AP Physics Enosburg Falls High School Mr. Bushey. Week 6: Work, Energy, Power


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1 AP Physcs Enosburg Falls Hgh School Mr. Bushey ee 6: or, Energy, Power Homewor! Read Gancol Chapter AND/OR Read Saxon Lessons 1, 16, 9, 48! Read Topc Summary Handout! Answer Gancol p.174 Problems 1, 4, 7, 17, 18, 19, 0, 1, 3, 5! Answer Prnceton Reew p.79 AP Set multple choce, ree response questons Supplemental Reew / Study Group Focus! Reew Key Concepts Handout! Schaum s Outlnes, Chapter 6 Problems ! Schaum s Outlnes, Chapter 7 Problems ! Sole problems rom assocated Saxon lessons
2 or and orenergy Theorem or or s dened as the scalar product o orce and dsplacement. or s done when a nonzero net orce acts on a mong object. I a orce F moes an object oer a dstance d, wor can be ound by usng the ormula: F! " F! " d!!!! F " d I F and d are parallel, the wor s! F d cos".! Fd. I they are not parallel, then the wor s From ths ormula, one can conclude that or the wor to be done, the ollowng condtons must be met: # The object must moe (.e. d $ 0). A orce can be exerted on an object wth no wor done, e.g. pushng on a wall, holdng up 100 lb, etc. # I F and d are perpendcular, no wor s done (cos" = 0). For example, carryng somethng s not wor because the angle between the orce (upwards) and dsplacement (horzontal) s equal to 0. # The SI unt o wor s Joule (J), and 1 J = 1 N x 1 m. # I the angle between orce and dsplacement s less than 90 0, cos" >0, and the wor s poste. I 90 o <" < 180 o, cos" <0, and the wor s negate. Frcton always does negate wor because t always acts n the drecton opposte (180 o ) to the moton o the object. or ether changes the elocty o an object or counteracts the wor done by an opposng orce. For example, a car s mong at a constant speed, the engne does wor on the car to counteract the wor done by rcton n the opposte drecton. I an object s mong wth constant elocty, the net orce actng on t s zero. In ths case, the net amount o wor done on t must also be zero ( net =F net d).
3 Energy Energy s one o the most undamental concepts o Physcs. It s a ery abstract concept that does not hae a sngle denton. Energy s a characterstc o a system. Energy n a system ges t the capablty to perorm some operaton. hen wor s done on a system or by a system, the energy o the system changes. In other words, the system gans or loses energy through the process o wor. I poste wor s done, the system gans energy. I wor s negate, the system loses energy. I no wor s done, the total energy o the system does not change. There are many types o energy. Knetc energy (K or E ) s the energy o moton. It depends only upon the square o the speed and the mass o the object. K! 1 m or  Energy Theorem An object s mong wth an acceleraton a oer a dstance d. Then accordng to the ormula,! " a d : F! " ( ) d. m Solng or F d ges F! F! m F d! # ( ) d! The wor done on the object s gen by m 1 1! F d! ( # )! m # m! K # K Or, we can rewrte t as! K # K! $ K Ths s the netc energy wor theorem: The change o netc energy o a body s equal to the wor o all the orces actng upon t.
