BI TP VT L I CNGBi 1: Cho phng trnh chuyn ng ca cht im l :x = Acos ty = Bsin tz = 0Trong A, B, l cc hng s.Hy tm phng trnh qu o, vn tc v gia tc ca cht im.HD : qu o l mt ellip c cc bn trc l A v B nm trong mt phng xOy :z = 0= - A sin t + B cos t= - A 2cos t - B 2sin t = - 2: gia tc hng vo tm ellip v t l vi bn knh . Bi2.T cao h = 25m mt vt c nm theo phng nm ngang vi vn tc ban u vo = 15m/s. Hy xc nh :a) Qu o ca vtb) Thi gian chuyn ng ca vt t khi nm cho ti khi chm t.c) Gia tc ton phn, gia tc tip tuyn, gia tc php tuyn ca vt khi chm t.d) Bn knh cong ca qu o khi vt chm t.HD : qu o parapola) y = b) t == 2,26sc) a=g=9,8m/s2 ; at= =8,112m/s2 ; an= = 5,6m/s2d) R = =122,7m)Bi 3. Mt qu bng c nm vi vn tc ban u vo theo mt gc nghing so vi b mt ca mt mt phng nghing. Mt phng nghing lm vi mt phng ngang mt gc .a) Hy xc nh khong cch dc theo mt phng nghing t im nmcho ti khi qu bng chm mt phng nghing theo vo, g,, .b) Vi gc no khi nm th khong cch trn t gi tr cc i.( S : a) ( )[tan( +) - tan ]b))Bi 4.Mt cu th bng r b phm li khi c gng nm bng vo r ca i bn v c hng hai qu nm pht. Theo phng nm ngang t tm ca r n im nm pht l 4,21m v cao ca r l 3,05m tnh t mt sn. Trong ln nm pht th nht cu th nm qu bng theo mt gc 35o so vi phng nm ngang vi vn tc ban u vo=4,88m/s2. Khi bt u ri khi tay cu th th qu bng cao 1,83m so vi mt sn. Ln nm ny qu bng khng lt vo r. Gi s b qua sc cn ca khng kh.a) Hi cao cc i m qu bng t c.b) xa qu bng t c theo phng nm ngang khi ri chm t.c) Trong ln nm pht th hai cao ban u v gc nghing ca qu bng khi nm cng vn gi nguyn nh trong ln nm u tin tc l 1,83m v 35o. Ln ny qu bng i vo tm r. Hi vn tc ban u ca qu bng ln ny l bao nhiu?d) cao cc i ca qu bng t c trong ln nm th hai.(S : a) 2,32m; b) 3,84m; c) 8,65m/s; d) 3,09m. )Bi 5. Mt cht im chuyn ng trn qu o trn bn knh bng 50m. qung ng i c trn qu o c cho bi cng thc : s = -0,5t2+10t+10 (m) Tm gia tc php tuyn, gia tc tip tuyn v gia tc ton phn ca cht im lc t = 5(s).(S : at = -1m/s2; an = 0,5m/s2; a = 1,12m/s2)Bi 6.Mt vt A c t trn mt mt bn nm ngang. Dng mt si dy, mt u buc vo A cho vng qua rng rc v u kia ca si dy buc vo vt B sao cho vt B ri khng ma st thng ng t trn xung. Cho bit mA= 2kg, h s ma st gia A v mt bn l k=0,25; gia tc chuyn ng ca h l a= 4,9m/s2. Hy xc nh :a) Khi lng mB. b) Lc cng ca dy.( S : a/ mB= 3kg; b/ T=14,7N)Bi 7. Mt vt trt t nh mt mt phng nghing hp vi mt phng nm ngang mt gc . H s ma st gia vt v mt phng nghing l k, vn tc ban u ca vt bng 0. Vt trt ht mt phng nghing sau thi gian t. Tnh chiu di l ca mt phng nghing.(S : l =1/2.g(sin -kcos )t2 )Bi 8. Cho hai vt A v B c mc nh hnh di. Vt A c t nm trn mt phng nghing c h s ma st k=0,2. B qua khi lng ca rng rc v dy. Cho bit mA=1kg, lc cng ca si dy T=9,91N; g = 9,8m/s2; = 30o. Hy tnh gia tc ca h.( S : a=3,24m/s2) Bi 9. Cho hai vt m1 v m2 nh c mc hnh bn vi m1=m2=1kg. B qua ma st, khi lng ca cc rng rc v dy. Xc nh gia tc ca cc vt m1, m2 v lc cng ca si dy. Cho g = 9,8m/s2. ( S : a1= = g = 3,92m/s2; a2 = g =1,96m/s2T= m2g 5,9N )Bi 10. Trn mt toa tu khi lng M c hai vt m1 v m2 c mc nh hnh di. Cho bit h s ma st gia vt m1 v mui tu l k. B qua khi lng ca dy v rng rc cng nh ma st gia chng. Tc dng mt lc ytheo phng nm ngang lm cho toa tu chuyn ng trn ng ray. B qua lc ma st ln gia toa tu v ng ray. Hi lc yphi c ln bng bao nhiu cho khi toa tu chy m hai vt m1 v m2 vn ng yn so vi toa tu?Bi gii : gip cc bn c th hiu r vic p dng cc nh lut ng lc hc Niu-tn trong cc h qui chiu khc nhau cng nh vai tr ca lc qun tnh chng ta s gii bi ton ny trong hai h qui chiu khc nhau :- H qui chiu ng yn (chng hn nh sn ga) - H qui chiu chuyn ng (toa tu)1- Gii bi ton trong h qui chiu ng yn :(sn ga)y l mt h qui chiu qun tnh (gn ng), ta nhn thy vt m1 chuyn ng vi gia tc(l gia tc ca con tu) cn vt m2 ng yn khng chuyn ng theo phng thng ng (trong thc t m2 chuyn ng theo phng nm ngang vigia tc nhng ta khng quan tm n chuyn ng ny ca m2 m ch quan tm chuyn ng ca m2 theo phng thng ng!). Ta qui c chn chiu dng ca trc ta nm ngang hng t tri qua phi.*Xt vt m1 :C hai lc tc dng ln m1 : lc cngca si dy hng t tri qua phi, lc Fms = km1g hng t phi qua tri. Hiu ca hai lc ny gy ra chuyn ng c gia tcca vt m1 nn theo nh lut II Niu-tn ta c: T-km1g = m1a (1)*Xt vt m2 :Theo phng thng ng vt m2 chu tc dng ca hai lc : lc cngca si dy hng thng ng ln trn v trng lng ca n 2 hng thng ng xung di. V theo u bi th m2 ng yn theo phng ny nn theo nh lut II Niu-tn ta c:T-P2 = 0 hay T= P2 = m2g.Thay gi tr T= m2g vo (1), ta tm c a =Lc ytc dng ln h gm toa tu v hai vt m1, m2 v gy cho h gia tcnn := (m1+m2+M)T , ln ca lc yl : F = (m1+m2+M) (2)2- Gii bi ton trong h qui chiu chuyn ng : (toa tu)V toa tu chuyn ng thng vi gia tc lnn y l mt h qui chiu khng qun tnh. Trong h qui chiu ny v hnh thc ta cng c th p dng nh lut II Niu-tn nhng khi trong cc lc tc dng ln vt ta phi k thm c lc qun tnh.*Xt vt m1 :Trong h qui chiu ny vt m1 ng yn. Cc lc tc dng ln vt m1 gm :- Lc cngca si dy hng sang phi.- Lc ma st Fms=km1g hng sang tri.- Lc qun tnh qt= -m1 hng sang tri.V vt ng yn nn tng ca cc lc ny phi bng 0, tc l ;T-km1g-m1a = 0 T a =(3) *Xt vt m2 :Trong h qui chiu ny m2 ng yn. Theo phng nm ngang m2 chu tc dng ca lc qun tnh m2hng t phi qua tri lm cho m2 p st vo thnh toa. Phn lc ca thnh toa s trit tiu lc qun tnh ny nn theo phng ngang m2 ng yn. Theo phng thng ng m2 chu hai lc : lc cnghng ln trn v trng lng 2= m2 hng xung di. Hai lc ny trit tiu nhau nn :T= P2 = m2g Thay gi tr ny ca T vo (3) ta tm c :a = *Xt h gm toa tu khi lng M v cc vt m1, m2. Lc tc dng ln h gm :- Lc ytc dng ln toa tu.- Lc qun tnh -(m1+m2+M)a tc dng ln tt c cc vt ca h.V trong h qui chiu ny cc vt ca h u ng yn nn theo nh II Niu-tn ta phi c :-(m1+m2+M)a+F = 0.Hay t : F = (m1+m2+M)a = (m1+m2+M)(4)So snh (4) v (2) ta thy chng ta thu c cng mt kt qu khi gii bi ton ny trong hai h qui chiu khc nhau.Bi 11. Trn mt mt phng nghing lm vi mt phng nm ngang mt gc c hai vt c khi lng m1 v m2 (m2>m1) c ni vi nhau bng mt si dy khng co dn. H s ma st gia m1 v m2 vi mt phng nghing l k1 v k2 (k1< k2). Hy xc nh :a) gc ti thiu ca mt phng nging cho h hai vt c th bt u chuyn ng xung pha dui.b) gia tc a ca hai vt v sc cng T ca si dy ni khi h chuyn ng trn mt phng nghing c = 45o.p dng vi m1=2kg; m2 = 8kg; k1= 0,2; k2 = 0,4; g =10m/s2.(S: a) tan gh = 0,36 19o8b) a = gsin -gcos ( ) = 4,525m/s2 )Bi 12. Mt h gm hai vt c khi lng mA v mB c ni vi nhau bng mt si dy khng co dn vt qua mt rng rc nh hnh bn. Mt phng nghing c gc nghing l . H s ma st gia mA v mt phng nghing l k . a) Hy tm iu kin cho mA chuyn ng xung pha di. b) Tm iu kin cho mA chuyn ng ln pha trn. c) Tm iu kin cho h ng yn.(S : a)< sin -kcos ;b)> sin +kcos ;c) sin -kcos < < sin +kcos )Bi 13.Trn mt t mt ngi ng ln mt ci cn th ch s ca cn l 50 (50kg).a) Khi trong thang my ang i ln vi gia tc a=1m/s2 ci cn ch bao nhiu?b) Khi thang my i xung vi gia tc bng bao nhiu th ch s ca ci cn l 0 (tnh trng khng trng lc)(S: a) 54; b) a= g = 9,8m/s2 )Bi 14. Mt vt c khi lng m c ko trn mt mt phng nghing c gc nghing l , vi vn tc khng i bi mt si dy ni. H s ma st gia vt v mt phng nghing l k. Hy xc nh gc hp bi si dy v mt phng nghing lc cng dy l nh nht. Tnh gi tr lc cng dy lc .(S: = arctgk; Tmin= (sin +kcos ) )Bi 15. Mt vt khi lng m1 chuyn ng ti va chm vi vt th hai ng yn c khi lng m2=1kg. Bit rng sau va chm vt th nht truyn cho vt th hai mt lng x=36% ng nng ban u ca mnh. Coi va chm l hon ton n hi. Hy tnh m1.(S: m1= 9kg hay m1= 1/9 kg)Bi 16.Mt con lc n c trng lng P c ko ra khi phng thng ng mt gc =90o sau con lc c th ri. Hy tnh sc cng T ca dy khi con lc i qua v tr cn bng.(S: T= 3P ) Bi 17. T nh ca mt bn cu bn knh l R ngi ta bung tay cho mt vt trt xung pha di theo b mt ca bn cu. Hi cao no so vi so vi mt t vt bt u ri khi b mt ca bn cuBi giiXt mt v tr bt k ca vt khi n cn tip xc vi mt bn cu nh hnh trn. Khi vt chu tc dng ca hai lc :- Phn lcca mt bn cu. Lc ny hng theo phng ni tm O ca bn cu vi vt v c chiu hng t trong ra ngoi bn cu.OORRv00mg mFLLvvv- Trng lng mg ca vt hng theo phng thng ng t trn xung di. Lc ny c th phn tch thnh hai lc thnh phn:+ Thnh phn php tuyn mgcos .+ Thnh phn tip tuyn mgsin l thnh phn lc trc tip lm cho vt chuyn ng trt xung di theo b mt bn cu. Trong l gc lc php tuyn lm vi phng thng ng.Do theo phng php tuyn ca qu o hai lc mgcos vngc chiu nhau nn khi tng hp li ta c lc (mgcos -N). Chnh lc ny ng vai tr l lc hng tm lm cho vt chuyn ng theo qu o trn. Do vy, ta c :mgcos -N = mvi v l vn tc tc thi ca vt ti thi im m ta ang xt. T :N= mgcos - m = m(gcos - ) (1)T (1) ta c nhn xt sau : khi vt cng trt xung pha di th gc cng tng lm cho thnh phn gcos cng gim trong khi thnh phncng tng do v cng ln (v cng ln do thnh phn mgsin cng tng) n mt lc no th gcos =v khi N= 0. Lc ny vt bt u ri khi mt cu (v rng khi ri khi bn cu vt khng cn tip xc vi mt bn cu nn n khng cn chu tc dng ca phn lc N, tc l N = 0)Khi vt bt u ri khi bn cu, ta c :mgcos = mgcos =cos = (2)Mun tnh c gc khi vt bt u ri khi bn cu ta phi tnh c vn tc tc thi v ca vt khi . Mun vy, ta c th p dng nh lut bo ton c nng v vt chuyn ng trong trng trng lc l mt trng th. Gi h l khong cch tnh theo phng thng ng t nh ca bn cu n v tr m vt bt u ri khi bn cu. Theo nh lut bo ton c nng, ta c :1/2. mv2= mgh v2= 2gh Thay gi tr ny vo (2) ta tm c :cos =(3)Mt khc, t hnh v ta tnh c :cos =(4)T (3) v (4) ta c :=2h = R-h h = R/3Cui cng, vt bt u ri khi bn cu cao :H = R h = R R/3 = 2R/3Bi 18. Mt khc g c khi lng 1,5 kg tip xc vi mt l xo b nn t chn mt mt phng nghing c gc nghing 30o (im A). Khi l xo c th lng cho bung ra n y khc g chuyn ng dc mt phng nghing . im B cch A 6m dc theo mt phng nghing th khc g c vn tc 7m/s v khng lin kt vi l xo na. Cho bit h s ma st ng gia khc g v mt phng nghing l OORRv00mg mFLLvvvk = 0,5. B qua khi lng ca l xo. Hy tnh th nng bin dng d tr trong l xo lc ban u. Cho g = 9,8m/s2.(S: U =119J ) Bi 19. Mt h gm hai vt c khi lng m1= 12kg v m2 = 4kg c ni vi nhau bng mt si dy v c vt qua mt rng rc nh hnh bn. Ban u vt m2 nm sn nh cn m1 nm cao 2m. bung tay cho m1 ri xung di. Hy xc nh vn tc ca m1 khi n chm nn nh. B qua ma st v khi lng ca dy v rng rc.(S: 4,4m/s ) Bi 20. Mt thang my c khi lng 1 tn, i ln nhanh dn u vi vn tc ban u bng khng v gia tc l 2 m/s2. Tnh:a) Cng ca thang my thc hin c trong 5 giy u tin. b) Cng sut trung bnh v cng sut cc i sau 10 giy u tin. (S: a) 50kW=Jb) CS t.bnh=Cng ton b/thi gian=20kW; CSmax=F.vmax=40kW) Bi 21. Mt ng c c cng sut l 3 m lc (HP) (1HP=736W). Hiu sut ca my l 75%. ng c dng nng mt vt ln cao vi vn tc khng i l 3m/pht. Tnh khi lng ti a ca vt c nng.(S: 3380kg)Bi 22.Vn ng vin chy xe p trn ng vng xic l mt ng trn tm O v bn knh R. Tm vn tc ti thiu v0 ngi i qua im cao nht ca ng trn m khng b ri xung.

