Posted by: PrudensOptimus

Can someone distribute their knowledge of the above topics? I have a test on these tomorrow.

Thanks.

Posted by: turin

WORK: Work is the transfer of mechanical energy
ENERGY: Energy is the potential to do work. It is conserved when you account for all of its forms, in a closed inertial system.
POWER: Power is the time rate at which work is done.
MOMENTUM: Momentum is the quantity of motion (or ability to provide impulse). It is conserved both linearly and rotationally in a closed system in an inertial frame.



WORK can be calculated in two basic ways:

1) Integrate the dot product of the force with the position displacement over the path. This has a simplification that comes up in first semester physics: W = Fd, where W is the work, F is the force in the direction of travel (assumed to be constant, and here, basically treated as a scalar, not a vector), d is the distance traveled. Note that W can be negative (the object had to do work in order to move) if the force is in the oposite direction of the travel.

2) Calculate the change in mechanical energy. Usually this will be simpler, so, if you are given a bunch of energy info, then you probably want to use this one. W = DeltaKE + DeltaU = KE_final - KE_initial + U_final - U_initial, where W is the work (done to the object), KE is the kinetic energy (1/2 m v^2), U is the potential energy (mgh for gravity, 1/2 k x^2 for a spring). Note that W can be negative for the same reason as before (it could also be positive). This indicates that the system is not conservative (usually this means that friction is present). If the work is zero, then the system is either conservative, or work is being done to the object to compensate for the work that it is doing. I would expect you to encounter the former case rather than the later.


Energy will probably just be the sum of the kinetic energies and potential energies. There are tons of details here. Basically, kinetic energy is translational and rotational (1/2 m v^2 and 1/2 I w^2). Potential will probably be gravitational or elastic (mgh or 1/2 k x^2). Remeber that the h is arbitrary for gravitational, but the x IS NOT ARBITRARY FOR THE ELASTIC POTENTIAL. IT IS THE DISTANCE FROM THE EQUILIBRIUM POSITION. It provides a resotring force, whereas gravity does not (usually, unless you're talking about a pendulum). Like i said, there are a lot of details for energy.


To calculate power, divide the amount of work done by the time it took to do the work. Also, there is thrust, which is the force of the exhaust gas multiplied by the velocity that it is flying out of the jet engine. I feel like I'm missing something important here, but I'll move on.


The rate of change of momentum of an object is defined to be the net (or resultant) force on an object (a generalization of Newton's second law). The change in momentum of an object is defined to be the impulse on an object. Momentum, force, and impulse are all VECTOR QUANTITIES. It is easy to screw up on a calculation when you use the magnitudes of these quantities without considering their direction.

Impluse can be calculated in two ways (notice the analogy to work):

1) It is the integral of the force (as a function of time) over time. This sometimes simplifies to I = Ft, where I is the impulse, F is the force (assumed to be constant), t is the duration of the force. Note that, since F is a vector, impulse will be a vector. (For work, F loses all of its vector like qualities except one: it can cause a negative result.)

2) Calculate the change in momentum: I = p_final - p_initial. Usually, it is safe to say that p = mv (for nonrelativistic particles, so unless otherwise stated in the problem, use this formula).


To help you organize:

energy is to work is to power (not vectors, scalars)
as
momentum is to impulse is to force (vectors)

Work is what breaks the conservation of energy.

Impulse is what breaks the conservation of momentum.

That is not to say that whenever you have work or impulse, that energy and momentum are not conserved. If you consider work or impulse inside the system, then energy and momentum are conserved. If you consider work and energy as acting on the system from outside, then energy and momentum can break conservation in the system.

Example for energy:
If we consider a mass falling due to gravity, then we can say that gravity is doing work on the object because it is applying a downward force to the object, and the object is moving downward. The force is mg, the distance it travels is y, therefore the work done on the object is mgy (Fd). This is manifest in the fact that the object is speeding up as it falls (and thus gaining kinetic energy). Now, consider the same situation, but include the gravitational potential in the system. In this view, the increase in kinetic energy is due to the decrease in gravitational potential energy. Thus, the energy is conserved. In the first point of view, we treated gravity as external to the system. In the second point of view, we included the potential energy of gravity as part of the system.

