Kinetic Energy and Work

Transcription

1 PH 1-3A Fall 010 Knetc Energy and Work Lecture Chapter 7 (Hallday/Resnck/Walker, Fundamentals o Physcs 8 th edton)

2 Chapter 7 Knetc Energy and Work In ths chapter we wll ntroduce the ollowng concepts: Knetc energy o a movng object Work done by a orce Power In addton we wll develop the work-knetc energy theorem and apply t to solve a varety o problems Ths approach s alternatve approach to mechancs. It uses scalars such as work and knetc energy rather than vectors such as velocty and acceleraton. Thereore t smpler to apply.

3 m m Knetc Energy: We dene a new physcal parameter to descrbe the state o moton o an object o mass m and speed v mv We dene ts knetc energy K as: K = We can use the equaton above to dene the SI unt or work (the joule, symbol: J ). An object o mass m = 1kg that moves wth speed v = 1 m/s has a knetc energy K = 1J Work: (symbol W) I a orce F s appled to an object o mass m t can accelerate t and ncrease ts speed v and knetc energy K. Smlarly F can decelerate m and decrease ts knetc energy. We account or these changes n K by sayng that F has transerred energy W to or rom the object. I energy t transerred to m (ts K ncreases) we say that work was done by F on the object (W > 0). I on the other hand. I on the other hand energy ts transerred rom the object (ts K decreases) we say that work was done by m (W < 0)

4 Problem 5. A ather racng hs son has hal the knetc energy o the son, who has hal the mass o the ather. The ather speeds up by 1.0 m/s and then has the same knetc energy as the son. What are the orgnal speeds o (a) the ather and (b) the son? We denote the mass o the ather as m and hs ntal speed v. The ntal knetc energy o the ather s K = 1 K son and hs nal knetc energy (when hs speed s v = v m/s) s K = K son. We use these relatons along wth denton o knetc energy n our soluton. (a) We see rom the above that K = 1 K whch (wth SI unts understood) leads to ( 1.0 m/s) mv = m v +. The mass cancels and we nd a second-degree equaton or v : 1 v 1 v = 0. The postve root (rom the quadratc ormula) yelds v =.4 m/s. (b) From the rst relaton above bk = 1 Ksong, we have ( /) mv = m vson and (ater cancelng m and one actor o 1/) are led to v son = v = 4.8 m s.

5 Fndng an epresson or Work: m m Consder a bead o mass m that can move wthout rcton along a straght wre along the -as. A constant orce F appled at an angle φ to the wre s actng on the bead We apply Newton's second law: F = ma We assume that the bead had an ntal velocty v and ater t has travelled a dsplacement d ts velocty s v. We apply the o thrd equaton o knematcs: v v = ad We multply both sdes by m/ o m m m m F m v vo = ad = d = Fd = Fcos φd K = vo m m K = v The change n knetc energy K K = Fd cos ϕ Thus the work W done by the orce on the bead s gven by: W = F d = Fd cosϕ W = Fd cosϕ W = F d

6 F A F m m C W = Fd cosϕ W = F d F B The unt o W s the same as that o K.e. joules Note 1: The epressons or work we have developed apply when F s constant Note : We have made the mplct assumpton that the movng object s pont-lke Note 3: W > 0 0 < φ < 90, W < 0 90 < φ < 180 Net Wor k: I we have several orces actng on a body (say three as n the pcture) there are two methods that can be used to calculate the net work Wnet Method 1: Frst calculate the work done by each orce: WA by orce FA, WB by orce FB, and WC by orce FC. Then determne Wn et = W A + W B + WC Method : Calculate rst F = F + F + F ; Then determne = F d net A B C Wnet

7 Acceleratng a Crate A 10kg crate on the latbed o a truck s movng wth an acceleraton a = +1.5m/s along the postve as. The crate does not slp wth respect to the truck, as the truck undergoes a dsplacement s = 65m. What s the total work done on the crate by all the orces actng on t? N

8 Work Problem 11: A 100kg car s beng drven up a 5.0 hll. The rctonal orce s drected opposte to the moton o the car and has a magntude o k = N. The orce F s appled to the car by the road and propels the car orward. In addton to those two orces, two other orces act on the car: ts weght W, and the normal orce N drected perpendcular to the road surace. The length o the hll s m. What should be the magntude o F so that the net work done by all the orces actng on the car s +150,000 J?

