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by: Alverta Blick III


Alverta Blick III
GPA 3.66

Raymond Chastain

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Raymond Chastain
Class Notes
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This 37 page Class Notes was uploaded by Alverta Blick III on Tuesday October 13, 2015. The Class Notes belongs to PHYS 2101 at Louisiana State University taught by Raymond Chastain in Fall. Since its upload, it has received 30 views. For similar materials see /class/222994/phys-2101-louisiana-state-university in Physics 2 at Louisiana State University.




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Date Created: 10/13/15
Physzl l Hummurkz shum39un Spring 11 n tth th t however 1 P438 Eqn t t tt fer A t a r eos 40 0 the tnne 1t takes for the ball to hi the wall 15 Al 22 071 15 s v 25 0 eos 40 0 a The vemcal dutance ls Ay vn stnenne gtz 25 051nm 06015 79 811152 12 0 m b mmal value Vt Vn eos 40 0 19 2 ms e The vemcal eonnponent becomes using Eq 42 vy vn stnenegtds 0511140 0 9 801154 so mS d Since v gt 0 when the ball bus the Wall1thas not reachedthe hxghest point yet 2 We denote the two forces F and i Accordmg to Newton39s secondlaw 501 a 1n umt vector notauon i12oNfano1 12712 05m 30 own52 12 0 cos 3o 0 ms 75 00 mSz 1o 4mSz ThereforE 20 0 N 2 2 00kg is UUmsz 2 00kg7104ms2 732 0Nr 20 EN b The magnitude of 1 15 F2 1 1ya 732 0Nfe2o 8N2 38 2N y W21 e The angle that i makes Wxth the posmve x axxs 15 0 tan a FanF 720 8 Ne32 0 N 0 656 Consequent1y the angle 15 e then 33 0 or 33 0 180 213 Since both the x andy eonnponents are negatwe the eoneet result 15 213 An a1tematwe answens 147 213 e350 A1quot 17 cmd0 From the freerbody dAagram shown on the nght we a e 0 ZFW FceosayeFAeoss w FAsln5chm 7275 To solve for 751113 rst eonnpute w With FA 220 N Fa 170N and a 47 we get F 5 1054 A505 Ww 883 2 a 28 0 FE 170 N Subsmutmg the value mto the second force equahon we nd F F stnewcstng 22oms1n47 0 G70Nsm28 o 241N 4 The solutions to parts a and b have been combined here The freebody diagram is shown below with the tension of the string T the force of gravity mg and the force of the air F Our coordinate system is shown Since the sphere is motionless the net force on it is zero and the x and the y components of the equations are Tsin iF0 Tcos 67mg0 where 639 37 We answer the questions in the reverse order Solving T cos 6397 mg 0 for the tension we obtain Tmg cos 6 30 x 101 kg 98 msz cos 370 37 X 10 3 N Solving T sin 67F 0 for the force ofthe air F Tsin 6 37 gtlt10 3 N sin 370 22 gtlt10 3 N 5517 The freebody diagram of the problem is shown to the right Since the acceleration of the F V y block is zero the components of the Newton s A X second law equation yield Timgsin60 T 7 FN mg cos 639 0 lugging where T is the tension in the cord and F N is the 0 normal force on the block 9 m mgcose a Solving the rst equation for the tension in the string we nd T mg sin 6 85kg98mszsin 300 421v b We solve the second equation in part a for the normal force F N FN mgcos6 85 kg98 msz cos 30 72 N c When the cord is cut it no longer exerts a force on the block and the block accelerates The x component equation of Newton s second law becomes 7mgsin639 ma so the acceleration becomes a g sin6 98 ms2sin30 49 msz The negative sign indicates the acceleration is down the plane The magnitude of the acceleration is 49 ms Note The normal force F on the block must be equalto mg c055 so that the block xs m contact V th the surface ofthe melme at all tme When the cord xs cut the bloekhas an acceleration gems wheh m the hmxt a gt90 becomes 1 s 5 42 a The term deceleration means the aeeelehatmh veetons m the dAreetxon opposite to y acceleration he a 2 4 ms2 Newton39s Secondlaw leads to Temg ha m g a wheh yields m 7 3 kg forthe mass Repeatmg the above eomputatmh nowto solve forthe tension with a 2 4 ms2 ml of 5 course T 89N eomputatmh ths 15 to be expected 7 5 47 The heeebody dAagram not to scale forthe block s shown 5 the below F 15 the normal force exerted by the oor and mg t force ofgravxty 39 c055 ma the block anda 15 the x eonnponent of b as accelerauon We obtsnn 218m52 aFosSf120N05250a m 7 5 00kg Thxs 15 us accelerauon promdedxt remams 1n eontaet wth the oor Assummg 1t does we nd the value ofFNand1fFN15 posmve then the assumption 15 tme but 1fFN15 negauve then the b1oek leaves the oor They eonnponent ofNewton39s secondlaw becomes FN stna mg 0 FNmgeF stns 5 00 kg9 80 nns e 12 omstn2s 0 43 9 N Henee the blockremams onthe oor andxts accelerauon 15 a 218 nns2 b Ifos the mmmum force forwhmh the blockleaves the oor then F 0 and they 5 00kg9 80 nnf Fstnaemgw 113 116N stns E 1 t 1 a a F2055 116Neos25m 210m51 m 