1. THE WORK OF A FORCE, THE PRINCIPLE OF WORK AND ENERGY & SYSTEMS OF PARTICLES
Today’s Objectives:
Students will be able to:
Calculate the work of a force.
Apply the principle of work and energy to a particle or system of particles.
In-Class Activities:
• Check Homework
• Reading Quiz
• Applications
• Work of A Force
• Principle of Work And Energy
• Concept Quiz
• Group Problem Solving
• Attention Quiz

2. 1. What is the work done by the force F? A) F s B) –F s
READING QUIZ
1. What is the work done by the force F?
A) F s B) –F s
C) Zero D) None of the above. s s1 s2 F
2. If a particle is moved from 1 to 2, the work done on the particle by the force, FR will be
A) B)
C) D)
Anwers:
1. C
2. A

3. APPLICATIONS
A roller coaster makes use of gravitational forces to assist the cars in reaching high speeds in the “valleys” of the track.
How can we design the track (e.g., the height, h, and the radius of curvature, r) to control the forces experienced by the passengers?

4. APPLICATIONS (continued)
Crash barrels are often used along roadways for crash protection.
The barrels absorb the car’s kinetic energy by deforming.
If we know the velocity of an oncoming car and the amount of energy that can be absorbed by each barrel, how can we design a crash cushion?

5. WORK AND ENERGY
Another equation for working kinetics problems involving particles can be derived by integrating the equation of motion (F = ma) with respect to displacement.
By substituting at = v (dv/ds) into Ft = mat, the result is integrated to yield an equation known as the principle of work and energy.
This principle is useful for solving problems that involve force, velocity, and displacement. It can also be used to explore the concept of power.
To use this principle, we must first understand how to calculate the work of a force.

6. WORK OF A FORCE (Section 14.1)
A force does work on a particle when the particle undergoes a displacement along the line of action of the force.
Work is defined as the product of force and displacement components acting in the same direction. So, if the angle between the force and displacement vector is q, the increment of work dU done by the force is
dU = F ds cos q
By using the definition of the dot product and integrating, the total work can be written as ò r2 r1
U1-2 =
F • dr

7. WORK OF A FORCE (continued)
If F is a function of position (a common case) this becomes ò = s2 s1
F cos q ds
U1-2
If both F and q are constant (F = Fc), this equation further simplifies to
U1-2 = Fc cos q (s2 - s1)
Work is positive if the force and the movement are in the same direction. If they are opposing, then the work is negative. If the force and the displacement directions are perpendicular, the work is zero.

8. WORK OF A WEIGHT
The work done by the gravitational force acting on a particle (or weight of an object) can be calculated by using
- W (y2 − y1) = - W Dy
- W dy =
U1-2 = ò y2 y1
The work of a weight is the product of the magnitude of the particle’s weight and its vertical displacement. If Dy is upward, the work is negative since the weight force always acts downward.

9. The work of the spring force moving from position s1 to position s2 is
WORK OF A SPRING FORCE
When stretched, a linear elastic spring develops a force of magnitude Fs = ks, where k is the spring stiffness and s is the displacement from the unstretched position.
The work of the spring force moving from position s1 to position s2 is
= 0.5 k (s2)2 – 0.5 k (s1)2
k s ds
Fs ds
U1-2 s2 s1
= ò
If a particle is attached to the spring, the force Fs exerted on the particle is opposite to that exerted on the spring. Thus, the work done on the particle by the spring force will be negative or
U1-2 = – [ 0.5 k (s2)2 – 0.5 k (s1)2 ] .

10. SPRING FORCES
It is important to note the following about spring forces.
1. The equations above are for linear springs only! Recall that a linear spring develops a force according to
F = ks (essentially the equation of a line).
2. The work of a spring is not just spring force times distance at some point, i.e., (ksi)(si). Beware, this is a trap that students often fall into!
3. Always double check the sign of the spring work after calculating it. It is positive work if the force put on the object by the spring and the movement are in the same direction.

11. PRINCIPLE OF WORK AND ENERGY
(Section 14.2 & Section 14.3)
By integrating the equation of motion,  Ft = mat = mv(dv/ds), the principle of work and energy can be written as
 U1-2 = 0.5 m (v2)2 – 0.5 m (v1)2 or T1 +  U1-2 = T2
U1-2 is the work done by all the forces acting on the particle as it moves from point 1 to point 2. Work can be either a positive or negative scalar.
T1 and T2 are the kinetic energies of the particle at the initial and final position, respectively. Thus, T1 = 0.5 m (v1)2 and T2 = 0.5 m (v2)2. The kinetic energy is always a positive scalar (velocity is squared!).
So, the particle’s initial kinetic energy plus the work done by all the forces acting on the particle as it moves from its initial to final position is equal to the particle’s final kinetic energy.