Example for momentum:
This one is a bit more tricky. I will use a bit more obscure example, because I can't think of a more reasonable one. A ball hits a wall and bounces back. Its velocity has changed, and therefore its momentum has changed. Does this violate the conservation of momentum. No. Actually, the wall recoiled ever so slightly, but, since it is rigidly connected to the earth, and the earth is so huge, then the recoil was negligible. So, the change in mometum of the ball was mV, and the change in the momentum of the earth was Mv, and mV must equal Mv, but, since M >> m, then v << V.

Another example for momentum would be the falling object again. Its downward momentum increases because the force of gravity puts an impulse on it of Ft = mgt. Thus, it seems like the impulse breaks the conservation of momentum. But actually, the earth's momentum is again the compensation. The earth moves upward to meet the object as it falls. The impulse on the earth is -mgt. Since M >> m, then, again, the change in velocity isn't noticible.

Posted by: turin

I forgot to mention rotational stuff:

angular moment of inertia and angular velocity is to angular momentum is to torque
as
mass and velocity is to momentum is to force

(Usually, when you just hear "momentum" and "velocity" without a pre-qualifier, that refers to the linear version)

Symbolically:

Iw : L : �„ :: mv : p : F

Angular momentum is conserved, separately from linear momentum (in the absence of a torque applied from outside the system).

Don't forget your centripetal acceleration equations:

a_c = m (v^2) / r = m (w^2) r

Posted by: PrudensOptimus

Wow, looks yummy, thanks for your share of knowledge, Turin.


I tried to integrate Work, ∫ΣF dx, where x is displacement, replacing F with ma, I end up getting δKE∫1/dx dx. Does ∫1/dx dx give you ln|x| + C, or something else? I'm not sure. And what is the bottom and top index of the Work Integral? Is it the initial position and the final posistion?

Posted by: PrudensOptimus

When do you use W_net = �KE?

I saw some questions where W = �KE + �PE.

Or was that just for non conservative works? If it is for non conservative works only, then that means in cases when W = �KE, the W must mean for conservative?

Posted by: PrudensOptimus

What does it mean:

"It must be emphasized that all the forces acting on a body must be included in equation 6-10 either in the potential energy term on the right (if it is a conservative force), or in the work term W_NC, on the left(but not in both!)"

EQ 6-10 : W_NC = �KE + � PE.

Posted by: turin

quote:

Originally posted by PrudensOptimus
I tried to integrate Work, ∫ΣF dx, where x is displacement, replacing F with ma, I end up getting δKE∫1/dx dx.

I'm assuming that you mean you tried to use the integrate idea to find work. Don't forget that force is a vector, so it is a dot product of the force with the displacement. Also, one subtle issue, work is <i>usually</i> associated with a particular force (i.e. friction), not the resultant, so you usually will not have that summation in the integrand. I'm really sorry, but I don't quite follow what you are trying to do.

Is this what you were trying to do?

∫Fdx = ∫madx = m∫(dv/dt)dx = m∫(dv/dx)(dx/dt)dx
= m∫(dv/dx)vdx = m∫vdv = (m/2)[v² - v₀²] = KE - KE₀ = ÃŽâ€Â�KE

Posted by: turin

quote:

Originally posted by PrudensOptimus
When do you use W_net = �KE?

I saw some questions where W = �KE + �PE.

Or was that just for non conservative works? If it is for non conservative works only, then that means in cases when W = �KE, the W must mean for conservative?

You probably use this notion when you are considering the work being done by the potential energy in the problem. If you include the potential energy in the problem, then the system is conservative. If you define the system without the potential energy (i.e. call gravity an external force), then the system is not conservative.