500kg 8 565 clockwxse b1oekA mgg e mg Ma where M mass ofthe system 24 0 kg Ths yxelds an aeee1enanon of a gmge mom11 63 nns2 3 AE Next we analyze the forces ust on b1oek c mgg e T an Thus the tension 15 Tmcg2m4 mM817 N 9 578 We take xuph111forthe quot121 0 kg box and xnghtward for the m 3 0 kg box so The uphill force on m 15 F andthe downhdl forces on 1t are Tand ng Sm a where 937 The only honzonta1 39 L L we nd FeTeWhg Sm a m wheh can be addedto obtam Femlg Sm a mmga Ths yxelds the aeeelenatmn a 12 e 1098sm37 153m 10 3 0 Thus the tensmn ls Tma 3 01534 s N 610 a The freerbody dAagram for the block s shown on the nght wahF being the force apphedto the block F the normal force othe oor on the block mg the force ofgravxty and f the force offncuon We take the 7 duecuon to be honzontal to the nght and the y duecuon to be up The equatmns for the x andthe y eonnponents ofthe force accordmg to Newto secondlaw are 2 t n39s ma y Fstnaw emg 0 NowMN and the second equation gives F mg 7 FsmS wheh ytelds f kOng Fstne This expresslonls substttuted form the rst equation to obtam F 05 semmgeF Sm a ma so the aeeelenahonts 5eos 51k stn 5y kg m a If 1 0 500 and uh 0 500 then the magmde of f has amaxlmum value of A AF 0 500xquot 0 500m Sm 20 049mg on the otherhand Feose 0 500mgc0520 0470ng Therefore nose qm andthe block remams stationary with a 0 b If h 0400 and k 0 300 then the magmtude of f has amaxlmum value of A h F 0 400Wg e 0 500m Sm 20v 0 332g In thls ease Fcose 0 500mg c0520 047cm gtfm Therefore the aeeeleratloh ofthe bloekrs F 7505 9 M 5m 571kg m 0 5009 80 msz50520 0 300srh 20 e0 name 80 ms 17 rhs2 r16 We use coordlnates auol Welghtrcomponents as hareateolrh Flg 5r18see Sample Problem 577 from the prewous ehapter a 1n thrs srtuauoh we take f to porht uphlll auolto be equal to rts maxrrhurh value rh wheh caserm AFN apphes where so 0 25 Applymg Newton39s seeohollaw to the bloek ofmass m Wg 8 2 kg m the x andy dlrectlons prooluees FmemgsrheerLM rm 0 F 7mg eos a wheh wrth a 20V leaols to Fm mgsrneetl cosS85N b Now we take f to porht downth auolto be equal to rts maxrrhurh value h wheh easerm MFNapplles where so 0 25 Applyrhg Newton39s secondlaw to the block ofmass m Wgr 8 2 kg h the r andy drreetrohs prooluees Fm2 rmgsrheeW ma 0 F mg eos a 0 wheh wrth a 20v leaols to m mgsln 5 a 2055 45N exact39 urreol uphlll but srhee we quote our results here to two slgmflcant gures 46 le a good ehoughquot auswer e Flnally we are deallng wrth klnetl fnctlon porhtrhg downhlll so that Femgsrhsek ma0 0 F 7 mg eos alohg wlth MNwhere so 0 15 bhhgs us to F mgsln51cos 39N 12 o 79 a Ifwe ehoose downhlll posrtwe theh Newton39s law gwes mgsrnaer T Wu fertd t 1 t fnmzm Now Frmum es utussloos pun A 14W Fmm4 5055 usmg Eq 642 apphes to b1oekd and fa tFm was t 1 n strarghtforward1y solved forthe tensxon T 13 N b 51m1ar1y ndng the value ofa 15 strmghtforward a gmsrnae 11s mgm4 W 1 6 nns2 13 o 45 a At the top the hxghest pomt 1n the erreu1ar mohon the seat pushes up on the student 1 556N 1 W667N The seat 15 pushmg up wrth a force that 15 srna11erthanthe student39s wexght and we say the student expenences a deerease n h15 apparentwexghtquot at the hxghest pomt Thus he feels hght quot M n 1owest pomt ofthe an 15 now choosmg upward as the posmve dueetron The Fems whee1 15 steaddy rotatmgquot so the value vaR 15 the same as 1n part a Thus F 2 N KW111N667 N778N R c mv R to444N39T hrFr Hh Mk V pomt we have We FNmv R whreh 1eads to N 5 N 7444 N 223 N d Sunuar1y the nonna1 force at the 1owest pont 15 now foundto be F 5 444 N kN 14 o 51 The Seerbody dagrarn forthe aup1ane ofmass m1s shown below We note that E 15 the force of aerodynarnre ha and a pomts nghtwards n the gure We a1so note that 151 v2 R 480 kmh where v 133 ms 2 v F sm 5 m7 e s mg where a 40 Ehmmatmg mass eom these equauons leads to r 1 1d t t whmh yrelds R e Vzg tan 15 o 5 M upper sthhg oh the ball i rs the tehsroh force othe 1ower strrhg and m rs the mass ofthe ball Note that the tehsrohrh the upper sthhg rs greater than the tehsroh m the lower sthhg It must b ahee the downward pull ofgrawty ahdthe force othe 1ower strrhg smug Venglh t 70 m ezcl a We take the x ddrecuon to be le ward toward the eehter of the erreuhrr orht and y un rd 39 law rs mv2 TLEOSSTLEOSS i where we the speed ofthe ball and R rs the radws ofxts orbrt They eorhpoheht rs T sin 7 pg o thug TFTrngsme w th m quot1 t equxlateral a 30 0 Thus 2 1350NW874N 5m 30 0 b The net force has magmtude Fmx TTcos 35 0N874Nyeos3o 0 37 9 N e The radms othe path 15 R 170 m2tan 30 0 147 m Usmg A Wine we shdthat the speed ofthe ball 15 RFmLsu 7 W445 134kg m a The duecuon of Fm 15 le ward rad1allymward circumference othe annular path dmded by 271 therefore