Bi gii:Ti im cao nht, gn h quy chiu vi xe (hqc l phi qun tnh), ta c cc lc tc dng: trng lc mg v lc ly tm FL. xe khng ri: FL = m.v2/R >= mg (1).vi v c th tnh c t nh lut bo ton c nng:m.v02/2 = m.v2/2 +2mgR. Suy ra: v2=v02 - 4gR. Th vo (1), ta c: v02 >= 5gR.Bi 23. Mt qu cu c khi lng l 1 kg treo vo u mt si dy buc c nh vo trn nh. a qu cu lch khi phng thng ng mt gc 600 ri bung ra khng vn tc u. Tnh vn tc ca qu cu khi n v tr to vi v tr cn bng mt gc 300. Tnh lc cng ca dy cng ti v tr .Cho bit chiu di ca dy l 1m. (S: 2,7m/s; 16N)Bi 24. Tnh khi tm ca vt hnh trn c khot mt hnh trn nh pha trong c bn knh bng na hnh trn ln v ca hnh vung cnh l ng knh hnh trn ln v cng khot mt hnh trn nh pha trong (theo hnh v). OORRv00mg mFLLvvvBi 25.Mt vin n khi lng m=10g bay vi vn tc 600 m/s.Sau khi xuyn thng mt bc tng, vn tc ch cn 200 m/s. Tm bin thin xung lng v bin thin ng lng ca vinn. Tnh lc cn trung bnh m bc tng tc dng vo vin n cho bit thi gian m vin n xuyn qua tng l 1/1000 s.Bi 26.Sau va chm n hi ca hai qu cu c khi lng bng nhau, c hai c cng vn tc sau l 10 m/s. Cho bit trc va chm qu cu th hai ng yn. Tnh gc to bi phng chuyn ng ca qu cu th hai so vi phng chuyn ng ca qu cu th nht trc khi va chm. Tnh vn tc ca qu cu th nht trc va chm.Bi 27. Mt ngi ng gia gh Giukpski cm trong tay hai qu t, mi qu khi lng m=10kg. Khong cch t qu t n trc quay l 0,2m. Gi s ban u gh quay vi vn tc w 1 = 4,1vng/s. Hi vn tc gc ca gh thay i nh th no nu ngi dang tay ra khong cch t mi qu t n trc l 0,75m. Cho bit mmen qun tnh ca ngi v gh (khng k qu t) i vi trc quay l IO = 2,5kgm2.(S : w 2 = 0,984 1vng/s.)Bi 28. Mt rng rc bn knh R, khi lng M. Trn rng rc c qun mt si dy mt u treo mt vt nng khi lng m. Hy tnh :a/ Gia tc ri ca vt nng.b/ Sc cng T ca si dy.c/ Vn tc ca vt nng khi n ri c mt on s.(S : a/ a = 2mg/(2m+M) ; b/ T = mMg/(2m+M) ; c/ v =)Bi 29. Mt h gm hai vt m1 v m2 c mc nh hnh bn. Khi lng ca rng rc l M v bn knh R. H s ma st gia m1 v mt bn l k. Hy xc nh gia tc chuyn ng ca h v cc lc cng T1, T2 ca cc on dy. Cho m1=1kg; m2 =2kg; M = 2kg; k = 0,1; g =10m/s2.( S : a == 4,75m/s2T1 = m1(kg+a) = 5,75NT2 = m2(g-a) = 10,5N )Bi 30. Cho hai rng rc ging ht nhau c khi lng m v bn knh R. Hai vt m1 v m2 c mc nh hnh v. Si dy ni khng co dn v b qua khi lng ca dy.a/ Gi s ban u cc vt ng yn. Hy xc nh gia tc chuyn ng ca cc vt m1, m2.b/ Tm iu kin cho m1 ri t trn xung v ko m2 ln.c/ Tm cc sc cng T1, T2, T3 ca cc on dy.Bi giiChn h trc ta Ox c chiu dng hng thng ng t trn xung di nh hnh trn. Gi ta ca rng rc th nht l xo. Ta c nhn xt l xo khng i trong qu trnh h chuyn ng. Gi ta ca rng rc th hai l x2. Ta cng c nhn xt l chuyn ng ca m2 ging ht nh chuyn ng ca rng rc th hai v chng c ni vi nhau bng mt on dy khng co dn. Gi ta m1 l x1.Ta ln lt vit phng trnh chuyn ng ca cc vt trong h.* Xt vt m1 :Vt m1 chu tc dng ca hai lc ngc chiu nhau : m1 hng xung di (m1g > 0), lc cng 1 ca si dy hng ln trn (T1< 0).Vy theo nh lut II phng trnh chuyn ng ca m1 l :m1g T1 = m1a1 (1)* Xt chuyn ng quay ca rng rc th nht :Rng rc chu tc dng ca hai mmen lc : T1R v T2R ngc chiu nhau, do phng trnh chuyn ng ca rng rc th nht l :(T1-T2)R = I 1= I(a1/R) =mR2(a1/R) =ma1Rhay : T1-T2 = ma1(2)* Xt chuyn ng quay ca rng rc th hai (lu rng rc 2 bao gi cng quay ngc chiu vi rng rc 1)Tng t rng rc 1, ta c phng trnh chuyn ng :(T3-T2)R = I 2 = I(a2 /R) =mR2(a2 /R) =ma2RHay : T3-T2 =ma2 (3)* Lu l rng rc th hai va tham gia c chuyn ng quay v c chuyn ng tnh tin. V vt m2 ni vi rng rc th hai bng mt si dy khng co dn nn ta c th coi rng rc th hai v vt m2 nh l mt vt vi khi lng l (m+m2) cng tham gia chuyn ng tnh tin.Lc tc dng ln (m+m2) gm (m+m2)g hng xung di v (T2+T3) hng ln trn. Vy phng trnh chuyn ng l :(m+m2)g - (T2+T3) = (m+m2) a2 (4)* n y ta tm c bn phng trnh (1),(2),(3) v (4) nhng ta li c ti nm n s : a1, a2, T1, T2, T3 do c th gii c bi ton ta cn phi tm thm mt phng trnh na. Ta c nhn xt l hai rng rc khng chuyn ng hon ton c lp i vi nhau v chng c vt qua bi cng mt si dy c di khng i.Biu thc xc nh chiu di l ca si dy l :x2 + (x2 xo) + (x1 xo) + R + R = lhay 2x2 + x1 2xo +2 R = lly o hm bc hai theo thi gian v rng xo, R, l l hng s, ta i n phng trnh :2 =0 hay 2a2 + a1 = 0hay a1 = -2a2 (5)* Cui cng ta gii h gm nm phng trnh (1),(2),(3),(4) v (5) nh sau :(1)+(2)m1g T2 = (m1+ m)a1 =(m+2m1)a1hay - T2 =(m+2m1)a1 m1gthay a1 = -2a2 t (5) vo phng trnh trn, ta c :- T2 = -(m+2m1) a2 m1g (*)(3) + (4) (m+m2)g 2T2 =(3m+2m2)a2 (**)Thay 2T2 t (*) vo (**), ta c :(m+m2)g 2(m+2m1)a2 2m1g =(3m+2m2)a2hay (m+m2-2m1)g =(3m+2m2)a2+2(m+2m1)a2=(3m+2m2+4m+8m1)a2 = (7m+2m2+8m1)a2T ta tnh c :a2 = T (5) a1 = -2a2 nn :a1=-2a2 = Mun cho m1 chuyn ng xung pha di, tc l a1>0, th cn tha mn iu kin :8m1 > 4(m+m2) hay 2m1>(m+m2) (1) T1 = m1(g-a1) = m1[g- ]= m1[=T1=(*)T2 = (m+2m1)a2 +m1g = (m+2m1)+ m1gT2 =(3)T3= T2 + ma2 = T3= Bi 31.Mt khi trc c c khi lng l M v c bn knh R c th quay khng ma st chung quanh trc ca n theo phng nm ngang. Ngi ta treo hai vt c khi lng bng nhau v bng m nh hai si dy nh qun quanh khi tr v th cho chng ri khng vn tc ban u. Hy xc nh :a/ Gia tc ca cc vt.b/ Lc cng ca mi si dy.c/ Vn tc gc ca khi tr khi hai vt ri c mt on h.(S : a/ a = 4mg/(M+4m) ; b/ T = Mmg/(M+4m) ; c/ = )Bi 32. Cho mt h nh hnh v. Rng rc l mt rng rc kp ng tm c bn knh ln lt l R v R/2. Cho bit mmen qun tnh ca rng rc l I. Tm iu kin cho m1 chuyn ng i xung. Vi iu kin hy tnh gia tc gc ca rng rc v lc cng trn cc oan dy.(S : m1>2m2 = g ;T1= m1g[1- ] ; T2= m2g[1+ ])Bi 33.Mt thanh mnh khi lng m c chiu di l L c th quay khng ma st quanh trc O nm ngang i qua u thanh. Trn trc O cn treo mt si dy chiu di l khng co dn. u kia ca si dy c vt nng khi lng m. B qua khi lng ca dy treo. Ko qu cu sao cho dy lch mt gc no so vi thanh (dy vn phi cng) ri th tay. Hi chiu di l ca dy treo qu cu phi bng bao nhiu sau khi va chm vi thanh th qu cu dng li. Coi va chm gia qu cu vi thanh l hon ton n hi.(S : l=)Bi 34. cao h trn mt mt phng nghing lm vi mt phng nm ngang mt gc , ngi ta th cho mt hnh xuyn, c khi lng M c cc bn knh ngoi v trong ln lt l R1 v R2, ln khng trt vi vn tc ban u bng khng. Cho h s ma st ln ca hnh xuyn vi mt phng nghing v mt phng ngang l . Hy tnh vn tc ca hnh xuyn khi n ln n mt phng nm ngang v qung ng BC m n tip tc ln trn mt phng nm ngang cho n khi dng li.(S : v=2 ;) Bi 35. Mt thanh ng cht c chiu di l ang v tr thng ng th b xung. Hy xc nh :a/ Vn tc di ca nh thanh khi n chm t.b/ V tr ca im M trn thanh sao cho khi M chm t th vn tc ca n ng bng vn tc chm t ca mt vt ri t do t v tr M.Bi gii a/ Khi thanh xung c th xem thanh quay quanh im O vi vn tc gc .Khi thanh v tr thng ng th thanh c th nng (thay thanh bng cht im nm ti khi tm G cch O mt on l/2)U =mglKhi chm t th th nng ca thanh bin hon ton thnh ng nng quay ca thanh :Kquay=I 2 =( ml2) 2 = ml2 2 =mglT : = Vn tc di ca nh thanh c tnh theo cng thc v = l :v = l = b/ Ta bit rng vt ri t do cao h khi chm t th c vn tc l v= . Ap dng cng thc ny vi im M c cao xM ;vM =Theo u bi := xM = xMT tm c :xM =lBi 36. T nh mt bn cu bn knh R ngi ta bung tay cho mt vin bi ln khng trt trn b mt bn cu. Hi vin bi ri khi mt cu cao no so vi mt t. B qua nh hng ca ma st.( S : h = )Bi 37: T mt t mt vt c khi lng m (kg), c bn vi vn tc ban u V0(m/s) , hp vi phng nm ngang mt gc . Hy xc nh:a. Thi gian chuyn ng ca vt.

