51. We apply Newton s second law (specifically, Eq. 52). F = ma = ma sin 20.0 = 1.0 kg 2.00 m/s sin 20.0 = 0.684N. ( ) ( )


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1 51. We apply Newon s second law (specfcally, Eq. 5). (a) We fnd he componen of he foce s ( ) ( ) F = ma = ma cos 0.0 = 1.00kg.00m/s cos 0.0 = 1.88N. (b) The y componen of he foce s ( ) ( ) F = ma = ma sn 0.0 = 1.0 kg.00 m/s sn 0.0 = 0.84N. y y (c) In unveco noaon, he foce veco s F = F ˆ + F ˆj = (1.88 N) ˆ + (0.84 N)j ˆ. y 5. We apply Newon s second law (Eq. 51 o, equvalenly, Eq. 5). The ne foce appled on he choppng block s F F F ne = 1 +, whee he veco addon s done usng unveco noaon. The acceleaon of he block s gven by a = F + F m d1 /. (a) In he fs case F1 + F = ( 3.0N) ˆ + ( 4.0 N) ˆj + ( 3.0N) ˆ + ( 4.0N) ˆj = 0 so a = 0. (b) In he second case, he acceleaon a equals F + F m 1 (( ) + ( ) ) + ( ) + ( ) ( ) 3.0N ˆ 4.0 N ˆj 3.0 N ˆ 4.0N ˆj = = (4.0m/s ) ˆj..0kg (c) In hs fnal suaon, a s F + F m 1 (( ) + ( ) ) + ( ) + ( ) ( ) 3.0N ˆ 4.0 N ˆj 3.0 N ˆ 4.0N ˆj = = (3.0m/s ). ˆ.0 kg 55. We denoe he wo foces F + F = ma, so F = ma F. 1 1 b F and 1 (a) In un veco noaon F 1 = 0. 0 Ng $ and F. Accodng o Newon s second law, a = 1.0 sn 30.0 m/s ˆ 1.0 cos 30.0 m/s =.00 m/s ˆ 10.4m/s ˆj. Theefoe, ( ) ( ) ˆj ( ) ( )
2 F = (.00kg) (.00 m/s ) ˆ + (.00 kg) ( 10.4 m/s ) ˆj ( 0.0 N) ˆ = 3.0 N ˆ 0.8 N ˆj. ( ) ( ) (b) The magnude of F s F = F + F = ( 3.0 N) + ( 0.8 N) = 38. N. y (c) The angle ha F makes wh he posve as s found fom an θ = (F y /F ) = [( 0.8 N)/( 3.0 N)] = 0.5. Consequenly, he angle s ehe 33.0 o = 13. Snce boh he and y componens ae negave, he coec esul s 13. An alenave answe s = To solve he poblem, we noe ha acceleaon s he second me devave of he poson funcon, and he ne foce s elaed o he acceleaon va Newon s second law. Thus, dffeenang ( ) = wce wh espec o, we ge d d = , = The ne foce acng on he pacle a = 3.40 s s d F = m ˆ = (0.150) [ (3.40) ] ˆ = ( 7.98 N) ˆ To solve he poblem, we noe ha acceleaon s he second me devave of he poson funcon; s a veco and can be deemned fom s componens. The ne foce s elaed o he acceleaon va Newon s second law. Thus, dffeenang 3 ( ) = wce wh espec o, we ge d = = d , 4.0 Smlaly, dffeenang y( ) = + wce wh espec o yelds dy d y = , = 18.0 (a) The acceleaon s ˆ ˆ d ˆ d y a = a + j ˆj ( 4.0 ) ˆ ( 18.0) ˆ a = y j. + = +
3 A = s, we have a = ( 1.8) ˆ + ( 18.0) ˆj wh a magnude of a = = + = a ( 1.8) ( 18.0) 4. m/s. Thus, he magnude of he foce s (b) The angle F o a = F / m makes wh F = ma = = (0.34 kg)(4. m/s ) 8.37 N. + s a 1 y m/s θ = an = an 47.0 o 133. = a 1.8 m/s We choose he lae ( 133 ) snce F s n he hd quadan. (c) The decon of avel s he decon of a angen o he pah, whch s he decon of he velocy veco: ˆ ˆ d ˆ dy ˆ v( ) = v j j ( ) ˆ ( ) ˆ + vy = + = + j. A = s, we have v( = s) = ( 3.88 m/s) ˆ + ( 5.0 m/s) ˆj. angle v makes wh + s Theefoe, he v 1 y m/s θv = an = an = 55.3 o 15. v 3.88 m/s We choose he lae ( 15 ) snce v s n he hd quadan (a) The slope of each gaph gves he coespondng componen of acceleaon. Thus, we fnd a = 3.00 m/s and a y = 5.00 m/s. The magnude of he acceleaon veco s heefoe a = (3.00 m/s ) + ( 5.00 m/s ) = 5.83 m/s, and he foce s obaned fom hs by mulplyng wh he mass (m=.00 kg). The esul s F = ma =11.7 N. (b) The decon of he foce s he same as ha of he acceleaon: θ = an 1 [( 5.00 m/s )/(3.00 m/s )] = We esolve hs hozonal foce no appopae componens. (a) Newon s second law appled o he as poduces F cosθ mg sn θ = ma.
4 Fo a = 0, hs yelds F = 5 N. (b) Applyng Newon s second law o he y as (whee hee s no acceleaon), we have F F sn θ mg cos θ = 0 whch yelds he nomal foce F N = N. N 54. Applyng Newon s second law o cab B (of mass m) we have a = Eo! g = 4.89 m/s. Ne, we apply o he bo (of mass m b ) o fnd he nomal foce: F N = m b (g + a) = 17 N We apply Newon s second law fs o he hee blocks as a sngle sysem and hen o he ndvdual blocks. The + decon s o he gh n Fg (a) Wh m sys = m 1 + m + m 3 = 7.0 kg, we apply Eq. 5 o he moon of he sysem n whch case, hee s only one foce T = + T $. Theefoe, 3 3 T3 = msysa 5.0 N = (7.0 kg) a whch yelds a = m/s fo he sysem (and fo each of he blocks ndvdually). (b) Applyng Eq. 5 o block 1, we fnd ( )( ) T1 = m1a = 1.0kg m/s = 11.N. (c) In ode o fnd T, we can ehe analyze he foces on block 3 o we can ea blocks 1 and as a sysem and eamne s foces. We choose he lae. ( ) ( )( ) T = m1 + m a = 1.0 kg kg m/s = 34.9 N Usng Eq. 78 (and Eq. 33), we fnd he wok done by he wae on he ce block: W = F d = (10 N) ˆ (150 N) ˆj (15 m) ˆ (1 m) ˆj = (10 N)(15 m) + ( 150 N)( 1 m) J. = 70. (a) Usng noaon common o many veco capable calculaos, we have (fom Eq. 78) W = do([0.0,0] + [0, (3.00)(9.8)], [ º]) = J. (b) Eq (along wh Eq. 71) hen leads o v = (1.31 J)/(3.00 kg) = m/s.
5 734. Fom Eq. 73, we see ha he aea n he gaph s equvalen o he wok done. Fndng ha aea (n ems of ecangula [lengh wh] and angula base hegh] aeas) we oban [ 1 W = W0 + W 4 + W4 + W 8 = ( ) J = 5 J. < < < < < < < < (a) The compesson of he spng s d = 0.1 m. The wok done by he foce of gavy (acng on he block) s, by Eq. 71, c h W1 = mgd = ( 0. 5 kg) 9.8 m / s (0.1 m) = 0. 9 J. (b) The wok done by he spng s, by Eq. 7, W = 1 1 kd = ( N / m) (0.1 m) =. J. (c) The speed v of he block jus befoe hs he spng s found fom he wokknec enegy heoem (Eq. 715): whch yelds K = 0 1 mv = W + W 1 v ( )( W + W ) ( )(0.9 J 1.8 J) m 0.5 kg 1 = = = 3.5 m/s. ' (d) If we nsead had v = 7 m/s, we evese he above seps and solve fo d. Recallng he heoem used n pa (c), we have mv = W + W = mgd kd 1 whch (choosng he posve oo) leads o mg + m g + mkv d = k whch yelds d = 0.3 m. In ode o oban hs esul, we have used moe dgs n ou ' nemedae esuls han ae shown above (so v = m/s = m/s and v =.94 m/s) Thee s no acceleaon, so he lfng foce s equal o he wegh of he objec. We noe ha he peson s pull F s equal (n magnude) o he enson n he cod. (a) As ndcaed n he hn, enson conbues wce o he lfng of he canse: T = mg. Snce F = T, we fnd F = 98 N.