12. PRINCIPLE OF WORK AND ENERGY (continued)
Note that the principle of work and energy (T1 +  U1-2 = T2) is not a vector equation! Each term results in a scalar value.
Both kinetic energy and work have the same units, that of energy! In the SI system, the unit for energy is called a joule (J), where 1 J = 1 N·m. In the FPS system, units are ft·lb.
The principle of work and energy cannot be used, in general, to determine forces directed normal to the path, since these forces do no work.
The principle of work and energy can also be applied to a system of particles by summing the kinetic energies of all particles in the system and the work due to all forces acting on the system.

13. WORK OF FRICTION CAUSED BY SLIDING
The case of a body sliding over a rough surface merits special consideration.
Consider a block which is moving over a rough surface. If the applied force P just balances the resultant frictional force k N, a constant velocity v would be maintained.
The principle of work and energy would be applied as
0.5m (v)2 + P s – (k N) s = 0.5m (v)2
This equation is satisfied if P = k N. However, we know from experience that friction generates heat, a form of energy that does not seem to be accounted for in this equation. It can be shown that the work term (k N)s represents both the external work of the friction force and the internal work that is converted into heat.

14. Find: The distance s when the block stops. Plan:
EXAMPLE
Given: When s = 0.6 m, the spring is not stretched or compressed, and the 10 kg block, which is subjected to a force of F= 100 N, has a speed of 5 m/s down the smooth plane.
Find: The distance s when the block stops.
Plan:
Since this problem involves forces, velocity and displacement, apply the principle of work and energy to determine s.

15. EXAMPLE (continued)
Solution:
Apply the principle of work and energy between position 1
(s1 = 0.6 m) and position 2 (s2). Note that the normal force (N) does no work since it is always perpendicular to the displacement.
T1 + U1-2 = T2
There is work done by three different forces;
1) work of a the force F =100 N;
UF = 100 (s2− s1) = 100 (s2 − 0.6)
2) work of the block weight;
UW = 10 (9.81) (s2− s1) sin 30 = (s2 − 0.6)
3) and, work of the spring force.
US = (200) (s2−0.6)2 = -100 (s2 − 0.6)2

16. EXAMPLE (continued)
The work and energy equation will be
T1 + U1-2 = T2
0.5 (10) (s2 − 0.6) (s2 − 0.6) − 100(s2 − 0.6)2 = 0
 (s2 − 0.6) − 100(s2 − 0.6)2 = 0
Solving for (s2 − 0.6),
(s2 − 0.6) = { ± ( – 4×(-100)×125)0.5} / 2(-100)
Selecting the positive root, indicating a positive spring deflection,
(s2 − 0.6) = 2.09 m
Therefore, s2 = 2.69 m

17. A) -[0.5 k(4 in)2 - 0.5 k(2 in)2] B) 0.5 k (2 in)2
CONCEPT QUIZ
1. A spring with an un-stretched length of 5 in expands from a length of 2 in to a length of 4 in. The work done on the spring is _________ in·lb .
A) -[0.5 k(4 in) k(2 in)2] B) 0.5 k (2 in)2
C) -[0.5 k(3 in) k(1 in)2] D) 0.5 k(3 in) k(1 in)2
2. If a spring force is F = 5 s3 N/m and the spring is compressed by s = 0.5 m, the work done on a particle attached to the spring will be
A) N · m B) – N · m
C) N · m D) – N · m
Answers:
1. C
2. D

18. GROUP PROBLEM SOLVING
Given: Block A has a weight of 60 lb and block B has a weight of 40 lb. The coefficient of kinetic friction between the blocks and the incline is mk = Neglect the mass of the cord and pulleys.
Find: The speed of block A after block B moves 2 ft up the plane, starting from rest.
Plan:
1) Define the kinematic relationships between the blocks.
2) Draw the FBD of each block.
3) Apply the principle of work and energy to the system of blocks. Why choose this method?

19. GROUP PROBLEM SOLVING (continued)
Solution:
1) The kinematic relationships can be determined by defining position coordinates sA and sB, and then differentiating. sA sB
Since the cable length is constant:
2sA + sB = l
2DsA + DsB = 0
When DsB = -2 ft => DsA = 1 ft
and 2vA + vB = 0
=> vB = -2vA
Note that, by this definition of sA and sB, positive motion for each block is defined as downwards.