R 0 94271 0 15 m The angle that the cord makes wnh the honmta1 15 how easle 0 a eosquotRL osquot015 m0 90 m 80 a pull ofgravxtymg Thus mg 0 40 N sms Note that we are usmg Tfortensxon not for the penod b The honzonta1 eonnponent ofthat tension nnust supply the eentnpets1 force Eq 6718 so we tune for one revolution 0 940 49 1 9 s 17 6 55 Forthe puekto remam at rest the magnitude ofthe tension force Tofthe cord must that keeps the puek m its eneular ohtnt so T szV Thus Mg szV We solve for the speed Phys 2101 Homework 1 Solution Spring 11 These solutions use the parameter values from the problems printed in the book not those that appear in your personal homework assignment The logic used to get to the answer is the same however 1 a Denoting the travel time and distance from San Antonio to Houston as T and D respectively the average speed is D 55 kmhT290 kmhT2 avgl f which should be rounded to 73 kn h s 725 kmh b Using the fact that time distancespeed while the speed is constant we nd D D avg DZ m 683kmh which should be rounded to 68 anh S c The total distance traveled 2D must not be confused with the net displacement zero We obtain for the twoway trip 2D 2 70 kmh savg 725 kmh 683 kInh d Since the net displacement vanishes the average velocity for the trip in its entirety is zero e In asking for a sketch the problem is allowing the student to arbitrarily set the distance D the intent is not to make the student go to an Atlas to look it up the student can just as easily arbitrarily set T instead of D as will be clear in the following discussion We brie y describe the graph with kilometersperhour understood for the slopes two contiguous line segments the rst having a slope of55 and connecting the origin to t1 x1 T2 SST2 and the second having a slope of 90 and connecting t1 x1 to T D where D 55 90T2 The average velocity from the graphical point of view is the slope of a line drawn from the origin to T D The graph not drawn to scale is depicted below x 2P220 a Taking derivatives of xt 12t2 7 2t3 we obtain the velocity and the acceleration functions Va 24 6t2 and at 24 7 12t with length in meters and time in seconds Plugging in the value I 3 yields x3 54 m b Similarly plugging in the value I 3 yields v3 18 ms c Fort 3 a3 712 msz d At the maximum x we must have v 0 eliminating the t 0 root the velocity equation reveals t 246 4 s for the time of maximum x Plugging t 4 into the equation for x leads to x 64 m for the largest x value reached by the particle e From d we see that the x reaches its maximum at t 40 s f A maximum v requires a 0 which occurs when t 2412 20 s This inserted into the velocity equation gives vmax 24 ms g From f we see that the maximum ofv occurs at t 2412 20 s h In part e the particle was momentarily motionless at t 4 s The acceleration at that time is readily found to be 24 7 124 724 msz i The average velocity is defined by Eq 22 so we see that the values ofx at t 0 and t 3 s are needed these are respectively x 0 and x 54 m found in part a Thus 18ms 3 0 3 P222 We use Eq 22 average velocity and Eq 27 average acceleration Regarding our coordinate choices the initial position of the man is taken as the origin and his direction of motion during 5 min S t S 10 min is taken to be the positive x direction We also use the fact that Ax vAt when the velocity is constant during a time interval At39 a The entire interval considered is At 8 7 2 6 min which is equivalent to 360 s whereas the subinterval in which he is moving is only At 8 5 3min 180 s His position at t 2 min is x 0 and his position at t 8 min is x vAf 22180 396 m Therefore 396 m 0 avg 110 m s 360 s V b The man is at rest at t 2 min and has velocity v 22 ms at t 8 min Thus keeping the answer to 3 significant figures 22 m s 0 am 000611 m s2 360 s c Now the entire interval considered is At 9 7 3 6 min 360 s again whereas the sub interval in which he is moving is At 9 5 4min 240 5 His position at t 3 minis x 0 and his position at t 9 min is x vAt 22240 528 m Therefore 528 m 0 vavg 147 ms 360 s d The man is at rest at t 3 min and has velocity v 22 ms at t 9 min Consequently Llan 22360 000611 ms2 just as in part b e The horizontal line near the bottom of this xvst graph represents the man standing at x 0 for 0 S t lt 300 s and the linearly rising line for 300 S t S 600 s represents his constantvelocity motion The dotted lines represent the answers to part a and c in the sense that their slopes yield those results X The graph of vvst is not shown here but would consist of two horizontal steps one at v 0 for 0 S t lt 300 s and the next at v 22 nVs for 300 S t S 600 s The indications ofthe average accelerations found in parts b and d would be dotted lines