4 or and orenergy Theorem 1. A box, ntally at rest, s pulled a dstance o 5 m across a loor by a horzontal orce o 3 N. At the end o the 5 m, the netc energy o the box s 87 J. a. hat was the net wor done on the box? Accordng to the orenergy theorem, the net wor done on the box s equal to the change n the netc energy. net! " E! 87J b. How much wor was done by the 3 N orce? = F d = (3 N)(5 m) = 115 J c. How much wor was done by rcton? The net wor s equal to the sum o the wor done by the appled orce and the wor done by the rcton. net = A + = net A =87 J 115 J = 8 J d. How much wor was done by the gratatonal orce? The gratatonal orce s perpendcular to the dsplacement. Thus, the gratatonal orce does no wor. g = 0 J.. Bll carres a 35 N pacage rom the ground up to the th loor o a 15 m hgh oce buldng. How much wor s done by Bll on the pacage? The orce Bll s applyng to the pacage s drected up. He must be applyng a orce drected up to the pacage that s equal to the pacage s weght. He s also mong up. Thereore, # = 0. F = 35 N d = 15 m =? = F d cos# = F d = (35 N)(15 m)(cos 0º) = 55 J
5 3. A 4 g brc sldes a dstance (d) o 5m along an cy slope o nclnaton angle o 30 o. The coecent o netc rcton s! = 0.0. hat wll be the speed o the brc at the end o the slope? " There are three orces actng on the brc. The wor done by graty = m g d cos(90  "), as the angle enclosed by G and the slope s 90  ". Note that cos(90  ") = sn" (trgonometry), thereore, g = m g d sn ". Note that d sn" = h where h s the heght o the slope. Thus, the wor s equal to m g h. The normal orce encloses 90 o wth the dsplacement, so n = F n d cos(90 o ) = 0. The wor o the rcton s r = !#F n #d = ! m g d#cos("). The wor o all orces s = m g d (sn " ! cos ") and ths wll be equal to the netc energy K = 1/ m. m mgd(sn" $! cos ") % gd(sn" $! cos ") % % gd(sn" $! cos ") (9.81m/s )(5 m)(0.500 (0.0)(0.866)) % 5.7 m/s % $ 4. A g bloc s accelerated rom rest along a horzontal, smooth surace by a orce o 5 N oer a dstance o 6 m. a. Determne the acceleraton o the bloc and the nal elocty o the bloc. F 5 N a % % %.50 m/s m g $ % a # ( x o & ' & '& ' $ 0 m/s %.50m/s 6 m % 30 m /s % % 30 m /s 5.48 m/s
6 b. Calculate the wor done by the acceleratng orce. o! "! "! " $ F d Cos # $ 5 N 6 m Cos 0 $ 30 N % m $ 30.0 J c. Use the orenergy theorem to determne the nal elocty o the bloc. Compare your answer to the answer obtaned n part (a). $ & KE $ KE  KE KE $ ' KE $ 30 J ' 0 J $ 30 J 1! " g $ 30 J 30 J 1 g $ $ 30 m /s $ 5.48 m/s d. Fnd the nal elocty o the bloc t has an ntal elocty o + m/s. $ & KE $ KE  KE 1 1 $ m ( m J = g g m/s 30J = 1 g ( 4 J! " (! "! "! " 34 J $ $ $ 1 g 34 m /s 5.83 m/s e. I the bloc s mong wth a elocty o 8 m/s, what magntude retardng orce s needed to brng the bloc to rest oer a dstance o 5 m? & KE $ KE  KE F $ $ $ 1 1 m  m 1 1! g"! 0 m/s " ! g"! 8 m/s" 64 J & KE $ & x % Cos # 64 J $ (5 m)(1) $ 1.8 N
7 5. Hooe s law states that F!" x. a. Fnd the wor requred to compress the sprng through a dstance x. Because the orce s not constant throughout the dstance moed, we need to ntegrate the product o the dstance and the orce to determne the wor. Also the orce we exert s opposte n drecton to the orce exerted by the sprng, so the orce we exert s F! x. x x x x 1 1! # F dx! 0 0 # x dx!! x "! x 0 The wor requred s then 0 1! x. b. hat s the potental energy o a sprng whch was ntally at zero and was compressed to a pont x? hen the sprng has been compressed ts elocty s zero, so there s no netc energy, all o the energy s then potental energy. 1 or s equal to the change n the energy. Thus, Usprng! x.
