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Chap 12.6-10, 13.1-6
Principle of Work and Energy - Sections 14.1-3
Today’s Objectives:
Students will be able to:
a) Calculate the work of a force.
b) Apply the principle of work
and energy to a particle or
system of particles.
Engr222 Spring 2004 Chapter 14
In-Class Activities:
• Reading quiz
• Applications
• Work of a force
• Principle of work and
energy
• Concept quiz
• Group problem solving
• Attention quiz
1
Reading Quiz
F
1. What is the work done by the force F ?
s1
A) F s
B) –F s
C) Zero
D) None of the above
s2
s
2. If a particle is moved from 1 to 2, the work done on the
particle by the force, FR will be
A)
C)
s2
s1
s2
s1
ΣFt ds
B) −
ΣFn ds
D) −
s2
s1
s2
s1
ΣFt ds
ΣFn ds
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, ρ) to control the forces experienced by
the passengers?
Engr222 Spring 2004 Chapter 14
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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 typical velocity of an
oncoming car and the amount of energy
that can be absorbed by each barrel, how
can we design a crash cushion?
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.
Engr222 Spring 2004 Chapter 14
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Work of a Force
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 θ, the increment of work dU
done by the force is
dU = F ds cos θ
By using the definition of the dot product
and integrating, the total work can be U =
1-2
written as
r2
r1
F • dr
Work of a Force - continued
If F is a function of position (a common case) this becomes
s2
U1-2 = F cos θ ds
s1
If both F and θ are constant (F = Fc), this equation further
simplifies to
U1-2 = Fc cos θ (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.
Engr222 Spring 2004 Chapter 14
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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
y2
U1-2 =
y1
- W dy = - W (y2 - y1) = - W ∆y
The work of a weight is the product of the magnitude of
the particle’s weight and its vertical displacement. If
∆y is upward, the work is negative since the weight
force always acts downward.
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
s2
s2 is
U1-2 = Fs ds = k s ds = 0.5k(s2)2 - 0.5k(s1)2
s1
s1
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 = – [ 0.5k (s2)2 – 0.5k (s1)2 ].
Engr222 Spring 2004 Chapter 14
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Spring Forces
It is important to note the following about spring forces:
1. The equations just shown 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).
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.
Principle of Work and Energy
By integrating the equation of motion, Ft = mat = mv(dv/ds), the
principle of work and energy can be written as
U1-2 = 0.5m(v2)2 – 0.5m(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.
Engr222 Spring 2004 Chapter 14
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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.
Example
Given:A 0.5 kg ball of negligible size is fired
up a vertical track of radius 1.5 m using
a spring plunger with k = 500 N/m.
The plunger keeps the spring
compressed 0.08 m when s = 0.
Find: The distance s the plunger must be pulled back and
released so the ball will begin to leave the track when
θ = 135°.
Plan: 1) Draw the FBD of the ball at θ = 135°.
2) Apply the equation of motion in the n-direction to
determine the speed of the ball when it leaves the
track.
3) Apply the principle of work and energy to determine s.
Engr222 Spring 2004 Chapter 14
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Example - continued
Solution:
1) Draw the FBD of the ball at θ = 135°.
N
t
n
45°
W
The weight (W) acts downward through the
center of the ball. The normal force exerted
by the track is perpendicular to the surface.
The friction force between the ball and the
track has no component in the n-direction.
2) Apply the equation of motion in the n-direction. Since the
ball leaves the track at θ = 135°, set N = 0.
=>
+
Fn = man = m (v2/ρ) => W cos45° = m (v2/ρ)
=> (0.5)(9.81) cos 45° = (0.5/1.5)v2 => v = 3.2257 m/s
Example - continued
3) Apply the principle of work and energy between position 1
(θ = 0) and position 2 (θ = 135°). Note that the normal force
(N) does no work since it is always perpendicular to the
displacement direction.
0.5m (v1
and
)2
T1 + U1-2 = T2
– W ∆y – (0.5k(s2)2 – 0.5k (s1)2) = 0.5m (v2)2
v1 = 0, v2 = 3.2257 m/s
s1 = s + 0.08 m, s2 = 0.08 m
∆y = 1.5 + 1.5 sin 45° = 2.5607 m
=> 0 – (0.5)(9.81)(2.5607) – [0.5(500)(0.08)2 – 0.5(500)(5 + 0.08)2]
= 0.5(0.5)(3.2257)2
=> s = 0.179 m = 179 mm
Engr222 Spring 2004 Chapter 14
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Concept Quiz
1. A spring with an unstretched 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 (2 in)2
B) - [0.5 k(4 in)2 - 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. 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
2 kg
2 kg
Group Problem Solving
Given: Block A has a weight of 60 lb and
block B has a weight of 10 lb. The
coefficient of kinetic friction between
block A and the incline is µk = 0.2.
Neglect the mass of the cord and pulleys.
Find: The speed of block A after it moves 3 ft down 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.
Engr222 Spring 2004 Chapter 14
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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
2∆sA + ∆sB = 0
∆sA = 3ft => ∆sB = -6 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.