gvtgtt v yd dsin 22. sin 0020 b. Tm xa m vt c th t c.

gvt v xd d cos . sin 2. cos200 c. cao cc i m vt c th t c. gVygVt vat gt V yHHH2sin . sin2.. sin2 20max020 max d. Vct vn tc ti thi im chm t.

02020 02020202 2) sin 2 sin ( ) cos () sin ( ) cos (v v v v vgt v v v v v v v vd dy dx d dy dx d + + + + e. Vct vn tc ti thi im t bt k k t lc nm.

20202 2) sin ( ) cos (A Ay Ax A Ay Ax Agt v v v v v v v v + + + f. Gi s gcc th thay i c . Hy xc nh gc vt c th t c tm xa cc i v tnh gi tr cc i .

020200 max45 1 2 sin2 sin cos . sin 2. cos gvgvt v xd dg. Phng trnh qu o ca vt.

x tg xvgvx gvx v t gt v yvxt t v x. .cos . 2 cos2 cos. sin2.. sincos. cos22 202 202002000 + h. Ti thi im tA (s) k t lc bt u nm hy xc nh gia tc tip tuyn, gia tc php tuyn, bn knh cong qu o.

2020020200) sin ( ) cos (cos. . cos) sin ( ) cos (. sin. . sin .AAx AnAAAy Att g v vvgvvg g at g v vt g vgvvg g a + + 21 T ( ) [ ] cos . .) sin ( cos02 / 320202 2v gt g v vavRRvaAnn + i. Mmenngoi lc tc dng ln vt i vi im nm ti thi im vt t cao cc i. mggVMmg x P r P r M P r MH.sin . cos .. cos . .2sin . .20 ,_

+ j. Mmen ng lng ca vt i vi im nm ti v tr vt t cao cc i. cos . .2sin. . sin . .02 20maxV mgVV m y mV r L V m r LHx H H yrx00VmaxyHVHPyrx00VH Hnh v at an ak. Mmenngoi lc tc dng ln vt i vi im nm ti thi im t k t lc nm.mg t Vmg x P rP r MP r MAA. . cos .. cos . .2sin . .0

,_

+ l. M men ng lng i vi im nm ti thi im t (s )k t lc nm.2. cos 0. . cos . . . cos .20 00 0 00 0 0 0AAttttAttLLtmg V L Ldt t mg V dt t mg V L L dt M dL MdtdLA A A A Bi 38:Mt vt ri t do i c 10m cui cng ca qung ng trong khong thi gian t1 = 0,25s Cho g = 9,8m/s2. Tnh:a. Vn tc ca vt khi chm t.b. cao t vt bt u ri.c. Nu t cao ny ngi ta nm thng ng mt vt khc th phi nm vi vn tc bng bao nhiu v phi theo hng no vt ri xung ti mt t chm hn (v nhanh hn ) vt ri t do mt khong t2= 1s.a. Ti im chm t) 1 (2 2.2 20d dd dt g t gt V h y + Ti B :) 2 (22BBt gS h y ) 3 ( 25 , 0 25 , 01 d B B dt t s t t tT (1) v (2) ta c :) 4 (2 22 2Sgt gtB d+ Thay (3) vo (4) ta c :( ) s t t s t g gtd d d d2066 , 4 20 6125 , 0 9 , 4 2 25 , 02 2 + + s m gt Vd d/ 225 , 41 2066 , 4 . 8 , 9 b.mt ghd71 , 862.2 Pyrx00VA Hnh v at an ac. + + + 02 2222 02 201 1211 01 1) 1 (2.) 1 (2.V t tgtt V h yV t tgtt V h yd ddd dd ddd dBi 3 9:Mt v lng sau khi quay c mt pht th thu c vn tc 700 vng/pht. Tnh gia tc gc ca v lng v s vng m v lng quay c trong mt pht y nu chuyn ng ca v lng l nhanh dn u.Vn tc gc ca v lng t = 700v ng/pht = 700.2/60 (rad/s), sau thi gian t = 1pht = 60s.M = . t Gia tc gc: ( )2/ 22 , 1360014006060 / 1400s radt .Gc quay c sau thi gian t= 1 pht l:( ) rad t2 260 . 22 , 1 .21.21 Do vy s vng quay c trong 1 pht l:vng4.2n 2 t .Bi 40 . Mt bnh xe c bn knh R = 10cm lc u ng yn, sau quay xung quanh trc ca n vi gia tc gc bng 3,14 rad/s2. Hisau giy th nht:a) Vn tc gc v vn tc di ca mt im trn vnh bnh?b) Gia tc php tuyn, gia tc tip tuyn v gia tc ton phn ca mt im trn vnh bnh?c) Gc gia gia tc ton phn v bn knh ca bnh xe (ng vi cng mt im trn vnh bnh?)Bi gii:a. Sau giy th nht, vn tc gc v vn tc di ca mt im trn vnh bnh l:( )( ) s m R vs rad t/ ... ./ ... . b. Gia tc tip tuyn c gi tr khng i v gia tc php tuyn :( )( )2 2 2n2ts m 986 0 1 0 14 3 R as m 314 0 1 0 14 3 R a/ , , . , ./ , , . , . - Gia tc ton phn bng:( )2 2n2ts m 03 1 a a a / , + . c. Gc gia gia tc ton phn a v bn knh l tha mn:03 1314 0aat,,sin = 17046.Bi 41 . Chu k quay ca mt bnh xe c bn knh 50cm l 0,1 giy. Tm:a) Vn tc di v vn tc gc ca mt im vnh bnh;b) Gia tc php tuyn ca mt im nm gia mt bn knh.Bi gii:a. Vn tc di v vn tc gc ca mt im trn vnh bnh: ( )( ) s radRvs mTRv/ 8 , 625 , 04 , 31/ 4 , 311 , 05 , 0 . 2 2 b. Gia tc php tuyn (gia tc hng tm) ca mt imnm gia mt bn knh:( )2 2 2 2ns m 986 2 5 0 8 62 2 R r a / / , . , / . . Hnh v at an aBi 42 : Cho ba qu cu nh khi lng bng nhau m = 0,1 kg c ni bi cc si dy khng dn, khi lng khng ng k c cng chiu di l = 0,5m, dy quay u trong mt phng nm ngang xung quanh trc quay i qua 0 vi vn tc gc =100 rad/s . Tnh sc cng ca tng on dy.( bn knh ca qu cu khng ng k ) Vit phng tr nh chuyn ng cho tng vt3 3 32 2 2 21 1 1 1a m T Pa m T T Pa m T T P + + + + +Chiu ln phng chuyn ng ( Hng tm)23 322 2 221 1 1. 3 .. 2 ..l m ma Tl m ma T Tml ma T Tnnn Theo iu kin u bi : 2 3 1 2; T T T T Thay vo trn : 232 2 222 2 21. 3 .. . . 5 . 3 . . 2 .6 . 5 . . l m Tm l l m l m Tm l l m ml T + + Bi 43 : Mt t chuyn ng chm dn u vi vn tc ban u V0= 54 km/h , trn on ng c dng cung trn bn knh R = 800m. Khi i c on ng S = 800 m th vn tc ca n l V= 18 km/h.a. Tnh thi gian chuyn ng ca t khi i ht on ng .b. Tr s v phng gia tc ton phn ca t ti thi im u v thi im cui ca qung ng..c. Gia tc gc, vn tc gc ca t ti thi im t = 2s k t lc bt u chuyn ng vo on ng .Bi 44 : Cho mt cht im chuyn ng trn tm 0 bn knh R ngc chiu (cng chiu ) kim ng h . Hy biu din cc vct: Vn tc, gia tc tip tuyn, gia tc php tuyn, gia tc ton phn, vn tc gc, gia tc gc,vct ng lng, vct m men ng lng ca cht im ti mt thi im t , khi cht im chuyn ng chm dn v nhanh dn.Bi 45 : Mt qut my quay u vi vn tc gc = 900 vng/pht. Sau khi ngt mch qut quay chm dn u c N = 75 vng th dng hn. Tm :a. Thi gian t lc ngt mch n khi dng hnb. Tr s gia tc ton phn ti mt im nm cch trc quay mt khong r = 10cm ti thi im t1= 5s k t lc ngt mch.Bi 46 : Mt vt nm ngang p vo bc tng thng ng cch im nm S = 6,75 m. im cao ca im va chm thp hn so vi im nm mt on h = 1m, cho g = 9,8m/s2. Tnh :a. Vn tc ban u ca vtb. Bn knh cong qu o ti thi im t =0,3s k t lc nmc. Tr s v phng ca vn tc ti im va chm.d. Mmen ngoi lc i vi im nm ti thi im vt va chm tng.1P2P3P2T2T3T1T1T00m3m1m2me.Mmen ng lng i vi im nm ti thi im vt chm tng.Bi 4 7 :Cho mt h c hc nh hnh v : Cho m1 = 1 kg , m2 = 3 kg . Rng rc l mt a trn c khi lng m3 =2 kg, gc = 300, h s ma st gia vt m1 v mt phng nghing k = 0,1 . Cho dy khng dn khi lng khng ng k . Hy tnh gia tc chuyn ng ca h v sc cng ca dy.Bi 4 8 . Cho mt h c hc nh hnh v .Hnh tr c c khi lng m1 = 300 g m2 = 400 g.Ni vi nhau bi si dy khng dn, khi lng khng ng k , xem dy khng trt trn rng rc. Ly g = 10 m/s2.Hy xc nh gia tc ca h v sc cng ca dy .Bi 49 . Cho rng rc l mt a trn c khi lngm1 = 100 g, quay xung quanh mt trc nm ngang i qua tm O. Trn rng rc c cun mt si dy khng dn, khi lng khng ng k , u kia ca dy treo mt vt nng c khi lng m2 = 50 g . vt nng t do chuyn ng .Tm gia tc ca vt nng v sc cng ca dy . Ly g = 10 m/s2Gii:Vit phng tr nh chuyn ng cho tng vt: I Ma m T PT +2 2 2 21m2mP1T2T1m2m1m2mmmChiu ln phng chuyn ng: 2 22 1 2212 2 2 2a mTra r mI rTa m T P Thay vo trn ta c :N Ts mm mg ma a amm g m25 , 025 . 1 , 0) / ( 522)2(21 222 212 2 + + Bi 50. Trn mt tr rng khi lng m = 1kg, ngi ta cun mt si dy khng gin c khi lng v ng knh nh khng ng k. u t do ca dy c gn trn mt gi c nh. tr ri di tc dng ca trng lng. tnh gia tc ca tr v sc cng ca dy treo.Tr chuyn ng tnh tin va chuyn ng quay.Gi T l sc cng dy. Vit cc phng tr nhchuyn ng cho vt ta c: (**).(*)22ma TRa mRmR I TR Ima T mg a m T PT + T (*) v (**) ta c: ( )( )' NmgTs mga ma mg9 , 428 , 9 . 12/ 9 , 428 , 9222Bi 51 . Mt a trn, tr rng, qu cu c, c khilng m , bn knh R, quay quanh trc i qua tm vi vn tc gc 0vng/pht. Tc dng ln vt mt lc hm tip tuyn vi vnh a ( tr, qu cu) v vung gc vi trc quay. Sau t phtth vt dng li. Tmgi tr ca mmen lc hm i vi trc quay .p dng nh l v mmen ng lng : ) (60 .2 . . . . 00 0 0NmtItItIM MtL - Mmen qun tnh ca a: 22mRI - Mmen qun tnh ca tr rng: 2mR I - Mmen qun tnh ca qu cu c: 252mR I Bi 52 .Mt a trn c khilng m = 3kg , bn knh R = 0,6m , quay quanh trc i qua tm a vi vn tc gc 6000 vng/pht. Tc dng ln a mt lc hm tip tuyn vi vnh a v vung gc vi trc quay. Sau 2 pht tha dng li, tm ln ca lc hm tip tuyn.PTmNtR mFtmRtIM MtLRMF F R Mh h hhh h h60 . 120 . 214 , 3 . 2 . 600 . 6 , 0 . 3. 2. .. 2. 0.0 020 Bi 53.T cao h = 0,7 m trn mt phng nghing, ngi ta cho mt qu cu c, mt a trn, mt tr c, mt vnh trn, mt tr rng, c cng bn knh, ln khng trt trn mt phng nghing Bit = 300,600, 450, ly g = 9,8 m/s2. Hy xc nh : a. Vn tc di ca cc vt cui mt phng nghing. gia tc khi tm ca cc vt. b. Thi gian chuyn ng ca vt khi i ht mt phng nghing . (coi vn tc ban u ca cc vt u bng khng).c. Tm gi tr ca lc ma st gia vt v mt phng nghing. d. Nu gc nghing thay i, h s ma st khng i th gc nghing phi bng bao nhiu cc hnh ln khng trt.e. Tm gica h s ma st sao cho s ln khng xy ra.BI GIIa. * Vn tc di cui mt phng nghing : Vt ln khng trt th lc ma st l lc ma st ngh, cng ca lc ma st bng 0,const W A dS NN 0p dng nh lut bo ton c nng ta c: ( )I mRR mghVRV I mVmghI mVmgR R h mg+ + + + +2222 2 2 2. 2. 2.2 2 2- a trn, tr c:3. 42112222 2ghmRmghRVmRI