8 Conserate Forces and Potental Energy Conserate Forces Denton: A orce s called conserate the wor t does on an object depends only on the ntal and nal postons o the object and s ndependent o the path taen between those postons. Graty s one such conserate orce. Near Earth s surace, the wor done by graty on an object o mass m depends only on the change n the object s heght h. In the case o conserate orces, we assgn a numberthe potental energy, Uto each conguraton o the system. The zero leel o the potental energy s arbtrary; t can be assgned to any poston. I the place wth zero potental energy s chosen, the potental energy at any pont A s the wor done by the orce when the body moes rom the pont A to the pont 0 wth zero potental energy. So, U(A) = A0. The wor s addte. I the body moes rom A to B and then to pont 0, then! " B A0 AB 0, U ( A)! " U ( B) or! U ( A) # U ( B). AB AB Examples o Potental Energy The Gratatonal Potental Energy Graty s an example o a conserate orce. The wor done on an object o mass m when t s rased heght h at constant elocty by an external orce s mgh. U g = mgh s called the gratatonal potental energy o the object. The only physcally sgncant quantty s the change n potental energy, and not the absolute alue o the potental energy. For ths reason, n problem solng, we usually choose a conenent pont (table top leel, ground leel, sea leel, etc.) and set the potental energy at ths leel to be zero. All heghts are measured rom ths leel. Then, the object s below the zero leel, ts potental energy s negate. It s mportant to stress agan that only the change n potental energy s physcally sgncant. Gratatonal Potental Energy: U g = mgy Change n Gratatonal Potental Energy: $ U = U  U g g g o = mgh  mgh o
9 The Elastc Potental Energy Sprngs, rubber bands, and other sprngy objects also exert conserate orces. At relately small elongatons, the orce exerted by a sprng s proportonal to the amount x by whch the sprng has been stretched, Fspr! " x, (The  sgn ndcates that the orce s drected n the opposte drecton to the stretch.) s the constant o proportonalty and s called the sprng s sprng constant. Ths proportonalty between the orce F and the amount o stretchng (or compresson) x s nown as Hooe s Law. The wor done by a sprng wth a gen sprng constant depends only on x, so an assocated potental energy s U x x 1 spr ( )!. Note that x = 0 s at the equlbrum (unstretched) poston o the sprng. Nonconserate orces Unle conserate orces, the wor done by nonconserate orces depends on the path the object moes and not just on the ntal and nal ponts. Frcton s an example o a nonconserate orce. A nonconserate orce s oten called a dsspate orce.
10 Conserate Forces and Potental Energy 1. A 5 g bowlng ball s carred to the top o a tower that s 45 m hgh. The bowlng ball s released rom rest and allng reely to the ground. a. hat orce s requred to lt the bowlng ball to the top o the tower at constant elocty? F 0 N  Net!! FAppled F!! m g= (5g)(10 m/s)! 50 N Appled F App = 50 N t = 50 N b. How much wor s done by the ltng orce to lt the bowlng ball rom the ground to the top o the tower? " #" #! FNet d! 50 N 45 m! 50 J c. hat s the Gratatonal Potental Energy o the bowlng ball on the top o the tower? GPEtop!! $ 50 J d. How do the GPE and the KE o the bowlng ball change the bowlng ball alls rom the top o the tower to the ground? As the bowlng ball alls, t loses heght and ts GPE decreases. Because o the gratatonal orce, the bowlng ball accelerates (g = 10 m/s ), ts speed ncreases, and ts KE ncreases. e. Use the equatons o moton to nd the speed o the bowlng ball at the moment t hts the ground. hat s the KE o the bowlng ball at the nstant t stres the ground? Compare t wth the answer rom part (c). d! % 45 m ;! 0 m/s ; a! % 10 m/s %! a d o 0 " #" # % (0 m/s)! % 10 m/s % 45 m! 900 m /s E!! 900 m /s 30 m/s 1 1! m! " 5 g#" 30 m/s #! 50 J The answer s the same as n part (c).