Group Problem Solving - continued
2) Draw the FBD of each block.
T
WA
y
2T
x
A
5
4
3
µNA
NA
B
WB
Sum forces in the y-direction for block A (note that there is no
motion in this direction):
Fy = 0: NA –(4/5)WA = 0 => NA = (4/5)WA
Engr222 Spring 2004 Chapter 14
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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) + ((3/5)WA – 2T – µNA)∆sA
+ (WB – T)∆sB = (0.5mA(vA2)2 + 0.5mB(vB2)2)
vA1 = vB1 = 0, ∆sA = 3ft, ∆sB = -6 ft, vB = -2vA, NA = (4/5)WA
=> 0 + 0 + (3/5)(60)(3) – 2T(3) – (0.2)(0.8)(60)(3) + (10)(-6)
– T(-6) = 0.5(60/32.2)(vA2)2 + 0.5(10/32.2)(-2vA2)2
=> vA2 = 3.52 ft/s
Note that the work due to the cable tension force on each block
cancels out.
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) - 1.24 lb·ft
B) 0 lb·ft
C) 1.24 lb·ft
D) 2.48 lb·ft
N
B
2. If a spring force is F = 5s3 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
Engr222 Spring 2004 Chapter 14
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Textbook Problem 14-
Announcements
Engr222 Spring 2004 Chapter 14
12
Power and Efficiency - Section 14.4
Today’s Objectives:
Students will be able to:
a) Determine the power
generated by a machine,
engine, or motor.
b) Calculate the mechanical
efficiency of a machine.
In-Class Activities:
• Reading quiz
• Applications
• Define power
• Define efficiency
• Concept quiz
• Group problem solving
• Attention quiz
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
Engr222 Spring 2004 Chapter 14
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Applications
Engines and motors are often rated in
terms of their power output. The power
requirements of the motor lifting this
elevator depend on the vertical force F that
acts on the elevator, causing it to move
upwards.
Given the desired lift velocity for the elevator, how can
we determine the power requirement of the motor?
Applications - continued
The speed at which a vehicle 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
jeep, knowing the power transmitted by the engine to the
wheels?
Engr222 Spring 2004 Chapter 14
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Power
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.
Power - continued
Using scalar notation, power can be written
P = F • v = F v cos θ
where θ 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 .
Engr222 Spring 2004 Chapter 14
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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
ε = (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
ε = (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.
Solving Problems
• 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 θ).
• 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.
Engr222 Spring 2004 Chapter 14
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Example
Given:A sports car has a mass of 2 Mg and an engine efficiency
of ε = 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 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 power required for this motion.
4) Use the engine’s efficiency to determine input power.
Example - 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 = (10,000 + 1.2v2) N
Engr222 Spring 2004 Chapter 14
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Example - 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) = 209.6 kW
5) The power developed by the engine (prior to its frictional
losses) is obtained using the efficiency equation.
Pi = Po/ε = 209.6/0.65 = 322 kW
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 264 ft/s. The thrust developed by a jet engine is 1000 lb.
The power developed by the engines is
A) (1000 lb)(140 ft/s)
C) (1000 lb)(140 ft/s) cos 10
Engr222 Spring 2004 Chapter 14
B) (2000 lb)(140 ft/s) cos 10
D) (2000 lb)(140 ft/s)
18
Group Problem Solving
Given: A 50-lb load (B) is hoisted by the pulley
system and motor M. The motor has an
efficiency of 0.76 and exerts a constant
force of 30 lb on the cable. Neglect the
mass of the pulleys and cable.
Find: The power supplied to the motor when the load has been
hoisted 10 ft. The block started from rest.
Plan: 1) Relate the cable and block velocities by defining
position coordinates. Draw a FBD of the block.
2) Use the equation of motion or energy methods to
determine the block’s velocity at 10 feet.
3) Calculate the power supplied by the motor and to the
motor.
Group Problem Solving - continued
Solution:
1) Define position coordinates to relate velocities.
sm
Here sm is defined to a point on the cable. Also sB
sB is defined only to the lower pulley, since the block
moves with the pulley. From kinematics,
sm + 2sB = l
=> vm + 2vB = 0
=> vm = -2vB
Draw the FBD of the block:
2T
Since the pulley has no mass, a force
balance requires that the tension in the
B
lower cable is twice the tension in the upper
WB = 50 lb cable.
Engr222 Spring 2004 Chapter 14
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Group Problem Solving - continued
2) The velocity of the block can be obtained by
applying the principle of work and energy to the
block (recall that the block starts from rest).
+
T1 + U1-2 = T2
0.5m(v1)2 + [2T(s) – wB(s)] = 0.5m (v2)2
0 + [2(30)(10) - (50)(10)] = 0.5(50/32.2)(v2)2
=> v2 = vB = 11.35 ft/s
Since this velocity is upwards, it is a negative velocity in
terms of the kinematic equation coordinates.
