Introduction and Mathematical Concepts

Transcription

1 Basic Physics I Seleced Soled Problems from Cunell & Johnson Texbook 7 h Ediion Wiley Compiled by Associae Professor Dr. Jaafar Janan aka Dr JJ Faculy of Applied Sciences, Uniersiy Teknologi MARA, Shah Alam, Selangor, Malaysia hp://drjj.uim.edu.my HP: ; jjnia@salam.uim.edu.my Inroducion and Mahemaical Conceps. Vesna Vuloic suried he longes fall on record wihou a parachue when her plane exploded and she fell 6 miles, 551 yards. Wha is his disance in meers? We use he facs ha 1 mi = 580 f, 1 m = 3.81 f, and 1 yd = 3 f. Wih hese facs we consruc hree conersion facors: (580 f)/(1 mi) = 1, (1 m)/(3.81 f) = 1, and (3 f)/(1 yd) = 1. By muliplying by he gien disance d of he fall by he appropriae conersion facors we find ha d 6 mi 580 f 1 mi 1 m 3.81 f 551 yd 3 f 1 yd 1 m 3.81 f m 3. Bicycliss in he Tour de France reach speeds of 34.0 miles per hour (mi/h) on fla secions of he road. Wha is his speed in (a) kilomeers per hour (km/h) and (b) meers per second (m/s)? a. To coner he speed from miles per hour (mi/h) o kilomeers per hour (km/h), we need o coner miles o kilomeers. This conersion is achieed by using he relaion km = 1 mi (see he page facing he inside of he fron coer of he ex). b. To coner he speed from miles per hour (mi/h) o meers per second (m/s), we mus coner miles o meers and hours o seconds. This is accomplished by using he conersions 1 mi = 1609 m and 1 h = a. Muliplying he speed of 34.0 mi/h by a facor of uniy, (1.609 km)/(1 mi) = 1, we find he speed of he bicycliss is mi mi km Speed = h h 1 mi km 54.7 h b. Muliplying he speed of 34.0 mi/h by wo facors of uniy, (1609 m)/(1 mi) = 1 and (1 h)/(3600 s) = 1, he speed of he bicycliss is mi mi 1609m 1h m Speed = h h 1 mi 3600s s 4. Azelasine hydrochloride is an anihisamine nasal spray. A sandard size conainer holds one fluid ounce (oz) of he liquid. You are searching for his medicaion in a European drugsore and are asked how many milliliers (ml) here are in one fluid ounce. Using he following conersion facors, deermine he number P a g e 1

2 of milliliers in a olume of one fluid ounce:,, and. Muliplying an equaion by a facor of 1 does no aler he equaion; his is he basis of our soluion. We will use facors of 1 in he following forms: 1 gal 18 oz m 1 1 gal 1, since 1 gal = 18 oz 3 3, since m 3 = 1 gal 1 ml m 1, since 1 ml = 10 6 m 3 The saring poin for our soluion is he fac ha Volume = 1 oz Muliplying his equaion on he righ by facors of 1 does no aler he equaion, so i follows ha 1 gal m 1 ml Volume 1 oz oz 9.6 ml 18 oz 1 gal m Noe ha all he unis on he righ, excep one, are eliminaed algebraically, leaing only he desired unis of milliliers (ml). 5. The mass of he parasiic wasp Caraphracus cinus can be as small as. Wha is his mass in (a) grams (g), (b) milligrams (mg), and (c) micrograms (μg)? When conering beween unis, we wrie down he unis explicily in he calculaions and rea hem like any algebraic quaniy. We consruc he appropriae conersion facor (equal o uniy) so ha he final resul has he desired unis. a. Since grams = 1.0 kilogram, i follows ha he appropriae conersion facor is. Therefore, b. Since milligrams = 1.0 gram, kg g 1.0 kg g c. Since micrograms = 1.0 gram, g mg 1.0 g 5 mg P a g e

3 3 510 g g 1.0 g g 8. The olume of liquid flowing per second is called he olume flow rae Q and has he dimensions of [L] 3 /[T]. The flow rae of a liquid hrough a hypodermic needle during an injecion can be esimaed wih he following equaion: The lengh and radius of he needle are L and R, respeciely, boh of which hae he dimension [L]. The pressures a opposie ends of he needle are P and P 1, boh of which hae he dimensions of [M]/{[L] [T] }. The symbol ή represens he iscosiy of he liquid and has he dimensions of [M]/{[L][T]}. The symbol π sands for pi and, like he number 8 and he exponen n, has no dimensions. Using dimensional analysis, deermine he alue of n in he expression for Q In he expression for he olume flow rae, he dimensions on he lef side of he equals sign are [L] 3 /[T]. If he expression is o be alid, he dimensions on he righ side of he equals sign mus also be [L] 3 /[T]. Thus, he dimensions for he arious symbols on he righ mus combine algebraically o yield [L] 3 /[T]. We will subsiue he dimensions for each symbol in he expression and rea he dimensions of [M], [L], and [T] as algebraic ariables, soling he resuling equaion for he alue of he exponen n. We begin by noing ha he symbol and he number 8 hae no dimensions. I follows, hen, ha n R P P Q 3 1 L or 8 L T n M L LT M L L T n LT n LT L T L 3 L T n L L T n L L or L or L L L L n Thus, we find ha n = The deph of he ocean is someimes measured in fahoms. Disance on he surface of he ocean is someimes measured in nauical miles. The waer beneah a surface recangle 1.0 nauical miles by.60 nauical miles has a deph of 16.0 fahoms. Find he olume of waer (in cubic meers) beneah his recangle. The olume of waer a a deph d beneah he recangle is equal o he area of he recangle muliplied by d. The area of he recangle = (1.0 nauical miles) (.60 nauical miles) = 3.1 (nauical miles). Since 6076 f = 1 nauical mile and m =, he conersion facor beween nauical miles and meers is P a g e 3