connecting the steps at the appropriate t values the slopes of the dotted lines representing the values of aavg 4 P233 The problem statement see part a indicates that a constant which allows us to use Table 21 a We take x0 0 and solve x vot at2 Eq 215 for the acceleration a 2x 7 vot12 Substituting x 240 m v0 560 kmh 1555 nVs and t 200 s we nd 2 240 1555n 200 a m 25H S 356n s2 200s or a 356 msz The negative sign indicates that the acceleration is opposite to the direction of motion of the car The car is slowing down b We evaluate v v0 at as follows v1555ms 356ms2 200s843ms which can also be converted to 303 kmh 5 P239 a h trarrrrsthe areaquot h lt h lt 12base x herght Thus the absolute value ofthe duplacement for one tram 1240 rhs5 s100 rh andthat ofthe othertram xs 1230 rhs4 s so rh ht r m The mmal gap between the trams was 200 rh and accordmg to our duplacement eorhputatrorrs the gap has narrowed by 160 rh Thus the answer rs 200 715040 rh for the duranoh ofthe matron between 1aunehmg an 2 47 We hegleet ar resrstahee d 79 8 rhs1 we take downwardto be the 7y dArecuon We use the t t s Handingquot so 1 Graph A Graph 5 Graph c Graph D Graph E Graph F Graph G Graph H RRFL a At the hrghest porht the veloerty ofthe ball vamshes Takmgyu 0 we set v 0 m vg Zgy Zgy Vn b 0 Applyrhg Eq 2715 to the ehtrre motion the use and the falloftota1umetgt 0 we have Vn yvnzeEgz 217 which using our result from part a produces t 64 s It is possible to obtain this without using part a s result39 one can nd the time just for the rise from ground to highest point from Eq 216 and then double it c SI units are understood in the x and v graphs shown In the interest of saving space we do not show the graph of a which is a horizontal line at 798 msz y 10 40 20 l a 5 3c 7 249 We neglect air resistance which justi es setting a 7g 798 ms2 taking down as the 7y direction for the duration of the motion We are allowed to use Table 21 with Ay replacing Ax because this is constant acceleration motion We are placing the coordinate origin on the ground We note that the initial velocity of the package is the same as the velocity of the balloon v0 12 nVs and that its initial coordinate is yo 80 m a We solve y y0 v0t gt2 for time with y 0 using the quadratic formula choosing the positive root to yield a positive value for t v0 v3 2gyo 12w122 29880 t 54s g 98 b If we wish to avoid using the result from part a we could use Eq 216 but if that is not a concern then a variety of formulas from Table 21 can be used For instance Eq 211 leads to v v0 igt 12 7 9854 7 41 ms Its nal speed is 41 Ms 8 278 In this solution we make use of the notation xt for the value of x at a particular t Thus xt 50t 10t2 with SI units meters and seconds understood a The average velocity during the rst 3 s is given by x3 x0 503 1032 0 g At 3 80ms b The instantaneous velocity at time t is given by v dxdt 50 20t in SI units At t 30 s v 50 2030 110 nVS x v g t Z t Figure A Frgure E e The rnstantaneous aeeeleratron at tnne t is gwen bya dvdt 20 rns2 It is eonstant so the aeeeleratton at anyttrne 15 20 rns2 d and E Wth T t A t A x U to t 3 Os X 240 m Its slope is the average veloetty dunng the rst 3s of rnotaon The dashed hne nut me mo nt ugt t I ttt M w M 9 3 28 Many ofthe operattons are done emetently on rnost modern graphtea1 ea1eu1ators thetr bulltrm usmg veetor rnantputatton and rectangular e po1ar shortcuts quot 1n thts so1utton we employ the tradduonay rnethods sueh as Eq 36 a The magnrtude of a tsa4 o nn2 93 0 m e 5 o m b The angle between a and the 47 ants 5 tanquot 73 0 m4 0 m sznskwtse from the ants de ned by i e The rnagnttude ofh 15 545 0m28 o m 2 10m d The angle between 5 and the 7 ants xs tan 8 0 nno 0 nn 53 737 The veetorts 37 e ab4 0m5 0 m 3e30 m8 onn 10nn5 orny The magmtude ofthxs veetons 15E1J10 nn2 5 0 m 11nn we mundto two sxgm cant gures 1n our results 0 The angle between the Vector desenbed 1n pan 3 andthe x axxs 15 tanquot5 0 m10 m 27 ye t J veetons 751 2 0 nn2 11nn 2 11nn thch 15 mterestmgly the same result as 1n pan Y e exaetl faet that a m h The angle between the Vector descnbedm part g andthe x axxs 15 tanquot11 m2 0 m 80 1 V i 11 veetons 112751 72 0m211m2 11nn desenbed 1n part 1 and the 7 dArecuon are tanquot 711 m 2 0 m 80 and 180 80 260 pan n 1 t atnnnt thxs result Since 575 711675 they pomt 1n opposxte anurparallel dArecuons the angle between 180 them 15 10 3 36 We applyEq 3730 anqu 3723 unanan at hens nneennn at As It ens m T annpnnent n17 nan ennnan 0139 Imam a 5x0 apy 7120 12 smce an otherter ms vamsh due to the fact that neither 2 nor E have any 2 eomponents Consequent1y we obtam 5 014 eye 5 02 012 2 012 b I b 12be ayby yxelds 3 02 0 5 04 0 26 c a1 30 20 