Also, for instance, when you have an object sliding along some surface, and there is friction, then the friction will slow it down. Since the object slows down, it will lose kinetic energy, but there really isn't any potential energy to account for (the energy changes into thermal energy, the mechanism is called heat). So, friction usually makes the system non-conservative. To calculate the work done by friction, it is usually easiest to use the equation that you have up there. You would use the first equation if there was no change in any potential energy. If, say, you have a block sliding down an inclined plane, you would need to use the second equation, because the gravitational potential energy is part of the mechanical energy, and thus, it will increase the kinetic energy.

Posted by: turin

quote:

Originally posted by PrudensOptimus
What does it mean:

"It must be emphasized that all the forces acting on a body must be included in equation 6-10 either in the potential energy term on the right (if it is a conservative force), or in the work term W_NC, on the left(but not in both!)"

EQ 6-10 : W_NC = �KE + � PE.

It looks like they are using the subscript, "NC," to indicate that they are strictly talking about "Non-Conservative" work. That means that there is no potential energy that you can associate with it. Friction, I'm pretty sure, will contribute to the W_NC. Gravity, and spring force stuff, goes into the �PE on the right.

It's a little tricky, though, because it depends on how you define your system. I would say that it really doesn't matter, as long as you stick with the same defined system through your calculations.

The first law of thermodynamics says that �E = Q - W, where E is the energy in the system (accounting for ALL forms), Q is the heat transfered into the system, and W is the work done by the system. Maybe you will see the first law written as �E = W - Q. In this case, you just change the directions of W and Q. The sign just tells you which way the work and heat is going (into the system or out of the system). If you've seen this thermo stuff, then that W_NC almost definitely means the W or -W in the first law of thermodynamics, assuming Q = 0, and all the energy is considered as kinetic and potential energy.



Sorry, I said "second" law when I meant "first" law. I have editted the mistake.
I changed some other stupid statements as well. I hope I didn't screw you up on your test.
Oh hell, I don't really know what to say about this. You would probably do best to ignore the thermodynamics part; I'm confusing myself with it.

Posted by: PrudensOptimus

quote:

Originally posted by turin
I'm assuming that you mean you tried to use the integrate idea to find work. Don't forget that force is a vector, so it is a dot product of the force with the displacement. Also, one subtle issue, work is <i>usually</i> associated with a particular force (i.e. friction), not the resultant, so you usually will not have that summation in the integrand. I'm really sorry, but I don't quite follow what you are trying to do.

Is this what you were trying to do?

∫Fdx = ∫madx = m∫(dv/dt)dx = m∫(dv/dx)(dx/dt)dx
= m∫(dv/dx)vdx = m∫vdv = (m/2)[v² - v₀²] = KE - KE₀ = ÃŽâ€Â�KE

When I replaced F with ma, and a with (v^2 - v₀^2)/2ÃŽâ€Â�x, isn't m((v^2 - v₀^2)/2) suppose to be treated as constants? and thus getting ÃŽâ€Â�KE * ∫1/ÃŽâ€Â�x dx?

Posted by: PrudensOptimus

m�ç(dv/dx)vdx --- how did you get to that part?

Posted by: turin

quote:

Originally posted by PrudensOptimus
When I replaced F with ma, and a with (v^2 - v₀^2)/2ÃŽâ€Â�x, isn't m((v^2 - v₀^2)/2) suppose to be treated as constants?

F is the integrand, and it can be a function of the position of the particle, in general. If you replace it by ma, then the a must be a function of the position of the particle, in general. You cannot replace this by a constant, in general.

Posted by: turin

quote:

Originally posted by PrudensOptimus
m�ç(dv/dx)vdx --- how did you get to that part?