4 6076 f m m 1 nauical mile 1 f 1 nauical mile The area of he recangle of waer in m is, herefore, (nauical miles) m 1 nauical mile m Since 1 fahom = 6 f, and 1 f = m, he deph d in meers is 16.0 fahoms The olume of waer beneah he recangle is 6 f 1 fahom m 1 f 1 = m ( m ) ( m) = 10. A spring is hanging down from he ceiling, and an objec of mass m is aached o he free end. The objec is pulled down, hereby sreching he spring, and hen released. The objec oscillaes up and down, and he ime T required for one complee up-and-down oscillaion is gien by he equaion, where k is known as he spring consan. Wha mus be he dimension of k for his equaion o be dimensionally correc? The dimension of he spring consan k can be deermined by firs soling he equaion T m/ k for k in erms of he ime T and he mass m. Then, he dimensions of T and m can be subsiued ino his expression o yield he dimension of k. Algebraically soling he expression aboe for k gies k 4 m/ T. The erm 4 is a numerical facor ha does no hae a dimension, so i can be ignored in his analysis. Since he dimension for mass is [M] and ha for ime is [T], he dimension of k is Dimension of k M T P a g e 4

5 KINEMATICS IN ONE DIMENSION 4. An 18-year-old runner can complee a 10.0-km course wih an aerage speed of 4.39 m/s. A 50-year-old runner can coer he same disance wih an aerage speed of 4.7 m/s. How much laer (in seconds) should he younger runner sar in order o finish he course a he same ime as he older runner? The younger (and faser) runner should sar he race afer he older runner, he delay being he difference beween he ime required for he older runner o complee he race and ha for he younger runner. The ime for each runner o complee he race is equal o he disance of he race diided by he aerage speed of ha runner (see Equaion.1). The difference in he imes for he wo runners o complee he race is 50 18, where Disance 50 and 18 (.1) Aerage Speed50-yr-old Aerage Speed18-yr-old Disance The difference in hese wo imes (which is how much laer he younger runner should sar) is Disance Aerage Speed Aerage Speed 50-yr-old 3 3 Disance 18-yr-old m m 64 s 4.7 m/s 4.39 m/s 5. The Space Shule raels a a speed of abou m/s. The blink of an asronau s eye lass abou 110 ms. How many fooball fields (lengh = 91.4 m) does he Shule coer in he blink of an eye? The disance raeled by he Space Shule is equal o is speed muliplied by he ime. The number of fooball fields is equal o his disance diided by he lengh L of one fooball field. The number of fooball fields is m / s s x Number = 9.1 L L 91.4 m 6. The hree-oed sloh is he slowes moing land mammal. On he ground, he sloh moes a an aerage speed of m/s, considerably slower han he gian oroise, which walks a m/s. Afer 1 minues of walking, how much furher would he oroise hae gone relaie o he sloh? AND In 1 minues he sloh raels a disance of x s 60 s = s = (0.037 m/s)(1 min) = 7 m 1 min P a g e 5

6 while he oroise raels a disance of x 60 s = = (0.076 m/s)(1 min) = 55 m 1 min The oroise goes farher han he sloh by an amoun ha equals 55 m 7 m = 8 m 7. A ouris being chased by an angry bear is running in a sraigh line oward his car a a speed of 4.0 m/s. The car is a disance d away. The bear is 6 m behind he ouris and running a 6.0 m/s. The ouris reaches he car safely. Wha is he maximum possible alue for d? In order for he bear o cach he ouris oer he disance d, he bear mus reach he car a he same ime as he ouris. During he ime ha i akes for he ouris o reach he car, he bear mus rael a oal disance of d + 6 m. From Equaion.1, ouris d (1) and bear d 6 m () Equaions (1) and () can be soled simulaneously o find d. Soling Equaion (1) for and subsiuing ino Equaion (), we find bear d 6 m ( d 6 m) d / d ouris ouris Soling for d yields: bear 6 m 1 d ouris 6 m 6 m d bear 6.0 m/s m/s ouris 5 m 8. In reaching her desinaion, a backpacker walks wih an aerage elociy of 1.34 m/s, due wes. This aerage elociy resuls because she hikes for 6.44 km wih an aerage elociy of.68 m/s, due wes, urns around, and hikes wih an aerage elociy of m/s, due eas. How far eas did she walk? AND Le wes be he posiie direcion. The aerage elociy of he backpacker is x x x x w e where w and e w e w e w e Combining hese equaions and soling for x e (suppressing he unis) gies P a g e 6

