Ground Rules. PC1221 Fundamentals of Physics I. Introduction to Energy. Energy Approach to Problems. Lectures 13 and 14. Energy and Energy Transfer

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1 PC11 Fundamentals o Physics I Lectures 13 and 14 Energy and Energy Transer A/Pro Tay Seng Chuan 1 Ground Rules Switch o your handphone and pager Switch o your laptop computer and keep it No talking while lecture is going on No gossiping while the lecture is going on Raise your hand i you have question to ask Be on time or lecture Be on time to come back rom the recess break to continue the lecture Bring your lecturenotes to lecture Introduction to Energy Energy Approach to Problems The concept o energy is one o the most important topics in science while it is not the only important topic Every physical process that occurs in the Universe involves energy and energy transers or energy transormations Giving a lecture now is an eample o the transer and transorm o energy but it is more complicated (chemical energy rom the ood I ate this morning is used to create the sound energy in my voice now) The energy approach to describing motion is particularly useul when the orce is not constant An approach will involve Conservation o Energy This could be etended to biological organisms, technological systems and engineering situations 3 4

2 Systems Work A system is a small portion o the Universe A valid system may be a single object or particle be a collection o objects or particles be a region o space vary in size and shape Energy in a system is conserved. 5 The work, W, done on a system by an agent eerting a constant orce on the system is the product o the magnitude, F, o the orce, the magnitude r o the displacement o the point o application o the orce, and cos where is the angle between the orce and the displacement vectors. W = F r cos F F cos r 6 Work, cont. F F cos W = F r cos r The displacement is that o the point o application o the orce A orce does no work on the object i the orce does not move through a displacement The work done by a orce on a moving object is zero when the orce applied is perpendicular to the displacement o its point o application 7 Did you do any work i you were repelling along a rope down a helicopter? Did the gravity do work on you? 8

3 Did you do any work i you were running on a level ground? F F cos r Eample. In a baseball game, the catcher stops a 90-mph pitch. What can you say about the work done by the catcher on the ball? 1. catcher has done positive work. catcher has done negative work 3. catcher has done zero work Why did you eel tired ater the run? 9 180º F Δr The orce eerted by the catcher is opposite in direction to the displacement o the ball, so the work is negative. Or using the deinition o work (W = F (Δr)cos( ), because = 180º, then W < 0. Note that because the work done on the ball is negative, its speed decreases. 10 Eample. A bo is being pulled up a rough incline by a rope connected to a pulley. How many orces are doing work on the bo? Any orce not perpendicular to the motion will do work: N does no work T does positive work does negative work mg sin does negative work T displacement mg N Work in Pushing a Block The normal orce, n, and the gravitational orce, m g, do no work on the object cos = cos 90 = 0 The orce F does do work on the object 1

4 More About Work The system and the environment must be determined when dealing with work The environment does work on the system Work by the environment on the system The sign o the work depends on the direction o F relative to r Work is positive when projection o F onto r is in the same direction as the displacement Work is negative when the projection is in the opposite direction Units o Work Work is a scalar quantity The unit o work is a joule (J) 1 joule = 1 newton. 1 meter J = N m Work Is An Energy Transer I the work is done on (by) a system and it is positive, energy is transerred to the system I the work done on (by) the system is negative, energy is transerred rom the system I a system interacts with its environment, this interaction can be described as a transer o energy across the system boundary (e.g., heat generated due to riction) This will result in a change in the amount o energy stored in the system 15 Scalar Product o Two Vectors The scalar product o two vectors is written as A. B (read as A dot B) It is also called the dot product A. B = A B cos is the angle between A and B 16