11 . Use the equatons o moton or ree all to determne the heght aboe the ground and the potental energy o the bowlng ball at t = s ater t s released. 1 h # h0 $ 0t $ at 1 # 45m $ 0 m/s s $ % 10m/s s h # 5 m! "! "! "! " GPE! "! "! " # mgh # 5 g 10m/s 5 m # 150 J g. Calculate the elocty and the KE o the bowlng ball at tme t = s ater t s released. KE # $ a t 0! "! " # 0 m/s $ % 10 m/s s # % 0 m/s # 1 m 1 (5 g) 0 m/s 1000 J! " # % # h. Calculate the sum o the KE and GPE o the bowlng ball at ths tme and compare t wth ts ntal potental energy and wth ts nal netc energy at the moment t hts the ground. GPE + KE = 1000J + 150J = 50J = GPE top = KE. The bowlng ball hts the ground and compresses the ground a dstance o 0.00 m. hat s the magntude o the aerage orce exerted by the ground on the bowlng ball as t comes to rest? Ignorng the small change n GPE as the ground s compressed, the Change n KE s equal to the wor done by the orce exerted by the ground: & KE # KE % KE # 0 J  50 J # 50 J o or Done # & KE F d = F( d % d ) # & KE F( % 0.00 m % 0 m) # % 50 J % 50 J F # # 113,000 N % 0.00 m
12 Accordng to the orenergy Theorem, All _ Forces!" K Conseraton o Energy and Power or o all orces can be separated nto two components, wor o conserate orces and wor o nonconserate orces.! # _ All _ Forces Consearate _ Forces Non _ Conserate Forces Further, Conserate _ Forces! $" U Then the wor energy theorem becomes Non _ Conserate _ Forces $ " U! " K Non _ Conserate _ Forces! " K # " U Ths leads to the concept o the conseraton o mechancal energy. I there s no nonconserate orce n the system, Non _ Conserate _ Forces! 0 and 0! " K # " U. In the absence o dsspate orces n a mechancal system, the mechancal energy E mech = K + U s a constant. For example, E mech n tme t = t ntal s the same as E mech at tme t = t nal. The wor energy theorem ges the result " Emech! " K # " U! 0 In an solated system where only conserate orces cause energy to be transormed, the mechancal energy o the system s constant. I no nonconserate orces are actng n a system, then 0! " K # " U. hen there are orces such as rcton present, we can stll nclude these orces and consere total energy tang nto account the heat generated as a result o wor o dsspate orces such as rcton.
13 !!"!" For Frcton, nc " # d " $! Nd Ths wor taes away mechancal energy and transers ths detracted amount nto heat. From the wor energy theorem, $! Nd " % K & % U 0 " % & % &! K U Nd 0 " % Emech & % Ethermal Thermal energy s one o the orms o nonmechancal nternal energy stored nsde o the bodes. Internal energy wll be coered n the ollowng unts o the course. The conseraton o energy wth ncluson o nternal energy becomes 0 Emech " % & % E nt I there are external orces appled to an solated system as a whole then, the wor energy theorem s or,!!!"!" F # d " % E & % E ext mech nt Fext "% E Total All o ths can by summed up n the ollowng equaton: Total nal mechancal energy = Total ntal + Any wor done mechancal energy  Any losses due to nonconserate orces or K & U " K & U & $ app loss
14 Power Power s the rate (how ast) at whch wor s done. or Done Power!. Tme Spent Perormng or P! t The unt o power s the "att," whch s equal to a Joule/second. The "horsepower" s the Englsh system unt o power; t s equal to 746.
15 Conseraton o Energy and Power 1. An electrcal motor lts a 575 N box 0 m straght up by a rope n 10 s. hat power s deeloped by the motor? F = 575 N the amount o orce the rope must apply to the object to lt t up at constant elocty s equal to the object s weght F = 575 N P! t d = 0 m! F d t = 10 s P =? Fd P! t (575 N)(0 m) 3 P!! 1.15" 10 10s. A bloc sldes down a rctonless nclned plane o heght h = 1 m, mang angle # wth the horzontal. At the bottom o the plane, the bloc contnues to moe on a lat surace wth a coecent o rcton $ = How ar does the mass moe on the lat surace? e apply the law o Conseraton o energy. The equaton s % E! % K & % U The elocty o the bloc s zero both n the begnnng and the end. Thus, % K! 0 e also hae % E!! ' $ m g d rcton The change n GPE s % U! ' mgh Puttng t all together % E! % K & % U ' $ m g d! ' mgh h 1m d!!! 3.3m $ A cyclst approaches the bottom o a hll at a speed o 11 m/s. The hll s 6 m hgh. Ignorng rcton, how ast s the cyclst mong at the top o the hll? Assume that he doesn t peddle and gnore ar resstance. Snce there s no rcton, the total mechancal energy s consered. Thus, % U & % K! % U! mgh ' 0! mgh and % K! m ' m, Substtute these nto the energy conseraton equaton 1 1 mgh & m ' m! 0! ' gh Thus,! ' gh! (11m/s) ' (9.81m/s )(6 m) = 3.8m /s! 1.81m/s.