20. GROUP PROBLEM SOLVING (continued)
2) Draw the FBD of each block. y x NA
mNA 2T WA A
60
30 B NB
mNB T WB
Sum forces in the y-direction for block A (note that there is no motion in y-direction):
Fy = 0: NA – WA cos 60 = 0
NA = WA cos 60
Similarly, for block B:
NB = WB cos 30

21. GROUP PROBLEM SOLVING (continued)
3) Apply the principle of work and energy to the system (the blocks start from rest).
T1 + U1-2 = T2
[0.5mA(vA1)2 + .5mB(vB1)2] + [WA sin 60– 2T – mNA]DsA
+ [WB sin 30– T + mNB]DsB = [0.5mA(vA2) mB(vB2)2]
where vA1 = vB1 = 0, DsA = 1ft, DsB = -2 ft, vB = -2vA,
NA = WA cos 60, NB = WB cos 30
=> [0 + 0] + [60 sin 60– 2T – 0.1(60 cos 60)] (1)
+ [40 sin 30 – T + 0.1(40 cos 30)] (-2)
= [0.5(60/32.2)(vA2) (40/32.2)(-2vA2)2]

22. GROUP PROBLEM SOLVING (continued)
Again, the Work and Energy equation is:
=> [0 + 0] + [60 sin 60– 2T – 0.1(60 cos 60)] (1)
+ [40 sin 30 – T + 0.1(40 cos 30)] (-2)
= [0.5(60/32.2)(vA2) (40/32.2)(-2vA2)2]
Solving for the unknown velocity yeilds
=> vA2 = ft/s
Note that the work due to the cable tension force on each block cancels out.

23. ATTENTION QUIZ
1. What is the work done by the normal force N if a 10 lb box is moved from A to B ?
A) lb · ft B) lb · ft
C) lb · ft D) lb · ft
2. Two blocks are initially at rest. How many equations would be needed to determine the velocity of block A after block B moves 4 m horizontally on the smooth surface?
A) One B) Two
C) Three D) Four
Answers:
1. B
2. C (Work & Energy, length of cord equation, velocity of blocks equation)
2 kg

24. POWER AND EFFICIENCY Today’s Objectives: Students will be able to:
Determine the power generated by a machine, engine, or motor.
Calculate the mechanical efficiency of a machine.
In-Class Activities:
• Check Homework
• Reading Quiz
• Applications
• Define Power
• Define Efficiency
• Concept Quiz
• Group Problem Solving
• Attention Quiz

25. 1. The formula definition of power is ___________. A) dU / dt B) F  v
READING QUIZ
1. The formula definition of power is ___________.
A) dU / dt B) F  v
C) F  dr/dt D) All of the above.
2. Kinetic energy results from _______.
A) displacement B) velocity
C) gravity D) friction
Answers:
1. D
2. B 25

26. APPLICATIONS
Engines and motors are often rated in terms of their power output. The power output of the motor lifting this elevator is related to the vertical force F acting on the elevator, causing it to move upwards.
Given a desired lift velocity for the elevator (with a known maximum load), how can we determine the power requirement of the motor? 26

27. APPLICATIONS (continued)
The speed at which a truck can climb a hill depends in part on the power output of the engine and the angle of inclination of the hill.
For a given angle, how can we determine the speed of this truck, knowing the power transmitted by the engine to the wheels? Can we find the speed, if we know the power?
If we know the engine power output and speed of the truck, can we determine the maximum angle of climb of this truck ? 27

28. P = dU/dt = (F • dr)/dt = F • (dr/dt) = F • v
POWER AND EFFICIENCY
(Section 14.4)
Power is defined as the amount of work performed per unit of time.
If a machine or engine performs a certain amount of work, dU, within a given time interval, dt, the power generated can be calculated as
P = dU/dt
Since the work can be expressed as dU = F • dr, the power can be written
P = dU/dt = (F • dr)/dt = F • (dr/dt) = F • v
Thus, power is a scalar defined as the product of the force and velocity components acting in the same direction. 28

29. POWER
Using scalar notation, power can be written
P = F • v = F v cos q
where q is the angle between the force and velocity vectors.
So if the velocity of a body acted on by a force F is known, the power can be determined by calculating the dot product or by multiplying force and velocity components.
The unit of power in the SI system is the Watt (W) where
1 W = 1 J/s = 1 (N · m)/s .
In the FPS system, power is usually expressed in units of horsepower (hp) where
1 hp = 550 (ft · lb)/s = 746 W .