The velocity of the cable coming into the motor (vm) is
calculated from the kinematic equation.
vm = - 2vB = - (2)(-11.35) = 22.70 ft/s
Group Problem Solving - continued
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 = (30)(22.70) = 681 (ft ·lb)/s
The power supplied to the motor is determined using the
motor’s efficiency and the basic efficiency equation.
Pi = Po/ε = 681/0.76 = 896 (ft ·lb)/s
Converting to horsepower
Pi = 896/550 = 1.63 hp
Engr222 Spring 2004 Chapter 14
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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) 8.82 lb
B) 400 lb
C) 600 lb
D) 4.64 x 106 lb
Textbook Problem 14-
Engr222 Spring 2004 Chapter 14
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Announcements
Potential Energy and Conservation of Energy
Sections 14.5-6
Today’s Objectives:
Students will be able to:
a) Understand the concept of
conservative forces and
determine the potential
energy of such forces.
b) Apply the principle of
conservation of energy.
Engr222 Spring 2004 Chapter 14
In-Class Activities:
• Reading quiz
• Applications
• Conservative force
• Potential energy
• Conservation of energy
• Concept quiz
• Group problem solving
• Attention quiz
22
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
increase.
D) does not change.
Applications
The weight of the sacks resting on this platform causes
potential energy to be stored in the supporting springs.
If the sacks weigh 100 lb and the equivalent spring constant
is k = 500 lb/ft, what is the energy stored in the springs?
Engr222 Spring 2004 Chapter 14
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Applications - continued
When a ball of weight W is dropped (from rest) from a height
h above the ground, the potential energy stored in the ball is
converted to kinetic energy as the ball drops.
What is the velocity of the ball when it hits the ground? Does
the weight of the ball affect the final velocity?
Conservative Force
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. In other words, the work done by the
force F in a closed path (i.e., from A to B and then back to A)
equals zero.
z
F · dr = 0
B
F
This means the work is conserved.
A
A conservative force depends
only on the position of the
particle, and is independent of its
velocity or acceleration.
y
x
Engr222 Spring 2004 Chapter 14
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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 textbook. 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.
V total = V gravity + V springs
Potential Energy
Potential energy is a measure of the amount of work a
conservative force will do when it 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 (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.
Engr222 Spring 2004 Chapter 14
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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.
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.
Ve (where ‘e’ denotes an
elastic spring) has the distance
“s” raised to a power (the
result of an integration) or
Ve =
1 2
ks
2
Notice that the potential
function Ve always yields
positive energy.
Engr222 Spring 2004 Chapter 14
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Conservation of Energy
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 + V1 = T2 + V2 = 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.
Example
Given: The girl and bicycle
weigh 125 lbs. She moves from
point A to B.
Find: The velocity and the
normal force at B if the velocity
at A is 10 ft/s and she stops
pedaling at A.
Plan: Note that only kinetic energy and potential energy due
to gravity (Vg) are involved. Determine the velocity at B
using the conservation of energy equation and then apply
equilibrium equations to find the normal force.
Engr222 Spring 2004 Chapter 14
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Solution:
Example - continued
Placing the datum at B:
TA + VA = TB + VB
1 125
1 125 2
2
(
)(10) + 125(30) = (
)v
2 32.2
2 32.2 B
VB = 45.1 ft
s
Equation of motion applied at B:
v2
Fn = man = m ρ
NB − 125 =
125 ( 45.1) 2
32.2 50
N B = 283 lb
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!
Engr222 Spring 2004 Chapter 14
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Group Problem Solving
Given: The mass of the collar is 2 kg and
the spring constant is 60 N/m. The
collar has no velocity at A and the
spring is un-deformed at A.
Find:
The maximum distance y the collar
drops before it stops at Point C.
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).
Group Problem Solving - continued
Solution:
Notice that the potential energy at C has two parts (Tc = 0).
Vc = (Vc)e + (Vc)g
Placing the datum for gravitational potential at A yields a
conservation of energy equation with the left side all zeros.
Since Tc equals zero at points A and C, the equation becomes
60
0 + 0 = 0 + [ ( (.75) 2 + y 2 − .75) 2 − 2(9.81) y ]
2
Note that (Vc)g is negative since point C is below the datum.
Since the equation is nonlinear, a numerical solver can be
used to find the solution or root of the equation. This solving
routine can be done with a calculator or a program like Excel.
The solution yields y = 1.61 m.
Engr222 Spring 2004 Chapter 14
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Group Problem Solving - continued
Also notice that since the velocities at A and C are zero, the
velocity must reach a maximum somewhere between A and C.
Since energy is conserved, the point of maximum kinetic energy
(maximum velocity) corresponds to the point of minimum
potential energy.
By expressing the potential energy at any given position as a
function of y and then differentiating, we can determine the
position at which the velocity is maximum (since dV/dy = 0 at
this position). The derivative yields another nonlinear equation
which could be solved using a numerical solver.
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) Don’t pick this one.
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) 14.7 m/s.
D) 21 m/s.
Engr222 Spring 2004 Chapter 14
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Textbook Problem 14-
Engr222 Spring 2004 Chapter 14
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