This is the chain rule in action:

Velocity, at a particular instant of time, has a value. You can make a convenient rule for this value which will accept, as its input, the time, and will give, as its output, the value of the velocity. This rule is the time derivative of the position of the particle, evaluated at the specific time input into the rule. In other words, this is the function, f(t):

v = f(t)

Velocity, at a particular point in space, has a value (because we're talking about a trajectory). I couldn't tell you off the top of my head what exactly that rule is, or where it comes from, without mentioning energy, but lets see what happens when we differentiate this rule (the function, g(x)) with respect to time:

v = g(x)


Now for the chain rule:

d[g(x)]/dt = d[g(x(t))]/dt

Here, I have just explicated the time dependence. Now, using the chain rule:

= g'(x(t))(dx/dt) = (dg/dx)(dx/dt)

But, going back to the first rule for assigning a value to v, you can see that:

(dx/dt) = v

so that:

d[g(x)]/dt = (dg/dx)v

But, g(x) also equals v at this point on the trajectory, so:

d[v]/dt = (dv/dx)v

But,

dv/dt = a

Thus,

a = v(dv/dx).


This is usually done a bit sloppily, referring to the velocity itself as an inherrent function of position and time. I think that doing so causes confusion. It is just the chain rule, but try to remember that we're talking about functions that return a value that is v, not v as a function in itself. After having said this, my notation is probably a bit sloppy. To try to make the point a bit more clear:

We usually talk about v as a function of t, and we denote it:

v = v(t)

This is the f(t).

What we want to end up with is an expression with a, v, and x, without t hanging around and bothering us. What we do in terms of the math is to start with an integral of a function of t over the variable x. This is messy. We could express x in terms of t, but we choose t in terms of x. We use the chain rule to get the time derivative of f(t) in terms of a function, g(x), and its derivative. So, I should probably give it to you like this:

h(x) = g(x)(dg/dx)

where, h(x) gives the acceleration at the point, x:

a = h(x)

and g(x) gives the velocity at the point x

v = g(x).

So:

df/dt = g(x)(dg/dx).

It should be emphasised that this does not generally happen with the chain rule that the function just kind of pops out of the derivative. This occurs in this case because of the relationship between the two choices of independent variable that we're playing around with to each other.

physiology

bridge

The kilowatt-hour is a unit of energy used by electrical utilities.

The btu per hour (often erroneously shortened to btu) is a unit of power used by the heating, ventillation, and cooling industry (HVAC).

A horsepower is a unit of power sufficient to raise 33,000 pounds 1 foot every 1 minute (550 lbs, 1 ft, 1 sec) equivalent to roughly 745.70 W

Summary

• bullet

Problems

practice

1.        A typical adult in the United States consumes something like 2000 dietetic calories of food per day. Determine the average power generated by such an adult (assuming he or she is not gaining or losing weight).

o        We have to assume here that all the food energy consumed goes into work on some level (mechanical or metabolic). This is true only so long as the person is not gaining or losing weight, which occurs when the energy consumed is greater than or less than the work done. Start the problem by converting the units -- convert the energy from calories to joules and convert the time from days to seconds.

o        With something between 700 and 800 watts in a horsepower, this corresponds to a rate of energy conversion somewhere between one-seventh and one-eighth of a horsepower.

2.        Determine the cost of operating a 7000 Btu, room-sized air conditioner in New York City for the duration of the summer. Assume that electricity costs 14¢ per kilowatt·hour and that the air conditioner will run about 10 hours a day for 80 days. Solution ...

o        In the United States, the Btu is often confused with the Btu per hour (Btu/h). This is partially the fault of the heating, ventilation, and air conditioning industries (HVAC). The rate at which energy is transformed from one form to another or transferred from one place to another is called power. Power is stated in units of energy divided by time. Furnaces and air conditioners are rated according to their heating and cooling power; that is, how quickly they can add or subtract heat from a room or home. American appliance manufacturers often quote the power of their devices in "Btu" when they really mean "Btu/h". Most certainly, this is just a form of shorthand and does not reveal any malicious intent on the part of the industry. However ...