5 Scalar Product, cont The scalar product is commutative A. B = B. A The scalar product obeys the distributive law o multiplication A. (B + C) = A. B + A. C 17 Dot Products o Unit Vectors î î ĵ ĵ kˆ kˆ 1 î ĵ î kˆ ĵ kˆ 0 Using component orm with A and B: A A B B A B î A î B A B y y ĵ A ĵ B A y z z kˆ B kˆ y A B z z 18 Work Done by a Varying Force Work Done by a Varying Force, cont Assume that during a very small displacement,, F is constant For that displacement, W ~ F For all o the intervals, W F i Force Displacement lim 0 i i F F d Thereore, W The work done is equal to the area under the curve o orce versus displacement i F d Force Displacement 19 0

6 Work Done By Multiple Forces I more than one orce acts on a system and the system can be modeled as a particle, the total work done on the system is the work done by the net orce W Wnet F d i Work Done by Multiple Forces, cont. I the system cannot be modeled as a particle, then the total work is equal to the algebraic sum o the work done by the individual orces W net W by individual orces 1 Eample. Four orces are acting 5 N on an object along a horizontal plane. The orce with N is acting along the plane, 4 N with 30º rom the horizontal line, 5 N perpendicular to the plane, and 3 N 60º rom the horizontal line. What is the net work done o all these orces or the ball to travel 4 N N 3 N a horizontal distance o 5 m 5 m shown in the igure? The net horizontal orce is + 4 cos 30º - 3 cos 60º = 3.96 N The net work done = 3.96 N 5 m = 19.8 J Or: The net work done is the sum o the workdone by the individual orces: cos 30º 5-3 cos 60º 5 = 19.8 J 3 Hooke s Law The orce eerted by the spring is F s = - k is the position o the block with respect to the equilibrium position ( = 0) k is called the spring constant or orce constant and measures the stiness o the spring This is called Hooke s s Law 4

7 Hooke s Law, cont. When is positive (spring is stretched), F is negative When is 0 (at the equilibrium position), F is 0 When is negative (spring is compressed), F is positive Hooke s Law, inal The orce eerted by the spring is always directed opposite to the displacement rom equilibrium F is called the restoring orce I the block is released it will oscillate back and orth between and 5 6 Spring Combination A spring can be stretched a distance o 60 cm with an applied orce o 1 N. I an identical spring is connected in series with the irst spring, how much orce will be required to stretch this series combination a distance o 60 cm? 60 cm 60 cm Here, the springs are in series, so each spring is only stretched 30 cm thus only hal the orce is needed. But also, because the springs are in a row, the orce applied to one spring is transmitted to the other spring (like tension in a rope). So the overall applied orce o 0.5 N is all that is needed. The combination o two springs in series behaves like a weaker spring!! Spring Combination What i the two springs are arranged in parallel? 60 cm 60 cm Each spring is still stretched 60 cm, so each spring requires 1 N o orce. But because there are two springs, there must be a total o N o orce! Thus, the combination o two parallel springs behaves like a stronger spring!!

8 Work Done by a Spring Identiy the block as the system Calculate the work as the block moves rom i = - ma to = 0, or i = ma to =0 0 1 Ws Fd k d k i ma ma The total work done as the block moves rom ma to ma is zero Spring with an Applied Force Suppose an eternal agent, F app, stretches the spring The applied orce is equal and opposite to the spring orce F app = -F s = -(-k) = k Work done by F app is equal to ½ k ma 9 30 Eample. I it takes 4.00 J o work to stretch a Hooke's-law spring 10.0 cm rom its unstressed length, determine the etra work required to stretch it an additional 10.0 cm J m k k 800 N m To stretch the spring to 0. m rom 0.1 m requires an additional work o 1 W J 1.0 J 31 Kinetic Energy Kinetic Energy is the energy o a particle due to its motion K = ½ mv K is the kinetic energy m is the mass o the particle v is the speed o the particle the epression ½ mv can be derived using F = ma, and A change in kinetic energy is one possible result o doing work to transer energy into a system 3