9 m m Work-Knetc Energy Theorem We have seen earler that: K K = W. We dene the change n knetc energy as: K = K K. The equaton above becomes th e work-knetc energy theorem K = K K = Wnet net Change n the knetc net work done on energy o a partcle = the partcle The work-energy theorem: when a net eternal orce does work W on an object, the knetc energy o the object changes rom ntal value KE 0 to the nal value KE, the derence between the two values beng equal to the work W = K K 0 = 1/mv 1/mv 0 The work-knetc energy theorem holds or both postve and negatve values o I W > 0 K K > 0 K > K net I W < 0 K K < 0 K < K net W net

10 Work and Knetc Energy In a crcular orbt the gravtatonal orce F s always perpendcular to the dsplacement s o the satellte and does no work KE = constant W>0 KE ncreases In an ellptcal orbt, there can be a component o the orce along the dsplacement Work s done W<0 KE decreases

11 The Work-Energy Theorem and Knetc Energy Problem : A 0.075kg arrow s red horzontally. The bowstrng eerts an average orce o 65 N on the arrow over a dstance o 0.90 m. Wth what speed does the arrow leave the bow?

12 The Work-Energy Theorem and Knetc Energy The speed o a hockey puck decreases rom to 44.67m/s n coastng 16m across the ce. Fnd the coecent o knetc rcton between the puck and the ce.

13 The work-energy theorem deals wth the work done by the net eternal orce. The work-energy theorem does not apply to the work done by an ndvdual orce. I W>0 then KE ncreases; W<0 then KE decreases; W=0 then KE remans constant. Downhll Skng: A 58kg sker s coastng down a 5 slope. A knetc rcton orce k =70N opposes her moton. Near the top o the slope, the sker s speed s v 0 =3.6m/s. Ignorng ar resstance, determne the speed v at a pont that s dsplaced 57m downhll.

14 Problem 15. A 1.0 N orce wth a ed orentaton does work on a partcle as the partcle moves through dsplacement d = (.00ˆ 4.00 ˆj kˆ ) m. What s the angle between the orce and the dsplacement the charge n the partcle's knetc energy s (a) +30.0J and (b) -30.0J? Usng the work-knetc energy theorem, we have K = W = F d = Fd cosφ In addton, F = 1 Nand d = (.00 m) + ( 4.00 m) + (3.00 m) = 5.39 m. (a) I K = J, then 1 K J φ = cos = cos = 6.3. Fd (1.0 N)(5.39 m) (b) K = 30.0 J, then 1 K J φ = cos = cos = 118 Fd (1.0 N)(5.39 m)

15 B A Work Done by the Gravtatonal Force: Consder a tomato o mass m that s thrown upwards at pont A wth ntal speed v gravtatonal orce F ( ) o. As the tomato rses, t slows down by the g ( ) ( ) so that at pont B ts has a smaller speed v. The work W A B done by the gravtatonal orce on the g tomato as t travels rom pont A to pont B s: W A B = mgd cos180 = mgd g The work W B A done by the gravtatonal orce on the ( ) g tomato as t travels rom pont B to pont A s: W B A = mgd cos 0 = mgd g

16 B A. m Work done by a orce n Ltng an object: Consder an object o mass m that s lted by a orce F orm pont A to pont B. The object starts rom rest at A and arrves at B wth zero speed. The orce F s not necessarly constant durng the trp. The work-knetc energy theorem states that: We also have that K = K K = 0 W = 0 There are two orces actng on the object: The gravtatonal orce that lts the W a g net object. Wnet Wa ( A B) Wg ( A B) ( A B) = Wg ( A B ) ( A B) = cos180 = - ( A B) = Work done by a orce n Lowerng an object: In ths case the object moves rom B to A g K = K K = W F g net and the appled orce F = + = 0 W mgd mgd W mgd ( B A) = cos 0 = ( ) ( ) W mgd mgd a W B A = W B A = mgd a g

17 Problem 17. A helcopter lts a 7 kg astronaut 15 m vertcally rom the ocean by means o a cable. The acceleraton o the astronaut s g/10. How much work s done on the astronaut by (a) the orce rom the helcopter and (b0 the gravtatonal orce on her? Just beore she reaches the helcopter, what are her (c) knetc energy and (d) speed? (a) We use F to denote the upward orce eerted by the cable on the astronaut. The orce o the cable s upward and the orce o gravty s mg downward. Furthermore, the acceleraton o the astronaut s g/10 upward. Accordng to Newton s second law, F mg = mg/10, so F = 11 mg/10. Snce the orce F and the dsplacement d are n the same drecton, the work done by F s W F (7 kg)(9.8 m/s )(15 m) mgd = Fd = = = J whch (wth respect to sgncant gures) should be quoted as J. (b) The orce o gravty has magntude mg and s opposte n drecton to the dsplacement. Thus the work done by gravty s Wg = mgd = = 4 (7 kg)(9.8 m/s )(15 m) J whch should be quoted as J (c) The total work done s W = J J = J. Snce the astronaut started rom rest, the work-knetc energy theorem tells us that ths (whch we 3 round to J ) s her nal knetc energy. (d) Snce K = 1 mv, her nal speed s v K ( = = m 7 kg 3 J) = 5. 4 m / s.

