motor from rest until the distance is 10 m. The motor M has an efficiency of 0.76 and exerts a constant force of 30 N. Find the power supplied to the motor at that instant. Neglect the mass of the pulleys and cable.
v=?
No!
F = 30 N
(const)
50N
2F= 2(30)
Energy
Approach
s = 10 m
(start from rest)
F = 30 N
(const)
50 N
50 N

73. A car has a mass of 2 Mg and an engine efficiency of e = 0. 65
A car has a mass of 2 Mg and an engine efficiency of e = The car uniformly accelerates at 5 m/s2, starting from rest. During that constant acceleration, the wind outside creates a drag resistance on the car of FD = 1.2v2 N, where v is the velocity in m/s. Find the engine output input when t=4 s. a x
Constant
acceleration:

74. Summary Make sure you write FBD (no FBD, no score)
Scalar Equation (Only 1 unknown)
Equation itself is not hard to solve, but calculating work may be more difficult than you thought.

75. Recommended Problem M3/144 M3/155 M3/160 M3/168 H14/93 , H 14/96