7 x e e m/s / m/s 1 / w xw m/s /.68 m/s 6.44 km 0.81 km 1 / The disance raeled is he magniude of x e, or 0.81 km. 9. A bicyclis makes a rip ha consiss of hree pars, each in he same direcion (due norh) along a sraigh road. During he firs par, she rides for minues a an aerage speed of 7. m/s. During he second par, she rides for 36 minues a an aerage speed of 5.1 m/s. Finally, during he hird par, she rides for 8.0 minues a an aerage speed of 13 m/s. (a) How far has he bicyclis raeled during he enire rip? (b) Wha is her aerage elociy for he rip? AND a. The oal displacemen raeled by he bicyclis for he enire rip is equal o he sum of he displacemens raeled during each par of he rip. The displacemen raeled during each par of he rip is gien by Equaion.: x. Therefore, 60 s x1 (7. m/s)( min) 9500 m 1 min 60 s x (5.1 m/s)(36 min) m 1 min 60 s x3 (13 m/s)(8.0 min) 600 m 1 min The oal displacemen raeled by he bicyclis during he enire rip is hen x 9500 m m 600 m m 4 b. The aerage elociy can be found from Equaion.. 4 x.6710 m 1min 6.74 m/s, due norh min 36 min 8.0 min 60 s 10. A golfer rides in a golf car a an aerage speed of 3.10 m/s for 8.0 s. She hen ges ou of he car and sars walking a an aerage speed of 1.30 m/s. For how long (in seconds) mus she walk if her aerage speed for he enire rip, riding and walking, is 1.80 m/s? The ime rip o make he enire rip is equal o he ime car ha he golfer rides in he golf car plus he ime walk ha she walks; rip = car + walk. Therefore, he ime ha she walks is walk = rip car (1) The aerage speed rip for he enire rip is equal o he oal disance, x car + x walk, she raels diided by he ime o make he enire rip (see Equaion.1); rip x car x rip walk P a g e 7

8 Soling his equaion for rip and subsiuing he resuling expression ino Equaion 1 yields The disance raeled by he car is x walk walk walk car x x car walk walk () car rip x car car, and he disance walked by he golfer is. Subsiuing hese expressions for x car and x walk ino Equaion gies walk carcar walkwalk rip car The unknown ariable walk appears on boh sides of his equaion. Algebraically soling for his ariable gies walk The ime ha he golfer spends walking is carcar rip rip car walk walk carcar ripcar 3.10 m/s8.0 s 1.80 m/s8.0 s 1.80 m/s 1.30 m/s rip walk 73 s 1. A spriner explodes ou of he saring block wih an acceleraion of +.3 m/s, which she susains for 1. s. Then, her acceleraion drops o zero for he res of he race. Wha is her elociy (a) a = 1. s and (b) a he end of he race? a 0 / 0 (Equaion.4) o find he spriner s final elociy a he end of he acceleraion phase, because her iniial elociy ( 0 m/s, We can use he definiion of aerage acceleraion since she sars from res), her aerage acceleraion a, and he ime ineral 0 are known. 0 a. Since he spriner has a consan acceleraion, i is also equal o her aerage acceleraion, so a.3 m/s Her elociy a he end of he 1.-s period is a 0 0 m/s.3 m/s 1. s.8 m/s 0 b. Since her acceleraion is zero during he remainder of he race, her elociy remains consan a.8 m/s. 13. A moorcycle has a consan acceleraion of.5 m/s. Boh he elociy and acceleraion of he moorcycle poin in he same direcion. How much ime is required for he moorcycle o change is speed from (a) 1 o 31 m/s, and (b) 51 o 61 m/s? P a g e 8

9 Since he elociy and acceleraion of he moorcycle poin in he same direcion, heir numerical alues will hae he same algebraic sign. For conenience, we will choose hem o be posiie. The elociy, acceleraion, and he ime are relaed by Equaion.4: a. Soling Equaion.4 for we hae a. 0 0 (+31 m/s) (+1 m/s) a +.5 m/s 4.0 s b. Similarly, 0 (+61 m/s) (+51 m/s) 4.0 s a +.5 m/s 14. For a sandard producion car, he highes road-esed acceleraion eer repored occurred in 1993, when a Ford RS00 Eoluion wen from zero o 6.8 m/s (60 mi/h) in 3.75 s. Find he magniude of he car s acceleraion. AND The magniude of he car's acceleraion can be found from Equaion.4 ( = 0 + a) as m/s 0 m/s a 8.18 m/s 3.75 s 15. A runner acceleraes o a elociy of 4.15 m/s due wes in 1.50 s. His aerage acceleraion is m/s, also direced due wes. Wha was his elociy when he began acceleraing? AND The iniial elociy of he runner can be found by soling Equaion.4 ( = 0 + a) for 0. Taking wes as he posiie direcion, we hae 0 a ( 4.15 m/s) ( m/s )(1.50 s) = m/s Therefore, he iniial elociy of he runner is 3.19 m/s, due wes. 16. An auomobile sars from res and acceleraes o a final elociy in wo sages along a sraigh road. Each sage occupies he same amoun of ime. In sage 1, he magniude of he car s acceleraion is 3.0 m/s. The magniude of he car s elociy a he end of sage is.5 imes greaer han i is a he end of sage 1. Find he magniude of he acceleraion in sage AND The elociy of he auomobile for each sage is gien by Equaion.4: 0 a. Therefore, a 0 m/s + a and a P a g e 9

