Application: Work. 8.1 What is Work?

Transcription

1 Applcto: Work 81 Wht s Work? Work, the physcs sese, s usully defed s force ctg over dstce Work s sometmes force tmes dstce, 1 but ot lwys Work s more subtle th tht Every tme you exert force, t s ot the cse tht y work s doe (eve though t my feel lke tht to you! Why? Well, the work/eergy equto sys tht work doe (by the et force o object equls the object s chge ketc eergy More smply: 1 Work wll be force tmes dstce ll of the pplctos we cosder Work = Chge Ketc Eergy (81 Ths mes tht f object s ketc eergy does t chge, the o work hs bee doe o the object whether or ot force hs bee exerted I prtculr, force wll do work oly f the force hs compoet the drecto tht the object moves Here s the dstcto: f you push o bg box d t moves the drecto you push, the work s ccomplshed d t equls force tmes dstce If the box s very bg d whe you push t othg hppes (t does ot move, the o work s doe If force s ppled to the box cert drecto d the box moves, but ot the drecto you push (mybe your fred s pushg o other sde but some other drecto, the work s doe, but the mout depeds o the gle the box moves reltve to the drecto the box moves So g the swer s ot smply force tmes dstce However, to keep thgs smple, the cotext of ths secto we wll ssume the drecto the force s ppled d the drecto of the moto of the object re oe d the sme I ths cse, ssumg tht the force s costt, W = Work = force Dstce = F x The uts used to mesure work vry depedg o the system you re d they re probbly less fmlr to you th the uts for velocty d ccelerto Tble 81 lsts the uts for three commo systems System Force Dstce Work Tble 81: Uts of work Brtsh pouds feet foot-pouds (ft-lbs cgs dyes cetmeters ergs SI (tertol Newtos meters joules EXAMPLE 811 Clculte the work doe lftg 17 lb object 22 feet (eg, me wlkg from the frst to the thrd floor Lsg? Well, Work = force Dstce = =, 2 ft-lbs

2 mth 11 pplcto: work 2 Tht ws esy Now the key thg to otce s tht uder our ssumptos, work s product Oe of the ssumptos s tht costt force s ppled But most forces re ot costt Cosder the followg stuto EXAMPLE 812 Suppose we wt to host lekg bucket vertclly Becuse of the lek, the force ppled to lft the bucket decreses, so tht F(x = 6 (1 x2, x 5 5 where x s mesured feet Fd the work doe lftg the bucket So how do we clculte the work doe f the force vres? Well, remember tht Rem sums volve products So we wll use the subdvde d coquer strtegy oce more Ulke the erler cses there s o turl fgure to drw whose re, volume, or rc legth we re tryg to clculte Ths tme we must pply the theory Geerl Stuto Assume tht F(x s vrble but cotuous force tht s fucto of the posto x d tht t s ppled over tervl [, b] Fd the work doe As usul, let P = {x, x 1,, x } be regulr prtto of [, b] to equl wdth subtervls of legth x Now f the tervls re short eough, sce the force s cotuous, the force wll be erly costt o ech tervl, though ts vlue wll vry from tervl to tervl So let W deote the work doe o the th subtervl Sce the legth of the th subtervl s x, the W F(x x Cosequetly, ddg up the peces of work o ech subtervl, Totl Work = W F(x x (82 Notce tht we ow hve Rem sum volvg the force fucto! To mprove the pproxmto we do the stdrd thg: We let the umber of subdvsos get lrge d tke the lmt We fd Totl Work = lm W = lm F(x x = F(x dx We re cert tht ths lmt of the Rem sums exsts d s, fct, the defte tegrl becuse we ssumed tht the force F(x s cotuous fucto of the posto THEOREM 81 (Work Formul If F(x s cotuous force tht s fucto of the posto x tht s ppled over tervl [, b], the the work doe over the tervl s Work = F(x dx Stop d Step Bck There re couple of thgs I wt you to otce Frst, Theorem 81 mouts to syg tht work s the re uder the force curve Tht s probbly ot how you would frst thk of t, but tht s wht the theorem sys! Worktex Verso: Mtchell-215/1/89:5:21

