PC1221 Fundamentals o Physics I Lectures 13 and 14 Energy and Energy Transer Dr 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 2 1. I vectors A and B (collinear) pointing in the same direction are added, the resultant has a magnitude equal to 4.0. I B is subtracted rom A, the resultant has a magnitude equal to 8.0. What is the magnitude o A? A. 2.0 B. 3.0 C. 4.0 D. 5.0 E. 6.0 A and B are parallel. A + B = 4 (1) A - B = 8 (2) (1) + (2) 2 A = 12, so A = 6. (I A and B are opposite, A can be -2, and B can be 6.) 3 2. The site rom which an airplane takes o is the origin. The -ais points east; the y-ais points straight up. The position and velocity vectors o the plane at a later time are given by 6ˆ 3. r = (1. 61 10 i+ 9. 14 10 ˆ)m j andv = + 224ˆm i s The plane is most likely a. just touching down. b. in level light in the air. c. ascending. d. descending. e. taking o. The y value is not equal to 0 meaning the airplane is above the ground. The airplane is cruising toward east at an altitude o about 9 km. 4
3. I P = 6.0 N, what is the magnitude o the orce eerted on block 1 by block 2? A. 6.4 N 3 B. 5.6 N 2 C. 4.8 N 1 6 D. 7.2 N P 3.0 kg 2.0 kg E. 8.4 N F2 F2 F1 F1 5.0 kg Let F1 be the orce in between block 1 and block 2, and F2 be the orce in between Block 2 and Block 3. Let a be the acceleration. 6 - F1 = 2a (1) F1 - F2= 3a (2) F2 = 5a (3) Substitute (3) in (2), F1-5a= 3a, i.e., F1= 8a (4) Substitute (4) in (1), 6-8a=2a, i.e., 6 = 10a, so a= 0.6 m/s 2. Using (1), F1 = 6-2a = 6-20.6 = 6-1.2 = 4.8N 5 4. A roller-coaster car has a mass o 500 kg when ully loaded with passengers. The car passes over a hill o radius 15 m, as shown. At the top o the hill, the car has a speed o 8.0 m/s. What is the orce o the track on the car at the top o the hill? 8.0 m/s A. 7.0 kn up R B. 7.0 kn down C. 2.8 kn down D. 2.8 kn up mg 15 m E. 5.6 kn down mg R = m v 2 /r, so R = mg - m v 2 /r = 5009.8 500 8 2 /15 = 2766.66 N = 2.8 kn up 6 SM2 Students are to attend a seminar this Thursday at 2pm, LT32. Attendance is compulsory. Class monitor is to take attendance at 1:40pm. 7 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 8
Introduction to Energy 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 ood I ate this morning to the sound energy in my voice now) Energy Approach to Problems 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 9 10 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. 11 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 Δr o the displacement o the point o application o the orce, and cos θ, where θ is the angle between the orce and the displacement vectors. W = F Δr cos θ F θ F cosθ Δr 12
Work, cont. W = F Δr cos θ F θ F cosθ 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 Δr 13 Did you do any work i you were repelling along a rope down a helicopter? 14 Did you do any work i you were running on a level ground? F θ F cosθ Δr Why did you eel tired ater the run? Take up Physics major biophysics. This course will give you interesting knowledge. 15 Eample. Batman, whose mass is 80.0 kg, is dangling on the ree end o a 12.0-m rope, the other end o which is ied to a tree limb above. He is able to get the rope in motion as only Batman knows how, eventually getting it to swing enough that he can reach a ledge when the rope makes a 60.0 angle with the vertical. How much work was done by the gravitational orce on Batman in this maneuver? The orce o gravity on Batman is down. Only his vertical displacement contributes to the work gravity does. His original y- coordinate below the tree limb is -12 m. His change in elevation (the distance traveled) is Δy = -12 cos (60 ) m (-12 m) = 6 m. The work done by gravity is 80 kg 9.8 m/s 2 6 m = 4704 J 16
Work in Pushing a Block The normal orce, n, and the gravitational orce, m g, do no work on the object cos θ = cos 90 = 0 The orce F does do work on the object 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 Δr Work is positive when projection o F onto Δr is in the same direction as the displacement Work is negative when the projection is in the opposite direction 17 18 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 a system and it is positive, energy is transerred to the system I the work done on 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 19 20
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 θ θ is the angle between A and B 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 21 22 Dot Products o Unit Vectors Work Done by a Varying Force î î ĵ î = ĵ ĵ = kˆ kˆ = 1 = 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 23 Assume that during a very small displacement, Δ, F is constant For that displacement, W ~ F Δ For all o the intervals, W F Δ i Force Displacement 24
Work Done by a Varying Force, cont Work Done By Multiple Forces lim Δ 0 Δ = 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 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 25 26 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 27 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 28
