Sections 14.1-14.3

1
THE WORK OF A FORCE, THE PRINCIPLE OF WORK
AND ENERGY SYSTEMS OF PARTICLES

• Todays 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
READING QUIZ
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 cars
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
7
WORK OF A FORCE (continued)
If F is a function of position (a common case)
this becomes
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
The work of a weight is the product of the
magnitude of the particles weight and its
vertical displacement. If Dy is upward, the work
is negative since the weight force always acts
downward.
9
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.
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 particles 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 particles 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 Nm.
In the FPS system, units are ftlb.
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.
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
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.
16
EXAMPLE (continued)
The work and energy equation will be T1
?U1-2 T2 0.5 (10) 52 100(s2 - 0.6)
49.05(s2 - 0.6) - 100(s2 - 0.6)2 0 ? 125
149.05(s2 - 0.6) - 100(s2 - 0.6)2 0
Solving for (s2 - 0.6), (s2 - 0.6) -149.05
(149.052 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
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 _________
inlb . A) -0.5 k(4 in)2 - 0.5 k(2 in)2
B) 0.5 k (2 in)2 C) -0.5 k(3 in)2 -
0.5 k(1 in)2 D) 0.5 k(3 in)2 - 0.5 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) 0.625 N m B) 0.625 N m C) 0.0781 N
m D) 0.0781 N m
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 0.1. 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
Since the cable length is constant 2sA sB
l 2DsA DsB 0 When DsB -2 ft gt DsA 1
ft and 2vA vB 0 gt 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.
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
mNADsA WB sin 30? T mNBDsB 0.5mA(vA2)2
0.5mB(vB2)2
where vA1 vB1 0, DsA 1ft, DsB -2 ft, vB
-2vA, NA WA cos 60?, NB WB cos 30?
gt 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)2 0.5(40/32.2)(-2vA2)2

22
GROUP PROBLEM SOLVING (continued)
Again, the Work and Energy equation is gt 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)2 0.5(40/32.2)(-2vA2)2
Solving for the unknown velocity yeilds gt vA2
0.771 ft/s
Note that the work due to the cable tension force
on each block cancels out.
23
ATTENTION QUIZ
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
24
POWER AND EFFICIENCY

• Todays 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
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
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?
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 ?
28
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.
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
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 0.8. 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
EXAMPLE (continued)
Solution
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 - 490.5 50 (3) ? T 320.3
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 motors efficiency and the basic
efficiency equation.
Pi Po/e 3844/0.8 4804 W 4.8 kW
35
CONCEPT QUIZ
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 10 D) (2000 lb)(260
ft/s)
36
GROUP PROBLEM SOLVING
Given A sports car has a mass of 2000 kg and an
engine efficiency of e 0.65. 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 engines 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 cars velocity at
t 4 s. 3) Determine the output power required
for this motion. 4) Use the engines 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
38
GROUP PROBLEM SOLVING (continued)
3) The constant acceleration equations can be
used to determine the cars 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 cars velocity
Po (Fc)(vx ) 10,000 (1.2)(20)2(20)
209.6 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
ftlb/s. Find the combined friction force on the
tires. A) 8.82 lb B) 400 lb C) 600 lb
D) 4.64 x 106 lb
40
CONSERVATIVE FORCES, POTENTIAL ENERGY AND
CONSERVATION OF ENERGY

• Todays 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
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.
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 ?
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)
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.
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
Vg is positive if y is above the datum and
negative if y is below the datum. Remember, YOU
get to set the datum.
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.
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
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 Awhy?).
52
EXAMPLE (continued)
Solution
Similarly, the potential and kinetic energies at
A will be VA 0.5 (30) (2 1)2, TA
0.5 (2) 42
53
CONCEPT QUIZ
1. If the work done by a conservative force on a
particle as it moves between two positions is 10
ftlb, the change in its potential energy is
_______ A) 0 ftlb. B) -10 ftlb. C) 10
ftlb. 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!
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) 32 0 0.5 (800)
(vB)2 - 800(9.81) (h - 20) (1)
2) Find the required velocity of the coaster at B
so it doesnt leave the track.
56
GROUP PROBLEM SOLVING (continued)
Now using the energy conservation, eq. (1), the
minimum h can be determined. 0.5 (800) 32 0
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) 32 0
0.5 (800) (vC)2 - 800(9.81) (24.5 - 14) ? VC
14.66 m/s
? NC 16.8 kN
57
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!
58
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