3 mth 11 pplcto: work Secod, whe I ws wrtg these otes, I smply cut-d-psted the erler mterl o rc legth d chged the few words here d there But t s the sme subdvde d coquer strtegy tht you hve see severl tmes You should be ble to crete such rgumets for yourself ow EXAMPLE 81 (Retur to the leky bucket Retur to the lekg bucket The force ppled ws cotuous d t ws ppled o the F(x = 6 (1 x2, 5 d t ws ppled o the tervl [, 5] where x s mesured feet So by Theorem 81 the work doe lftg the bucket s 5 Work = F(x dx = (1 x2 dx 5 = 6 (x x 5 15, ( 125, = , = 25 ft-lbs Pretty strghtforwrd! 82 Work Doe Emptyg Tk H b The followg tk problems volve pumpg lquds from oe heght to other d determg the mout of work requred to do t Here s the geerl questo Gve tk cotg lqud betwee heghts d b, how much work s requred to pump the lqud to heght H (Note: H my or my ot be the heght of the top of the tk See Fgure 81 To solve the problem we do the usul thg We subdvde d coquer Use regulr prtto of the tervl [, b] o the vertcl y-xs to equl wdth subtervls, ech of wdth y The subtervls re used to crete lyers (slces of lqud, ech of wdth y See Fgure 81 Wht propertes of the lyer re mportt determg the work doe lftg the lyer out? Tke momet before redg o We eed y y Fgure 81: A tk cotg lqud betwee levels d b The lqud s to be moved to heght H Also show s represettve lyer of the lqud t heght y C we determe how much work s requred to lft ths lyer to heght H? Wht wll ths work deped o? the volume of the lyer; the desty of the lyer; the dstce the lyer s moved The weght of the lyer s the force here (eg, weght s mesured pouds d s the desty of the lqud tmes the volume So the work to move the th lyer to heght H s W = Work to move the th lyer to heght H = Weght Dstce the lyer s moved = (Desty Volume Dstce the lyer s moved (8 OK, t s tme to use bt of lyss d otto we sw the secto o volume Let s deote the desty by D ( uts such s lbs/ft Worktex Verso: Mtchell-215/1/89:5:21

4 mth 11 pplcto: work As usul, let y deote the th prtto pot of the tervl [, b] The th lyer of lqud s pproxmtely s pproxmtely cylder (see Fgure 81 so ts volume s V = Are of the bse heght = Cross-sectol re heght = A(y y, where A(y deotes the the cross-sectol re of the tk t heght y d the heght of the lyer s y Flly, sce the th lyer s t heght y d hs to be moved to heght H, the the th lyer s moved dstce of H y Puttg ths ll together, we c rewrte the work to move the th lyer to heght H (8 s W = (Desty Volume Dstce the lyer s moved = D V (H y D[A(y y](h y (8 But ow you kow how the rest of the story goes To estmte the totl work we dd up the work to move ech lyer for ech subtervl, Totl Work = W D[A(y y](h y (85 We ow hve Rem sum 2 To mprove the pproxmto we do the stdrd thg: We let the umber of subdvsos get lrge d tke the lmt We fd 2 Tke secod to compre (82 d (85 The tke other to tcpte the rest of ths rgumet Totl Work = lm W = lm D[A(y y](h y = D A(y[H y] dy We re cert tht ths lmt of the Rem sums exsts d s, fct, the defte tegrl s log s we ssume tht the cross-sectol re A(y s cotuous fucto of y THEOREM 821 (Work Formul for Emptyg Tk Assume the cross-sectol A(y s cotuous fucto of the posto y d tht the desty of the cotets s costt D If the cotets of the tk to be moved le the tervl [, b], the the work doe to move ths mterl to heght H s Work = D A(y[H y] dy Cuto: The tk my ot be full, the cotets my be moved to heght H bove the tk, or the etre tk my ot be empted If the tk s beg flled from source t heght H (ether t the bottom of or below the tk, the the cotets must be moved to ech lyer heght y betwee d b so the dstce moved s y H rther th H y EXAMPLE 822 Here s smple frst exmple A bove-groud bckyrd swmmg pool hs the shpe of crculr cylder wth rdus of 1 ft d depth of 8 ft Assume the depth of the wter the pool s 5 ft Fd the work doe emptyg the pool by pumpg the wter over the top edge of the pool Note: The desty of wter s 625 lbs/ft See Fgure 82 Soluto We pply Theorem 821 Be ttetve to the dfferet heghts The lqud les betwee d 5 feet but hs to be moved to heght of H = 8 feet The Worktex 8 5 Fgure 82: The pool s flled to depth of 5 feet d the wter s to be pumped out over the top edge t heght 8 feet Verso: Mtchell-215/1/89:5:21

