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Chapter 3 Energy (Spring 09)

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What 4-letter word in the English language upsets more people than any other? Especially college students

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Work Work is defined as being the product of a force and the distance the force moves in the direction of the force. – Movement must happen in the direction of the force. – The force and the motion cannot be perpendicular. – No work is done in holding something. – No work is done in carrying something parallel to the ground. – No work is done in doing physical science homework unless a pencil or pen is pushed along the paper.

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Work Work = force x distance moved W = F d (put on formula sheet) What work is done when a force of 10 N is used to push an object 3 m? Note that the weight or mass of the object is not important, only the force necessary to move it. DataFormulaWorkAnswer W=? F=10N d=3m W = F dW=10N3m=30 joules

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The joule 1 joule = 1 newton 1 meter A joule is the work done in pushing with a force of one newton for a distance of one meter. James Prescott Joule was an English brew- master. We will hear more of him later. 3q

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Work against gravity Work = force x distance When we lift something, the force necessary to lift it is the weight of the object – w=mg (force) Therefore, the work necessary to raise an object a height h is – W=mgh (formula sheet) = work – h is the height the object is raised (or lowered). The weight is w=mg, don’t confuse weight and work. Weight is a force that can do work.

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Calculate the amount of work for a 100 kg person to climb stairs that are 10 m high. Data W=? m=100 kg h=10 m

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How fast can you work? Power is the rate of doing work. Power = work/time Formula sheet! If 30 joules of work is done in 6 seconds, what is the power? DataFormulaWorkAnswer W=30 j t=6 s P=? P = W/tP = 30j/6s5 ?

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Units of power One watt is one joule per second. Watt = joule/sec 5 watts = 5 w

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What is my power if my mass is 70 kg and I can climb stairs that are 10 m high in 3 seconds? DataFormulaWorkAnswer P=? m=70kg h=10m t=3 s P = W/t W=F·d F=mg d=h P = W/t=F d/t= mgh/t P=70kg 9.8m/s 2 10m/3s 2286 W One horsepower is 746 watts so 2286 W is 3.1 horsepower. This calculation is necessary for your pre-lab next week (measure your horsepower). Note that this says nothing about how steep the stairs are, only their height. 2q

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Suggestions to improve the class. Clicker doesn't work Slow down Present more challenging material Room too hot/cold More class interaction Too much talking in the back of the room Give more CPS questions

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If I have a lot of energy I can do a lot of work. Energy is the ability (capacity) to do work. Energy comes in many forms: – Heat energy – Chemical energy Food Gasoline Dynamite – Potential energy – Kinetic energy

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Potential energy Potential energy is energy due to the position of an object. Examples of potential energy include: – A compressed spring – A nail near a magnet – A car on a hill Gravitational potential energy

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Gravitational Potential Energy Energy = work done To have one joule of energy, I must do 1 joule of work to raise the object. The units of work are also joules. The force of gravity is mass times the acceleration of gravity W = mg

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Gravitational P.E. (cont) P.E. = Work = force distance W = f d f = Weight = W = mg d = distance = height = h P.E. = mgh (Formula for gravitational potential energy) This applies only on the surface of the Earth.

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Gravitational P.E. (cont) What is the P.E. of a 20 kg block, 2 m above the floor? DataFormulaWorkAnswer P.E. = ? m = 20 kg h = 2 m g = 9.8m/s 2 P.E. = mghP = 20kg 9.8m/s 2 2m392 j

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Kinetic Energy Kinetic energy is the energy something has because it is moving. Energy of motion K.E. = ½ mv 2 Where m = mass of the object v = velocity of the object

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Kinetic energy What is the K.E. of a 70 kg person running 10 m/s? DataFormulaWorkAnswer K.E. = ? m = 70 kg V= 10 m/s K.E. = ½ mv 2 = ½ 70kg(10m/s) 2 3500 j

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Energy – a review Energy – the ability to do work Potential energy P.E. = mgh Kinetic energy K.E. = ½ mv 2

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Conservation of Energy Energy may be changed from one type to another but the total amount of energy remains the same.

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More on conservation of energy Total energy is always conserved We will be especially interested in situations in which mechanical energy is conserved – This means no energy is given to heat or friction When mechanical energy is conserved, the sum of the potential and the kinetic energies will remain constant. There are many types of problems we can solve using this knowledge. This will be the idea that will be worth more points on the next two tests than any other topic.

