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Work
When you hold something, are you exerting a
force on the object? Yes.
When you hold something, are you doing
work?
If you set the object on a table, does the table
exert a force on the object? Yes. Does the
table do any work? No.
When you hold something, then, you are not
doing work either.
Work
What can you do that the table can’t do? You
can lift the object up - which is work!
We define the concept of WORK as the
exertion of force through a distance.
There is one more point about work: in tether
ball, does the pole do work while it exerts a
force (via the rope) on the ball while the
ball moves around the pole? There is both
force and distance.
Work
No, the pole does not do work in the tether
ball case. This leads to the following
formal definition of work:
In doing work, the direction of the force has to
be parallel (or anti-parallel) to the distance
moved. We write this this way:
Work = F  s
where the dot indicates the cosine of the angle
between F and s: Work = F s cos(qFs) .
Work
Work = F  s
Although F and s are vectors (with magnitude
and direction), Work is a scalar
(magnitude only).
Can we have positive and negative work? If
the Force and distance are parallel, the
amount of work is positive, but if the two
vectors are anti-parallel, then the work is
negative.
More general definition of work
Work = F  s
If the force varies over the distance, then we
must break the distance into small pieces,
and use the average force on each piece to
find the work on that piece, then add up all
the small works to find the total work:
W =  Fi cos(qi) Dsi
If we carry this to the limit using calculus, we
get:
Work = sisf F cos(q) ds .
Energy
We can now define the concept of energy:
Energy is the capacity to do work (in ideal
circumstances).
We all know that we can do work: exert a
force through a distance. But to do that
requires food. Thus we convert the energy
in food into work. The same thing happens
when we burn coal to generate heat which
can be converted into electricity which can
be converted into lots of useful work.
Conservation of Energy
A Natural Law
Many such examples as we just saw lead us to
propose a natural law. Remember that a
natural law is a statement of how nature
seems to work - it is not “derived” from
anything more basic, it is observed to fit the
results of observations (experiments).
Energy can neither be created nor
destroyed (that is, energy is conserved).
However, it can be transformed from one
form into another.
Conservation of Energy
The equation that comes from this law of
conservation of energy is:
 Energiesinitially =  Energiesfinally .
Our job now is to find out how the amount of
energy in different forms relates to the
various parameters associated with that
form. That is, we need to derive formulas
for various kinds of energy.
Units
The units of energy (and work) are:
Nt*m = Joule.
A British unit of energy is the BTU (British
Thermal Unit). 1 BTU = 1054 Joules
(This is the energy necessary to heat one pound of water 1 oF.)
Another unit of energy is the calorie.
1 calorie = 4.186 Joules
(This is the energy necessary to heat one gram of water 1 oC.)
However, the calorie we refer to when we eat
is really a kilocalorie = 4186 Joules.
Positive and Negative
Can we have energies that are negative?
First, can we have negative money? Yes - it’s
called debt. You need to earn money to pay
off the debt and reach up to zero.
In the same way, some energies can be
negative - we need to gain some energy to
reach what we define as zero energy.
Forces and Energies
Since Energy is the capacity to do work, and work is
force thru a distance, then we need to consider
every force to see what kind of energy is
associated with it.
But from Newton’s Second Law, ΣF = ma, is there
an energy associated with the “ma” term as well?
Since “ma”, mass times acceleration, relates to the
motion, is there energy associated with motion?
Can a moving object do work (that is, exert a force
through a distance)?
Forms of Energy
Kinetic Energy
Energy of motion, called Kinetic Energy:
should depend on mass and speed of object.
Your car has energy when it is moving.
The wind has energy when it is moving, and
we can convert this wind energy into
electric energy via windmills.
Potential Energies
Energy of position, called Potential Energy:
should depend on why that position has
energy.
The water stored behind a dam has energy due to it’s
height above the base of the dam. We can use this
to run a hydroelectric station. The energy in food
is due to the molecular binding of the atoms in the
food. The same is true for coal, oil and gas.
There is also energy stored in the nucleus of atoms
- nuclear energy.
Some other forms
• Heat: should depend on temperature, type
and amount of material. We burn coal to
get heat to turn water into steam and use the
steam pressure to get work (or electricity).
• Light: should depend on type and intensity
of the light.
• Sound: should depend on type and
intensity of the sound.
Kinetic Energy
getting a formula
If we start with a single net force doing work
on an object so that the object picks up
speed (and no other energy is involved), we
have: Work = sisf F cos(q) ds .