10 Since he magniude of he car's elociy a he end of sage is.5 imes greaer han i is a he end of sage 1,.5. Thus, rearranging he resul for, we find ( a1 ) a 1.5a1 1.5(3.0 m/s ) 4.5 m/s 17. A car is raeling along a sraigh road a a elociy of m/s when is engine cus ou. For he nex wele seconds he car slows down, and is aerage acceleraion is. For he nex six seconds he car slows down furher, and is aerage acceleraion is. The elociy of he car a he end of he eigheensecond period is +8.0 m/s. The raio of he aerage acceleraion alues is. Find he elociy of he car a he end of he iniial wele-second ineral. According o Equaion.4, he aerage acceleraion of he car for he firs wele seconds afer he engine cus ou is a1 1f 10 1 (1) and he aerage acceleraion of he car during he nex six seconds is a f 0 f 1f () The elociy 1f of he car a he end of he iniial wele-second ineral can be found by soling Equaions (1) and () simulaneously. Diiding Equaion (1) by Equaion (), we hae Soling for 1f, we obain a1 ( ) / ( ) a ( ) / ( ) 1f f 10 f 1f f 1f 1 1f a a ( a / a ) a a ( a / a ) 1 1 f f f 1.50(1.0 s)(+8.0 m/s) (6.0 s)( 36.0 m/s) 1.50(1.0 s) 6.0 s m/s 18. A fooball player, saring from res a he line of scrimmage, acceleraes along a sraigh line for a ime of 1.5 s. Then, during a negligible amoun of ime, he changes he magniude of his acceleraion o a alue of 1.1 m/s. Wih his acceleraion, he coninues in he same direcion for anoher 1. s, unil he reaches a speed of 3.4 m/s. Wha is he alue of his acceleraion (assumed o be consan) during he iniial 1.5-s period? AND During he firs phase of he acceleraion, a1 1 P a g e 10

11 During he second phase of he acceleraion, = (3.4 m/s) (1.1 m/s )(1. s) =.1 m/s Then.1 m/s a1 1.4 m/s 1.5 s 19. In geing ready o slam-dunk he ball, a baskeball player sars from res and sprins o a speed of 6.0 m/s in 1.5 s. Assuming ha he player acceleraes uniformly, deermine he disance he runs. AND The aerage acceleraion of he baskeball player is a /, so m/s x a 1.5 s 4.5 m 1.5 s 0. A car is drien by a large propeller or fan, which can accelerae or decelerae he car. The car sars ou a he posiion x = 0 m, wih an iniial elociy of +5.0 m/s and a consan acceleraion due o he fan. The direcion o he righ is posiie. The car reaches a maximum posiion of x = +1.5 m, where i begins o rael in he negaie direcion. Find he acceleraion of he car. The car has an iniial elociy of 0 = +5.0 m/s, so iniially i is moing o he righ, which is he posiie direcion. I eenually reaches a poin where he displacemen is x = +1.5 m, and i begins o moe o he lef. This mus mean ha he car comes o a momenary hal a his poin (final elociy is = 0 m/s), before beginning o moe o he lef. In oher words, he car is deceleraing, and is acceleraion mus poin opposie o he elociy, or o he lef. Thus, he acceleraion is negaie. Since he iniial elociy, he final elociy, and he displacemen are known, Equaion.9 0 ax can be used o deermine he acceleraion. Soling Equaion.9 for he acceleraion a shows ha 0 0 m/s 5.0 m/s a 1.0 m/s x 1.5 m 1. A VW Beele goes from 0 o 60.0 mi/h wih an acceleraion of +.35 m/s. (a) How much ime does i ake for he Beele o reach his speed? (b) A op-fuel dragser can go from 0 o 60.0 mi/h in s. Find he acceleraion (in m/s ) of he dragser. The aerage acceleraion is defined by Equaion.4 as he change in elociy diided by he elapsed ime. We can find he elapsed ime from his relaion because he acceleraion and he change in elociy are gien. a. The ime ha i akes for he VW Beele o change is elociy by an amoun = 0 is (and noing ha m/s = 1 mi/h) P a g e 11