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 29 30 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 2 = = ( ) = Ws Fd k d k i ma 2 ma The total work done as the block moves rom ma to ma is zero 31 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 2 ma 32
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. 1 4.00 J = ( 0.100 m ) 2 2 k k = 800 N m To stretch the spring to 0.200 m requires 1 Δ W = ( 800 )( 0.200 ) 2 4.00 J = 12.0 J 2 33 Kinetic Energy (F ) Kinetic Energy is the energy o a particle due to its motion K = ½ mv 2 K is the kinetic energy m is the mass o the particle v is the speed o the particle the epression ½ mv 2 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 34 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 2 2 2 2 i dv ma d = m d = mv dv dt 35 Work-Kinetic Energy Theorem The Work-Kinetic Energy Principle states ΣW = K K i = ΔK 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 2 36
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 Δ = ΔK = ½ mv 2 0 (the vertical normal orce has no eect) 37 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.) 38 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 39 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 40
Eample. A 2.00-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 0.350. (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 2 = ½ m v 2. k 500 So, v = = 0.05 = 0.79 m/s m 2 41 (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 2 - μmg = ½ m v 2. In terms o initial sum and inal sum o energies, you can also treat it as ½k 2 = μmg + ½ m v 2 ½ 500 0.05 2 = 0.35 2 9.8 + ½ 2 v 2 v = 0.53 m/s 42 Eample. The ball launcher in a pinball machine has a spring that has a orce constant o 1.20 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 2 = ½ 1.20 N/(10-2 m) (5 cm 10-2 ) 2 = 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 2 0.0087 m + ½ 0.1 kg v 2 = 0.15 J v = 1.68 m/s 43 Eample. You need to move a heavy crate by sliding it across a lat loor with a coeicient o sliding riction o 0.2. 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 less, regardless o the weight o the crate or the displacement. 44
Normal Push orce F1 µ= 1 µ= 1 mg F1 = µmg Work done = F1 d What i you pull at 0,, i.e., in parallel with the surace? Normal orce mg F2 cos 30 = µ (mg F2 sin 30 ) F2 cos 30 + µ F2 sin 30 = µ mg F2 (0.866 + 0.5) = µ mg Pull F2 F2 = 0.732 µmg = 0.732 F1 s1 Slide 45 s1 scitaysc, 10/2/2007 Work done = F2 d = 0.732 F1 d 45 Ways to Transer Energy Into or Out o A System More Ways to Transer Energy Into or Out o A System Work transer by applying a orce and causing a displacement o the point o application o the orce, e.g., you push a book Mechanical Waves allow a disturbance to propagate through a medium, e.g., sound, earthquake, tsunami Heat is driven by a temperature dierence between two regions in space, e.g., hot tea 46 Matter Transer matter physically crosses the boundary o the system, carrying energy with it, e.g., use o gasoline in car Electrical Transmission transer is by electric current, e.g., use o electrical hair dryer Electromagnetic Radiation energy is transerred by electromagnetic waves, e.g., use o light bulb. 47
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 Power The time rate o energy transer is called power The average power is given by W P = Δt when the method o energy transer is work 48 49 Instantaneous Power Units o Power The instantaneous power is the limiting value o the average power as Δt approaches zero lim W dw P = Δ t 0 = Δt dt This can also be written as dw dr P = = F = F v dt dt The SI unit o power is called the watt 1 watt = 1 joule / second = ((kg m/s 2 ) m ) / s = 1 kg. m 2 / 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) = 3.6 10 6 J Energy transerred in Not 1 kilo-watt 1 hr at a constant rate Power = orce velocity 50 51 per hour
Eample. An energy-eicient light bulb, taking in 28.0 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 10 000 h and its purchase price is $17.0, whereas the conventional bulb has lietime 750 h and costs $0.420 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. kwh kwh Energy and the Automobile The concepts o energy, power, and riction help to analyze automobile uel consumption About 67% o the energy available rom the uel is lost in the engine About 10% is lost due to riction in the transmission, drive shat, bearings, etc. About 6% goes to internal energy and 4% to operate the uel and oil pumps and accessories This leaves about 13% to actually propel the car 52 53 Friction in a Car Friction in a Car, cont The magnitude o the total riction orce (ƒ t ) is the sum o the rolling riction (ƒ r ) orce and air resistance (ƒ a ) ƒ t = ƒ r + ƒ a At low speeds, rolling riction predominates At high speeds, air drag predominates 54 55