,_

+ - Qu cu c:7. 10521252222ghmRmghRV mR I

,_

+ - Vnh trn, tr rng:ghmRmghRV mR I 222. 22* Gia tc khi tm ca cc vt :hVSVahS V S a V V. 2sin .2;sin, 0 ; 22 20202 - a trn, tr c: 3sin 22 . 3sin . 43. 4 ghghaghV -Qu cu c: sin . .752 . 7sin . 107. 10ghghaghV - Vnh trn, tr rng:2sin .2sin . ghgha gh V h RRmmmb. Thi gian chuyn ng ca vt :aVt V at at V V + ) 0 ( ;0 0- a trn, tr c: 2sin .33sin 2;3. 4ghtgaghV

-Qu cu c: 2sin . 5. 14sin . .75;7. 10ght g aghV - Vnh trn, tr rng:2sin .. 42sin .;ghtga gh V c. Lc ma st gia hnh tr v mt phng nghingI mRmg IRI mRmgRIFRImmgaRIm a PRa IFRaI I F Rma F Pmsms msms+

,_

+

,_

+

,_

+ 222 2222sin . . sin . sin .. sin ... . .sin . -Qu cu ng cht: sin .72522mg F mR Ims - a trn, tr c : sin .3122mg FmRIms - Vnh trn, tr rng:2sin .2 mgF mR Ims d. Vt ln khng trt: N k F F Ft ms. mstr msngh- Qu cu ng cht: k tg mg k mg Fms27cos . . sin .72 - a trn, tr c : k tg mg k mg Fms3 cos . . sin .31 bt ng thc cho ta xc nh gii hn trn ca gc nghing0 03 k tgVt ln khng trt- Vnh trn, tr rng: k tg mg k mg Fms2 cos . . sin .21 e. Gi tr ca h s ma st sao cho s ln khng xy ra:iu kin vt khng ln:N k F F Ft ms. mstr msnghBi 54 . C hai hnh tr, mt bng nhm (c), mt bng ch (rng) cng c th t nh mt mt phng nghing. Chng c cng bn knh R = 6cm v cng khi lng m = 0,5kg. Mt cc hnh tr c qut sn ging nhau. Hi: a) Vn tc tnh tin ca cc hnh tr cui mt phng nghing c khc nhau khng?b) Mmen qun tnh ca mi hnh tr;Bi 55 :Mt bao ct c khi lngM, c treobi si dy khng dn chiu di l, khi lng khng ng k. Mt vin n c khi lng m bay theo phng ngang (h.v). Hi ti v tr thp ca bao ct M MmM RRmmm RRmmmmmth vn tc b nht ca vin n phi bng bao nhiu khi vin n cm vo bao ct, th c bao ct v vin n chuyn ng quay trn trong mt phng thng ng quanh im treo.Bi gii:p dng nh lut bo ton ng lng theo phng ngang:m V m MvmV M mv V m M mvA AAminmin). ( ). (). (+ + + p dng nh lut bo ton c nng (thes lc l P A S d TT ; 0 )B AW W Chn gc th nng ti A bng 0(*)2) (2 ) (2) (min2 2min B AV m Ml g m MV m M ++ + +Vit phng tr nh chuyn ng cho vt ti B:

lV m Ma m M T PBn2) () (+ + +Mun VBmin th ti B sc cng T= 0 glm Ml g m MVB++ ) (. ) (2

Thay vo (*) glm m Mv gl VA5) (5min min+ Bi 56: Cho mt h c hc nh hnh v m1 = 400g ,m2 = 200g , rng rc l mt a trn c khi lng m3 = 100g . Gi m2 chm t th m1 cch mt t mt khongh1 = 2m. Cho dy khng dn , khi lng khng ng k .a. Hy xc nh gia tc chuyn ng ca h v sc cng ca cc on dy. b. Tnh cao cc i m m2 c th t c. Bi 57:Mt vt nh trt khng ma st t nh mt mt cu c bn knh R = 1,2m. Mt cu t trn mt t, ly g = 9,8m/s2. Xc nh :a. V tr vt bt u ri khi mt cu so vi mt t. b. Vn tc ca vt khi chm t. BI GII:a. V tr vt bt u ri khi mt cu so vi mt t. Gi s ti B vt bt u ri khi mt cu1m1h2m3mmmmmmm Rh BAAmR00m RRmmmmmAMmBAVPTp dng nh lut bo ton c nng: gim th nng ca vt bng tng ng nng ca vt nn ta c(*)2 22 2gVhmVh mgB B Tnh VB: Vit phng trnh ng lc hc cho vt ti B, chiu ln phng hng tm ta cRmVma N PBn2cos . Ti B vt ri khi mt cu nn mt cu khng cn tc dng ln vt (N=0)(**) ) ( ) (22h R g VRmVRh RmgBB Thay (**) vo (*) ta c: 3 2 2RhR hh + V tr vt bt u ri khi mt cu so vi mt t: R h R352 b. Vn tc ca vt khi chm t: p dng nh lut bo ton c nng: Chn gc th nng ti mt t bng 0gR VVgRmV mVR mg B42 2 . 32352 2 352 2 2

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+ +Bi 58.Mt a tr n ng cht khi lngm1 =100kg, bn knh R = 1,5m, quay khng ma st quanh mt trc thng ng i qua tm vi vn tc gc 10 vng/pht. Mt ngi c khi lng m2= 50kg ng mp a v i dn vo tm a dc theo phng bn knh. Xc nh:a. Vn tc gc ca a khi ngi ng tm a.b. Cng m ngi thc hin khi ngi i t mp a vo tm a .(Coi ngi l mt cht im)Bi gii:a. Vn tc gc ca a khi ngi vo tm a:p dng nh lut bo ton mmen ng lng cho h ngi a:2212 2 2122211 1 12.2 R mI LR mR mI L

,_

+ m :2 2 1 1 2 1 I I L L 112 112122211212mm 2 mR m 5 0R m R m 5 0II ++ . ,. ,22 1010060 2 1002+ .. (vng/pht)b. Cng m ngi thc hin khi ngi i t mp a vo tm a( )) 2 (42 2. 22 2 2 221 221 122 12212221222121 122 21 2 m m mRAAR m R m R m I IW W Wd d d + Bi 59:Mt thanh ng cht thit din mnh, chiu di l (m), quay xung quanh trc nm ngang i qua mt u ca thanh. Lc u thanh v tr nm ngang , th thanh chuyn ng t do. Tm gia tc gc v vn tc gc ca thanh khi thanh i qua v tr hp vi phng thng ng mt gc v khi thanh i qua v tr cn bng. Ly g = 9,8m/s2l Rh BAAmmmR00m* Tm gia tc gc ca thanh:Mmen qun tnh i vi trc quay C i qua u thanh:3 4 122 2 2ml ml mlIc + Phng tr nh chuyn ng quay:) / (. 2sin . . 33sin .2.22s radlgmlPlI P r Mc * Tm vn tc gc ca thanh:p dng nh lut bo ton c nng, chn gc th nng ti D (WtD =0 )C nng bo ton : (WA = WB )) / (cos 32cos .2cos16.2.3 2cos .22 22 2 2s radlgmglmgl mglmlml ll mg I mgh mglc ,_

+ ,_

+ Bi 60 : Cho mt h c hc c gn vo thang my nh hnh v. Thang my chuyn ng i ln vi gia tc a0 = 2m/s2. Cho :m1 = 2kg, m2 = 1kg, m3 = 1,5kg, m4 = 5kg . Dy khng dn khi lng khng ng k, xem dy khng trt trn rng rc . Tnh : a. Gia tc chuyn ng ca cc vt i vi mt t.b. Sc cng ca cc on dy. c. p lc ca m2 ln m1.Bi 61 : Cho mt h c hc c gn vo thang my (hnh v). Thang my chuyn ng i ln (hoc i xung) vi gia tc a0 (m/s2) Cho : m1 (kg) > m2 (kg), Dy khng dn khi lng khng ng k, xem dy khng trt trn rng rc, ly g =10m/s2. a. Tnh gia tc chuyn ng ca cc vt i vi thang my. b. Tnh sc cng ca si dy.lhPc 0000lhPc 0000DAB0am2m4m1m3* Thang my chuyn ng i ln: - Gi gia tc ca vt i vi thang my l: aVit phng trnh chuyn ng cho tng vt:2 2 2 21 1 1 1T P a mT P a m + + chiu ln phng chuyn ng ca tng vt ( )( )0 2 2 20 1 1 1a a m P Ta a m T P+ Theo iu kin ca u bi :T T T 2 1Thay vo v cng hai v ta c : + 0 2 2 20 1 1 1a m a m P Ta m a m T P+ 0 2 1 2 1 2 1) ( ) ( a m m a m m P P + ) / () ).( (22 10 2 1s mm ma g m ma++

) ( ) (0 1N a a g m T + * Thang my chuyn ng i xung:Gi gia tc ca vt i vi thang my l: aVit phng trnh chuyn ng cho tng vt:2 2 2 21 1 1 1T P a mT P a m + + chiu ln phng chuyn ng ca tng vt ( )( )0 2 2 20 1 1 1a a m P Ta a m T P + Theo iu kin ca u bi :T T T 2 1Thay vo v cng hai v ta c : + 0 2 2 20 1 1 1a m a m P Ta m a m T P + 0 2 1 2 1 2 1) ( ) ( a m m a m m P P + + ) / () ).( (22 10 2 1s mm ma g m ma+

) ( ) (0 1N a a g m T Bi 62 . Cho mt h c hc c gn vo thang my (hnh v). Thang my chuyn ng i ln (hoc i xung) vi gia tc a0 (m/s2). Cho : m1 (kg)< m2 (kg).Dy khng dn khi lng khng ng k, xem dy khng trt trn rng rc, ly g =10m/s2. a. .Tnh gia tc chuyn ng ca cc vt i vi thang myb. Tnh sc cng ca si dy. * Thang my chuyn ng i ln: - Gi gia tc ca vt i vi thang my l: aVit phng trnh chuyn ng cho tng vt:1m2m0a1T2T2P1P1m2m0a1T2T2P1P2 2 2 21 1 1 1T P a mT P a m + + chiu ln phng chuyn ng ca tng vt ( )( )0 2 2 20 1 1 1a a m T Pa a m P T + Theo iu kin ca u bi :T T T 2 1Thay vo v cng hai v ta c : + 0 2 2 20 1 1 1a m a m T Pa m a m P T + 0 1 2 2 1 1 2) ( ) ( a m m a m m P P + ) / () ).( (22 10 1 2s mm ma g m ma++