5 mth 11 pplcto: work 5 cross-sectos re crcles of rdus r = 1 feet So A(y = πr 2 = 1π (I most problems the cross-sectos wll vry So 5 Work = D A(y[H y] dy = 625 1π[8 y] dy = 625π (8y y2 5 2 [( = 625π 25 2 = 17, 1875π ft-lbs ] EXAMPLE 82 (A revso Suppose the swmmg pool Exmple 822 bove ws -groud so tht the top of the pool ws t groud level How would the work tegrl chge? Would the work doe emptyg the pool chge? Soluto Three thgs chge the set-up The lqud ow les betwee 8 d feet d hs to be moved to heght of H = feet The cross-sectos re the sme Work = D A(y[H y] dy = 625 1π[ y] dy 8 = 625π ( y2 2 8 = 625π [ 92 ] + 2 The work should d does rem the sme = 17, 1875π ft-lbs EXAMPLE 82 (A tk formed by rotto A tk s formed by rottg the rego betwee y = x 2, the y-xs d the le y = the frst qudrt roud the y xs The tk s flled wth ol wth desty 5 lbs/ft ( Fd the work doe pumpg the ol to the top of the tk (b Fd the work doe pumpg the ol to the top of the tk f there s oly 1 foot of ol the tk (c Suppose the tk s empty d s flled from hole the bottom to depth of feet Fd the work doe Soluto Fgure 8 shows sketch of the tk wth represettve crosssecto The crculr cross-secto formed by rottg the pot (x, y bout the y-xs hs rdus r = x So the cross-sectol re s s fucto of y A(y = πr 2 = πx 2 = πy, r = x (x, y Fgure 8: The crculr cross-secto formed by rottg the pot (x, y bout the y-xs hs rdus r = x Worktex Verso: Mtchell-215/1/89:5:21

6 mth 11 pplcto: work 6 ( We pply Theorem 821 Work = D A(y[H y] dy = 5 = 5 π(y[ y] dy ( π y y 2 dy = 5π (2y 2 y [( = 5π 2 6 ] = 16π ft-lbs (b If there s oly 1 foot of ol the tk, the oly chge s tht the upper lmt of tegrto s ow 1 sted of (Remember the lmts of tegrto represet the upper lower levels of the cotet of the tk So 1 Work = D A(y[H y] dy = 5 π(y[ y] dy = 5π (2y 2 y 1 [( = 5π 2 1 ] = 25π ft-lbs (c The key observto whe tk s flled from the bottom s tht ech lyer must be moved or lfted from groud level to ts fl heght H = y Sce the tk s oly flled to depth of feet, the lmts of tegrto re (bottom to ft So ths tme Work = D A(y[H y] dy = 5 π(y[y ] dy ( y = 5π [( = 5π 2 1 ] = 5π ft-lbs YOU TRY IT 81 How would the tegrl d work chge f the tk were full d the ol ws pumped to heght feet bove the top of the tk? (Aswer: 52 π ft-lbs EXAMPLE 825 (A more complcted problem A udergroud hemsphercl tk wth rdus 1 ft s flled wth ol of desty 5 lbs/ft Fd the work doe pumpg the ol to the surfce f the top of the tk s 6 feet below groud Soluto It wll be esest to set up the equto of the hemsphere f we thk of the top of the tk t heght d the pump the ol to heght of 6 feet See Fgure 8 The cross-sectos re crcles We wll be ble to determe the cross-sectol re oce we determe the rdus of the cross-secto The sem-crcle s prt of the crcle of rdus 1 cetered t the org whch hs equto x 2 + y 2 = 1 The rdus of cross-secto s the x-coordte of the pot (x, y tht les o the 6 1 Fgure 8: The tk wth ts posto re-mged d represettve lyer Worktex Verso: Mtchell-215/1/89:5:21