50 40 3 a 13 5020 90 4046 d Several approaches are available In this solution we will construct a 5 unitvector and dot it take the scalar product of it with a In this case we make the desired unitvector by 13 201 40 a T l l 1202 402 We therefore obtain ab a g 3020 5040 258 1202 402 11 352 Ifwe wish to use Eq 35 directly we should note that the angles for QR and Squot are 100 2500 and 310 respectively if they are measured counterclockwise from the x axis a Using unitvector notation with the unit meter understood we have 100 cos250 i100sin250 quot01 Q120cos100 i120sin100 3 a R 800 cos 250 800sin250 g 900cos310 i900sin310 13QR 100 mil63 111 b The magnitude ofthe vector sum is 100 m2 163 m2 102 m c The angle is tan 1 163 m100 m m 9240 measured counterclockwise from the x axis 12 357 The three vectors are l 1 30 302012 12 20 4032012 J20 301012 a Since 072 073 0i l033930k we have 511 12923 30130320120 103012 0 306030 m2 b Using Eq 3 30 we obtain a x a 10 602012 Thus 511512 x513 30i302012 10i602012 301840 52 m3 c We found 62 62 inpart a Use oqu 330 then leads to 11x572c73 30i302012x0i 103012 11i903012m2 13 48 Our coordinate system has pointed east and pointed north The rst displacement is a 483 kmi and the second is 2 966 km a The net displacement is as 7 756 483 kmi 966 km which yields WAC lquot483 km2 966 km2 l08gtlt103 km b The angle is given by 966 km 483 km We observe that the angle can be alternatively expressed as 6340 south of east or 2660 east of south 6 tan 1 634 c Dividing the magnitude of FAC by the total time 225 h gives 483 kmi 966 km avg 225 h with a magnitude ing i 215 kinh2 429 kinh2 480 kmh d The direction of 17an is 2660 east of south same as in part b In magnitudeangle notation we would have 17an 480 kmh 4 634 215 kmhi 429 kin11 e Assuming the AB trip was a straight one and similarly for the BC trip then lFABl is the distance traveled during the AB trip and lid is the distance traveled during the BC trip Since the average speed is the total distance divided by the total time it equals 483 km 966 km 225 h 644 kmh 14 414 We adopt a coordinate system with pointed east and pointed north the coordinate origin is the agpole We translate the given information into unitvector notation as follows f 400 mi and 170 100 ms3 7 400 111 and 17 100 msi a Using Eq 42 the displacement A is A 7 70 400 mi400 111 with a magnitude 1A 1 K 400 m2 400 m2 566 m b The direction of A is 6tan 1 amp tan 1 450 or 135 Ax 400 m Since the desired angle is in the second quadrant we pick 13 5 45 north of due west Note that the displacement can be written as A 17 r70 566 4 135 in terms of the magnitude angle notation c The magnitude of 7an is simply the magnitude of the displacement divided by the time At 300 s Thus the average velocity has magnitude 566 m300 s 189 ms d Eq 48 shows that 7 points in the same direction as A ie 135 45 north of due west avg e Using Eq 415 we have am V 3 0333 ms2i 0333 ms2 The magnitude of the average acceleration vector is therefore equal to 1an i 0333 ms22 0333 ms22 0471 msz f The direction of an is 0333 ms2 0333 ms2 Since the desired angle is now in the rst quadrant we choose 45 and 2 avg 0tan 1 J45 or 135 points north of due east 15 426 a Using the same coordinate system assumed in Eq 422 we solve fory h hy0 v0 sin60t gt2 which yields h 518 m foryo 0 v0 420 ms 1 600 and t 550 s b The horizontal motion is steady so vx V0 v0 cos 1 but the vertical component of velocity varies according to Eq 423 Thus the speed at impact is v loan cosan2 unsrnan gt 27 4rns e We use Eq 4724 wrth v 0 andy may H Ea 67Sm 16 431 A rjease pornt We wrrte 3347 0 for the angle measured fromx srnee the angle grven m the p1ane39s speed at the moment ofrelease a We use Eq 4722 to nd Vn yeyn vn 5m 5 ze gz 07730 rn vn srne37 055 00 979 80 rns 5 00 s2 whreh yre1ds Vn 202 rns b The hunzunta1 arstaneetrave1eurs XVutc05 at 202 ms5 00 seose37 0v 806 rn e The x eornponent ofthe ve1oerty Just before rrnpaet 15 Vt Vneosat 202 rnseose37 0v 161 rns d The y eornponent of the ve1oerty Just before rrnpaet 15 v Vn 5m 50gt 202 ms 5m 737 0 e 9 80 m525 00 s171ms 17 4102 We assume the ball39s rnrtra1 ve1oerty 15 perpendmularto the p1ane ofthe net We ehoose coordmates so that xuyu 0 3 0 rn and Vx gt 0 note that v3 0 a To barely e1ear the net we have 1 2 7 1 2 2 yeyn vnyzeagz 2 24m 3 0rn 0739 8ms 2 whreh grves t 0 39 s for the urne rt 15 passmg over the net Thrs 15 plugged rnto the xrequauon to yre1dthe rnrnrrnurn rnrtra1 ve1oerty Vt 8 0 rn0 39 s 20 3 rns b We require y 0 and nd tfromy y0 voyt gt2 This value I 4230 m98ms2 078 s is plugged into the x equation to yield the maximum initial velocity vx 170 m078 s 217 n s 18 462 a The circumference is c 27W 