Transcript chapter 3 - UniMAP Portal

CHAPTER 3
KINETICS OF A PARTICLE: WORK AND
ENERGY
3.1 The Work of a Force
Today’s Objectives:
Students will be able to:
a) Calculate the work of a force.
b) Apply the principle of work
In-Class Activities:
and energy to a particle or
• Check homework, if any
system of particles.
• Applications
• Work of a force
• Principle of work and
energy
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?
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.
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 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 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 q ds
s1
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.
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 =
 - W dy = - W (y2 - y1) =
- W Dy
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.
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 ] .
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). 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.
3.2 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.
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.
3.3 Principle of Work and Energy for a System of Particles
The principle of work and energy can be extended to include a system of particles
isolated within an enclosed region of space. Symbolically, the principle looks like
 T1 +  U1-2 =  T2
In works, this equations states that
~ The system’s initial kinetic energy ( T1) plus the work done
by all the external and internal force acting on the particles of
the system ( U1-2) is equal to the system’s final kinetic
energy ( T2).
Note that although the internal force on adjacent particles occur in equal but opposite
collinear pairs, the total work done by each of these forces will, in general, not cancel
out since the paths over which corresponding particles travel will be different. There
are two important exceptions to this rule which often occur in practice.
- When particles are contained within the boundary of a translating rigid body.
- When particles are connected by inextensible cables. In this cases, adjacent particles
exert equal but opposite internal forces that have components which undergo the
same displacement, and therefore the work of these forces cancels.
• Special class of problem involving work of friction caused by sliding.
We note also that equation “ T1 +  U1-2 =  T2” can be applied to problems
involved sliding friction; however it should be realized that the work of the
resultant friction force is not represented by µkNS; instead, this term represents
both the external work of friction (µkNS’) and internal work µkN(S – S’) which
is converted into various forms of internal energy, such as heat.
3.4 Power and Efficiency
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:
• Check homework, if any
• Applications
• Define power
• Define efficiency
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?
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 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 .
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.
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 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.
EXAMPLE
Given:A sports car has a mass of 2 Mg 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 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
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/e = 209.6/0.65 = 322 kW
3.5 Conservative Force And Potential Energy
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.
In-Class Activities:
• Check homework, if any
• Applications
• Conservative force
• Potential energy
• Conservation of energy
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?
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
B
F
 F · dr = 0
This means the work is conserved.
A
y
A conservative force depends
only on the position of the
particle, and is independent of its
velocity or acceleration.
x
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 texts. 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.
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
1 2
=
Ve
ks
2
Notice that the potential
function Ve always yields
positive energy.
3.6 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.
EXAMPLE (continued)
Solution:
Placing the datum at B:
TA + VA = TB + VB
1 125
1 125 2
(
)(10) 2 + 125(30) = (
)v B
2 32.2
2 32.2
VB = 45.1 ft
s
Equation of motion applied at B:
2
v
 Fn = man = m r
125 (45.1) 2
N B - 125 =
32.2 50
N B = 283 lb