Using Newton’s Second Law, with the force
simply directed always along the direction
of motion: F = ma = m dv/dt we get:
Work=sisf m dv/dt ds =vivf m ds/dt dv and
using v = ds/dt we get: Work= vivf m v dv
or finally, Work= ½mvf2 - ½mvi2 = DKE
Kinetic Energy
an alternative derivation
If we let an object fall, it gains speed. It also
gains what we call kinetic energy. By the
Conservation of Energy law, the amount of
work going into the object (from gravity)
will equal the amount of energy the object
has (kinetic): F s cos(q) = mg h (1). But if
an object falls a distance h with an
acceleration of g, how fast is it going?
Kinetic Energy – alternate derivation
KE(m,v) = mgh (The amount of kinetic energy,
which depends on the quantities mass and speed in
this case equals the amount of work done by
gravity, mgh).
From our motion equations, v = vo + gt and
h = ho + vot + ½gt2 or in this case (ho=0,
vo=0): h = ½gt2, or t = (2h/g)1/2 so v = gt =
g(2h/g)1/2, or v = (2hg)1/2, or h = ½v2/g; thus
mgh = mg½v2/g = ½mv2 = KE .
Kinetic Energy - formula
KE = ½mv2 . Note that the kinetic energy
depends on m (the more mass the more
kinetic energy) and on v2 (if you double the
speed, you quadruple the kinetic energy).
Also note that the kinetic energy must always
be either zero or positive - it can’t be
negative. (This is like cash in the money
analogy.)
Kinetic Energy - considerations
You’ve probably heard the expression: “speed
kills”. This comes from the fact that KE
depends on the square of the speed. If you
double your speed, you quadruple the
amount of energy of the object. And
remember that energy is the capacity to do
work - for either good or bad. Uncontrolled
energy can exert large forces through
significant distances - which can be very
dangerous!
Kinetic Energy - considerations
Note that the difference between (1 m/s)2 and
(2 m/s)2 is 3 m2/s2, whereas the difference
between (99 m/s)2 and (100 m/s)2 is 199
m2/s2 . What this indicates is that it takes
more and more energy to move faster and
faster. This explains why there is so little
difference between first and tenth in a speed
race between trained athletes!
Gravitational Potential Energy
If we let the force of gravity act on an object as it
moves, gravity is exerting a force through a
distance and may add or subtract energy from
the object. We can work with this near the
earth (where gravity is constant):
Work = sisf Fapplied cos(q) ds
To lift something up, the applied force is
opposite the force of gravity, so
Work = hihf (-Fgravity) dh = hihf (+mg) dh
= mg Dh = DPEgravvity , or PEgravity = mgh .
h and Dh
The formula for PEgravity really should be
DPEgr = mgDh instead of PEgr = mgh. We often
use h when we mean Dh, as in the case of “what is
your height?”. What we mean is, “what is the
difference in height between the top of your head
and the bottom of your feet?”. We often assume
that we are measuring from the floor, ground, or
some other standard position. If you use (or see)
PEgr = mgh, be sure to interpret it as DPEgr = mgDh,
and know where the “standard” position of h=0 is.
Gravitational Potential Energy
PEgravity = mgh
The greater the height, the more potential
energy there is - that is, if the object is
allowed to fall, the gravity will exert a force
through a distance and cause work.
We can say that the object has energy due to
its height (position). We call this a
potential energy.
Gravitational Potential Energy
more general form
Farther from the earth’s surface, gravity changes
with height, and we need a more general
formula:Work = sisf Fapplied cos(q) ds
To lift something up, the applied force is opposite
the force of gravity, so
Work = hihf (-Fgravity) dh = rirf (+GMm/r2) dr
= -GMm/rf + GMm/ri = DPE , or
PEgravity = -Gm1m2/r12 .
PE versus DPE
We have a similar case here as we did with the
near-earth PE: when should we use PE and
when should we use DPE ?
If we use PE = mgh, we are assuming that there
is some standard position where h=0 (usually
the ground, the floor, or table top).
If we use PE = -GM1m2/r12 , we are assuming
that the “standard r12” is a position that makes
PE = 0; to have this, the “standard r12” would
have to be infinity, since 1/infinity = 0.
PEgravity Considerations
In the simpler formula near the earth’s
surface, PEgravity = mgh both m and g are
positive numbers, but h is a height
measured from some point that you
determine. It can be the ground, but doesn’t
have to be. Note that h can be either
positive or negative since it is possible to
be below ground level.
PEgravity Considerations
In the more general form,
PEgravity = -Gm1m2/r12 , the PE is always
negative, with the highest (least negative)
value being when PE=0 or r12 goes to
infinity.
Note in particular, that while the force of
gravity goes as r122, the PE of gravity goes
only as r12.
Potential Energy for spring
DPEspring = -sis Fspring ds where Fspring = -kx
Recall that the minus sign in the PE equation
comes from the fact that to store energy, we
need to provide a force opposite that of the
spring. This relation then gives:
DPEspring = -sis -kx dx = ½kx2 - ½kxi2
If we choose xi to be the equilibrium position
and call this zero, then we have:
PEspring= ½kx2
Note that this PE does depend on position (x).
