Transcript Energy - Edublogs

Work, Energy and Power
AP style
Energy
Energy: the currency of the universe.
Everything has to be “paid for” with
energy.
Energy can’t be created or destroyed, but
it can be transformed from one kind to
another and it can be transferred from
one object to another.
How do you know an
object has
mechanical energy
(kinetic / potential)?
If it can change itself
or change its
environment, then it
certainly has
energy.
• Doing WORK is one way to transfer
energy from one object to another.
Work = Force x displacement
W = Fd
• Unit for work is Newton x meter. One
Newton-meter is also called a Joule, J.
Work = Force x displacement
• Work is not done unless there is a
displacement.
• If you hold an object a long time, you
may get tired, but NO work was done.
• If you push against a solid wall for
hours, there is still NO work done.
• For work to be done, the displacement of the
object must be in the same direction, as the
applied force. They must be parallel.
• If the force and the displacement are
perpendicular to each other, NO work is done
by the force.
So, using vector multiplication,
W=F•d
(In many university texts, as well as the AP
test, the displacement is represented by “s”
and not “d”
W=F•s
An object is subject to a force given by
F = 6i – 8j as it moves from the position r = -4i + 3j
to the position r = i + 7j
What work was done by this force?
First find the displacement, s
s = Dr =
rf – ro =
(i + 7j) - (-4i + 3j) =
5i + 4j
Then, do the dot product W = F · s
(6i – 8j) · (5i + 4j) =
30 – 32 = -2J of work
• In lifting a book, the force exerted by your
hands is upward and the displacement is
upward- work is done.
• Similarly, in lowering a book, the force
exerted by your hands is still upward, and
the displacement is downward.
• The force and the displacement are STILL
parallel, so work is still done.
• But since they are in opposite directions,
now it is NEGATIVE work.
• On the other hand, while carrying a
book down the hallway, the force from
your hands is vertical, and the
displacement of the book is horizontal.
• Therefore, NO work is done by your
hands.
Your Force
Vertical component of d
• So,….while
climbing stairs or
walking up an
incline, only the
vertical
component of
the displacement
is used to
calculate the
work done in
moving the
object from the
bottom to the
top.
Horizontal component of d
Example
How much work is done to
carry a 5 kg cat to the top of
a ramp that is 7 meters long
and 3 meters tall?
W = Force x displacement
Force = weight of the cat
d = height NOT length
W = mg x h
W = 5 x 10 x 3
W = 150 J
3m
Example
Displacement = 20 m
A boy pushes a
lawnmower 20 meters
across the yard. If he
pushed with a force of
200 N and the angle
between the handle
and the ground was
50 degrees, how
much work did he do?
F cos q
q
F
W = (F cos q )d
W = (200 cos 50) 20
W = 2571 J
NOTE: If while pushing an object, it is
moving at a constant velocity,
the NET force must be zero.
So….. Your applied force must be exactly
equal to any resistant forces like friction.
How much work do you do to
carry a 30 kg cat from one
side of the room to the other
if the room is 10 meters
long?
ZERO, because your Force is
vertical, but the displacement
is horizontal.
• Energy and Work have no direction
associated with them and are therefore
scalar quantities, not vectors.
Power
The rate at which work is done.
1. Power = Work ÷ time
Unit for power = J ÷ s
= Watt, W
What is a Watt in “fundamental units”?
Since work is also the energy transferred or
transformed, “power” is the rate at which energy
is transferred or transformed.
2. P = Energy ÷ time
This energy could be in ANY form: heat, light,
potential, chemical, nuclear
Since NET work = D K,
3. P = DK ÷ t
And yet another approach:
P = W ÷ t = (F·d) ÷ t = F · (d ÷ t)
P=F·v
Kinetic Energy
the energy of motion
K = ½ mv2
Where does K = ½ mv2 come from??
Did your amazing teacher just arbitrarily
make that equation up??
Hmmm…
The “Work- Kinetic Energy
Theorem”
Worknet = DK
Fnet •d = DK = ½ mvf2 - ½ mvo2
You are supposed to be able to
use Fnet = ma, kinematics equations and Wnet = Fnet •d to
derive the work-kinetic energy theorem…..