9 Kinetic Energy, cont Calculating the work by integration: W F d ma d i i W v v i mvdv 1 1 W mv mv i dv ma d = m d = mv dv dt 33 Work-Kinetic Energy Theorem The Work-Kinetic Energy Principle states W = K K i = K In the case in which work is done on a system and the only change in the system is in its speed, the work done by the net orce equals the change in kinetic energy o the system. Kinetic energy possessed by an object o mass m and velocity v is K = ½ mv 34 Work-Kinetic Energy Theorem Nonisolated System The normal and gravitational orces do no work since they are perpendicular to the direction o the displacement W = F = K = ½ mv 0 (the vertical normal orce has no eect) 35 A nonisolated system is one that interacts with or is inluenced by its environment, e.g., push a block on a rough surace An isolated system would not interact with its environment (e.g., push a block on a riction-less surace.) The Work-Kinetic Energy Theorem can also be applied to nonisolated systems. In that case there will be a transer o energy across the boundary o an object (e.g., rom the block to the surace where heat is generated.) 36

10 Internal Energy The energy associated with an object s temperature is called its internal energy, E int In this eample, the surace is the system The riction does work and increases the internal energy o the surace 37 Potential Energy Potential energy is energy related to the coniguration o a system in which the components o the system interact by orces Eamples include: elastic potential energy stored in a spring gravitational potential energy, e.g., you stand on top o Bukit Timah Hill electrical potential energy, e.g., you switch on electricity to iron your shirt 38 Eample. A.00-kg block is attached to a spring o orce constant 500 N/m as shown in igure. The block is pulled 5.00 cm to the right o equilibrium and released rom rest. Find the speed o the block as it passes through equilibrium i (a) the horizontal surace is rictionless and (b) the coeicient o riction between block and surace is (a) When the block is pulled back to equilibrium position, the potential energy in spring will be ully converted to the kinetic energy in the block when the spring returns to original length, i.e., ½k = ½ m v. k 500 So, v = = 0.05 = 0.79 m/s m 39 (b) I there is iction on the surace, the potential energy on the spring is irst wasted on the work done by riction. The remaining energy is converted to the kinetic energy in the block, i.e., ½k - μmg = ½ m v. In terms o initial sum and inal sum o energies, you can also treat it as ½k = μmg + ½ m v ½ = ½ v v = 0.53 m/s 40

11 Eample. The ball launcher in a pinball machine has a spring that has a orce constant o 1.0 N/cm. The surace on which the ball moves is inclined 10.0 with respect to the horizontal. I the spring is initially compressed 5.00 cm, ind the launching speed o a 100-g ball when the plunger is released. Friction and the mass o the plunger are negligible. Initial energy stored in the spring = ½ k = ½ 1.0 N/(10 - m) (5 cm 10 - ) = 0.15 J This 0.15 J is used to (i) move the ball up the slope o 10 or a vertical distance o 5 cm sin( 10) = 0.87 cm, and (ii) provide the ball with a muzzle speed o v. So we have 0.1 kg 9.8 m/s m + ½ 0.1 kg v = 0.15 J v = 1.68 m/s 41 V Eample. You need to move a heavy crate by sliding it across a lat loor with a coeicient o sliding riction o 0.. You can either push the crate horizontally or pull the crate using an attached rope. When you pull on the rope, it makes 30 angle with the loor. Which way should you choose to move the crate so that you will do the least amount o work? How can you answer this question without knowing the weight o the crate or the displacement o the crate? Normal orce Normal orce When pushing the crate with a orce parallel to the ground, the orce o riction acting to impede its motion is proportional to the normal orce acting on the crate in this situation, the normal orce is equal to the crate s weight. When pulling the crate with a rope angled above the horizontal, the normal orce on the crate is less than its weight the orce o riction is thereore reduced. To keep the crate moving across the loor, the applied orce in the parallel direction must be greater than or equal to the orce o riction pulling on the rope thereore requires a smaller parallel applied orce. The work done in moving an object is equal to the product o the displacement through which it has been moved and the orce component parallel to the direction o motion. The applied orce component parallel to the ground is smaller when pulling the crate with the rope - thus, the work done to move the crate with the rope must be lesser, regardless o the weight o the crate or the displacement. 4 µ= 1 Push F1 mg Normal orce To overcome the riction: F1 = µmg Work done = F1 d What i you pull at 0,, i.e., in parallel with the surace? To overcome the riction: Normal orce Pull F µ= 1 mg F cos 30 = µ (mg F sin 30 ) F cos 30 + µ F sin 30 = µ mg F ( ) = µ mg 1 F = 0.73 µmg = 0.73 F1 1 Work done = (F cos 30 ) d = 0.73 F d = F1 d The mechanical advantage is F1 F F 0.73F Conservation o Energy Energy is conserved This means that energy cannot be created or destroyed I the total amount o energy in a system changes, it can only be due to the act that energy has crossed the boundary o the system by some method o energy transer 44