18 Problem 4. A cave rescue team lts an njured spelunker drectly upward and out o a snkhole by means o a motor-drven cable. The lt s perormed n three stages, each requrng a vertcal dstance o 10.0 m: (a) the ntally statonary spelunker s accelerated to a speed o 5.00 m/s; (b0 he s then lted at the constant speed o 5.00 m/s; nally he s decelerated to zero speed. How much work s done on the 80.0kg rescuee by the orce ltng hm durng each stage? We use d to denote the magntude o the spelunker s dsplacement durng each stage. The mass o the spelunker s m = 80.0 kg. The work done by the ltng orce s denoted W where = 1,, 3 or the three stages. We apply the work-energy theorem, K -K =W a +W g 1 (a) For stage 1, W1 mgd = K1 = mv, where v1 = m / s. Ths gves W1= mgd + mv1 = (80.0 kg)(9.80 m/s )(10.0 m) + (80.0 kg)(5.00 m/s) = J. (b) For stage, W mgd = K = 0, whch leads to W 3 = mgd = (80.0 kg)(9.80 m/s )(10.0 m) = J. 1 (c) For stage 3, W mgd = K = mv We obtan W3 = mgd mv1 = (80.0 kg)(9.80 m/s )(10.0 m) (80.0 kg)(5.00 m/s) = J.

19 Work done by a varable orce F( ) actng along the A orce F that s not constant but nstead vares as uncton o s shown n g.a. We wsh to calculate the work W that F does on an object t moves rom poston to poston. j, avg ( ) -as: We partton the nterval, nto N "elements" o length each as s shown n g.b. The work done by F n the j- th nterval s: j= 1 W = F Where F, j over the j-th element. W = F j, avg j= 1 j, avg the sum as 0, (or equvalently N ) N s the average value o F We then take the lmt o W = lm F = F( ) d Geometrcally, W s the area j avg between F( ) curve and the -as, between and (shaded blue n g.d) W N = F( ) d

20 Three dmensonal Analyss: In the general case the orce F acts n three dmensonal space and moves an object on a three dmensonal path rom an ntal pont A to a nal pont B The orce has th eo rm: F= F yz,, ˆ+ F yz,, ˆj+ F yzk,, ˆ ( ) ( ) ( ) ( y z) ( y z ) Pont s A and B have coordnates,, and,,, respectvely dw = F dr = F d + F dy + F dz B y z y z W = dw = F d + F dy + F dz y z A y z y z y z W = F d + F dy + F dz y z y z A z O path y B

21 Work-Knetc Energy Theorem wth a Varable Force: Consde a varable orce F() whch moves an object o mass m rom pont A( = ) dv to pont B( = ). We apply Newton's second law: F = ma = m We then dt dv multply both sdes o the last equaton wth d and get: Fd = m d dt dv We ntegrate both sdes over d rom to : Fd = m d dt dv dv d dv dv d = d = d = vdv Thus the ntegral becomes: dt d dt dt d dt m mv mv W = m vdv = v = = K K = K Note: The work-knetc energy theorem has eactly the same orm as n the case when F s constant! W = K K = K O. A m B. F() d -as

22 The Ideal Sprng Sprngs are objects that ehbt elastc behavor. It wll return back to ts orgnal length ater beng stretched or compressed. Equlbrum poston Relaed or unstraned length o the sprng For small deormatons, the orce F requred to stretch or compress a sprng obeys the equaton: F = k - dsplacement o the sprng rom ts unstraned length k sprng constant [N/m] unt A sprng that behaves accordng to the relatonshp F = k t s sad to be an deal sprng

23 Restorng Force Equlbrum poston Restorng orce Stretched poston To stretch or compress a sprng a orce F must be appled Newton s 3 rd Law: Every acton has an equal n magntude and opposte reacton. The reacton orce that s appled by the sprng to the agent that does the pullng or pushng s called restorng orce The restorng orce s always opposte to the dsplacement o the sprng