So…. do it!! Then you’ll see where the equation we call
“kinetic energy” comes from.
(Hint: start with Fnet = ma and use the kinematics equation
that doesn’t involve time.)
Potential Energy
Stored energy
It is called potential
energy because it
has the potential to
do work.
• Example 1: Spring potential energy in
the stretched string of a bow or spring
or rubber band. SPE = ½ kx2
• Example 2: Chemical potential energy
in fuels- gasoline, propane, batteries,
food!
• Example 3: Gravitational potential
energy- stored in an object due to its
position from a chosen reference point.
Gravitational potential energy
GPE = weight x height
GPE = mgh
• The GPE may be negative. For
example, if your reference point is the
top of a cliff and the object is at its
base, its “height” would be negative, so
mgh would also be negative.
• The GPE only depends on the weight
and the height, not on the path that it
took to get to that height.
Work and Energy
Often, some force must do work
to give an object potential or
kinetic energy.
You push a wagon and it starts
moving. You do work to
stretch a spring and you
transform your work energy
into spring potential energy.
Or, you lift an object to a certain
height- you transfer your
energy into the object in the
form of gravitational potential
energy.
Work = Force x distance =
change in energy
Example
How much more distance is required to stop
if a car is going twice as fast?
Fd = ½ mv2
The work done by the brakes = the change in the kinetic energy
With TWICE the speed, the car has
FOUR times the kinetic energy.
Therefore it takes FOUR times the stopping
distance.
The Work-Kinetic Energy
Theorem
NET Work done = D Kinetic Energy
Wnet = ½ mv2f – ½ mv2o
Varying Forces
The rule is… “If the Force varies, you must
integrate!”
If the force varies with displacement , in
other words Force is a function of
displacement, you must integrate to find
the work done.
If the force is a function of velocity, you must
integrate to find the power output.
Examples:
If F(x) = 5x3, what work is done by the force
as the object moves from x = 2 to x = 5?
If F(v) = 4v2, what power was developed as
the velocity changed from 3 m/s to 7 m/s?
Examples of Integration
An object of mass m is subject to a force given by F(x) = 3x3 N.
What is the work done by the force?
3 4
W   F x dx   3 x dx  x
4
3
Examples of Integration
At object, moving along the x-axis, has a velocity of 3m/s
when it passes the origin at t = 0s. It has an acceleration
given by a(t) = 1.5t m/s2. Find v(t) and x(t)
15
. 2
v (t )  a(t )dt   15
. tdt 
t  C 0.75t 2 3
2
0.75 2
x(t )  v (t )dt   (0.75t 3)dt 
t 3t  C 0.375t 2 3t 0
2
Examples of Integration
A particle of mass m moves along the y-axis as
y(t)=at4-bt2+2t, where a and b are constants, due
to a constant applied force, Fy(t) What is the
power, P(t) delivered by the force?
P = F·v
We need to find both v and F!
v t  4at 3 2bt 2
a(t )12at 2 2b
F  ma  m(12at 2 2b)
P  Fy  v y  m(12at 2b)(4at 2bt 2)
2
3
Examples of Definite Integration
An object is moving along the x-axis with a velocity
given by v(t)=-2t2+8t-6 where t is in seconds.
What was its displacement from t = 2s to t = 6s?
6
6
2
8
x (t )  v (t )dt   ( 2t 2 8t 6)dt   t 3  t 2 6t
3
2
2
2
6
2
2
8
2
8
(  6 3  6 2 66)(  2 3  2 2 62) 45.33m
3
2
3
2
Examples of Definite Integration
To stretch a NON linear spring a distance x requires
a force given by F(x) = 4x2 – x.
What work is done to stretch the spring 2 meters
beyond its equilibrium position?
2m
W 

0
2m
4 3 1 2
F  ds   (4 x  x )ds  x  x
3
2
0
2
2m
0
4 3 1 2 4 2 1
( 2  2 )( 0  0)8.667J
3
2
3
2
Examples of Definite Integration
A 2.88kg particle starts from rest at x = 0 and moves under
the influence of a single force, Fx= 6 + 4x -3x2 where Fx is
in Newtons and x is in meters. Find the power delivered to
the particle when it is at x = 3m.
P = F·v, but what is the velocity at x = 3m?
Hmmm….
W   F( x )dx
W  DK so.....
and
3
xf
1 mv 2  1 mv 2
F
(
x
)
dx

D
K

f
o

2
2
 (6 4 x 3 x
0
xo
3
 (64x 3x 2 )dx 6x 
0
4 2 3 3
x  x
2
3
3
0
2
)dx  DK  12 mv x2 3
4
3
(63  3 2  3 3 )0 9J  12 mv x2 3
2
3
v x  3 2.5m / s and Fx  3  64333 2  9N so P  F  v  92.5 22.5W