) ( ) (0 1N g a a m T + + * Thang my chuyn ng i xung:Gi gia tc ca vt i vi thang my l: aVit phng tr nh chuyn ng cho tng vt:2 2 2 21 1 1 1T P a mT P a m + + chiu ln phng chuyn ng ca tng vt ( )( )0 2 2 20 1 1 1a a m T Pa a m P T+ Theo iu kin ca u bi :T T T 2 1Thay vo v cng hai v ta c : + 0 2 2 20 1 1 1a m a m T Pa m a m P T+ 0 1 2 2 1 1 2) ( ) ( a m m a m m P P + + ) / () ).( (22 10 1 2s mm ma g m ma+

) ( ) (0 1N g a a m T + Bi 63. Mt vt c khi lng m(kg)chuyn ng trn sn thang my di tc dng ca lc F (N)theo phng ngang. H s ma st gia vt v sn l k. Thang my chuyn ng ln trn ( hoc chuyn ng xung di) vi gia tc ) / (20s m a. Ly g =9,8m/s2 a. Tnh gia tc ca vt i vi sn thang my ( Cho lc F)b. Tnh lc F tc dng ln vt ( Khi cho gia tc) F0aF0aPqtFNamSF1m2m0a1T2T2P1P1m2m0a1T2T2P1P* Thang my chuyn ng i ln:a. Vit phng tr nh chuyn ng cho vt trong h quy chiu khng qun tnh: ( )( )mF P K Fama mg K F P K F a m F Fa m F F F N Pqtqt mS mSqt ms+ + + + + + +) ( ;0 * Thang my chuyn ng i xung: Vit phng tr nh chuyn ng cho vt trong h quy chiu khng qun tnh: ( )mF P K FmF Fama mg k F P K F a m F Fa m F F F N PqtmSqt mS mSqt ms) () ( ;0 + + + + Bi 64 . Mt vt c khi lng m(kg) chuyn ng trn sn thang my vi gia tc a (m/s2) i vi sn, di tc dng ca lc F (N)theo phng ngang. H s ma st gia vt v sn l k. Thang my chuyn ng ln trn ( hoc chuyn ng xung di) vi gia tc ) / (20s m a. Tnh lc F tc dng ln vt. Ly g =9,8m/s2 * Thang my chuyn ng i ln:F0aPqtFNamSF

F0aF0aPqtFNamSFVit phng tr nh chuyn ng cho vt trong h quy chiu khng qun tnh: ( ) ) (0ma mg K a m F P K a m F a m F Fa m F F F N Pqt mSqt ms+ + + + + + + + * Thang my chuyn ng i xung: Vit phng tr nh chuyn ng cho vt trong h quy chiu khng qun tnh: ( ) ) ( ;0ma mg k a m F P K a m F a m F Fa m F F F N Pqt mSqt ms + + + + + + Bi 65 . Mt thanh ng cht c chiu di l = 5m ang v tr thng ng th b xung. g=10m/s2aa. Xc nh vn tc di ca nh thanh khi n chm t . b. Xc nh cao ca im M trn thanh sao cho khi im M chm t th vn tc ca n ng bng vn tc chm t ca vt ri t do t cao .Bi gii : a) v tr thng ng, ct c th nng wt = 2mgl. Khi ti mt t th th nng ny bin thnh ng nng quay ca ct v tr chm t2.2 IWd , trong I l mmen qun tnh ca thanh i vi trc qua u mt ca thanh: I = 32ml, l vn tc gc ca nh thanh lc chm t.p dng nh lut bo ton c nng: s m gl VlV ml I mgl/ 2 , 12 32 . 3 2.222 2 2 b) Gi x l cao ca im M khi thanh v tr thng ng. p dng cng thc tnh vn tc ca vt ri t do t cao x, ta c vn tc ca im M khi chm t: Mv=gx 2 (*).p dng nh lut bo ton nng lngxlgVxV mlmglI mglMM.3. 3.2 222 2 2 (**)Theo iu kin ca u bi ta c biu thc (*) bng (**):gxlgx 23. x = l32 = 10/3 =3,33m.Bi 66 . Mt si dy khng dn vt qua rng rc, mt u buc vt c khi lng m1 = 300g, u kia c ci vng khi lng m2 = 200gtrt c ma st trn dy. Gia tc ca vt m2 i vi dy l a/ = 0,5m/s2. B qua khi lng ca dy, xem dy khng trt trn rng rc. Hy xc nh a. Gia tc ca vt m1b. Lc ma st gia vng v dy. mx

F0aPqtFNamSFBi gii :- Phn tch lc nh hnh v. Chiu chuyn ng nh hnh v- Phng trnh chuyn ng ca h :) (1 2 2 2 21 1 1 1a a m a m F Pa m T Pms + + + - Chiu ln phng chuyn ng ta c :) 2 ( ) () 1 (1 2 21 1 1 1a a m F Pa m T Pms vlc ma st ng vai tr ko cng dy nn: Fms=T1Thay vo phng trnh trn ta c :) (1 2 1 21 1 1 1a a m T Pa m T P tr v vi v ta c: ( )( )1 12 12 2 11.a g mm ma m g m ma + + 1T T vBi 67 . Mt si dy khng dn khi lng khng ng k c cun trn hnh tr c c khi lng m (kg). Mt u dy c gn vo trn thang my, thang my chuyn ng thng ng ln trn (xung di) vi gia tc a0 (m/s2). Tnh : a. Gia tc ca khi tm hnh tr i vi thang my v i vi mt t b. Lc cng ca dy .* Thang my chuyn ng i ln:-Gia tc khi tm ca hnh tr i vi thangmy :Chiu ln phng chuyn ng ta c: 2.2.. .222a mra mrra ITraI I rTa m T F Pqt +( )( )00 03223.2a g am a a g m a ma mma mg+ + + - Gia tc khi tm ca hnh tr i vi mt t: ) 2 (31) (320 0 0 0a g a a g a a a + - Lc cng ca dy: 3) () (322 200a g ma gmamT+ + * Thang my chuyn ng i xung :-Gia tc khi tm ca hnh tr i vi thangmy : Chiu ln phng chuyn ng ta c: m0aTqtFaP0aTqtFaP

0am1m2m1P2P1TmsF0RPPPPP

2.2.. .222a mra mrra ITraI I rTa m T F Pqt

Thay vo trn ta c :( )( )00 03223.2a g am a a g m a ma mma mg - Gia tc khi tm ca hnh tr i vi mt t: ) 2 (31) (320 0 0 0a g a a g a a a + + + - Lc cng ca dy: 3) () (322 200a g ma gmamT Bi 68. Mt khu pho khi lng M = 450kg nh n theo phng nm ngang. n pho c khi lng m = 5kg, vn tc u nng v = 450m/s. Khi bn, b pho git v pha sau mt on s = 45cm. Tm lc hm trung bnh tc dng ln pho.Bi gii:Gi V l vn tc git li ca khu pho. Da vo nh lut bo ton ng lng p dng cho phng ngang ta c: 0 v m V M +. . vMmV. Lc hm khu pho sinh cng lm gim ng nng ca khu pho:2V M21S F . . hmT : ( ) N 12500450 45 0 2450 5M s 2v mS 2V MF2 22 2 2 . , ... . ...hm Bi 69 . Mt chic thuyn di l =4m, khi lng m = 100kg nm yn trn mt nc. Hai ngi c khi lng m1 = 60kg , m2 = 40kg ng hai u thuyn . Hi thuyn dch chuyn mt on bng bao nhiu so vi mt nc khi :a. Khi ngi c khi lng m1 n v tr ngi c khi lng m2b. Hai ngi cng i n gia thuyn vi cng vn tc.p dng nh lut bo ton ng lng theo phng ngang . Gi V1 vn tc ca ngi 1 i vi thuyn, V2 l vn tc ca ngi 2 i vi thuyn , V l vn tc ca thuyn i vi nc.- Khi ngi cha chuyn ng h ng yn K1 = 0- Khi ngi chuyn ng : a. Ngi th nht n v tr ca ngi th 2: ( ) ( ) V m m V V m K .2 1 1 2+ + mMVvlx1m2m0RPPPPPBo ton ng lng:( )1 1 2 12 1 1 1.0 ). (V m V m m mV m m V m V m + + + + V thi gian chuyn ng ca ngi v thuyn bng nhau nn :( ) mm m ml ms t V m t V m m mlS2 , 1100 40 604 . 60. . .2 111 1 2 1+ ++ + + +b. C hai ngi cng ra gia thuyn( ) ( ) ( )1 1 2 2 2 1 2 2 1 1 2V m V m V m m m mV V V m V V m K + + + + + + Ba o ton ng lng ( )m m mV m V mV V m V m V m m m K K+ + + + + 2 12 2 1 11 1 2 2 2 1 2 10Vthi gian chuyn ng ca thuyn v ngi bng nhau nn : ( )mm m ml m mSm m mt V m t V mt Vl lS2 , 0200 . 24 ). 40 60 (. 2). (. ..2 12 12 12 /2 22 /1 1+ + + +Bi 70:Mt thanh c khi lng m1 = 1kg, chiu di l = 1,5m c th quay t do xung quanh mt trc nm ngang i qua u mt ca thanh . Mt vin n c khi lng m2 = 0,1kg , bay theo phng nm ngang vi vn tc V = 400m/s ti xuyn vo u kia ca thanh v mc vo thanh.Tm vn tc gc ca thanh ngay sau khi vin n cm vo thanh Bi gii:Ti v tr thp nht mmen trng lc i vi trc quay bng 0 nn mmen ng lng bo tonA Addl ml m LV l mlVl m l m L .3.21 22 2222 122 1+ mmen ng lng bo ton L1=L2s radl m mV mV l m lm mdA d A/5 , 1 ). 1 1 , 0 . 3 (400 . 1 , 0 . 3). 3 (3.331 2222 1 2++ + Bi 71.Mt a trn ng chtbn knh R , khi lng m c th quay xung quanh trc nm ngang vung gc vi a v cch tm a mt onR/2 . a bt u quay t v tr ng vi v tr cao nht ca tm a vi vn tc ban u bng 0. Hy xc nh m men ng lngca a i vi trc quay khi a i qua v tr thp nht . Bi gii:Mmen qun tnh ca a i vi trc quay l:2m1mlA00R1m2mR rrPPPPP0 432 21222mR Rm mR I

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+ p dng nh lut bo ton nng lng:( )ILIII mgR2 2 212 22 gR mRgR mmgRI L234623 2 Bi 72. Mt con lc n trng lng P c treo bi si dy khng dn khi lng khng ng k, ko con lc ra khi phng thng ng mt gc = 900, sau con lc c th ri t do. Xc nh sc cng ca dy treo khi con lc i qua v tr cn bng.Bi gii:Khi qua v tr cn bng, hp lc gia sc cng ca dy v trng lc ca con lc to ra lc hng tm ca chuyn ng ny.lmva m P T2ht .mglmvT2+ p dng nh lut bo ton c nng ta c:mgl mv mgl mv 2212 2 T suy ra sccng T: P 3 mg 3 mglmgl 2T + Bi 73.Cho mt hc hc nh hnh v : Cho m1 = 1 kg , m2 = 3 kg . Rng rc hai nc c mmen qun tnh ivi trc quay i qua tm l I = 8.104kgm2, bn knh b r = 20cm , bn knh ln R =40cm. Chody khng dn khi lng khng ng k . Hy xc nh gia tc gc ca h v sc cng ca cc on dy . Bi 74. Hai bnh c th tch bng nhau cha cng mt cht kh. Nhit v p sut trong mi bnh l t1 =270C , p1 = 2,1.105N/m2 , t2 = 470C , p2 = 3,2.105N/m2 . M kho K sau khicn bng nhit khi kh t = 350C . Xc nh p sut v khi lng ring ca khi kh nhit . Cho = 2kg/kmolBi gii:Phng tr nh trng thi ca tng b nh : ) (2. 2.222112211 2 1222 222111 111mNTPTP TPVTTPTPRTm mV PRTV Pm RTmV PRTV Pm RTmV P

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+

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+ + Bi 75 :Mt bm ht kh c th tch V= 100cm3 , dng ht kh trong bnh c th tch V = 1,5.10-3m3 t p sut P0 = 105N/m2 n p sut P = 102N/m2. Hi phi ht bao nhiu ln? (Coi qu trnh bm thc hin chm c th coi l qu trnh ng nhit).Gii :Qu tr nh ng nhit ?lnlnln . ln) () () (000 120 2 2 10 1 1 0 + +