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Energy conservation- Changing P.E. to K.E. and K.E. to P.E. - - - - Case 1: A falling object 100% P.E at the top 90% P.E. 10%K.E. 80% P.E. 20%K.E 70% P.E. 30%K.E 60% P.E. 40%K.E 50% P.E. 50%K.Ehalf way down 40% P.E. 60%K.E 30% P.E. 70%K.E 20% P.E. 80%K.E 10% P.E. 90%K.E. 100%K.Eat the bottom

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P.E. to K.E. and K.E. to P.E. At the top – All P.E. 1/3 of the way from the top 2/3 P.E., 1/3 K.E. Half way down ½ K.E., ½ P.E. 1/3 of the way from the bottom, 1/3 P.E., 2/3 K.E. At the bottom – All K.E.

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Potential to Kinetic What is the velocity of an object after falling 10m? v=? h= 10m Initial state = Potential energy Final state = kinetic energy PrincipleP.E. = K.E.

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You will be asked to add one additional step in the solution of these problems. That step is the principle. In the principle you must give the state of the energy of the system in its initial state and its final state. For this problem the principle is: Potential Energy is changed to Kinetic Energy. mgh = ½ mv 2

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Potential to Kinetic What is the velocity of an object after falling 10m? v=? h= 10m Initial state = Potential energy Final state = kinetic energy PrincipleP.E. = K.E. mgh = ½ mv 2 3q

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Kinetic to Potential How high will an arrow go that is shot straight into the air with an initial velocity of 100 m/s? h = ?, v = 100 m/s Initial state = kinetic energy – In this problem the arrow has already been shot and is moving with a velocity of 100 m/s. – But it is still at zero height. final state = potential energy ½ mv 2 =mgh Solve the equation for h 3q

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The energy can be used to do work or work can be used to give energy. Work to Kinetic energy Work to Potential energy Kinetic energy to work Potential energy to work

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Work to energy and energy to work A 3 kg hammer moving 10 m/s hits a nail and drives it 2 cm into a board. What was the force on the nail? m = 3 kg v = 10 m/s d = 2 cm = 0.02 mF=? A falling hammer (K.E.) drives a nail (work)(force x distance) Principle K.E. = work ½ mv 2 =F·d 3q

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How fast will a 5 kg rocket be traveling if its motor has a force of 2000 n and it operates for 500 m horizontally? v=?, m=5kg, F=2000N, d=500m Principle: work = K.E. F·d=½ mv 2 v=? m=5kg F = 2000 n d = 500 m

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Benefits of groups (from Fall 2007) 26 students are currently in groups On Test one, these students received a total of 212 bonus points The 5 participants of one group each received 22 points (the max on this test (10%). The way to receive maximum benefit is to work together and help each other understand as the material is covered in class. If any one would like to join but doesn't have a group, send me an email or talk to me and I will help you find a group.

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Work to energy and energy to work A 3 kg hammer moving 10 m/s hits a nail and drives it 2 cm. What was the force on the nail? m = 3 kg v = 10 m/s d = 2 cm = 0.02 mF=? A falling hammer (K.E.) drives a nail (work)(force x distance) Principle K.E. = work ½ mv 2 =F·d

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How fast will a 5 kg rocket be traveling if its motor has a force of 2000 n and it operates for 500 m? v=?, m=5kg, F=2000N, d=500m Principle: work = K.E. F·d=½ mv 2 v=? m=5kg F = 2000 n d = 500 m

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Conservation of momentum Two types: Linear momentum and angular momentum Linear momentum = mass times velocity In the absence of outside forces, the total momentum of a group of objects remains unchanged (even if the objects collide). Useful when objects collide or when they separate (boy jumping from wagon, etc) Angular momentum = radius x mass x velocity Useful for systems, such as planets, comets, etc We will only do calculations with linear momentum.

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momentum before collision = momentum after m 1 is the mass of object one. v 1i is the initial velocity of object one. m 2 is the mass of object two. v 2i is the initial velocity of object two. v 1f is the final velocity of object one. v 2f is the final velocity of object two. m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f m1m1 v 1i m2m2 v 2i This needs to go on your formula sheet!

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Example one A 500 kg car traveling 20 m/s collides head-on with a 2000 kg truck traveling 10 m/s in the opposite direction. The two stick together after the collision. What is the direction and velocity of the two after the collision? m 1 = 500 kg v 1i = 20 m/s v 1f = v 2f = v f = ? m 2 = 2000 kg v 2i = -10 m/s DATA This minus sign is very important

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General equation m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f In our special case (v 1f = v 2f = v f ) m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f 500 kg20 m/s+ 2000 kg(-10 m/s)=(500+2000)kgv f 10,000 kg m/s-20,000 kg m/s = 2500 kg v f -10,000 kg m/s = 2500 kg v f v f = -10,000 kg m/s 2500 kg v f = -4 m/s Direction is that of the truck

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Another momentum example A 70 kg man stands on a 20 kg boat in the water and jumps with a velocity of 5 m/s. What is the recoil velocity of the boat? m 1 = 70 kg v 1i = 0 v 1f = 5 m/s m 2 = 20 kg v 2i = 0 v 2f = ?