12 Power The time rate o energy transer is called power The average power is given by W P t when the method o energy transer is work 45 Instantaneous Power The instantaneous power is the limiting value o the average power as t approaches zero P lim t 0 W t dw dt This can also be written as dw dr P F Fv dt dt Power = orce velocity 46 Eample. Mike applied 10 N o orce over 3 m in 10 seconds. Joe applied the same orce over the same distance in 1 minute. Who did more work? 1. Mike. Joe 3. Both did the same work Both eerted the same orce over the same displacement. Thereore, both did the same amount o work. Time does not matter or determining the work done. Eample. Mike perormed 5 J o work in 10 secs. Joe did 3 J o work in 5 secs. Who produced the greater power? 1) Mike produced more power ) Joe produced more power 3) both produced the same amount o power Because power = work / time, we see that Mike produced 0.5 W and Joe produced 0.6 W o power. Thus, even though Mike did more work, he required twice the time to do the work, and thereore his power output was lower.

13 Eample. Engine #1 produces twice the power o engine #. Can we conclude that engine #1 does twice as much work as engine #? No!! We cannot conclude anything about how much work each engine does. Given the power output, the work will depend upon how much time is used. There are three possibilities: less, equal, more. Eample. A typical 10-year-old pitcher can throw a baseball at 70 km/h, but only a ew proessional athletes have the etraordinary strength needed to throw a baseball at twice that speed. Why is it so much harder to throw the baseball only twice as ast? 1 mv Kinetic Energy =. 1 mv 1 So the baseball requires quadrupling ( m(v) ) the energy transerred to it by the proessional pitcher. The time require to deliver this energy is in one-hal the time because the velocity o the ball is doubled. So the proessional pitcher needs to produce eight times as much power as lim W dw X 4 that rom the 10-year-old kid ( P t 0 ). 50 t dt X 0.5 Units o Power The SI unit o power is called the watt 1 watt = 1 joule / second = ((kg m/s ) m ) / s = 1 kg. m / s 3 A unit o power in the US Customary system is horsepower 1 hp = 746 W Units o power can also be used to epress units o work or energy 1 kwh = (1000 W)(3600 s) = J Energy transerred in Not 1 kilo-watt 1 hr at a constant rate per hour 51 Eample. When you pay the electric company by the kilowatt-hour, what are you actually paying or? 1) current ) Voltage 3) energy 4) power 5) none o the above

14 Eample. An energy-eicient light bulb, taking in 8.0 W o power, can produce the same level o brightness as a conventional bulb operating at power 100 W. The lietime o the energy eicient bulb is h and its purchase price is $17.0, whereas the conventional bulb has lietime 750 h and costs $0.40 per bulb. Determine the total savings obtained by using one energy-eicient bulb over its lietime, as opposed to using conventional bulbs over the same time period. Assume an energy cost o $0.080 per kilowatt-hour. (Energy eicient bulb) kwh (Conventional bulb) kwh 53