16 4. A 100 g mass traelng wth a elocty o 15 m/s on a horzontal surace stres a sprng wth a sprng constant = 5 N/m. a. Fnd the compresson o the sprng requred to stop the mass the surace s rctonless. Snce there s no dsspate orce, we wll use the Law o Conseraton o Energy.! U "! K # ! U # x $ 0 # x! K # 0 $ m # $ m 1 1 x $ m # Sole or x: x # m 100 g x # (15m/s) # 67.1m 5 N/m b. Fnd the compresson o the sprng the surace s rough ( % # 0.4 0). The more general orm o the equaton ncludng nonconserate orces s! E # U "! K!!! E # F & d # $% m g x 1! U # x and! K # $ Puttng t all together we get: 1 m % $ m g x # x $ m 1 1 where and and $ # $ # $ % m g x 0.40(100 g)(9.81 m/s ) x ( 39.4 N) x x # (5 N/m) x # (.5 N/m) x 1 1 m # (100g)(15m/s) # 1150N 1 1 Smplyng (and leang o the unts to mae the quadratc equaton easer to read) $ 39.4x#.5x $ 1150 In standard orm ths quadratc equaton becomes.5x " 39.4x$ 1150 # 0 Solng by the quadratc ormula we obtan x# 4.8 m or x# $ 18 m The physcal soluton here s poste. Thus, x = 5 m.
17 5. An amusement par roller coaster car has a mass o 50 g. Durng the rde, t s towed to the top o a 30 m hll, where t s released rom rest and allowed to roll. The car plunges down the hll, then up a 10 m hll and through a loop wth a radus o 10 m. Assume that the tracs are rctonless. (Use g = 10 m/s.) a. hat s the Potental Energy o the car at the top o the 30 m hll? GPE # # Top o Hll #! 50 g"! 10 m/s "! 30 m" J m g h b. hat are the Knetc Energy and the speed o the car at the bottom o the 30 m hll? GPE # E # KE # KE Top o Hll Tot Bottom o Hll Bottom o Hll # 1 m J KE (75000 J) # # # # m 50 g 600 m /s 4 m/s c. hat are the Knetc Energy and the speed o the car at the top o the 10 m hll? GPE 10 m 10 m 10 m 10 m Tot 10 m Tot 10 m! "! "! " # m g h # 50 g 10 m/s 10 m # 5000 J GPE $ KE # E # J KE # E % GPE # J % 5000 J # J KE (50000 J) # # # # m 50 g 400 m /s 0 m/s d. I the hll maes an angle o 60 o wth the horzontal and the car taes 15 seconds to be towed up the hll, determne the length o the hll, the elocty o the car, the orce requred to tow the car up the hll, and the power o the motor pullng the car up the hll. o 30m # L(sn 60 ) 30m L # # o 34.64m sn 60 o F # t & sn 60 o (500 N)(sn 60 ) # 165 N = 00 N L 34.64m # # #.31m/s =.3m/s ' t 15s or Done 75000J P # # # 5000 J/s # 5000 ' t 15s or P # F!! = (m g)(sn 60 )( )=(50 g)(10 m/s )(sn 60 )(.31! " = (165 N).31m/s # 5001J/s # 5.0 m/s)
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