12 m / s 60.0 mi / h 0 m / s 0 1 mi / h a.35 m / s 11.4 s b. From Equaion.4, he acceleraion (in m/s ) of he dragser is m / s 60.0 mi / h 0 m / s 0 1 mi / h a 44.7 m / s s 0 s. (a) Wha is he magniude of he aerage acceleraion of a skier who, saring from res, reaches a speed of 8.0 m/s when going down a slope for 5.0 s? (b) How far does he skier rael in his ime? AND a. From Equaion.4, he definiion of aerage acceleraion, he magniude of he aerage acceleraion of he skier is m/s 0 m/s a 5.0 s m/s b. Wih x represening he displacemen raeled along he slope, Equaion.7 gies: x ( ) (8.0 m/s 0 m/s)(5.0 s) =.0 10 m 3. The lef enricle of he hear acceleraes blood from res o a elociy of +6 cm/s. (a) If he displacemen of he blood during he acceleraion is +.0 cm, deermine is acceleraion (in cm/s ). (b) How much ime does blood ake o reach is final elociy? We know he iniial and final elociies of he blood, as well as is displacemen. Therefore, Equaion.9 ax can be used o find he acceleraion of he blood. The ime i akes for he blood o reach i 0 final elociy can be found by using Equaion.7 a. The acceleraion of he blood is 1 x cm / s 0 cm / s a cm / s x.0 cm 1 b. The ime i akes for he blood, saring from 0 cm/s, o reach a final elociy of +6 cm/s is x.0 cm 0.15 s cm / s + 6 cm / s 0 P a g e 1

13 Newon s Laws of Moion 1. An airplane has a mass of and akes off under he influence of a consan ne force of. Wha is he ne force ha acs on he plane s 78-kg pilo? AND According o Newon s second law, he acceleraion is a = F/m. Since he pilo and he plane hae he same acceleraion, we can wrie F F or F m F PILOT m m m PILOT PILOT PLANE PLANE Therefore, we find N F PILOT 78 kg 93 N kg. A boa has a mass of 6800 kg. Is engines generae a drie force of 4100 N, due wes, while he wind exers a force of 800 N, due eas, and he waer exers a resisie force of 100 N due eas. Wha is he magniude and direcion of he boa s acceleraion? Newon s second law of moion gies he relaionship beween he ne force ΣF and he acceleraion a ha i causes for an objec of mass m. The ne force is he ecor sum of all he exernal forces ha ac on he objec. Here he exernal forces are he drie force, he force due o he wind, and he resisie force of he waer. We choose he direcion of he drie force (due wes) as he posiie direcion. Soling Newon s second law F ma for he acceleraion gies F 4100 N 800 N 100 N a 0.31 m/s m 6800 kg The posiie sign for he acceleraion indicaes ha is direcion is due wes. 3. In he amusemen park ride known as Magic Mounain Superman, powerful magnes accelerae a car and is riders from res o 45 m/s (abou 100 mi/h) in a ime of 7.0 s. The mass of he car and riders is. Find he aerage ne force exered on he car and riders by he magnes. F is equal o he produc of he According o Newon s second law, Equaion 4.1, he aerage ne force objec s mass m and he aerage acceleraion a. The aerage acceleraion is equal o he change in elociy diided by he elapsed ime (Equaion.4), where he change in elociy is he final elociy minus he iniial elociy 0. The aerage ne force exered on he car and riders is m/s 0 m/s F ma m kg N 7.0 s P a g e 13

14 4. During a circus performance, a 7-kg human cannonball is sho ou of an 18-m-long cannon. If he human cannonball spends 0.95 s in he cannon, deermine he aerage ne force exered on him in he barrel of he cannon AND The acceleraion is obained from x = a where 0 = 0 m/s. So Newon s second law gies a = x/ x 18 m F ma m 7 kg 900 N 0.95 s 5. A 15-g bulle is fired from a rifle. I akes s for he bulle o rael he lengh of he barrel, and i exis he barrel wih a speed of 715 m/s. Assuming ha he acceleraion of he bulle is consan, find he aerage ne force exered on he bulle. We can use he appropriae equaion of kinemaics o find he acceleraion of he bulle. Then Newon's second law can be used o find he aerage ne force on he bulle. According o Equaion.4, he acceleraion of he bulle is Therefore, he ne aerage force on he bulle is m/s 0 m/s a.8610 m/s s F ma (15 10 kg)( m/s ) 490 N 6. A 1580-kg car is raeling wih a speed of 15.0 m/s. Wha is he magniude of he horizonal ne force ha is required o bring he car o a hal in a disance of 50.0 m? AND The acceleraion required is m/s a.5 m/s x 50.0 m Newon's second law hen gies he magniude of he ne force as F = ma = (1580 kg)(.5 m/s ) = 3560 N P a g e 14

