ppt work energy
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THE WORK OF A FORCE, THE PRINCIPLE OF
WORK AND ENERGY & SYSTEMS OF PARTICLES
Todays Objectives:
Students will be able to:
1. Calculate the work of a force.2. Apply the principle of work and
energy to a particle or system of
particles.
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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 theparticle by the force, FRwill be
A) B)
C) D)
2
1
s
ts
F ds 21
s
ts
F ds 2
1
s
ns
F ds2
1
s
ns
F ds
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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, ) to control the forces experienced by thepassengers?
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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.
This principle is useful for solving problems that involve
force, velocity, and displacement. It can also be used to
explore the concept ofpower.
By substituting at = v (dv/ds) into Ft = mat, the result is
integrated to yield an equation known as theprinciple of work
and energy.
To use this principle, we must first understand how to
calculate the work of a force.
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WORK OF A FORCE (Section 14.1)
A force does workon a particle when the particle undergoes a
displacement along the line of action of the force.
Work is defined as theproduct offorceand displacement components acting in
the same direction. So, if the angle
between the force and displacement
vector is , the increment of work dU
done by the force is
dU = F ds cos
By using the definition of the dot product
and integrating, the total work can be
written asr2
r1
U1-2 = F dr
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WORK OF A FORCE
(continued)
Work ispositive if the force and the movement are in thesame direction. If they are opposing, then the work is
negative. If the force and the displacement directions are
perpendicular, the work is zero.
IfF is a function of position (a common
case) this becomes
s2
s1
F cos dsU1-2
If both F and are constant (F = Fc), this equation furthersimplifies to
U1-2 = Fc cos s2 - s1)
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WORK OF A WEIGHTThe work done by the gravitational force acting on a particle
(orweight 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.
Ify is upward, the work isnegative since the weight force
always acts downward.
y
W
r2
r1
U1-2 = F dr yWyyWWdykdzjdyidxjWy
y
r
r
2
1
2
1
12).()(
y1
y2
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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 thedisplacement from the unstretched position.
If a particle is attached to the spring, the force Fs exerted on theparticle 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 ] .
The work of the spring force moving from position s1
to position
s2 is
= 0.5 k (s2)2 0.5 k (s1)
2k s dsFs dsU1-2
s2
s1
s2
s1
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WORK OF A SPRING FORCE
r2
r1
U1-2 = F dr
2
)(
22)().()(
2
2
2
1
2
1
2
2
2
1
2
1
2
1
ssk
ksksdsksdsFkdzjdyidsiF
s
s
s
s
s
r
r
s
Note s here is measured
with respect to the
upstretched position.
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PRINCIPLE OF WORK AND ENERGY
(Section 14.2 & Section 14.3)
U1-2 is the work done by all the forces acting on the particle as itmoves from point 1 to point 2. Work can be either apositive or
negative scalar.
By integrating the equation of motion, Ft = mat = mv(dv/ds), theprinciple 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
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 apositive 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.
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PRINCIPLE OF WORK AND ENERGY
(continued)
The principle of work and energy cannotbe used, in general, to
determine forces directed normal to the path, since these forces
do no work.
Note that the principle of work and energy (T1 + U1-2 = T2) isnot 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 ajoule (J),
where 1 J = 1 Nm. In the FPS system, units are ftlb.
The principle of work and energy can also be applied to a system
of particlesby summing the kinetic energies of all particles in the
system and the work due to all forces acting on the system.
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The case of a body sliding over a rough surface merits special
consideration.
This equation is satisfied if P = kN. However, we know fromexperience that friction generates heat, a form of energy that doesnot seem to be accounted for in this equation. It can be shown thatthe work term (kN)s representsboth the external workof the
friction force and the internal workthat is converted into heat.
WORK OF FRICTION CAUSED BY SLIDING
The principle of work and energy would be
applied as
0.5m (v)2 + P s (kN) s = 0.5m (v)2
Consider a block which is moving over a
rough surface. If the applied forcePjust
balances the resultant frictional force kN,a constant velocity v would be maintained.
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Given: When s = 0.6 m, the spring is
not stretched or compressed,
and the 10 kg block, which issubjected 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.
EXAMPLE
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S2
S1
Fs
W
FNy2
y1
sin()=( y1-y2)/( s2-s1)
y1-y2
y2-y1= -(s2-s1) sin()
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Apply the principle of work and energybetween position 1
(s = 0.6 m) and position 2 (s). Note that the normal force (N)
does no work since it is always perpendicular to the
displacement.