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+ +

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+ + + + V VVPPnV VVnPPV VVP P P V V P V PV VVP P V V P V PV VVP P V V P V Pnn n nBi 76:Mt cht lng nguyn t ( a nguyn t, n nguyn t) th tch V1 = 0,5 lt, p sut P1 = 0,5 t. Nn on nhit n p sut P2, th tch V2. Sau gi nguyn th tch V2 lm lnh n nhit ban u. Khi p sut P3 = 1 t.a. V th ca qu trnh trn gin OPVb. Tm th tch V2, p sut P2 ?c. Tnh cng m khi kh nhn vo trong qu trnh trn?d. Tnh nhit m khi kh ta ra trong qu trnh trn?e. bin thin entropi ca qu trnh trn, Cho T1=300KBi gii: b- Qu trnh 1-3 ng nhit :Qu trnh 1-2 on nhit:atVVP P V P V P 32 , 125 , 05 , 0. 5 , 0 .57211 2 2 2 1 1 ,_

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c. Tnh cng m khi kh nhn vo trong cc qu trnh trn . JV P V PA 62 , 1915710 . 10 . 81 , 9 ). 5 , 0 . 5 , 0 25 , 0 . 32 , 1 (13 41 1 2 2d. Tnh nhit m khi kh ta ra trong cc qu trnh trn?Trng thi 1v trng thi 3 cng nhit nn bin thin ni nng ca qu trnh bng 0, theo nguyn l I :J A Q A Q Q A U 62 , 19 0 + Hay tnh theo cch khc: Trong qu tr nh on nhit Q=0 nn nhit ca khi kh ta ra trong cc qu trnh trn bng nhit m khi kh ta ra trong qu tr nh ng tchlitPPV V V P V P 25 , 015 , 05 , 0311 2 2 3 1 1 0VP123V1V2P2P1P3123P1P2P3P0VV1V2( ) ( )( ) ( ) J V P V PiQT T RTmV P RTmV PT TR i mT T CmQV62 , 19 10 . 10 . 81 , 9 . 5 , 0 . 5 , 0 25 , 0 . 32 , 12522 .. .3 41 1 2 21 3 2 2 1 1 1 13 2 3 2 + + + e. entropi: V entropi l hm trng thi nn bin thin entropi khng ph thuc vo qu trnh bin i m n ph thuc vo trng thi u v trng thi cuiHay s dng qu tr nh ng nhit: ) / ( ln12K JVVRmS Bi 77:Mt khi kh dn n on nhit sao cho p sut ca n gim t P1 = 2t n P2= 1t, V2=2lt, sau h nng ng tch n nhit ban u th p sut l P3 = 1,22t. a. V th v xc nh t s Cp/Cv. b. Nhit m khi kh nhn vo trong cc qu trnh trn?c. bin thin entropi ca qu trnh trn, cho T2=a. Qu tr nh on nhit 1-2:

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12212 211VVPPV P V PLy loga 2 v :4 , 16393 , 1 ln2 lnlnln:lnlnln . ln312131122 3 1 112211221 PPPPPPVVV P V PVVPPVVPP mb. Qu tr nh on nhit Q = 0 nn nhit m khi kh nhn c trong cc qu trnh trn bng nhit m khi kh nhn c trong qu tr nh ng tch ( ) ( )( ) ( ) J V P V PiQV PmRT RTmV PV PmRT V PmRT RTmV PT TR i mT T CmQV75 , 64 10 . 10 . 81 , 9 . 2 . 1 2 . 22 , 12322 .. .3 42 2 2 32 2 2 2 2 12 3 3 1 1 1 1 1 12 1 2 3 C. V entropi l hm trng thi nn bin thin entropi khng ph thuc vo qu trnh bin i m n ph thuc vo trng thi u v trng thi cui nn da vo qu trnh ng nhit 1-3 ta cK JPPTV PPPRmVVRmS / ln ln ln3122 23112 Bi 78:Cho 3kmol kh l tng n nguyn t thc hin mt chu trnh AB, CD ng nhit BC, DA ng tch , T1 = 1,5T2 = 450K ; V2 = 2V1.. a. Chng minh rng t s PA/PB = PD/PC 123P1P2P3P0VV1V2b. Tnh cng sinh ra, nhit m khi kh thc s nhn vo trong chu trnhc. Tnh hiu sut ca chu trnh.Bi gii: a. CDC DBAB APPVVV P V PPPVVV P V P 122 1122 1 CDBAPPPP b. Cng sinh ra, nhit thc s nhn vo trong chu trnh:JVVRTmVVRTmVVRTmA A ACD AB5 312121212110 . 92 , 255 , 15 , 0. 2 ln . 450 . 10 . 31 , 8 . 35 , 111 ln ln ln ,_

+ + ( )Ji iRTmT TR i mVVRTmQ Q QDA AB5 31 2 112110 . 85 , 133 1232 ln . 450 . 10 . 31 , 8 . 35 , 1 . 2 22 ln2.ln

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+ ,_

+ + + c. Hiu sut ca chu tr nh: % 4 , 1985 , 13392 , 25 QABi 79:3kmol kh O2 thc hin mt chu trnh CcN thun nghch gia ngun nng T1 = 600K v ngun lnh T2 = 300K. Bit t s P1/P3 = 20 trong chu trnh . Tnh : a. Hiu sut chu trnh.b. Nhit nhn vo ca ngun nng c. Nhit to ra cho ngun lnh d. Cng sinh ra trong mt chu trnha. Hiu sut chu trnh.% 50 112 TTb. Nhit nhn vo ca ngun nng211 1lnPPRTmQ(*) T phng tr nh trng thi ca qu trnh 2-3( )( ) ( ) ( ) JPPRTmQPTTP P P T P T5 5 , 3 3 5 , 3311 15 , 331123 2 312 21110 . 22 , 85 5 , 0 . 20 ln 600 . 10 . 31 , 8 . 3 5 , 0 . ln5 , 0 . . 1]1

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c. Nhit to ra cho ngun lnh( ) J Q Q Q A Q Q51 1 2 1 210 . 69 , 16 . 5 , 0 1 1.Q A md. Cng sinh ra trong mt chu trnhJ Q A5110 . 69 , 16 . Bi 80:cho 2kmol kh n nguyn t thc hin mt chu trnh thun nghch gm 3 qu trnh dn ng nhit , nn ng p, h nng ng tch vi nhit Tmax = 400K. Bit t s Vmax/Vmin = 2 .ACBT1T2P0VV1V2DTmaxV00PVmaxVminT1T21234PV0V1V4V2V3P1P3P2P4P 1 320VV1P2V2P1a. Hy v th . b. Tnh cng sinh ra, nhit nhn vo v hiu sut ca chu trnh.JRTmVVVVRTmVVRTmAVRT mP RTmV PV V PVVRTmA

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+

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+

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+ + 1212 ln . 400 . 10 . 31 , 8 . 21212 ln ln) ( ln3maxmaxmaxmaxminmaxminmaxmaxmaxmax2 max max 2max min 2minmaxmax Nhit nhn vo: Trong mt chu tr nh bin thin ni nng A A Q Q A U + 0Hiu sut l thuyt: % 50 5 , 0 1 5 , 0 . ; 1maxmaxminmax minmaxmaxminminmaxmin TVVT TTVTVTTc. So snh hiu sut thc t v hiu sut l thuyt chy theo chu trnh CacN thun nghch vi ngun nng v ngun lnh k trn Nhit m khi kh thc s nhn vo trong mt chu trnh:( )3 maxminmaxmax 31 122ln T TiR mVVRTmQ Q Q + + p dng qu tr nh ng p: KVVT TTVTV20021. 400maxminmax 3maxmax3min JVV i iRTmQ5 3maxminmax10 . 94 , 952 . 23232 ln . 400 . 10 . 31 , 8 . 2. 2.22 ln

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+

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+ % 4 , 1394 , 9584 , 1231 12 +Q QATTSuy ra : LT TT 504 , 13Bi 81:Mt khi kh ban u c th tch V1 = 0,39 m3 v p sut P1=1,55.105N/m2, c dn ng nhit sao cho th tch tng 10 ln. Sau kh c t nng ng tch trng thi cui p sut ca khi kh bng p sut ban u. Bit trong ton b qu trnh ny, nhit lng phi truyn cho khi kh l 1,5.106J.a. V qu trnh trn th OPV.b. Xc nh t s Potxng .c. Tnh bin thin ni nng, cng do khi kh sinh ra trong qu trnh trn.d. Tnh bin thin entropi trong qu trnh trn.b. Xc nh t s Potxng .Bi gii: ( )4 , 157 2510 . 55 , 1 . 39 , 0 . 910 . 61 , 13 . 210 . 61 , 132. 910 . 61 , 13 10 ln 10 . 55 , 1 . 39 , 0 10 . 5 , 1 10 ln 10 . 5 , 1 . 102) (210 . 5 , 12ln5551 15 5 61 161 1 1 1 1 36121 23 12 + + + iiiiV PJ V P V P V PiT TiR mJ TiR mVVRTmQ Q Q c. Tnh bin thin ni nng, cng do khi kh sinh ra trong qu trnh trn.TmaxV0PVmaxVmin123JV P QVVRTmQ A U5 6 561 121110 . 61 , 13 10 . 5 , 1101ln 10 . 55 , 1 . 39 , 010 . 5 , 1101ln ln + + + + Hay: Ni nng l hm trng thi khng ph thuc vo qu trnh bin i m n ph thuc vo trng thi u v trng thi cui nn: J V PiT TiR mU51 1 1 310 . 61 , 13 . 9 .2) (2 J U Q A A5 5 610 . 39 , 1 10 . 61 , 13 10 . 5 , 1 Bi 82:Mt cht kh l tng trng thi ban u p sut P0, c dn ng nhit ti th tch V2=3V1. Sau kh c nn on nhit tr v th tch ban u , p sut sau khi nn l P3= 31/3P0. Hya. Tm p sut sau khi dn P2 v xc nh kh l n nguyn t hay lng nguyn t, a nguyn t?b. ng nng trung bnh ca mt phn t kh trng thi cui so vi trng thi u thay i nh th no a. p sut P2 v bc t do:Qu trnh ng nhit 1-2:3. 3 .02 1 2 2 2 1 0PP V P V P V P Qu trnh on nhit 2-3:6234311 3 3. . 3 . 3 .33111 0311 3 101 3 2 2 + iiiV P V P VPV P V P b. ng nng trung bnh ca mt phn t kh trng thi cui so vi trng thi u thay i nh th no ?.13132 TTWWKTiW Da vo qu tr nh on nhit 2-3: 44 , 144 , 1 3. 3 . 313314103101321313 312 1

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WWPPPPTTP T P TBi 83 :Mt cht kh l tng trng thi ban u p sut P0 , c dn ng nhit ti th tch V2= 3V1 v p sut P2 . Sau kh c nn on nhit tr v th tch ban u , p sut sau khi nn l P3= 32/5P0. a. Hy xc nh kh l n nguyn t , lng nguyn t hay a nguyn t?.Qu trnh ng nhit 1-2:3. 3 .02 1 2 2 2 1 0PP V P V P V P Qu trnh don nhit 2-3:5257521 3 3. . 3 . 3 .35211 0521 3 101 3 2 2 + iiiV P V P VPV P V P Cht lng nguyn tb. ng nng trung bnh ca mt phn t kh trng thi cui so vi trng thi u thay i nh th no ?.131323TTWWKT W Da vo qu tr nh on nhit 2-3: 55 , 155 , 1 3. 3 . 313527205201321313 312 1

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WWPPPPTTP T P TBi 84 :Mt cht kh l tng trng thi ban u p sut P0, c dn ng nhit ti th tch V2= 3V1 v p sut P2 . Sau kh c nn on nhit tr v th tch ban u , p sut sau khi nn l P3= 32/3P0. a. Hy xc nh kh l n nguyn t hay lng nguyn t, a nguyn t?.Qu trnh ng nhit 1-2:3. 3 .02 1 2 2 2 1 0PP V P V P V P Qu trnh on nhit 2-3:3235321 3 3. . 3 . 3 .33211 0321 3 101 3 2 2 + iiiV P V P VPV P V P Cht n nguyn tb. ng nng trung bnh ca mt phn t kh trng thi cui so vi trng thi u thay i nh th no ?.131323TTWWKT W Da vo qu trnh on nhit 2-3: 08 , 208 , 2 3. 3 . 313325203201321313 312 1