15 7. A person wih a black bel in karae has a fis ha has a mass of 0.70 kg. Saring from res, his fis aains a elociy of 8.0 m/s in 0.15 s. Wha is he magniude of he aerage ne force applied o he fis o achiee his leel of performance? According o Newon's second law of moion, he ne force applied o he fis is equal o he mass of he fis muliplied by is acceleraion. The daa in he problem gies he final elociy of he fis and he ime i akes o acquire ha elociy. The aerage acceleraion can be obained direcly from hese daa using he definiion of aerage acceleraion gien in Equaion.4. The magniude of he aerage ne force applied o he fis is, herefore, 8.0 m/s 0 m/s F ma m 0.70 kg 37 N 0.15 s 8. An arrow, saring from res, leaes he bow wih a speed of 5.0 m/s. If he aerage force exered on he arrow by he bow were doubled, all else remaining he same, wih wha speed would he arrow leae he bow? AND From Equaion.9, 0 ax Since he arrow sars from res, 0 = 0 m/s. In boh cases x is he same so or 1 1 a x a a a x a a Since F = ma, i follows ha a = F/m. The mass of he arrow is unchanged, and or F1 F1 F 1 1 F F F 5.0 m/s 35.4 m/s P a g e 15

16 WORK AND ENERGY 8. A kg crae is being pushed across a horizonal floor by a force ha makes an angle of 30.0 below he horizonal. The coefficien of kineic fricion is Wha should be he magniude of ha he ne work done by i and he kineic fricional force is zero? The ne work done by he pushing force and he fricional force is zero, and our soluion is focused on his fac. Thus, we express his ne work as W P + W f = 0, where W P is he work done by he pushing force and W f is he work done by he fricional force. We will subsiue for each indiidual work using Equaion 6.1 [W = (F cos θ) s] and sole he resuling equaion for he magniude P of he pushing force. According o Equaion 6.1, he work done by he pushing force is W P = (P cos 30.0 ) s = P s The fricional force opposes he moion, so he angle beween he force and he displacemen is 180. Thus, he work done by he fricional force is W f = (f k cos 180 ) s = f k s, so Equaion 4.8 indicaes ha he magniude of he kineic fricional force is f k = µ k F N, where F N is he +y magniude of he normal force acing on he crae. The free-body diagram shows he forces acing on he crae. Since here is no acceleraion in he erical direcion, he y componen of he ne force mus be zero: P 30.0º F N f k +x F mg Psin N mg Therefore, F mg Psin 30.0 N I follows, hen, ha he magniude of he fricional force is The work done by he fricional force is f k = µ k F N = µ k (mg + P sin 30.0 ) W f = f k s = (0.00)[( kg)(9.80 m/s ) P]s = (0.100P + 196)s Since he ne work is zero, we hae W P + W f = Ps (0.100P + 196)s = 0 Eliminaing s algebraically and soling for P gies P = 56 N. P a g e 16

17 9. A husband and wife ake urns pulling heir child in a wagon along a horizonal sidewalk. Each exers a consan force and pulls he wagon hrough he same displacemen. They do he same amoun of work, bu he husband s pulling force is direced 58 aboe he horizonal, and he wife s pulling force is direced 38 aboe he horizonal. The husband pulls wih a force whose magniude is 67 N. Wha is he magniude of he pulling force exered by his wife? AND According o Equaion 6.1, he work done by he husband and wife are, respeciely, Husband W ( F cos ) s H H H W W W Wife W ( F cos ) s Since boh he husband and he wife do he same amoun of work, ( F cos ) s ( F cos ) s H H W W Since he displacemen has he same magniude s in boh cases, he magniude of he force exered by he wife is cos H cos 58 FW FH (67 N) 45 N cos W cos A 55-kg box is being pushed a disance of 7.0 m across he floor by a force whose magniude is 150 N. The force is parallel o he displacemen of he box. The coefficien of kineic fricion is 0.5. Deermine he work done on he box by each of he four forces ha ac on he box. Be sure o include he proper plus or minus sign for he work done by each force. AND The applied force does work The fricional force does work W P = Ps cos 0 = (150 N)(7.0 m) = where F N = mg, so W f = f k s cos 180 = µ k F N s W f = (0.5)(55 kg)(9.80 m/s )(7.0 m) = The normal force and graiy do no work, since hey boh ac a a 90 angle o he displacemen. 11. A 100-kg car is being drien up a 5.0 hill. The fricional force is direced opposie o he moion of he ca and has a magniude of. A force is applied o he car by he road and propels he car forward. In addiion o hese wo forces, wo oher forces ac on he car: is weigh and he normal force direced perpendicular o he road surface. The lengh of he road up he hill is 90 m. Wha should be he magniude of, so ha he ne work done by all he forces acing on he car is +150 kj? AND The ne work done on he car is W T = W F + W f + W g + W N P a g e 17