EXAMPLE
(continued)Solution:
T1 + U1-2 = T2
There is work done by three different forces;1) work of a the force F =100 N;
UF = 100 (s s1) = 100 (s 0.6)
2) work of the block weight= -Wy=-(s2-s1) sin()*-mg
UW = 10 (9.81) (s s1) sin 30 = 49.05 (s 0.6)
3) and, work of the spring force.
US
= - 0.5 (200) (s0.6)2 = -100 (s
0.6)2
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The work and energy equation will be
T1 + U1-2 = T2
0.5 (10) 52 + 100(s 0.6) + 49.05(s 0.6) 100(s 0.6)2 = 0
125 + 149.05(s 0.6) 100(s 0.6)2 = 0
EXAMPLE
(continued)
Solving for (s 0.6),(s 0.6) = {-149.05 (149.05
2 4(-100)125)0.5} / 2(-100)
Selecting the positive root, indicating a positive spring deflection,
(s 0.6) = 2.09 m
Therefore, s = 2.69 m
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CONSERVATIVE FORCES, POTENTIAL ENERGY
AND CONSERVATION OF ENERGY
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CONSERVATIVE FORCE
(Section 14.5)
A forceF 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 bythe forceF in a closed path (i.e., from A to B and then back to A)
is zero.
Thus, we say the work is conserved.
rF 0 d
x
y
z
A
BF
The work done by a conservative
force depends only on the positionsof the particle, and is independent of
its velocity or acceleration.
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CONSERVATIVE FORCE (continued)
A more rigorous definition of a conservative force makes
use of a potential function (V ).
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 andthe potential energy from springs or other elastic elements.
Vtotal = Vgravity + Vsprings
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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 ofpotential energy and how to calculate their magnitudes.
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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 ispositive if y is above the
datum and negative if y is
below the datum. Remember,
YOU get to set the datum.
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ELASTIC POTENTIAL ENERGY
Recall that the force of an elastic spring is F = ks.
Notice that the potential
function Ve always yields
positive energy.
k s22
1Ve
Ve (where e denotes an
elastic spring) has the distance
s raised to a power (the
result of an integration) or
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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 remainsconstant. 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
2211 VTVT = Constant
T1 stands for the kinetic energy at state 1 and V1 is thepotential energy function for state 1. T2 and V2represent these energy states at state 2. Recall, the
kinetic energy is defined as T = mv2.
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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 andC. Set the gravitational potential energy datum at point A
or point C (in this example, we choose point A).
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Similarly, the potential and kinetic energies at A will be
VA = 0.5 (30) (2 1)2, TA = 0.5 (2) 4
2
EXAMPLE
(continued)Solution:
Note that the potential energy at Chas two parts.
VC = (VC)e + (VC)gVC = 0.5 (30) (5 1)
2 2 (9.81) 1
The kinetic energy at C is
TC = 0.5 (2) v2
The energy conservation equation becomes TA + VA = TC + VC .[ 0.5(30) (5 1)2 2(9.81)1 ] + 0.5 (2) v2
= [0.5 (30) (2 1)2 ]+ 0.5 (2) 42
v = 5.26 m/s
Vg = W y
k s221Ve
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2211 VTWVT consernon
2211 VTVT
For a system of particles under conservative forces:
For a particle under conservative and non conservative
forces:
Important Notes:
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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.
EXAMPLE
mghkmvssFmv
VTWVT consernon
22
212
2
1
2211
2
1
2
1)(0
2
1
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S2
S1
y2y1
sin()=( y1-y2)/( s2-s1)
y1-y2
h=y1-y2= (s2-s1) sin()
Datum
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)sin()(21
2
1
)(02
1
12
2
12
2
2
12
2
1
ssmgsskmv
ssFmv
Below is the solution of the other approach in numbers:
T1 + U1-2 = T2
0.5 (10) 52 + 100(s 0.6) + 10 (9.81) (s s1) sin 30 100(s 0.6)
2 = 0
Both are the same.
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Example
Given: The 800 kg roller
coaster starts from
A with a speed of
3 m/s.
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 .
Find: The minimum height, h, of the hill so that the car
travels around inside loop at B without leaving the track. Alsofind the normal reaction on the car when the car is at C for this
height of A.
Plan:
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Example (continued)
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)
Solution:
manmg
NB 0
=
Equation of motion applied at B:
2v
mmaF nn
10800 (9.81) = 800
(vB)2
vB
= 9.905 m/s
2) Find the required velocity of the coaster at B so it doesntleave the track.
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Example (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
manmg
NC
=
NC = 16.8 kN
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
Equation of motion applied at B:
2vmFn
7NC+800 (9.81) = 800
14.662
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ATTENTION QUIZ
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.
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