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General equation m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f In our special case (v 1i = v 2i = 0 ) 0 =m 1 v 1f + m 2 v 2f - m 1 v 1f = m 2 v 2f -70 kg5 m/s = 20 kgv f v f = -350 kg m/s 20 kg v f = -17.5 m/s Direction is opposite the way the man jumped. The minus sign tells us the direction.

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Caloric A colorless, odorless liquid that is in most substances. When a substance burns the caloric in it is given off as heat This theory explained heat until the 18 th century.

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Benjamin Thompson An American who fled the U.S. in 1773 because of British sympathies. Became a military consultant (mercenary) Took the name of Count Rumford Made cannon Discovered that heat is a form of energy Set up the first public school for the children of his workers.

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James Prescott Joule An rich English beer maker who liked to do science experiments. James Prescott Joule showed how many joules (unit of energy) were in a calorie (unit of heat). 1 calorie = 4.2 joules (put on formula sheet) 1 calorie is the amount of heat necessary to raise the temperature of one gram of water one degree Celsius.

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Guess who Didn’t like grade school very well Got a doctor’s degree in physics but couldn’t get a job teaching (even in high school) so took a job in the post office (taking care of patents). Takes his girl friend on a trip and she gets pregnant. Child is taken by girl’s relatives, and is not heard of later Writes three papers in physics journals in 1905 that profoundly change three major areas of physics. Becomes most famous scientist of 20 th century

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Albert Einstein Theory of Brownian motion Theory of photons Theory of relativity Two postulates of special relativity 1) The speed of light is always measured to be the same even when two systems are moving with respect to one another. 2) The laws of physics apply in systems that are moving with respect to one another.

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Using these two postulates and some algebra the following results apply for the case in which two systems are moving with respect to one another with a relative speed close to speed of light The following happen (and have been proven to take place by experiments in the laboratory). o Length is measured to be different (shorter) o Mass is measured to be greater o Time is measured to be longer o Mass can be changed to energy o E = mc 2

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General Theory of Relativity The special theory that we have already discussed does not apply to systems that are accelerated or in a gravitational field. The general theory applies to accelerated systems or systems in a gravitational field. – The most significant result of the general theory we will note is that gravity affects light. Light is bent by gravity (stars) – We can see this when there is an eclipse of the sun.

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E = mc 2 E = Energy m = mass c = speed of light = 3 x 10 8 m/s (formula sheet) c 2 = 9 x 10 16 m 2 /s 2 What energy is obtained when 7 grams is changed to mass Data: m = 7 grams = 0.007 kg Data m =.007kg E=? Formula E=mc 2 =(.007 kg)(3 x 10 8 m/s) 2 = 6.3 x 10 14 joules

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The Energy Problem Energy is necessary for our life style. We are running out of the energy sources we use the most.

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Fossil Fuels The cheapest The easiest to use The most convenient The worst for the environment (Global warming and acid rain) Supplies will become limited in our lifetime.

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Energy Costs per kilowatt hour of electricity Natural gas3-4¢ Coal4-5¢ Wind4-5¢ Geothermal3-8¢ Hydropower4-10¢ Biomass6-8¢ Nuclear10-15¢ Solar thermal10-15¢ Photovoltaics20-30¢

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TypeInitial cost Operating cost Environmental cost Coal Oil Natural gas Low Lowest Low CO 2. Acid rain CO 2 CO 2,best ff for environ. Wind Solar energy Hydro power High Low Noise, kills birds Occupies space Needs a dam, rainfall Tide energy Geothermal Ocean thermal High Experi- mental High Unknown Very bad for fish, etc. Few places where possible unknown

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Solar generated hydrogen Burns without pollution – H 2 + O 2 H 2 O Could be used in cars and distributed at filling stations Car manufacturers are designing hydrogen cars

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Skateboard concept All electronic steering, brakes Tops plug-in to the fuel cell base. Cars without engines Each wheel has an electric motor that drives it

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Best possibilities for the future Conservation – Energy efficient cars/trucks – Better insulated houses Methane hydrate – Methane (natural gas) combined with water – Uncertain quantities – In difficult places to access Increased use of wind and solar energies Hydrogen powered cars/trucks Fusion

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Nuclear energy Good or Bad? Two types – Fission (to split apart) (Atomic bomb) Uses Uranium as a fuel What we now have All present nuclear power plants use fission – Fusion (to join together) (Hydrogen bomb) A possibility for the future Uses hydrogen as a fuel (sea water)

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Fission Fuel is uranium (which splits apart) Good points No CO 2 No radioactivity in normal operation In normal operation a nuclear power plant releases much less radioactivity than a coal burning plant. Bad points Possibility of an accident (Chernobyl) Disposing of spent (used) fuel

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Fusion Fuel is hydrogen (not radio-active) Good points Produces tremendous amounts of energy Very good for the environment Uses ocean water for fuel Bad points We don’t know how to do it yet.