30. e = (power output) / (power input)
EFFICIENCY
The mechanical efficiency of a machine is the ratio of the useful power produced (output power) to the power supplied to the machine (input power) or
e = (power output) / (power input)
If energy input and removal occur at the same time, efficiency may also be expressed in terms of the ratio of output energy to input energy or
e = (energy output) / (energy input)
Machines will always have frictional forces. Since frictional forces dissipate energy, additional power will be required to overcome these forces. Consequently, the efficiency of a machine is always less than 1.

31. PROCEDURE FOR ANALYSIS
• Find the resultant external force acting on the body causing its motion. It may be necessary to draw a free-body diagram.
• Determine the velocity of the point on the body at which the force is applied. Energy methods or the equation of motion and appropriate kinematic relations, may be necessary.
• Multiply the force magnitude by the component of velocity acting in the direction of F to determine the power supplied to the body (P = F v cos q ).
• In some cases, power may be found by calculating the work done per unit of time (P = dU/dt).
• If the mechanical efficiency of a machine is known, either the power input or output can be determined.

32. EXAMPLE
Given: A 50 kg block (A) is hoisted by the pulley system and motor M. The motor has an efficiency of At this instant, point P on the cable has a velocity of 12 m/s which is increasing at a rate of 6 m/s2. Neglect the mass of the pulleys and cable.
Find: The power supplied to the motor at this instant.
Plan:
1) Relate the cable and block velocities by defining position coordinates. Draw a FBD of the block.
2) Use the equation of motion to determine the cable tension.
3) Calculate the power supplied by the motor and then to the motor.

33. 1) Define position coordinates to relate velocities.
EXAMPLE (continued)
Solution:
1) Define position coordinates to relate velocities.
Datum SA SP
Here sP is defined to a point on the cable. Also sA is defined only to the lower pulley, since the block moves with the pulley. From kinematics,
sP + 2 sA = l
 aP + 2 aA = 0
aA = − aP / 2 = −3 m/s2 (↑) sB sm
Draw the FBD and kinetic diagram of the block: 2T WA A
mA aA = 33

34. EXAMPLE
(continued)
2) The tension of the cable can be obtained by applying the equation of motion to the block.
+↑ Fy = mA aA
2T − = 50 (3)  T = N
3) The power supplied by the motor is the product of the force applied to the cable and the velocity of the cable.
Po = F • v = (320.3)(12) = 3844 W
The power supplied to the motor is determined using the motor’s efficiency and the basic efficiency equation.
Pi = Po/e = 3844/0.8 = 4804 W = 4.8 kW

35. A) (1000 lb)(260 ft/s) B) (2000 lb)(260 ft/s) cos 10
CONCEPT QUIZ
1. A motor pulls a 10 lb block up a smooth incline at a constant velocity of 4 ft/s. Find the power supplied by the motor.
A) 8.4 ft·lb/s B) 20 ft·lb/s
C) 34.6 ft·lb/s D) 40 ft·lb/s
30º
2. A twin engine jet aircraft is climbing at a 10 degree angle at 260 ft/s. The thrust developed by a jet engine is 1000 lb. The power developed by the aircraft is
A) (1000 lb)(260 ft/s) B) (2000 lb)(260 ft/s) cos 10
C) (1000 lb)(260 ft/s) cos D) (2000 lb)(260 ft/s)
Answers: B D 35

36. GROUP PROBLEM SOLVING
Given: A sports car has a mass of 2000 kg and an engine efficiency of e = Moving forward, the wind creates a drag resistance on the car of FD = 1.2v2 N, where v is the velocity in m/s. The car accelerates at 5 m/s2, starting from rest.
Find: The engine’s input power when t = 4 s.
Plan:
1) Draw a free body diagram of the car.
2) Apply the equation of motion and kinematic equations to find the car’s velocity at t = 4 s.
3) Determine the output power required for this motion.
4) Use the engine’s efficiency to determine input power.