18 Rearranging his resul gies WT F f mg sin 5.0 s W T = Fs cos f s cos 180 mgs sin F N s cos J 3 = 54 N kg 9.80 m/s sin N 90 m 1. A kg arrow is fired horizonally. The bowsring exers an aerage force of 65 N on he arrow oer a disance of 0.90 m. Wih wha speed does he arrow leae he bow? AND The work done on he arrow by he bow is gien by W = Fs cos 0 = Fs This work is conered ino kineic energy according o he work energy heorem. Soling for f, we find ha 1 1 f 0 W m m W 65 N 0.90 m 0 m/s 39 m/s m kg f Two cars, A and B, are raeling wih he same speed of 40.0 m/s, each haing sared from res. Car A has a mass of, and car B has a mass of. Compared o he work required o bring car A up o speed, how much addiional work is required o bring car B up o speed? AND The work required o bring each car up o speed is, from he workenergy heorem,. Therefore, B f 0 W m ( kg) (40.0 m/s) 0 m/s J B f 0 W m ( kg) (40.0 m/s) 0 m/s J The addiional work required o bring car B up o speed is, herefore, 14. A figher je is launched from an aircraf carrier wih he aid of is own engines and a seam-powered caapul. The hrus of is engines is. In being launched from res i moes hrough a disance of 87 m and has a kineic energy of a lif-off. Wha is he work done on he je by he caapul? The work done by he caapul W caapul is one conribuion o he work done by he ne exernal force ha changes he kineic energy of he plane. The oher conribuion is he work done by he hrus force of he plane s engines W hrus. According o he work-energy heorem (Equaion 6.3), he work done by he P a g e 18

19 ne exernal force W caapul + W hrus is equal o he change in he kineic energy. The change in he kineic energy is he gien kineic energy of J a lif-off minus he iniial kineic energy, which is zero since he plane sars a res. The work done by he hrus force can be deermined from Equaion 6.1 [W = (F cos θ) s], since he magniude F of he hrus is N and he magniude s of he displacemen is 87 m. We noe ha he angle θ beween he hrus and he displacemen is 0º, because hey hae he same direcion. In summary, we will calculae W caapul from W caapul + W hrus = KE f KE 0. According o he work-energy heorem, we hae W caapul + W hrus = KE f KE 0 Using Equaion 6.1 and noing ha KE 0 = 0 J, we can wrie he work energy heorem as follows: Soling for W caapul gies caapul W KE F cos s caapul f W F cos s KE Work done by hrus Work done by hrus J.3 10 N cos 0 87 m.5 10 J f 15. When a kg golf ball akes off afer being hi, is speed is 41 m/s. (a) How much work is done on he ball by he club? (b) Assume ha he force of he golf club acs parallel o he moion of he ball and ha he club is in conac wih he ball for a disance of m. Ignore he weigh of he ball and deermine he aerage force applied o he ball by he club. AND a. The work-energy heorem gies b. From he definiion of work so W = (1/)m f (1/)mo = (1/)( kg)(41 m/s) = W = Fs cos 0 F = W/s = (38 J)/( m) = N 17. The hammer hrow is a rack-and-field een in which a 7.3-kg ball (he hammer ), saring from res, is whirled around in a circle seeral imes and released. I hen moes upward on he familiar curing pah of projecile moion. In one hrow, he hammer is gien a speed of 9 m/s. For comparison, a. caliber bulle has a mass of.6 g and, saring from res, exis he barrel of a gun wih a speed of 410 m/s. Deermine he work done o launch he moion of (a) he hammer and (b) he bulle. The work done o launch eiher objec can be found from Equaion 6.3, he work-energy heorem,. a. The work required o launch he hammer is P a g e 19

20 W m m m (7.3 kg) (9 m/s) 0 m/s J f 0 f 0 b. Similarly, he work required o launch he bulle is f 0 W m (0.006 kg) (410 m/s) 0 m/s. 10 J 1 1 P a g e 0

21 MOMENTUM 1. One aerage force has a magniude ha is hree imes as large as ha of anoher aerage force. Boh forces produce he same impulse. The aerage force ime ineral does he aerage force ac? acs for a ime ineral of 3. ms. For wha According o Equaion 7.1, he impulse J produced by an aerage force F is JF, where is he ime ineral during which he force acs. We will apply his definiion for each of he forces and hen se he wo impulses equal o one anoher. The fac ha one aerage force has a magniude ha is hree imes as large as ha of he oher aerage force will hen be used o obain he desired ime ineral. Applying Equaion 7.1, we wrie he impulse of each aerage force as follows: J F and J F Bu he impulses J 1 and J are he same, so we hae ha he magniudes of he forces gies F F. Wriing his resul in erms of 1 1 F F F or 1 F The raio of he force magniudes is gien as F1/ F 3, so we find ha F ms 9.6 ms F. A 6.0-kg person, sanding on a diing board, dies sraigh down ino he waer. Jus before sriking he waer, her speed is 5.50 m/s. A a ime of 1.65 s afer she eners he waer, her speed is reduced o 1.10 m/s. Wha is he ne aerage force (magniude and direcion) ha acs on her when she is in he waer? AND According o he impulse-momenum heorem, Equaion 7.4, F m m, where F is he ne f 0 aerage force acing on he person. Taking he direcion of moion (downward) as he negaie direcion and soling for he ne aerage force F, we obain m f kg 1.10 m/s ( 5.50 m/s) F 1.65 s +165 N The plus sign indicaes ha he force acs upward. P a g e 1