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Conservation of energy revisited This will be the problem on the chapter 3 and chapter 4 tests that will be worth the most points.

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Energy conservation- Changing P.E. to K.E. and K.E. to P.E. - - - - Case 1: A falling object 100% P.E at the top 90% P.E. 10%K.E. 80% P.E. 20%K.E 70% P.E. 30%K.E 60% P.E. 40%K.E 50% P.E. 50%K.Ehalf way down 40% P.E. 60%K.E 30% P.E. 70%K.E 20% P.E. 80%K.E 10% P.E. 90%K.E. 100%K.Eat the bottom

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P.E. to K.E. and K.E. to P.E. At the top – All P.E. 1/3 of the way from the top 2/3 P.E., 1/3 K.E. Half way down ½ K.E., ½ P.E. 1/3 of the way from the bottom, 1/3 P.E., 2/3 K.E. At the bottom – All K.E. It is very important to note that the total energy is always the same Total Energy = Potential Energy + Kinetic Energy (True when there is no friction and no work done). In the end, the energy will be all heat energy!

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Calculate the work necessary to raise an 10 kg rock 30 m high. (25 point problem) Datam = 10 kg, h = 30 m, Work=? W=F·d=mgh=10kg·9.8m/s 2 ·30m = 2940j What is the potential energy at this height (30m)? No calculation is necessary, the P.E. is 2940J. What is the kinetic energy at this height? No calculation is necessary, as the rock is sitting still, the K.E. is zero! Total energy = K.E. + P.E. = 2940j at all heights.

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What is the total energy at this height? No calculation is necessary, T.E. is 2940j. Calculate the P.E., the K.E. and the T.E when the rock is 10 m from the ground and when it just begins to touch the ground (the height is zero and the velocity is maximum). The total energy remains the same in all cases, 2940j, At ground level, h=0 so P.E. is zero, thus the K.E. is 2940j

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To calculate the P.E. at 10 m height, Data: m=10 kg, h=10m, P.E.=? P.E. = mgh = 10kg·9.8m/s 2 ·10m = 980j The K.E. at this point is: K.E. = T.E. – P.E. = 2940j – 980j = 1960j What is the heat energy produced when the rock hits the ground? No calculation is necessary, all the energy is changed to heat, so the answer is 2940j.

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The energy can be used to do work or work can be used to give energy. – The Principle Work to Kinetic energy Work to Potential energy Kinetic energy to work Potential energy to work Potential energy to Kinetic energy Kinetic energy to Potential energy

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A 3 kg hammer moving 10 m/s hits a nail and drives it 2 cm. What was the force on the nail? Data: m= 3kg, v= 10 m/s, d= 0.02 m, F=? Principle Initial = K.E Final = work ½ mv 2 =F·d

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How fast will a 5 kg rocket be traveling if its motor has a force of 2000 n and it operates for 500 m? Data V=?, m=5kg, F= 2000 N, d=500m Principle Initial state = work Final state = K.E. F·d = ½ mv 2 V= 632 m/s

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Potential Energy to Kinetic Energy What is the speed of a ball that has fallen 2 m without friction? In the beginning all the energy is P.E. – P.E. = mgh After falling 2 m, all the energy is Kinetic – K.E. = ½ mv 2 K.E. = P.E. ½ mv 2 = mgh v = √2gh v = √2·9.8 m/s 2 ·2 m = √39.2 m 2 /s 2 = 6.26 m/s

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Density – help for your pre-lab Density is the mass per unit volume. Density = mass/volume D=m/v

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Calculate the density of a wood cylinder of diameter 6 cm, length 10 cm and mass 250 grams. Data: d = 6 cm, (r = 3 cm), L = 10 cm and m = 250 g.

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Objectives Recognize the difference between the scientific and ordinary definitions of work. Define work by relating it to force and displacement. Identify.

Objectives Recognize the difference between the scientific and ordinary definitions of work. Define work by relating it to force and displacement. Identify.

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