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WWPPPPTTP T P TBi 85:Mt kmol kh l tng thc hin mt chu trnh P0 ABthun nghch nh h.v. BC,DA on nhit ; AB, CD ng p . Vi th tch VA = V0 , VB = 2V0 , VC = 16V0 , VD = 8V0 ; PA = P0 , PD = P0/32 . a. Cht kh l n nguyn t , a nguyn t , hay lng nguyn t ? D Cb. Hiu sut ca ng c chy theo chu trnh trn.0V0 2V0 8V0

Bi 86:Hai khi ng, khi lng mi khi l 850g, t tip sc nhit vi nhau trong hp cch nhit. Nhit ban u ca hai khi l 325K v 285K . Nhit dung ring ca ng l c =0,386 J/gKa. Nhit cn bng cui cng ca hai khi ng l bao nhiu ?b. bin thin entropi ca hai khi ng l bao nhiu ?Bi 87:Mt khi 0,1kg nc nhit 240K, c bin thnh hi nc nhit 373K . Tnh bin thin entropi trong qu tr nh bin i trn nu cho rng nhit dung ca nc v nc khng ph thuc vo nhit . p sut trong qu trnh bin i l p sut kh quyn . Nhit dung ring ca nc l 1,8.103J/kg ; ca nc l 4,18.103J/kg ; nhit nng chy ring ca nc l3,35.105J/kg ; nhit ho hi ring ca nc l 2,26.106J/kg . Bi 88: Cho 3kmol kh l tng n nguyn t thc hin mt chu trnh AB, CD ng nhit BC, DA ng tch, T1 = 1,5T2 = 450 (K); V2 = 2V1.. Tnh cng m khi kh sinh ra, nhit m khi kh thc s nhn vo trong chu trnh trn. Tnh hiu sut thc t ca chu trnh. a. Cng sinh ra, nhit thc s nhn vo trong chu trnh:JVVRTmVVRTmVVRTmA A ACD AB5 312121212110 . 92 , 255 , 15 , 0. 2 ln . 450 . 10 . 31 , 8 . 35 , 111 ln ln ln ,_

+ + ( )Ji iRTmT TR i mVVRTmQ Q QDA AB5 31 2 112110 . 85 , 133 1232 ln . 450 . 10 . 31 , 8 . 35 , 1 . 2 22 ln2.ln

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+ ,_

+ + + b. Hiu sut ca chu trnh: % 4 , 1985 , 13392 , 25 QA TNH ENTRPIBi 89 . Tnh bin thin entrpi khi h nng ng p 6,5 gam Hyr n th tch kh tng ln gp i.K JVVRi mVVCmSp/ 52 , 65 2 ln 10 . 31 , 822 5.210 . 5 , 6ln22ln331212++ Bi 90 .Tnh bin thin entrpi khi dn n ng nhit 10,5 gam kh Nit t th tch 2 lt n th tch 5 ltK JVVRmS / 86 , 225ln 10 . 31 , 8 .2810 . 5 , 10ln3312 Bi 91 . Mt kilmol kh a nguyn t c h nng ng tch, nhit tuyt i ca n c tng ln 1,5 ln. Tnh bin thin entrpi trong qu trnh .K JTT iR mTTCmSV/ 10 . 1 , 10 5 , 1 ln 10 . 31 , 8 . 3 . 1 ln2ln3 31212 Bi 92 . Tnh bin thin entrpi ca mt cht kh lng nguyn t khi h thay i t trng thi ban u c th tchC t m N P l V012 51 127 ; / 10 . 31 , 8 ; 3 sang trng thi hai c 2 52 2/ 10 . 6 ; 5 , 4 m N P l V , theo hai qu trnh dn ng p sau lm lnh ng tch.K JTV PPP iVV iRmPPCmVVCmS S Sv p/ 03 , 531 , 86ln255 , 1 ln27.30010 . 31 , 8 . 10 . 331 , 86ln2535 , 4ln27ln2ln22ln ln5 311 1121212122 1 ,_

+ ,_

+

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++ + + Bi 93 . Nn ng nhit mt khi kh xy t th tchK T at P l V 350 ; 2 ; 41 1 1 n th tch 1 241V V, sau lm lnh ng tch n p sut ban u. Hy tnh bin thin entrpi ca qu trnh bin i trn.P 1 320VV1P2V2P1P 1 320VV1P2V2P1K JTV PSP P V P V PPP iVVRmPPCmVVRmS S SV/ 88 , 1041ln .27.35010 . 81 , 9 . 2 . 10 . 425141ln441ln2ln ln ln4 311 11 2 1 2 1 1211221122 1

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+

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+ + + Bi 94 . Dn ng nhit mt khi kh lng nguyn t t th tchK T at P l V 350 ; 4 ; 21 1 1 n th tch 1 24V V, sau h nng ng tch n p sut ban u. Hy tnh bin thin entrpi ca qu trnh bin i trn.K JTV PSP P V P V PPP iVVRmPPCmVVRmS S SV/ 88 , 10 4 ln254 ln .35010 . 81 , 9 . 4 . 10 . 24 ln254 ln414ln2ln ln ln4 311 11 2 1 2 1 1211221122 1

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+

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+

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+ + + Hay tnh theo cch khc: entropi l hm trng thi nn bin thin entropi khng ph thuc vo qu trnh bin i m ph thuc vo trng thi u v trng thi cuiK JTV PVVRi mTTCmSp/ 88 , 10 4 ln22 5ln22ln11 11213++ Bi 94:Mt kmol kh lng nguyn t c h nng, nhit tuyt i ca n tng ln 1,5 ln. Tnh bin thin entropi nu qu trnh h nng l : a. ng tch . K JTT iR mTTCmSV/ 10 . 42 , 8 5 , 1 ln 10 . 31 , 8 .25. 1 ln2ln3 31212 .K JTTRi mTTCmSp/ 10 . 79 , 11 5 , 1 ln 10 . 31 , 822 5. 1 ln22ln3 31212++

Bi 95 :Mt khi kh lng nguyn t ban u c th tch V1 = 0,39 m3 v p sut P1 =1,55.105N/m2 nhit T1=300K , c dn ng nhit sao cho th tch tng 10 ln. Sau kh c t nng ng tch trng thi cui p sut ca khi kh bng p sut ban u. Hy tnh bin thin entrpi ca qu trnh bin i trn.V entropi l hm trng thi nn bin thin entropi khng ph thuc vo qu trnh bin i m n ph thuc vo trng thi u v trng thi cuiK JTV PVVRi mTTCmSp/ 10 . 6 , 1 10 ln22 5ln22ln311 11213++ CNG IN TRNG1E2E cos 2 E th1 M2 12 1EE E mE E EM+ ME MEcos 2cos 22 122212 122212 12 1E E E E HayE E E E E ThE E mE E EMM+ + + + 2E1EMMBI 96 :Cho hai in tch im q1 = 4.10-8Cv q2 = 4.10-8C (hai in tch tri du). t ti hai nh A v B ca tam gic u ABC cnh a = 20cm trong khng kh. Hy xc nh :a. Vct cng in trng v in th do hai in tch gy ra ti nh C. Theo nguyn ly chng cht( )) / ( 10 . 99 , 12 3 .6 , 010 . 310 . 9 3..3 30 cos 2..228921012 2 1 2 1m Vaqk EE E Eaqk E E E E ECCC + m Vbqbq kV V V 900 ) 2 1 (2 , 110 . 4 . 10 . 9)2(8 92 1 + + + b. Nu ti C ta t in tch qo = 2.10-7C tnh lc in trng tc dng ln q0.Lc in trng do h hai in tch tc dng ln in tch q0.( )Naq qF F F F Faq qF F7216 9020 9 01 2 120 92 110 . 81 , 25923.6 . 010 . 2 . 3 . 10 . 9 . 230 cos .... 10 . 9 . 2 30 cos 2..10 . 9 + c. Tnh cng ca lc in trng khi q0 dch chuyn t C ti H Cng ca lc in trng khi dch chuyn in tch q0 t C ti H: ) (0 H CV V q A A BCaqqHF1F2FA BCaq q2E1EE

0hQ, lM1E d2E dE ddqrmax( )( )( )( )) ( 10 . 72 ) 2400 600 .( 10 . 424006 , 010 . 4 4 . 10 . 9 . 2.. 26002 , 110 . 4 410 . 9.6 88 92 1892 1 2 1J AV q qakVV q qbkV V VHC + + + + + d. Hy xc nh trn ng thng AB mt im m ti vc t cng in trng tng hp bng khng.BI 97. Mt thanh mng AB c chiu dy l = 50cm, tch in tch Q = 4.10-6C phn b u, t trong chn khng. Hy xc nh: Vct cng in trng v in th do thanh gy ra ti im M nm trn ng trung trc ca thanh v cch thanh mt khong h = 60cm. v im N nm trn ng ko di ca thanh v cch u mt gn nht ca thanh mt khong R = 40cm.* Tnh cng in trng ti M:2 2 2 1.rdxlQk dE d E d E d E d + 1dE ln do d h dx tg h xhxtghrrhrdxlQk dE dE22 1cos1.coscos. cos. . 2 cos 2 ) / ( 10 . 87 , 946 , 0 2 , 016 , 010 . 4. 10 . 92..42.. . 2 sin.. 2.. cos. . 232 269220 22maxmaxm VhlhQkhl lh lQkh lQkh ld Qk dE EM++

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+ thanh 1/2* Tnh cng in trng ti N:Chia nh thanh ra thnh nhng phn t c chiu di dx mang in tch dq, cch im M mt khong x, gy ra ti M vc t cng in trng dE 2020. 4 ... 4 x ldx QxdqdE vt TonE d EM + Phng chiu: Do cc in tch im dq trn ton thanh gy ra ti M cc vect cng in trng c cng phng chiu nnE d o EM ngchiu cngph hMR Nl,Q QQ, l rN dxdqxdE

0QR01E dE d+ ln: ) / ( 10 . 906 , 0 4 , 014 , 016 . 010 . 410 . 91 110 . 936929m Vl r r lQkxdxlQEl rrM ,_

+

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+ +

* Tnh in th ti N: Chia nh thanh ra thnh nhng phn t c chiu di dx mang in tch dq, cch im M mt khong x, gy ra ti M mt in th dV r l rlQkxdxlQk VxdxlQkxdqdVl rr+ +ln. .... ) ( 10 . 904 , 01ln6 , 010 . 610 . 9369V V b. Mt nng lng in trng do thanh gy ra ti im M v N c. Nu ti M, N ta t mt in tch q = 2.10-7C tnh lc in trng tc dng ln in tch q.Cu 99 .Cho mt phn t vng dy trn (3/4 vng trn, mt na vng trn, 1/3 vng trn) bn knh R = 80cm, tch in tch Q =6.10-7C phn b u, t trong khng kh. Hy xc nh a. Vc t cng in trng do in tch gy ra ti tm 0 ca vng dy.b. inth do in tch gy ra ti tm 0 ca vng dy.c. Mt nng lng in trng ti tm 0 ca vng dy.d. Ti 0 t mt in tch q0 = - 2.10-8C tnh lc in trng tc dng ln in tch q0.* vng trnDo c tnh cht i xng ta ly 2 phn t i xng c chiu di dl mang in tchRd R QdlRQdq. . . 2.42 gy ra ti0 vc t cng in trng202 202 1. 4. . 2. 4 Rd QRdqdE dE Theo nguyn l chng cht in trng: cos 21 2 1dE dE E d E d E d + vt TonE d E 0 + Phng chiu: Do tng cp in tch im dq i xng gy ra ti 0 cc vct cng in trng c cng phng chiu nnE d o E ngchiu cngph0

+ ln: ) / ( 68 , 506 0228 . 0 . 14 , 310 . 4 . 210 . 9 . 2 ) 0 sin4(sin. 210 . 9 . 2 . cos. 210 . 9 . 22892940290m VRQdRQE

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* in th:RQk dlRQk dV VRdl QkRdqk dV dlRQdqR. .. 2... . 2...424202 2 vt ton - Tnh in th 1/3 vng trn :