37. GROUP PROBLEM SOLVING
(continued)
Solution:
1) Draw the FBD of the car.
The drag force and weight are known forces. The normal force Nc and frictional force Fc represent the resultant forces of all four wheels. The frictional force between the wheels and road pushes the car forward.
2) The equation of motion can be applied in the x-direction, with ax = 5 m/s2:
+ Fx = max => Fc – 1.2v2 = (2000)(5)
=> Fc = (10, v2) N

38. GROUP PROBLEM SOLVING
(continued)
3) The constant acceleration equations can be used to determine the car’s velocity.
vx = vxo + axt = 0 + (5)(4) = 20 m/s
4) The power output of the car is calculated by multiplying the driving (frictional) force and the car’s velocity:
Po = (Fc)(vx ) = [10,000 + (1.2)(20)2](20) = kW
5) The power developed by the engine (prior to its frictional losses) is obtained using the efficiency equation.
Pi = Po/e = 209.6/0.65 = 322 kW

39. ATTENTION QUIZ
1. The power supplied by a machine will always be _________ the power supplied to the machine.
A) less than B) equal to
C) greater than D) A or B
2. A car is traveling a level road at 88 ft/s. The power being supplied to the wheels is 52,800 ft·lb/s. Find the combined friction force on the tires.
A) lb B) 400 lb
C) 600 lb D) 4.64 x 106 lb
Answers: A C 39

40. CONSERVATIVE FORCES, POTENTIAL ENERGY AND CONSERVATION OF ENERGY
Today’s Objectives:
Students will be able to:
Understand the concept of conservative forces and determine the potential energy of such forces.
Apply the principle of conservation of energy.
In-Class Activities:
Check Homework
Reading Quiz
Applications
Conservative Force
Potential Energy
Conservation of Energy
Concept Quiz
Group Problem Solving
Attention Quiz

41. 1. The potential energy of a spring is ________
READING QUIZ
1. The potential energy of a spring is ________
A) always negative. B) always positive.
C) positive or negative. D) equal to ks.
2. When the potential energy of a conservative system increases, the kinetic energy _________
A) always decreases. B) always increases.
C) could decrease or D) does not change.
increase.
Answers:
1. B
2. A (Conservation of Energy Principal) 41

42. APPLICATIONS
The weight of the sacks resting on this platform causes potential energy to be stored in the supporting springs.
As each sack is removed, the platform will rise slightly since some of the potential energy within the springs will be transformed into an increase in gravitational potential energy of the remaining sacks.
If the sacks weigh 100 lb and the equivalent spring constant is k = 500 lb/ft, what is the energy stored in the springs?

43. APPLICATIONS (continued)
The boy pulls the water balloon launcher back, stretching each of the four elastic cords.
If we know the unstretched length and stiffness of each cord, can we estimate the maximum height and the maximum range of the water balloon when it is released from the current position ? 43

44. APPLICATIONS (continued)
The roller coaster is released from rest at the top of the hill. As the coaster moves down the hill, potential energy is transformed into kinetic energy.
What is the velocity of the coaster when it is at B and C?
Also, how can we determine the minimum height of the hill so that the car travels around both inside loops without leaving the track?

45. CONSERVATIVE FORCE (Section 14.5)
A force F is said to be conservative if the work done is independent of the path followed by the force acting on a particle as it moves from A to B. This also means that the work done by the force F in a closed path (i.e., from A to B and then back to A) is zero.
Thus, we say the work is conserved. ò r F = d x y z A B F
The work done by a conservative force depends only on the positions of the particle, and is independent of its velocity or acceleration.

46. CONSERVATIVE FORCE (continued)
A more rigorous definition of a conservative force makes use of a potential function (V) and partial differential calculus, as explained in the text. However, even without the use of the these mathematical relationships, much can be understood and accomplished.
The “conservative” potential energy of a particle/system is typically written using the potential function V. There are two major components to V commonly encountered in mechanical systems, the potential energy from gravity and the potential energy from springs or other elastic elements.
Vtotal = Vgravity + Vsprings

47. POTENTIAL ENERGY
Potential energy is a measure of the amount of work a conservative force will do when a body changes position.
In general, for any conservative force system, we can define the potential function (V) as a function of position. The work done by conservative forces as the particle moves equals the change in the value of the potential function (e.g., the sum of Vgravity and Vsprings).
It is important to become familiar with the two types of potential energy and how to calculate their magnitudes.

48. POTENTIAL ENERGY DUE TO GRAVITY
The potential function (formula) for a gravitational force, e.g., weight (W = mg), is the force multiplied by its elevation from a datum. The datum can be defined at any convenient location.
Vg = ± W y
Vg is positive if y is above the datum and negative if y is below the datum. Remember, YOU get to set the datum. 48

49. ELASTIC POTENTIAL ENERGY
Recall that the force of an elastic spring is F = ks. It is important to realize that the potential energy of a spring, while it looks similar, is a different formula.
k s2 2 1 V e =
Ve (where ‘e’ denotes an elastic spring) has the distance “s” raised to a power (the result of an integration) or
Notice that the potential function Ve always yields positive energy.