Problems
Having KE and PEgravity, we can start solving
some problems using the Conservation of
Energy.
Problem: How high will a ball go if it is
thrown with a speed of 25 m/s?
We could solve this problem using Newton’s
Second Law and the equations for constant
acceleration, or we could use Conservation
of Energy.
Tossing a ball up
Let’s try this problem using
Conservation of Energy:
We recognize that we have kinetic energy
(since we have motion), and we recognize
that we have gravitational potential energy
(since we have gravity); also vi=25 m/s;
hi=0 m (start from the ground); vf=0
(highest point). These are all related by the
Conservation of Energy:
Tossing a Ball Up
 Energiesinitially =  Energiesfinally .
KEi + PEi = KEf
+ PEf
(1/2)mvi2 + mghi = (1/2)mvf2 + mghf
(note that this is equivalent to saying DKE = DPE, or ½mvf2 – ½mvi2 = mg(hf-hi)
(1/2)*m*(25 m/s)2 + m*(9.8 m/s2)*(0 m) =
(1/2)*m*(0 m/s)2 + m*(9.8 m/s2)*hf .
Here we see that the mass cancels out, and we
have one equation in one unknown (hf):
hf = (1/2)*(25 m/s)2 / (9.8 m/s2) = 31.89 m.
Observations
We should note two things from this example:
Conservation of Energy is a scalar equation,
and so has no information about directions.
This makes it easier to solve, but gives less
information in the answer.
Conservation of Energy makes no mention of
time (only initial and final). This removes t
from the problem - making it easier but also
giving us less information in the answer.
Escape Speed
In the previous example, we threw something up that
went about 32 meters high. How fast would we
have to throw something to make it escape
from the earth altogether (if we continue to
neglect air resistance)?
We can use Conservation of Energy again, but
we need the more general form for potential
energy due to gravity.
To escape the earth, rf = infinity!
We start with ri = Rearth = 6.4 x 106m.
Escape Speed
 Energiesinitially =  Energiesfinally .
KEi + PEi = KEf
+ PEf
(1/2)mvi2 - Gmearthm/ri = (1/2)mvf2 - Gmearthm/rf
(note that this is equivalent to saying DKE = DPE, or ½mvf2 – ½mvi2 = -Gmem/rf - -Gmem/ri)
We see that m is in each term, so we cancel it.
(1/2)*(vi)2 - [(6.67x10-11 Nt*m2/kg2) *(6.0 x 1024 kg)/(6.4x106m)] =
(1/2)*(0 m/s)2 - [(6.67x10-11 Nt*m2/kg2) *(6.0 x 1024 kg)/(infinity)]
We again have one equation in one unknown (vi).
Escape Speed
We have used vf = 0 m/s since this is the
minimum speed we need at the end. We
could have more speed when we escape, but
we’re looking for the lowest speed for the
object to still escape; this means that both
the terms on the right side = 0.
Solving for vi = vescape = [2*G*Mearth /Rearth ]1/2
= 11,180 m/s = 25,000 mph.
Friction and Energy Loss
Can we use Conservation of Energy if we
have friction? What happens with friction?
We convert kinetic energy into heat!
What “formula” do we use for how much
energy is “lost” to friction, that is, how
much energy is converted from kinetic to
heat?
Friction
We start from the basic definition of energy:
the capacity to do work, where work =
Force thru a distance:
Elost = Ffriction * s. We still have Ffriction = mFc.
Where does this Elost go in the equation for
Conservation of Energy: on the initial or
final side? Is it a positive or negative
amount of energy?
Friction
Since some of the initial kinetic energy will go
(transform) into some heat, the Elost should be
a positive term if it is on the final side, or a
negative term if it is on the initial side.
 Energiesinitially =  Energiesfinally .
KEi + PEi = KEf + PEf + Elost
where Elost = + Ffriction*s = + mFcs .
Friction - example
Problem: If the coefficient of friction
between a block of wood and the concrete
floor is 0.50, how far will a block of wood
slide on the floor before coming to rest if it
starts with a speed of 10 m/s ?
We recognize this as a Conservation of
Energy problem with kinetic energy and
with friction (Elost).
Friction - an example
We are given: vi = 10 m/s; vf = 0 m/s; m = .5
We are looking for s (the distance of slide).
 Energiesinitially =  Energiesfinally .
KEi = KEf + Elost
where Elost = + Ffriction*s = + mFcs .
From  Fy = 0, we have Fc = mg. Therefore:
(1/2)*m*vi2 = (1/2)*m*vf2 + m*m*g*s .