2720 15 m 094 m b With T 60 sl200 0050 s the speed is v cT 094 m0050 s 19 ms This is equivalent to using Eq 435 c The magnitude ofthe acceleration is a vzr 19 ms2015 m 24 X 103 msz d The period ofrevolution is 1200 revmin 1 83 X 1071 min which becomes in SI units T 0050 s 50 ms 19 467 To calculate the centripetal acceleration of the stone we need to know its speed during its circular motion this is also its initial speed when it ies off We use the kinematic equations of projectile motion discussed in 46 to nd that speed Taking the y direction to be upward and placing the origin at the point where the stone leaves its circular orbit then the coordinates ofthe stone during its motion as a projectile are given by x vot and y gt2 since vuy 0 It hits the ground at x 10 m and y 720 m Formally solving the second equation for the time we obtain t l Z y g which we substitute into the first equation 9 i 52 2 v0 x 2y 10m 29mm 157 5 Therefore the magnitude of the centripetal acceleration is U2 157 2 a S160n 32 r 15 m 20 P468 We note that after three seconds have elapsed t2 7 t1 300 s the velocity for this object in circular motion of period T is reversed we infer that it takes three seconds to reach the opposite side ofthe circle Thus T 2300 600 s a Using Eq 435 V vT21t where vquot3002 4002 500 ms we obtain r 477 m The magnitude of the object s centripetal acceleration is therefore a vzr 524 msz b The average acceleration is given by Eq 4 15 A 17 41 3001 400 300400 200 ms2 i 267 msz 39 3 tz tl 500 200 J whrehrrnphes he 72 oo 72 572 3 33 rns2 represents F1 s nm x x U l t 1 2 x x x x kP g t t t t 5 a 7 x s s ope ons ahonzontal lme Thrs eonohtron rs true forgra hs 1 2 szerol p 3and7 he a ee1eratron at t1 s s posrtwe xfthe curvature ofthe graph rs eoneave whreh rs true for rs 1 e r only graph 1 out ofthe four rnentroned above Therefore the answer convex curvature Therefore the answens z e to the m t t t 50 the answer 5 6 d convex curvature Therefore the answerxs 4 e 22 1D kmematxes For eaeh partbeuowprekthe letteredpomts th r F r a the answers are E andE b Therefore the co wd mi 0 dt rnereasrng m pornts c andF rnrnon answer rsr Increase ofthe speed rneans that the steepness ofthe s1ope mereases at a eertam pomt Nut nnsmnn m 39hrFr the answets are E andE d wd mi 0 dt tneteastng tn potnts c andF Theref te the ephn n answ t ts F Increase ofthe speed ans that the steepness ofthe slope tneteases at a eettat pol wd F 4 4 wd T n tnt ts A potnt 0 Posttton and speed are both deeteastng tn potnt D Look above ferreason g Posttton ts deeteastng and speedts tneteastng lnpom h Aeeetetataon ts postttve 1f the eutvatttte othe positionrtlme plot ts concave Thatts true pt potnts 1313 andF t Aeeeletataon ts negattve xfthe eumatttte othe positionrtlme plot ts convex That ts tme pt potnts AB an Phys 2101 Homework 3 Solution Spring 11 These solutions use the parameter values from the problems printed in the book not those that appear in your personal homework assignment The logic used to get to the answer is the same however 1 l a From Table 21 we have v2 v3 2an Thus v JV 2an J24x107 ms2 2 36x10 ms20035 m 29gtlt107 ms b The initial kinetic energy is Kl mv 167 x10 27kg24 gtlt107 ms2 48 gtlt10 13 J The nal kinetic energy is l 2 l 27 7 2 713 Kf Emv E l67gtlt10 kg29gtlt10 ms 69gtlt10 J The change in kinetic energy is AK 69 X 10 13 J 748 X 10 13 J 21 X 10 13 J 2 F 713 We choose x as the direction of motion so a and I3 are negativevalued a Newton s second law readily yields F 85 kg 20 msz so that FlFll7gtlt102N b From Eq 216 with v 0 we have 37ms2 m34gtlt102m 0v 2an 3 Ax Altematively this can be worked using the workenergy theorem c Since I3 is opposite to the direction of motion so the angle between 1 and 67 Ax is 180 then Eq 77 gives the work done as W FAx 58 gtlt104 J d In thrs case Newton39s secondlaw yrelds F 85kge4 oms so that F F 34XIOZN s From Eq 2716 we now have 37rhs2 7 17x10 rh 24 Ums o The force f rs agam opposrte to the dArecuon ofmouon so the angle agam18 a so that 7r7leads o 7 Ax BXIO J promdes rhsrght rhto the eoh 7 The fact that thrs agrees wrth the result ofpart e eept of work FAX 3 P 7714 The forces are all eohstarrt so the tota1 work done by them rs grveh by W m Ax A We add the three veetors ndmg the x and y eorhpohehts ofthe net force Fersmso 0 F3eos35 0 e 3 00Ne4 00Nsm35 0 10 0Neos35 0 13N 7 F2505500 Fzsm350 MUNeos500 100Nsm350 The magmtude ofthe net force rs W W 3 82N The work done by the net force 15 W Fmd a 82N4 00m 15 3 J 3 HFquot whreh 15 expressed by Fm 4 F 717 We use I3 to denote the upward force exerted by the cable on the astronaut The force of the cable is upward and the force of gravity is mg downward Furthermore the