7 mth 11 pplcto: work 7 sem-crcle fourth qudrt (See Fgure 85 Thus, r = x = (1 2 y 2 Therefore the cross-sectol re s A(y = πr 2 = π[1 2 y 2 ] = π(1 y 2 We pply Theorem 821 Remember the lqud s pumped to heght H = 6 our re-cstg of the problem 1 1 y 12 y 2 Work = D A(y[H y] dy = 5 = 5 = 5π 1 1 π((1 2 y 2 [6 y] dy ( π 6 1y 6y 2 + y dy (6y 5y 2 2y + y 1 = 5π [ ( ] = 25, π ft-lbs Fgure 85: The rdus of the crosssecto t heght y s (1 2 y 2 YOU TRY IT 82 Set up the ew tegrl for ech modfcto of the exmple bove d determe the work requred ( How would the tegrl d work chge f the tk were oly 5 feet below groud? (Aswer: 875 π ft-lbs (b How would the tegrl d work chge f the top of the tk were t groud level? (Aswer: 125, π ft-lbs YOU TRY IT 8 Suppose tht tk s formed by rottg the rego the frst qudrt eclosed by y = (x + 2 2, the y-xs, d the x-xs bout the y-xs (see Fgure 86 ( Fd the work doe pumpg the ol to the top of the tk (Aswer: 5 π( y 2 2 ( y dy = 128 π ft-lbs (b Fd the work doe f the ol ws pumped to heght feet bove the top of the tk? (Aswer: 28 π ft-lbs (c Fd the work doe pumpg the ol to the top of the tk f there s oly 1 foot of ol the tk (Aswer: π ft-lbs EXAMPLE 826 (A more complcted problem A lrge gsole storge tk les feet udergroud It s the shpe of hlf-cylder (flt sde up ft rdus d 8 ft log If the tk s full of gs wth desty 6 lbs/ft, fd the work doe pumpg the gs to level 2 ft bove groud (x, y Fgure 86: The crculr cross-secto formed by rottg the pot (x, y bout the y-xs hs rdus r = x = y 2 Soluto It wll be esest to set up the equto of the hemsphere f we thk of the top of the tk t heght Sce the orgl tk ws feet udergroud d the cotets were to be pumped to 2 feet bove groud, our model we eed to pump the gs to totl of 5 feet bove groud See Fgure 87 The cross-sectos re rectgles of wdth 2x d legth 8 Sce x 2 + y 2 = o the sem-crcle, x = 16 y 2 So the cross-sectol re s A(y = 2x(8 = y 2 We pply Theorem 821 Remember the lqud s pumped to heght H = 5 our re-cstg of the problem Worktex Verso: Mtchell-215/1/89:5:21

8 mth 11 pplcto: work 8 5 x 2 + y 2 = 2 5 y x x x 2 + y 2 = 2 8 Fgure 87: Left: Thk of the tk top t groud level d the lqud beg pumped to heght H = 5 ft Rght: The wdth of the cross-secto t heght y s 2x = 2 2 y 2 Work = D A(y[H y] dy = 5 = 8 = (16 16 y 2 [5 y] dy ( 5 16 y 2 y 16 y 2 dy 16 y 2 dy 8 y 16 y 2 dy (86 The frst tegrl 16 y 2 dy represets the re of qurter crcle of rdus r = d so ts vlue (re s 2 π The secod tegrl (86 c be doe usg the substtuto u = 16 y 2 wth du = y dy Be creful to chge the lmts! 1 2 Cosequetly, (86 becomes 16 y 16 y 2 1 dy = = π Remember tht we hve ot yet developed tdervtve of 16 y 2, s we oted the secto o rc legth, so we hve to use the geometrc terpretto of the defte tegrl 16 here 2 u1/2 du = 1 u/2 Work = 16 y 2 dy 8 ( 6 = (π 8π ft-lbs = 6 y 16 y 2 dy YOU TRY IT 8 How would the tegrl d work chge f the tk the prevous problem hd the flt sde fcg dow? Ht: Put the ceter of the sem-crcle t the org The tk ow les bove the x-xs Wht heght s the gs pumped to ths tme? Now work out the swer (Aswer: 288π 512 ft-lbs YOU TRY IT 85 A tk the form of tructed coe s formed by rottg the segmet betwee (2, d (, roud the y-xs It s flled wth sludge (desty 8 lbs/ft If the sludge s pumped feet upwrds to tk truck, how much work ws requred? (Aswer: 96π ft-lbs YOU TRY IT 86 (Rotto bout o-xs le Ths tme tk the form of tructed coe s formed by rottg the segmet betwee (, d (2, roud the le x = It s flled wth sludge (desty 8 lbs/ft If the sludge s pumped feet upwrds to tk truck, how much work ws requred? (Aswer: 96π ft-lbs YOU TRY IT 87 (No-rotto A trgulr trough for cttle s 8 ft log The eds re trgles wth wth bse of ft d heght of 2 ft (but the vertex pots dow See Fgure 88 Fd the work doe by the cttle emptyg just the top foot of wter (desty 625 lbs/ft over the edge (Aswer: 5 ft-lbs Worktex 2 Verso: Mtchell-215/1/89:5:21 Fgure 88: Fd the work doe emptyg the top foot of the trough