Electric energy is sold by the kilowatt·hour while air conditioners are rated in Btu/h. This makes it extremely difficult to estimate the operating costs of these energy hungry appliances. For example, a 7000 Btu/h room air conditioner consumes energy at a rate of about ...

o        while electricity in the New York City metropolitan area averages about 14 Ãƒâ€šÃ‚¢ per kilowatt·hour. Thus, every ten hours of use costs ...

o        Given that summer days in NYC range from semitropical to subtropical to absolutely tropical, it's quite reasonable to assume 80 days of air conditioner use in a typical season. This brings the cost of cooling one room to ...

o        This cost may or may not be acceptable to an individual consumer for this particular use. That isn't the point. The point is that rating an air conditioner in a nonstandard unit adds one more step to the problem. Appliances should be rated in watts or kilowatts so that consumers would be able to make mental estimates of their operating costs more easily.

3.        The athlete in this video clip [] is performing a weightlifting maneuver known as the snatch. In this maneuver, the barbell must be lifted from the platform to a point above the head, with the arms and legs fully extended, in a single movement. The barbell must then be held motionless until the referees give the signal and then returned to the platform. In this particular video ...

o        the mass of the barbell is 77.5 kg (for comparison, the mass of athlete is 58 kg),

o        the disks on the barbell have a diameter of 450 mm, and

o        the video advances at 25 frames per second (with 71 frames total).

Determine the following quantities for the barbell this athlete is lifting as functions of time ...

d.        height

e.        velocity

f.         acceleration

g.        applied force

h.        work

i.         power

Most of this question is a review of mechanical concepts discussed in previous sections in this book. Only the last part deals with power. To begin this problem, you will need some sort of screen measuring tool. Many basic image editing applications have this function built into them. If your doesn't, you might like to try Wadruler [] for Windows or Freeruler [external link] for Mac OS X.

numerical

1.        A 64 kg student travels from the first floor to the fourth floor of a school (a height of 15 m).

a.        What total work did she do climbing the stairs?

b.        How long would this trip last if the student produced 480 W of power?

2.        A motorized winch is rated at 10.0 kW. At what maximum speed can this winch raise a mass of 27,500 kg?

3.        A pedaling cyclist turns a 17.5 cm crank arm at 200 rpm. (The crank arm distance is measured from one pedal to the axle.) Calculate the average force exerted on the pedals if the cyclist does work at the rate of 600 W.

4.        How fast must a cyclist climb a 12° hill to maintain a power output of 190 W? Ignore friction and assume the mass of the cyclist plus bicycle is 85 kg?

5.        The graph below shows the power output vs. time for an elevator motor in operation.

a.        What does the area under this curve represent?

b.        Calculate it.

6.        The world's most powerful laser (the Petawatt) went online at Lawrence Livermore National Laboratory (LLNL) in May of 1996. This laser produced a peak power of 1.25 petawatts (1.25 x 10¹⁵ W), ten times more power than the previous record holding laser (which was also built at LLNL) and 1200 times more powerful than the entire electrical generating capacity of the United States. Although it is incredibly powerful, the Petawatt is not particularly energetic. Pulses from the Petawatt typically last less than half a picosecond (10^-12 s). How long could an ordinary 60 W light bulb run on the energy delivered in one pulse of the Petawatt?

7.        A problem for Americans only: Show that one horsepower is equivalent to one pound of thrust at 375 mph.

Resources

• lasers
• Petawatt, Lawrence Livermore National Laboratory. December 1996.
• machines
• Terex Equipment Limited, Specifications on the most powerful trucks in production.