acceleration of the astronaut is a g 10 upward According to Newton s second law the force is given by F mgma 3 Fmgamg in the same direction as the displacement On the other hand the force of gravity has magnitude Fg mg and is opposite in direction to the displacement a Since the force of the cable 13 and the displacement a are in the same direction the work done by I3 is llmgd 11 72 kg98 ms215 m 10 10 WFFd ll64X104 le2X104J b Using Eq 77 the work done by gravity is Wg ng mgd 72 kg98 ms215 m l058X104 J z llgtlt 104 J c The total work done is the sum of the two works W net WF Wg ll64X104J l058X104Jl06X103lelX103J Since the astronaut started from rest the workkinetic energy theorem tells us that this is her nal kinetic energy 3 d Since K mv2 herfmalspeed is vJ w54 ms m 72 kg Note For a general upward acceleration a the net work done is W net WF Wg Fd ngmgad mgdmad Since W AK mv2 2 by the workkinetic energy theorem the speed of the astronaut would net be v JZad which is independent of the mass of the astronaut 5 F 723 The fact that the applied force Fa causes the box to move up a frictionless ramp at a constant speed implies that there is no net change in the kinetic energy AK 0 Thus the work one by FA must be equal to me negauye work done by gravity W o m we have W Smce the box rs d duplaced yemeany upward by h o 1 r w w 1 mmgh a 00 kg9 80mSz0150 m4411 o P 7727 From Eq 7725 we see that the work done by me sprrng force rs gwer by w 1m 7x 2 The fact that some eorsrmr re 90 Nem9 0x103 Nm k 360N 40m Blor39k a mchcrl to spring 39 0539 iv 1 1 e I Fx negative 0 O 0 O 39 7 a Whenthe block moves mm A 5 0 cmto x3 0 cm we have m 9 0x10 Nm0 050 m2 70 030 my 7 2 J b Mexmg from A 5 0 cmto X73 0 cm we have 9 0x103 Nm0 050 m2 770 030 quot0217 2 J cMovmg from x 5 0 ernto X75 0 ernwe have W 79 0810Z Nrn0 050 rn2 770 050 my 0 J a Mowng from x 5 0 ernto X39 0 ernwe have 9 0x103 Nrn0 050 rn2 770 090 my 725 I 7 P 7730 Hooke39s law and the work done by a spnng rs ducussedm the ehapter We apply the workkrnetre energy theorern m the form of AK WRWN to the pornts m Frgure 7735 at x 1 0 rn andx 2 0 rnrespeetwe1y The apphed work 1 rs that due to the eonstant force 15 4 JP10 mk10m2 0 P2 0 rn 142 0 rn2 quot a U Km Mm f Elm L AL U a Srrnultaneous solutron leads to P 8 0 N b Sxmdarly we nd It 8 0 Nm 8 P 7733 There are two forees wrth nonzero work dunng rnotron ofthe bloek the eonstant oree Thnmwr and tht rnrnmhmr to the dASpIacement ofthe bloek so therr works are both zero Forces Also the net force is drawn which is the sum of the applied constant force F and the spring force Fs Fnet F 39 kx The maximum speed will be achieved at the position where the net force becomes zero equilibrium position because up to that point the net work done on the block is positive and therefore increases block s kinetic energy It is so because the force F is a constant force and is NOT equal and opposite to the spring force at any moment what is usually the case when one stretches the spring at constant slow speed Here the work done by the net force is represented by the area of a triangle above the x axis A er that position x0 the work done by the net force is negative and it takes the same negative area to get to the position where kinetic energy and speed of the object becomes the same as the initial speed which is in this case zero a Using the gure above the block will stop at the distance that is twice the distance where the block has maximum speed The block has maximum speed when the net force is zero or where kxo F So the position where the block stops is x 2 x0 2 Fk 230 N50 Nm 012 m Another approach to nd the position of zero speed is to use workkinetic energy theorem The block is at rest at the initial and nal positions so the work done by the net force is zero because A K 0 so W WF WS Fx kx2 0 3 x 2Fk which is the same as above b The work done by the applied force during displacement x is WF F x 30 N0 12 m 036 c Because the net work is zero the work done by the spring force is equal in magnitude and opposite in sign to the work done by the applied force therefore M Wa 036 J d From the forcedisplacement plot we see that the block will have maximum speed at the position x0 9V2 Fk which is halfthe distance where the block stops Therefore x0 x 2 012 m2 006 m the lunetre n r A respeet to x andthen set the resulung expresslon equal to zero 8K Bx K erkx n F kx0 xFx whlehlsthe same as above a Uslng workrkmeu energy theorem the net Workls Wm AK K e K K andthe nal 2 2 2 Frge 5 LJF F 30 2 7 7 7009 1t 2k 2k 250Nm luneue energy atposruon x ls K P 7737 a We rstrnultrply the vertreal suns by the mass so that rt beeornes a graph ofthe the tnangular and reetangular areas m