0Q RQ R00QRQR001 2 0A BCa2aq 5qA BCa1q2qHRQk dlRQk dV VRdl QkRdqk dV dlRQdqR. 2 .. 3.2 .. . 3...323202 2 vt ton) ( 10 . 96 , 010 . 6. 10 . 9289V V Cu 100. Ti hai nh A,B ca tam gic vung ABC c cnh a = 60cm, b = 2a trongkhng kh , ta t cc in tch im c cng lnC q810 . 3. Hy xc nha. Cng in trng do h in tch gy ra ti Cb. Mt nng lng in trng ti im C.c. in th do h in tch gy ra ti C.d. Ti C t mt in tch q0 = - 2.10-8C tnh lc in trng tc dng ln in tch q0.Cu 101 . Tihai nh A,B ca tam gic u ABC c cnh a = 80cm trong khng kh , ta t hai intch c cng ln C q q q81 2 110 . 4 ; . 2 . a. Cng in trng do h in tch gy ra ti Cb. Mt nnglng in trng ti im C.c. in th do h in tch gy ra ti C.d. Ti C t mt in tch q0 = - 3.10-8 (C) tnh lc in trng tc dng ln in tch q0e. Hy xc nh cng ca lc in trng khi dch chuyn in tch q0 = - 3.10-8 (C) t im C tiim H Bi 102:Cho hai qu cu dn in ng tm c bn knh R1 = 20cm, R2 = 40cm, tch in u vi in tch tng ng l q1 = 9.10-9C, q2 = -3.10-9C, t trong khng kh. Xc nh in trng v in th ti cc im M, N, P. Cho OM =10cm, ON = 30cm, OP = 50cm.Bi 103. Cho mt dy dn mnh di 314cm, c un thnh vng trn tm 0, mang in tch Q = 4.10-6C phn b u , t trong khng kh. a. Hy xc nh cng in trng ti im M nm trn trc ca vng dy v cch tm vng dy mt khong h= 60cm.b. Mt nng lng in trng ti M.c. in th ti M.a. Cng in trng: Do c tnh cht i xng ta ly 2 phn t i xng c chiu di dl mang in tch dllQdq . , gy ra ti vc t cng in trng2 1E d E d E d + Do tnh cht i xng nn cos 01 2dE dE dE vt Ton1E d EM + Phng chiu: Do cc in tch im dqgy ra ti M cc vct cng in trng c cng phng chiu nn 1E d o EM ngchiu cngph + ln: ( ) ( )) / ( 10 . 34 , 455 , 0 6 , 06 , 0 . 10 . 410 . 9.. . ..10 . 932 / 32 2692 / 32 22029m VR h ll h Q kdlrhr lQERM++ b. Mt nng lng in trng: ( ) ( )) / ( 10 . 77 , 404 , 0 6 , 0 . 14 , 3 . 810 . 16 . 10 . 94 . 2 213 332 212 932 22 220m JR hh Q kE wM e++ 0hQMRE d1E d2E d rdl MhR0 Q

0N M P 2RI1I2R r0I0IR2R aI1I2I3aMc. in th ti M. Chia vng dy ra thnh nhng phn t c chiu di dl mang in tch dqm R m Rh RQk dlh R lQk dV Vr ldl Qkrdqk dV dllQdql21214 , 314 , 3 2. . ... .....2 202 2 ++ vt ton) ( 10 . 09 , 466 , 0 5 , 010 . 4. 10 . 932 269V V +T TRNGCu 104. Cho mt dng in thng di v hn c cng khng i I1 = 3A chy qua v mt dng in trn bn knh R = 10cm c cng I2 = 4A khng i chy qua (trng hp mt trong hai dng chy theo chiu ngc li),t cch nhau r = 5cm. C hai dng in t trong cng mt mt phng. a. Hy xc nh cm ng t do hai dng in gy ra ti tm 0 ca hai dng in trn.b. Hy xc nh mt nng lng t trng ti tm 0 ca dng in trn.Cho = 1, 0 = 4.10-7 H/m.a. Hy xc nh cm ng t do hai dng in gy ra ti tm 0 ca hai dng in trn.Theo nguyn l chng cht 2 1 0B B B + Do 2 1; B B c cng phng chiu nn( )TRIr R IB B B52 27 2 0 1 02 1 010 . 91 , 210 . 10410 . 15 . 14 , 3310 . 14 , 3 . 22 2 ,_

+ ++ + b. mt nng lng t trng ti tm 0 ca dng in trn.) / ( 10 . 71 , 3310 . 14 , 3 . 4 . 2) 10 . 91 , 2 (23 572 502m JBwm Cu 105. Cho hai dng in trn c cng cng I chy qua v ngc chiu nhau, ng tm bn knh R v 2R , cng nm trong mt mt phng (trng hp mt tronghai dng chy theo chiu ngc li), Hy xc nh:a. cm ng t do hai dng in gy ra ti tm 0 ca hai dng in trn. b. Hy xc nh mt nng lng t trng ti tm 0 ca dng in trn.Cho = 1, 0 = 4.10-7 H/m. Cu 106. Cho hai dng in trn ng tm c cng I v 2I chy qua c chiu nh (H.v)(trng hp mt trong hai dng chy theo chiu ngc li), c bn knh R v 2R, nm trong hai mt phng t nghing vi nhau mt gc 0 0 0 090 ; 60 ; 45 ; 30 . Hy xc nh a. cm ng t do hai dng in gy ra ti tm 0 ca hai dng in trn. b. Hy xc nh mt nng lng t trng ti tm 0 ca dng in trn. Cho = 1, 0 = 4.10-7 H/m.Cu 107. Hnh v biu din thit din ca ba dng in thng song song di v hnc cng bng nhau I1= I2 = I3 = 5A , i qua ba nh ca mt hnh vung cnh a = 50cm. Hy xc nh a. cm ng t do ba dng in gy ra ti im M.0IR2R2I 0 aI1I2I3aMIabRABCDBlVb. Ti M ta t mt dng in thng di v hn song song cng chiu vi dng in I2 trnv c cng I4 = 2A chy qua. Hy tnh t lc tc dng ln 50cm chiu di trn dng in I4. Cho : 1 ; mH7010 . 4 Cu 108. Hnh v biu din thit din ca ba dng in thng song song di v hnc cng bng nhau I1= I2 = I3 = 10A , i qua ba nh ca mt hnh vung cnh a = 50cm. Hy xc nh a. Cm ng t do ba dng in gy ra ti im M. b. Ti M ta t mt dng in thng di v hn song song cng chiu vi dng in I2 trn v c cng I4 = 2A chy qua, Hy tnh t lc tc dng ln 50cm chiu di trn dng in I4. Cho : 1 ; mH7010 . 4 Cu 109. Cho mt dng in thng di v hn, c cng I = 5A. B b gp tiO thnh mt gc = 600 ; 900; 1200 . Hy xc nh cm ng t, cng t trngdo dng in gy ra ti im M nm trn ng phn gic ca gc 0 v cch 0 mt khong 0M = 20cm nh (h.v ). Cho = 1, 0 = 4.10-7 H/m.

Cu 110.Cho mt on dng in c cng I = 8A chy qua, b b gp ti O thnh hai cnh ca mt hnh ch nht c chiu di a = 40cm , b = 60cm . Hy xc nh cm ng t do dng in gy ra ti im M nm trn nh ca hnh ch nht nh (h.v). Cho = 1, 0 = 4.10-7 H/m. Cu 111.Cho mt on dng in c cng I = 8A chy qua, b b gp ti O thnh hai cnh ca mt hnh tam gic u c cnh l a = 60cm . Hy xc nh cm ng t do dng in gy ra ti im H nm chnca ng cao ca tam gic u nh (h.v). Cho = 1, 0 = 4.10-7 H/m.Cu 112. Cho dng in thng di v hn c cng khng iI1 = 8A chy qua v dng in I2=4A chy trongkhung dy hnh ch nht theo chiu ABCD c cnh bng a = 30cm v b = 20cm. Cnh AB song song vi dng in, cch dng in mt khong R = 10cm. Hy xc nh t lc tc dng ln cnh AB ca khung. Cho = 1. 0 = 4.10-7 H/m.Cu 113 . Cho dng in thng di v hn c cng khng iI = 5A chy qua v mt khung dy hnh ch nht ABCD c cnh bng a = 20cm v b = 30cm. Cnh AB song song vi dng in , cch dng in mt khong R = 10cm. Hy xc nh tnh t thng gi qua khung dy. cho = 1. 0 = 4.10-7 H/ m.Cu 114 . Mt thanh dy dn thng di l = 20cm chuyn ng trong mt t trng u c cm ng t B = 0,1T. Bit 0R

IM00baMI1IRbaABCD2IIaHBaBlVrng: phng ca thanh, phng ng sc t trng v phng dch chuynlun vung gc vi nhau tng i mt. Tnh t thng qut bi dy dn trong chuyn dimt on s = 0,75mCu 115 . Mt khung dy hnh vung c cnh a = 50cm t trong t trng u c cm ng t B = 4,10-4 T. Mt phng ca khung hp vi vct cm ng t B mt gc 030 . Hy xc nh t thng gi qua din tch ca khung .Cu 116. Cho dng in thng di v hn chiu nh hnh v, c cng bin i tng dn chy qua v mt khung dy hnh ch nht ABCD c cnh bng a v b. Cnh AB song song vi dng in, cch dng in mt khong R. Hy xc nh chiu dng in cm ng chy trong khung Cu 117 . Mt thanh dy dn thng di l = 20cm chuyn ng vi vn tc V = 1,5m/s trong mt t trng uc cm ng t B = 0,1T. Bit rng: phng ca thanh, phng ng sc t trng v phng dch chuynlun vung gc vi nhau tng i mt.Tm sut in ng cm ng xut hin trn thanh dy dn . Cho = 1. 0 = 4.10-7 H/m. T thng qut bi thanh trong dch chuyn dx dx l B dS B dm. . 0 cos . .0 Sut in ng cm ng xut hin trn dy dn: V l Bdtdx l BdtdmC. .. . V dy dn khng khp kn nn sut in ng ) ( 10 . 3 5 , 1 . 2 , 0 . 1 , 0 . .2V V l B UC Cu 118 . Cho dng in thng di v hnc cng I1 = 5a chy qua v mt khungdy hnh ch nht ABCD c cnh bng a = 30cm v b = 20cm. Cnh AB song song vi dng in, cch dng in mt khongR = 10cm , cho = 1. a.Tnh t thng gi qua khung . b. Nu cho khung dy dch chuyn song song vi chnh n vi vn tc khng i ra xa dng in, hy xc nh chiu ca dng cm ng chy trong khung. c. Nu cho dng in c cng I2 = 2A chy qua khung dy c chiu t A n D, tnh t lc tc dng ln tng cnh ca khung. Bi 1 19 . Cho dng in thng di v hn c cng I = 5A chy qua, b b gp nh hnh v . Cho R = 10cm , 1 . a. Hy xc nh vct cm ng t do dng in gy ra ti tm 0 . b. Tnh mt nng lng t trng do dng in gy ra ti im 0.Bi 1 20 . Cho mt dng in c cng I = 5A chy trong mt ng dy in thng di v hn , bit rng 50cm chiu di ng dy c 1500 vng dy. Hy p dng nh RI00R

IRabAB CDI1ABCDRba0ab IIl v dng in ton phn, xc nh vct cm ng t ti mt im trong lng ng dy. cho 1 , 0= 4710 . H/mBi 1 21 . Mt thanh kim loi c chiu di l = 1m, quay trong t trng u c cm ng t B =4.10-3T, vi vn tc khng i30 vng/s =2.30rad/s, trc quay i qua u mt ca thanh, vung gc vi thanh v song song vi ng sc t trng. a. Tm hiu in th xut hin gia hai u thanh. b. Tm t thng qut bi thanh sau mt vng.Bi 1 22 . Cho mt dng in c cng I = 5A chy trong mt on dy b gp thnh mt hnh ch nht c cnh a = 40cm, b = 60cm t trong khng kh. Hy xc nh vct cm ng t do dng in gy ra ti tm 0 ca hnh vung. Bi 1 23 . cho mt dng in trn bn knh R = 20cm c cng I = 5A chy qua, hy xc nh : a. Vct cng t trng v vct cm ng t do dng in gy ra ti mt im M nm trn trc vng dy v cch tm vng dy mt khong h = 30cm, v ti tm 0 vng dy.b. Xc nh mt nng lng t trng ti hai im .Bi 1 24: Hnh v biu din tit din thng ca hai dng in thng di v hn c cng cng I1 = 10A chy qua, t cch nhau 20cm, P l im nm trn ng trung trc ca on ni gia hai dng in, v cch trung im 10cm. Hy xc nh : a. Vct cm ng t do hai dng in gy ra ti P. b. Ti P t mt dng in thng di l = 40cm, c cng I2 =5A chy qua, song song vi hai dng in trn v c chiu i vo. Hy xc nh t lc tc dng ln dng in I2 .Bi 1 25: Thit din thng ca hai dng in thng di v hn c cng I chy qua, t cch nhau mt khong d. a. Chng minh ln ca vct cm ng t ti im P nm cch u hai dng in v cch d mt khong h l B = ( )2 204 .. . 2d Rd I+ v xc nh phng chiu ca vct cm ng t.b. Xc nh mt nng lng t trng ti P. Cho 1

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