24 The orce F The Sprng Force: Fg.a shows a sprng n ts relaed state. In g.b we pull one end o the sprng and stretch t by an amount d. The sprng rests by eertng a orce the opposte drecton. F on our hand In g.c we push one end o the sprng and compress t by an amount d. Agan the sprng ressts by eertng a orce hand n the opposte drecton F n on our eerted by the sprng on whatever agent (n the pcture our hand) s tryng to change ts natural length ether by etendng or by compressng t s gven by the equaton: F = k Here s the amount by whch the sprng has been etended or compressed. Ths equaton s known as "Hookes law" k s known as "sprng constant" F = k

25 An object s attached to the lower end o a 100-col sprng that s hangng rom the celng. The sprng stretches by m. The sprng s then cut nto two dentcal sprngs o 50 cols each. As the drawng shows, each sprng s attached between the celng and the object. By how much does each sprng stretch?

26 O O (a) (b) Work Done by a Sprng Force Consder the relaed sprng o sprng constant k shown n (a) By applyng an eternal orce we change the sprng's length rom (see b) to (see c). We wll calculate the work W s done by the sprng on the eternal agent (n ths case our hand) that changed the sprng length. We O (c) assume that the sprng s massless and that t obeys Hooke's law We wll use the epresson: W = F( ) d = kd = k d W s k k = k = sprng ( s Qute oten we start wth a relaed = 0) and we ether stretch or compress the sprng by an amount ( = ± ). In ths case W s k =

27 Problem 9. The only orce actng on a.0 kg body as t moves along a postve as has an component F =-6 N, wth n meters. The velocty at = 3.0 m s 8.0 m/s. (a) What s the velocty o the body at =4.0 m? (b) At what postve value o wll the body have a velocty o 5.0 m/s? (a) As the body moves along the as rom = 3.0 m to = 4.0 m the work done by the orce s 6 3( ) 3 ( ) 1 J. W F d d = = = = = Accordng to the work-knetc energy theorem, ths gves the change n the knetc energy: 1 W = K = m v v d where v s the ntal velocty (at ) and v s the nal velocty (at ). The theorem yelds v W ( 1 J) = + v = + (8.0 m/s) = 6.6 m/s. m.0 kg (b) The velocty o the partcle s v = 5.0 m/s when t s at =. The work-knetc energy W =, so theorem s used to solve or. The net work done on the partcle s 3( ) the theorem leads to Thus, d d. 3 1 = m v v m.0 kg = ( v v ) + = ( (5.0 m/s) (8.0 m/s) ) + (3.0 m) = 4.7 m. 6 6 N/m

28 Power We dene "power" P as the rate at whch work s done by a orce F. I F does work W n a tme nterval t then we dene as the average power as: P avg W = t The nstantaneous po wer s dened as: P = dw dt Unt o P : The SI unt o power s the watt. It s dened as the power o an engne that does work W = 1 J n a tme t = 1 second A commonly used non-si power unt s the horsepower (hp) dened as: 1 hp = 746 W The klowatt-hour The klowatt-hour (kwh) s a unt o work. It s dened as the work perormed by an engne o power P = 1000 W n a tme t = 1 hour W 6 = Pt = = J The kwh s used by electrcal utlty companes (check your latest electrc bll)

29 Consder a orce F actng on a partcle at an angle φ to the moton. The rate dw F cosφd d at whch F does work s gven by: P = = = F cosφ = Fv cosφ dt dt dt P = Fv cosφ = F v v

30 Eample: An elevator cage has a mass o 1000kg. How many horsepower must the motor delver to the elevator t s to rase the elevator cage at the rate o.0m/s?

31 A car accelerates unormly rom rest to 7 m/s n 7.0s along a level stretch o road. Ignorng rcton, determne the average power requred to accelerate the car a) The weght o the car s N, and b) weght s N =86hp =114hp

32 Problem 48. (a) At a certan nstant, a partcle-lke obkect s acted on by a orce ˆ ˆ ˆ F = (4.0 N) (.0 N) j+ (9.0 N) k whle the objects velocty s v = (.0 m/ s) ˆ+ (4.0 m/ s) kˆ What s the nstantaneous rate at whch the orce does work on the object? (b) AT some other tme, the velocty conssts o only a y component. the orce s unchanged and the nstantaneous power s -1W, what s the velocty o the object? (a) we obtan P = F v = (4.0 N)(.0 m/s) + (9.0 N)(4.0 m/s) = 8 W. (b) wth a one-component velocty: v = vj. whch yelds v = 6 m/s. P= F v 1 W = (.0 N) v.