the graph for 0 gr 4 applled force Now addlng glves 42 J for the work alone b Countrng th to be 30 J at x Andatx9 omtheworkrs121 a Equatron 7710 along wrth Eq 771 leads to speedv o 5 rns at x 4 0 rn Returnrng to the veetorporntrng m the n dlrectlon at x4 0 rn 3 Now uslng the result ofpan b and Eq 7710 along wth Eq 771 we ndthe speedls 5 5 rns at F7 0 rn Although rt has enpeneneeol sorne oleeelerauon dunng the 0 st7lnterva1lts veloerty veetor sull pornts m the n dlrectlon 0 Flnally uslng the result ofpart e and Eq 7710 along wth Eq 771 we ndrts speedv 3 5 rn t cenamly has enpeneneeol a slgn eant arnount of oleeeleratron dunng the 0 g x g 9 rnterval nonetheless rts veloerty veetor strll pornts m the x dlrectlon 10 P 738 a Uslng the workrkmetl energy theorern K 1lt39J n257x2dx02 52 0e2 023I b For a variable endpoint we have K f as a lnction of x which could be diiTerentiated to nd the extremum value but we recognize that this is equivalent to solving F 0 for x F0 3 25 x20 Thus K is extremized at x 25 zl6 m and we obtain 5 2 1 3 ltfKlJ0 25 x dx02525 J25 26 J Recalling our answer for part a it is clear that this extreme value is a maximum 11 P749 We have a loaded elevator moving upward at a constant speed The forces involved are gravitational force on the elevator gravitational force on the counterweight and the force by the motor via cable The total work is the sum of the work done by gravity on the elevator the work done by gravity on the counterweight and the work done by the motor on the system W W VK Wm Since the elevator moves at constant velocity its kinetic energy does not change and according to the workkinetic energy theorem the total work done is zero that is W AK 0 The elevator moves upward through 54 In so the work done by gravity on it is W mggd 1200 kg980 ms254 m 635 X 105 J The counterweight moves downward the same distance so the work done by gravity on it is WC mcgd950 kg980 n s2 54 m 503gtlt105 J Since W 0 the work done by the motor on the system is Wm WE WC635gtlt105 J 503gtlt105 J 132gtlt105 J This work is done in a time interval of At 3 0 min 180 s so the power supplied by the motor to li the elevator is W 132 5 p m 2w At 180 s 74 X 102 W 12 F 751 a The object s displacement is 1 ampf 411 800 mi600 m200 mk Thus Eq 78 gives 5 comes oo mU 00mg 00 quot90 coma 00m3201 b The average powens gtveh by Eq 142 e The dASLance from the coordmate ongm to the ma position 15 g 00 m22 00 mZ5 00 m 515m d ties 00 m2 9 00 m2U 00 m2 949 m Theh scalar dot pmduetts amp 3 3 oo mxes oo m2 oo mX4 oo m5 oo mU 00 m12 o m2 Thus the angle between the two vectors is E E eos 782 M s 169 49 12m rm he Wmgd0 25 kg9 8m52012m0 291 b The work done by the spring is by Eq 726 1 2 1 2 W2 Ekd E 250 Nm 012 m 18 J c The speed v of the block just before it hits the spring is found from the workkinetic energy theorem Eq 715 AK0 lmvl2 W1 W2 2 which yields V1 I2Wl W2 2029 J 18 J 35 I m V 025 kg d If we instead had V 7ms we reverse the above steps and solve for d Recalling the theorem used in part c we have 1 2 I I 1 I2 O Emvl W1 W2 mgd Ek which choosing the positive root leads to d39 mgquotng2 mkvl392 k which yields 6139 023 m In order to obtain this result we have used more digits in our intermediate results than are shown above so VI 12048 ms 3471 ms and v 6942 ms 14 F 776 a The component of the force of gravity exerted on the ice block of mass m along the incline is mg sin 6 where 9sin391091 15 gives the angle of inclination for the inclined plane Since the ice block slides down with uniform velocity the worker must exert a force I3 uphill with a magnitude equal to mg sin 9 Consequently F mg sin 0 45 kg98 ms2013995 27 gtlt102 N m b Since the downhill displacement is opposite to F the work done by the worker is W1 27 xroszrs m 40x102 J c Since the displacement has a vertically downward component of magnitude 091 m in the same direction as the force of gravity we nd the work done by gravity to be W 45 kg9 8 msz0 91m4 0 x102 d Srnee F 15 perpendmular to the dArecuon ofmohon ofthe b1oek and 20590 0 work done by the norrna1 force 15 W3 yEq 777 e The resu1tant force FM 15 zero smce there 15 no aeee1eratron Thus rts workrs zero as ean be hacked by addmg the above resu1ts WW2 3 15 P 7754 From Eq 7732 we see that the area m the graph 15 equwa1ent to the work done We 4 0 ms where vn aW1d1 1g mvn215 J we have K KHVVHltXGVVIltXlt2VV2ltXG 401 so that K the lunatic energy when x 3 0 rn 15 foundto equal 12 I b 1 wme r r A m I workrkmeu energy theorern Kr Kz WRMX 74 730 so that the requrrernent K1 8 o J leads to x 40m E t ho1o1 unh1 x 1 0 rn At that 1oeatron the kmetn energy 15 110Wnltm 7151201181


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