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Most objects are not magnetic because the atoms are arranged so that the electrons spin in different, random directions, and cancel out each other. Magnets are different; the molecules in magnets are arranged so that the electrons spin in the same direction. This arrangement of atoms creates two poles in a magnet, a North-seeking pole and a South-seeking pole. Bar Magnet A magnet is labeled with North (N) and South (S) poles. The magnetic force in a magnet flows from the North pole to the South pole. This creates a magnetic field around a magnet. Have you ever held two magnets close to each other? They don’t act like most objects. If you try to push the South poles together, they repel each other. Two North poles also repel each other. Turn one magnet around and the North (N) and the South (S) poles are attracted to each other. The magnets come together with a strong force. Just like protons and electrons, opposites attract. These special properties of magnets can be used to make electricity. Moving magnetic fields can pull and push electrons. Some metals, like copper have electrons that are loosely held. They can be pushed from their shells by moving magnets. Magnets and wire are used together in electric generators. BATTERIES PRODUCE ELECTRICITY A battery produces electricity using two different metals in a chemical solution. A chemical reaction between the metals and the chemicals frees more electrons in one metal than in the other. One end of the battery is attached to one of the metals; the other end is attached to the other metal. The end that frees more electrons develops a positive charge and the other end develops a negative charge. If a wire is attached from one end of the battery to the other, electrons flow through the wire to balance the electrical charge. A load is a device that does work or performs a job. If a load––such as a lightbulb––is placed along the wire, the electricity can do work as it flows through the wire. In the picture above, electrons flow from the negative end of the battery through the wire to the lightbulb. The electricity flows through the wire in the lightbulb and back to the battery. ELECTRICITY TRAVELS IN CIRCUITS Electricity travels in closed loops, or circuits (from the word circle). It must have a complete path before the electrons can move. If a circuit is open, the electrons cannot flow. When we flip on a light switch, we close a circuit. The electricity flows from the electric wire through the light and back into the wire. When we flip the switch off, we open the circuit. No electricity flows to the light. When we turn a light switch on, electricity flows through a tiny wire in the bulb. The wire gets very hot. It makes the gas in the bulb glow. When the bulb burns out, the tiny wire has broken. The path through the bulb is gone. When we turn on the TV, electricity flows through wires inside the set, producing pictures and sound. Sometimes electricity runs motors—in washers or mixers. Electricity does a lot of work for us. We use it many times each day. HOW ELECTRICITY IS GENERATED A generator is a device that converts mechanical energy into electrical energy.  The process is based on the relationship between magnetism and electricity.  In 1831, Faraday discovered that when a magnet is moved inside a coil of wire, electrical current flows in the wire.  A typical generator at a power plant uses an electromagnet—a magnet produced by electricity—not a traditional magnet. The generator has a series of insulated coils of wire that form a stationary cylinder.  This cylinder surrounds a rotary electromagnetic shaft.  When the electromagnetic shaft rotates, it induces a small electric current in each section of the wire coil.   Each section of the wire becomes a small, separate electric conductor. The small currents of individual sections are added together to form one large current. This current is the electric power that is transmitted from the power company to the consumer. An electric utility power station uses either a turbine, engine, water wheel, or other similar machine to drive an electric generator or a device that converts mechanical or chemical energy to generate electricity. Steam turbines, internal-combustion engines, gas combustion turbines, water turbines, and wind turbines are the most common methods to generate electricity.  Most power plants are about 35 percent efficient. That means that for every 100 units of energy that go into a plant, only 35 units are converted to usable electrical energy. Most of the electricity in the United States is produced in steam turbines. A turbine converts the kinetic energy of a moving fluid (liquid or gas) to mechanical energy. Steam turbines have a series of blades mounted on a shaft against which steam is forced, thus rotating the shaft connected to the generator. In a fossil-fueled steam turbine, the fuel is burned in a furnace to heat water in a boiler to produce steam. Coal, petroleum (oil), and natural gas are burned in large furnaces to heat water to make steam that in turn pushes on the blades of a turbine. Did you know that coal is the largest single primary source of energy used to generate electricity in the United States? In 2005, more than half (51%) of the country's 3.9 trillion kilowatthours