50. CONSERVATION OF ENERGY (Section 14.6)
When a particle is acted upon by a system of conservative forces, the work done by these forces is conserved and the sum of kinetic energy and potential energy remains constant. In other words, as the particle moves, kinetic energy is converted to potential energy and vice versa. This principle is called the principle of conservation of energy and is expressed as 2 1 V T + =
= Constant
T1 stands for the kinetic energy at state 1 and V1 is the potential energy function for state 1. T2 and V2 represent these energy states at state 2. Recall, the kinetic energy is defined as T = ½ mv2.

51. EXAMPLE
Given: The 2 kg collar is moving down with the velocity of 4 m/s at A.
The spring constant is 30 N/m. The unstretched length of the spring is 1 m.
Find: The velocity of the collar when s = 1 m.
Plan:
Apply the conservation of energy equation between A and C. Set the gravitational potential energy datum at point A or point C (in this example, choose point A—why?).

52. EXAMPLE
(continued)
Solution:
Note that the potential energy at C has two parts.
VC = (VC)e + (VC)g
VC = 0.5 (30) (√5 – 1)2 – 2 (9.81) 1
The kinetic energy at C is
TC = 0.5 (2) v2
Similarly, the potential and kinetic energies at A will be
VA = 0.5 (30) (2 – 1)2, TA = 0.5 (2) 42
The energy conservation equation becomes TA + VA = TC + VC .
[ 0.5(30) (√5 – 1)2 – 2(9.81)1 ] (2) v2
= [0.5 (30) (2 – 1)2 ]+ 0.5 (2) 42
 v = 5.26 m/s

53. C) +10 ft·lb. D) None of the above.
CONCEPT QUIZ
1. If the work done by a conservative force on a particle as it moves between two positions is –10 ft·lb, the change in its potential energy is _______
A) 0 ft·lb. B) -10 ft·lb.
C) +10 ft·lb. D) None of the above.
2. Recall that the work of a spring is U1-2 = -½ k(s22 – s12) and can be either positive or negative. The potential energy of a spring is V = ½ ks2. Its value is __________
A) always negative B) either positive or negative.
C) always positive D) an imaginary number!
Answers:
1. C
2. C 53

54. GROUP PROBLEM SOLVING
Given: The 800 kg roller coaster starts from A with a speed of
3 m/s.
Find: The minimum height, h, of the hill so that the car travels around inside loop at B without leaving the track. Also find the normal reaction on the car when the car is at C for this height of A.
Plan:
Note that only kinetic energy and potential energy due to gravity are involved. Determine the velocity at B using the equation of equilibrium and then apply the conservation of energy equation to find minimum height h .

55. GROUP PROBLEM SOLVING (continued)
Solution:
1) Placing the datum at A:
TA + VA = TB + VB
 0.5 (800)
= 0.5 (800) (vB)2 − 800(9.81) (h − 20) (1)
2) Find the required velocity of the coaster at B so it doesn’t
leave the track.
Equation of motion applied at B: r 2 v m ma F n = å 10
800 (9.81) = 800
(vB)2
 vB = m/s
man mg
NB  0 = 55

56. GROUP PROBLEM SOLVING (continued)
Now using the energy conservation, eq. (1), the minimum h can be determined.
0.5 (800) = 0.5 (800) (9.905)2 − 800(9.81) (h − 20)
 h= 24.5 m
3) To find the normal reaction at C, we need vc.
TA + VA = TC + VC
 0.5 (800) = 0.5 (800) (vC)2 − 800(9.81) (24.5 − 14)
 VC = m/s
Equation of motion applied at B: r 2 v m F n = å 7
NC+800 (9.81) = 800
14.662 
man mg NC =
 NC = 16.8 kN 56

57. C) the same amount of work D) It is a mystery!
ATTENTION QUIZ
1. The principle of conservation of energy is usually ______ to apply than the principle of work & energy.
A) harder B) easier
C) the same amount of work D) It is a mystery!
2. If the pendulum is released from the
horizontal position, the velocity of its
bob in the vertical position is _____
A) 3.8 m/s. B) 6.9 m/s.
C) m/s. D) 21 m/s.
Answers:
1. B
2. A 57

58. End of the Lecture
Let Learning Continue