Friction - an example
(1/2)*m*vi2 = (1/2)*m*vf2 + m*m*g*s
We notice that there is an m in each term so it
cancels out!
(1/2)*(10 m/s)2 = 0 + (.5)*(9.8 m/s2)*s
This is one equation in one unknown (s):
s = (1/2)*(10 m/s)2 / (.5)*(9.8 m/s2) = 10.2 m.
Contact Force and Energy
Every force we have considered so far has an energy:
• gravity (potential energy due to gravity)
• friction (energy lost to heat)
• spring (potential energy)
What about contact force? Since the contact
force is not normally able to move an object
through a distance, it normally can do no
work.
Pulleys and Work
Recall the single pulley situation from Part 2:
In this case, the pull, P, was only half the weight, W.
However, in order to lift the weight up a distance, h,
we would have to pull the rope twice that distance!
Therefore the work done is
W = F*d = (½W)*(2h) = W*h,
P
which is the same work with or
without the pulley.
W
Power
We now know what Force and Energy are, but
what is Power?
The definition of Power is that it is the rate of
change of Energy: Power = DEnergy / Dt .
The units of power are: Joule/sec = Watt.
Another common unit is the horsepower, hp.
The conversion factor is: 1 hp = 746 Watts.
Power - example
What is your power output when you climb
stairs?
In this case, you are changing your potential
energy (mgh) in time, so … P = Dmgh / Dt
if your mass = 70 kg, gravity is 9.8 m/s2, and
you climb steps of height 10 meters in a
time of 20 seconds:
P = 70 kg * 9.8 m/s2 * 10 m / 20 sec = 343 W
or 343 W * (1 hp / 746 W) = .46 hp .
Power - example
What is your average power output per day?
If you eat 2000 Calories per day, (and
assuming you do not gain or lose weight),
that energy must be converted into energy
you use throughout the day.
P = 2,000 Calories / day =
[(2,000 Cal)*(4,186 joule/Cal)] / [(24 hours)*(60 min/hr)*(60 sec/min)]
= 97 Watts.
Power – another example
How powerful must a car engine be (on
average) if it is to accelerate a 2,000 kg car
from zero to 65 mph in 20 seconds?
This is a power question. The change in
energy is in the form of kinetic energy. We
should convert 65 mph into metric form:
vf = 65 mph * (1 m/s / 2.24 mph) = 29 m/s.
vi = 0 (starts at rest); Dt = 20 sec.
Power – another example
Power = DEnergy / Dt =
[ {½*m*vf2} – {½*m*vi2}] / t =
[(1/2)*(2,000 kg)*(29 m/s)2 - 0 ] / 20 sec =
42,000 Watts * (1 hp / 746 Watts) = 56 hp.
Note that this is the average power.
Force and Power
We know how force is related to energy, and
how energy is related to power. Can we
relate power to force?
Work = F s , Power = DWork /Dtime =
F Ds /Dt (but Ds/Dt = v), so Power = F v .
Power, like work, is a scalar.
Note that if F is constant, Power must go up
as speed (v) goes up! If Power is constant,
F must go down as v goes up.
Force and Power
Power = F v
At very low speeds, even a small power will
give a rather large force!
On cars with manual transmissions, you
normally don’t rev up the engine (high
power), and then pop the clutch! This causes
tremendous forces that can break the car!
A spreadsheet has been prepared showing Power and Force
versus speed, including air resistance, and it is available for
download on the web page.
What do you pay for:
Force, Energy, or Power?
What do you pay MLG&W (for example) for:
force, energy or power?
What do you pay the gas station for:
force, energy or power?
In both cases you pay for ENERGY!
Cost of Energy
What is the cost of energy?
We saw before that we could do work at the
rate of a couple 100 Watts, but that was
hard work!. If we worked for 40 hours a
week, how much useful work would we
perform? Work = Energy = Power * time.
The MKS unit of energy is the Joule, but this
is a very small unit. Another common unit
of energy is the Kilowatt*hour.
Cost of Energy
In terms of kilowatt-hours, if you worked at
the rate of 200 Watts for 40 hours, you
would do 8 KW-Hrs of work.
How much does the power company charge
for a KW-hr of energy?
Cost of Energy
In Memphis, MLG&W charges about 10 cents per
KW-hr. Thus, if you were to work for the power
company providing power, you would earn about 8
KW-hr/week * $.10 = $.80/week (80 cents per
week)!
As we see, energy is quite cheap! The reason our
utility bills are so high is that we use so much energy
- especially when we heat (or cool) things (like air
and water)!
More Examples
The computer homework program, Energy
and Power, Volume 2, #1, deals with
problems involving both energy and power.
We have two labs that also deal with energy:
Atwoods Machine and Hooke’s Law.