6 (b) To se 0.00 m, wo segmens of he cod (see Fg. 744) mus shoen by ha amoun. Thus, he amoun of sng pulled down a he lef end (hs s he magnude of d, he downwad dsplacemen of he hand) s d = m. (c) Snce (a he lef end) boh F and d ae downwad, hen Eq. 77 leads o W = F d = (98 N) (0.040 m) = 3.9 J. (d) Snce he foce of gavy F g (wh magnude mg) s oppose o he dsplacemen d c = m (up) of he canse, Eq. 77 leads o W = F d = (19 N)(0.00 m) = 3.9 J. g c Ths s conssen wh Eq snce hee s no change n knec enegy. 8. We denoe m as he mass of he block, h = 0.40 m as he hegh fom whch dopped (measued fom he elaed poson of he spng), and he compesson of he spng (measued downwad so ha yelds a posve value). Ou efeence pon fo he gavaonal poenal enegy s he nal poson of he block. The block dops a oal dsance h +, and he fnal gavaonal poenal enegy s mg(h + ). The spng poenal enegy s 1 k n he fnal suaon, and he knec enegy s zeo boh a he begnnng and end. Snce enegy s conseved K + U = K + U f f 0 = mg( h + ) + 1 k whch s a second degee equaon n. Usng he quadac fomula, s soluon s b g mg ± mg + mghk = k. Now mg = 19. N, h = 0.40 m, and k = 190 N m, and we choose he posve oo so ha > = 190 b gb gb g = 0.10 m The wok equed s he change n he gavaonal poenal enegy as a esul of he chan beng pulled ono he able. Dvdng he hangng chan no a lage numbe of nfnesmal segmens, each of lengh dy, we noe ha he mass of a segmen s (m/l) dy and he change n poenal enegy of a segmen when s a dsance y below he able op s
7 du = (m/l)g y dy = (m/l)gy dy snce y s negavevalued (we have +y upwad and he ogn s a he ableop). The oal poenal enegy change s z 1 mg 0 mg U = y dy = ( L L L 4) = mgl 3. L/ 4 The wok equed o pull he chan ono he able s heefoe W = U = mgl/3 = (0.01 kg)(9.8 m/s )(0.8 m)/3 = J We use Eq o oban and Eq. 78 o ge Eh = f d = (10 N)(5.0 m) = 50 J k W = Fd = (.0 N)(5.0 m) = 10 J. Smlaly, Eq gves W = K + U + E 10 = 35+ U + 50 h whch yelds U = 75 J. By Eq. 81, hen, he wok done by gavy s W = U = 75 J (a) Wh P = 1.5 MW = W (assumed consan) and =.0 mn = 30 s, he wokknec enegy heoem becomes The mass of he locomove s hen 1 W = P = K = mdv f v. P W 30s m = = v v 5m s 10 m s f bgc hb g b g b g c = (b) Wh abay, we use P = m v v ho solve fo he speed v = v() as a funcon of me and oban n SI uns (v n m/s and n s). kg. P vbg v bg bgc h = + = 10 + = m (c) The foce F() as a funcon of me s
8 n SI uns (F n N and n s). P Fbg = v bg = (d) The dsance d he an moved s gven by 30 1/ 3/ ( ) m. 0 0 d = v = + = + = 9 0
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