9 mth 11 pplcto: work 9 YOU TRY IT 88 A cup shped tk s obted by rottg the curve y = x bout the y-xs where x 2 ( Assume the tk s full of wter (desty 625 lbs/ft? How much work s doe emptyg the tk by removg the wter over the top edge of the tk? (Aswer: 6π ft-lbs? (b How much work would be doe rsg the wter 2 feet bove the tk s top? (c Suppose the depth of the lqud the tk s 1 foot Fd the work requred to pump the lqud to the top edge of the tk YOU TRY IT 89 ( A coe-shped reservor hs 1 foot rdus cross the top d 15 foot depth If the reservor hs 9 feet of ol (desty 5 lbs/ft t, how much work s requred to empty t by brgg the wter to the top of the reservor? (Frst fgure out the equto of the le tht determes the coe (b Sme questo wth the reservor beg completely full (Aswer: 1125π ft-lbs? YOU TRY IT 81 (Expdg Gs Boyle s lw sys tht the pressure (force exerted by gs s versely proportol to the volume A qutty of gs wth tl volume of 2 cubc feet d pressure of 1 pouds per squre foot expds to volume of 5 cubc feet Fd the work doe Ht: p = k/v, so frst solve for k Next, the work doe by expdg gs s Aswer: 2(l 5 l 2 ft lbs W = V1 V YOU TRY IT 811 (Extr Credt A hevy rope s 6 feet log d hs desty of 15 lbs/ft It s hgg over the edge of buldg 1 ft tll Fd the work doe pullg the rope to the top of the buldg Ht: Model your swer lke tk problem Thk of the rope sectos Wht force must be ppled to move ech secto of the rope? Wht dstce must ech secto be used? OK, set up the tegrl d do t! p dv 8 Problems 1 Assume 25 ft-lb of work s requred to stretch sprg feet beyod ts turl legth Fd the work doe stretchg the sprg from 2 to ft beyod ts turl legth Ht: Frst fd the sprg costt (Aswer: k = 5/9 lbs/ft d 1/ ft-lbs 2 Let R be the rego the frst qudrt bouded by the x-xs, the y-xs, d the curve y = x 2 Rotte the R roud the y-xs to form slo tk If the tk s flled wth whet (1 lbs per cu ft how much work s doe rsg the whet to the top of the slo (As: 6 π ft lbs Let R be the rego the frst qudrt bouded by the y-xs d the curves y = x 2 d y = 6 x Rotte the R roud the y-xs to form tk If the tk s flled wth hoey (9 lbs per cu ft how much work s doe rsg the hoey to the top of the tk Note: you wll hve to dvde the work to two peces! Be creful of the lmts (As: 276π ft lbs A beer bll wth rdus 1 foot s locted wth ts ceter t the org It hs tp 5 ft log stckg up out the top If beer hs desty of 6 pouds per cubc foot, fd the work doe emptyg the beer bll Ht: Locte the ceter of the bll t the org (As: 12π ft-lbs 5 Ats excvte chmber udergroud tht s descrbed s follows: Let S be the rego the fourth qudrt eclosed by y = x, y = 1, d the y-xs; revolve S roud the y-xs ( Set up d smplfy the tegrl for the volume of the chmber usg the shell method (For future referece: As: π/5 Worktex Verso: Mtchell-215/1/89:5:21

10 mth 11 pplcto: work 1 (b [From exm] Suppose tht the chmber coted sol whch weghed 5 lbs per cubc foot How much work dd the ts do rsg the sol to groud level? (As: 5π/6 ft lbs If you hve tme, set up ech of the followg tegrls whch we wll do wth Mple 6 A smll frm elevted wter tk s the shpe obted from rottg the rego the frst qudrt eclosed by the curves y = x2, y = 8, d the y-xs bout the y-xs ( Fd the work lost" f the wter (625 lbs/ft leks oto the groud from hole the bottom of the tk (Aswer: 65π/ ft-lbs (b Fd the work lost" f the wter leks oto the groud from hole the sde of the tk t heght 9 feet (Aswer: 175π/ ft-lbs 7 The le segmet betwee the pots (1, d (, 1 s rotted roud the y-xs to form tructed cocl gs storge tk If the tk hs oly 1 foot of gsole (desty 6 lbs/ft t, set up the tegrl for the work doe to pump ll of the gs to groud level webwork: Clck to try Problems 95 through 98 Use guest log, f ot my course Worktex Verso: Mtchell-215/1/89:5:21