of electricity used coal as its source of energy. Natural gas, in addition to being burned to heat water for steam, can also be burned to produce hot combustion gases that pass directly through a turbine, spinning the blades of the turbine to generate electricity. Gas turbines are commonly used when electricity utility usage is in high demand. In 2005, 17% of the nation's electricity was fueled by natural gas. Petroleum can also be used to make steam to turn a turbine. Residual fuel oil, a product refined from crude oil, is often the petroleum product used in electric plants that use petroleum to make steam. Petroleum was used to generate about three percent (3%) of all electricity generated in U.S. electricity plants in 2005. Nuclear power is a method in which steam is produced by heating water through a process called nuclear fission. In a nuclear power plant, a reactor contains a core of nuclear fuel, primarily enriched uranium. When atoms of uranium fuel are hit by neutrons they fission (split), releasing heat and more neutrons. Under controlled conditions, these other neutrons can strike more uranium atoms, splitting more atoms, and so on. Thereby, continuous fission can take place, forming a chain reaction releasing heat. The heat is used to turn water into steam, that, in turn, spins a turbine that generates electricity. Nuclear power was used to generate 20% of all the country's electricity in 2005. Hydropower, the source for almost 7% of U.S. electricity generation in 2005, is a process in which flowing water is used to spin a turbine connected to a generator. There are two basic types of hydroelectric systems that produce electricity. In the first system, flowing water accumulates in reservoirs created by the use of dams. The water falls through a pipe called a penstock and applies pressure against the turbine blades to drive the generator to produce electricity. In the second system, called run-of-river, the force of the river current (rather than falling water) applies pressure to the turbine blades to produce electricity. Geothermal power comes from heat energy buried beneath the surface of the earth. In some areas of the country, enough heat rises close to the surface of the earth to heat underground water into steam, which can be tapped for use at steam-turbine plants. This energy source generated less than 1% of the electricity in the country in 2005. Solar power is derived from the energy of the sun.  However, the sun's energy is not available full-time and it is widely scattered. The processes used to produce electricity using the sun's energy have historically been more expensive than using conventional fossil fuels. Photovoltaic conversion generates electric power directly from the light of the sun in a photovoltaic (solar) cell. Solar-thermal electric generators use the radiant energy from the sun to produce steam to drive turbines. In 2005, less than 1% of the nation's electricity was based on solar power. Wind power is derived from the conversion of the energy contained in wind into electricity. Wind power, less than 1% of the nation's electricity in 2005, is a rapidly growing source of electricity. A wind turbine is similar to a typical wind mill. Biomass includes wood, municipal solid waste (garbage), and agricultural waste, such as corn cobs and wheat straw. These are some other energy sources for producing electricity. These sources replace fossil fuels in the boiler. The combustion of wood and waste creates steam that is typically used in conventional steam-electric plants. Biomass accounts for about 1% of the electricity generated in the United States. THE TRANSFORMER - MOVING ELECTRICITY  To solve the problem of sending electricity over long distances, William Stanley developed a device called a transformer. The transformer allowed electricity to be efficiently transmitted over long distances. This made it possible to supply electricity to homes and businesses located far from the electric generating plant. The electricity produced by a generator travels along cables to a transformer, which changes electricity from low voltage to high voltage. Electricity can be moved long distances more efficiently using high voltage. Transmission lines are used to carry the electricity to a substation. Substations have transformers that change the high voltage electricity into lower voltage electricity. From the substation, distribution lines carry the electricity to homes, offices and factories, which require low voltage electricity. MEASURING ELECTRICITY Electricity is measured in units of power called watts. It was named to honor James Watt, the inventor of the steam engine. One watt is a very small amount of power. It would require nearly 750 watts to equal one horsepower. A kilowatt represents 1,000 watts. A kilowatthour (kWh) is equal to the energy of 1,000 watts working for one hour. The amount of electricity a power plant generates or a customer uses over a period of time is measured in kilowatthours (kWh). Kilowatthours are determined by multiplying the number of kW's required by the number of hours of use. For example, if you use a 40-watt light bulb 5 hours a day, you have used 200 watts of power, or 0.2 kilowatthours of electrical energy. See our Energy Calculator section to learn more about converting units.