22 3. A golfer, driing a golf ball off he ee, gies he ball a elociy of + 38 m/s. The mass of he ball is kg, and he duraion of he impac wih he golf club is. (a) Wha is he change in momenum of he ball? (b) Deermine he aerage force applied o he ball by he club. a. The change in momenum of he ball is he final momenum m f minus he iniial momenum m f, boh of which can be deermined. b. According o he impulse-momenum heorem, F = m f m 0, he ne aerage force F applied o he ball is equal o he change (m f m 0 ) in he ball s momenum, diided by he ime of impac. In his siuaion he ee upon which he ball is placed suppors is weigh, so he ne aerage force is FF, he aerage force ha he club applies o he ball. a. The change p in he ball s momenum is p m m m f 0 f kg 38 m/s 0 m/s 1.7 kg m/s b. Soling he impulse-momenum heorem for he aerage force gies kg38 m/s 0 m/s m f 0 F = 570 N s 4. A baseball ( ) approaches a ba horizonally a a speed of 40. m/s (90 mi/h) and is hi sraigh back a a speed of 45.6 m/s (10 mi/h). If he ball is in conac wih he ba for a ime of 1.10 ms, wha is he aerage force exered on he ball by he ba? Neglec he weigh of he ball, since i is so much less han he force of he ba. Choose he direcion of he incoming ball as he posiie direcion During he collision, he ba exers an impulse on he ball. The impulse is he produc of he aerage force ha he ba exers and he ime of conac. According o he impulse-momenum heorem, he impulse is also equal o he change in he momenum of he ball. We will use hese wo relaions o deermine he aerage force exered by he ba on he ball. The impulse J is gien by Equaion 7.1 as J = F, where F is he aerage force ha he ba exers on he ball and is he ime of conac. According o he impulse-momenum heorem, Equaion 7.4, he ne aerage impulse F is equal o he change in he ball s momenum; F m f m 0. Since we are ignoring he weigh of he ball, he ba s force is he ne force, so FF. Subsiuing his alue for he ne aerage force ino he impulse-momenum equaion and soling for he aerage force gies kg45.6 m/s kg40. m/s m m f 0 F N s where he posiie direcion for he elociy has been chosen as he direcion of he incoming ball. P a g e

23 5. A olleyball is spiked so ha is incoming elociy of m/s is changed o an ougoing elociy of 1 m/s. The mass of he olleyball is 0.35 kg. Wha impulse does he player apply o he ball? The impulse ha he olleyball player applies o he ball can be found from he impulse-momenum heorem, Equaion 7.4. Two forces ac on he olleyball while i s being spiked: an aerage force F exered by he player, and he weigh of he ball. As in Example 1, we will assume ha F is much greaer han he weigh of he ball, so he weigh can be negleced. Thus, he ne aerage force F is equal o F. From Equaion 7.4, he impulse ha he player applies o he olleyball is F m m f 0 Impulse Final Iniial momenum momenum f 0 m( ) (0.35 kg) ( 1 m/s) (+4.0 m/s) 8.7 kg m/s The minus sign indicaes ha he direcion of he impulse is he same as ha of he final elociy of he ball. 6. A space probe is raeling in ouer space wih a momenum ha has a magniude of. A rerorocke is fired o slow down he probe. I applies a force o he probe ha has a magniude of and a direcion opposie o he probe s moion. I fires for a period of 1 s. Deermine he momenum of he probe afer he rerorocke ceases o fire The impulse-momenum heorem (Equaion 7.4) saes ha he impulse of an applied force is equal o he change in he momenum of he objec o which he force is applied. We will use his heorem o deermine he final momenum from he gien alue of he iniial momenum. The impulse is he aerage force imes he ime ineral during which he force acs, according o Equaion 7.1. The force and he ime ineral during which i acs are gien, so we can calculae he impulse. According o he impulse-momenum heorem, he impulse applied by he rerorocke is f 0 J m m (7.4) The impulse is J F (Equaion 7.1), which can be subsiued ino Equaion 7.4 o gie F m m or m F m f 0 f 0 where m f is he final momenum. Taking he direcion in which he probe is raeling as he posiie direcion, we hae ha he iniial momenum is m 0 = kgm/s and he force is 6 F.0 10 N. The force is negaie, because i poins opposie o he direcion of he moion. Wih hese daa, we find ha he final momenum afer he rerorocke ceases o fire is mf F m0.010 N 1 s kg m/s kg m/s P a g e 3

24 7. A 46-kg skaer is sanding sill in fron of a wall. By pushing agains he wall she propels herself backward wih a elociy of 1. m/s. Her hands are in conac wih he wall for 0.80 s. Ignore fricion and wind resisance. Find he magniude and direcion of he aerage force she exers on he wall (which has he same magniude, bu opposie direcion, as he force ha he wall applies o her). The impulse ha he wall exers on he skaer can be found from he impulse-momenum heorem, Equaion 7.4. The aerage force F exered on he skaer by he wall is he only force exered on her in he horizonal direcion, so i is he ne force; F = F. From Equaion 7.4, he aerage force exered on he skaer by he wall is 46 kg 1. m/s 46 kg0 m/s mf m0 F 69 N 0.80 s From Newon's hird law, he aerage force exered on he wall by he skaer is equal in magniude and opposie in direcion o his force. Therefore, Force exered on wall = 69 N The plus sign indicaes ha his force poins opposie o he elociy of he skaer. P a g e 4