Aragtida tamarta shaqada-kinetic-ga waxay sheegaysaa in shaqada guud, ama dadaalka uu sameeyo xoog saafi ah oo ku jira shay, ay la mid tahay isbeddelka tamarta kinetic-ga ee shayga. Haddii xoogga saafiga ahi uu sameeyo shaqo togan (xoogga saafiga ah wuxuu ku jiraa jihada la mid ah barakicinta), markaa tamarta kinetic-ga ee shayga ayaa kordheysa. Taas beddelkeeda, haddii xoogga saafiga ahi uu sameeyo shaqo taban (xoogga saafiga ah wuxuu ku jiraa jihada ka soo horjeeda barakicinta), markaa tamarta kinetic-ga ee shayga ayaa hoos u dhacaysa.
W wadarta = EK 2 - EK 1 = ½ mv 2 2 - ½ mv 1 2
Haddii xoogagga muxaafidka ah oo keliya ay ku dhaqmaan shay, sida haddii shay xor ah uu dhaco , markaa xoogga saafiga ah wuxuu la mid yahay xoogga muxaafidka ah. Bayaanka aragtida tamarta shaqada-kinetic-ga waxaa loo rogi karaa shaqo dhammaystiran, ama shaqada ay qabato xoogga muxaafidka ah waxay la mid tahay isbeddelka tamarta kinetic-ga. Haddii xoogga muxaafidka ahi uu sameeyo shaqo togan (xoogga muxaafidka ah wuxuu ku jiraa jihada la mid ah barokaca), markaa tamarta kinetic-ga ee shayga way kordheysaa. Taas beddelkeeda, haddii xoogga muxaafidka ahi uu sameeyo shaqo taban (xoogga muxaafidka ah wuxuu ku jiraa jihada ka soo horjeeda barokaca), markaa tamarta kinetic-ga ee shayga way yaraanaysaa.
W c = EK 2 - EK 1 = ½ mv 2 2 - ½ mv 1 2
Shaqada ay ciidan muxaafid ah ku qabtaan shay waxay la mid tahay isbeddelka taban ee tamarta suurtagalka ah ee shayga. Haddii xoogga muxaafid ah uu sameeyo shaqo togan, tamarta suurtagalka ah way yaraanaysaa. Taas bedelkeeda, haddii xoogga muxaafid ah uu sameeyo shaqo taban, tamarta suurtagalka ah way kordheysaa.
W c = – (EP 2 – EP 1 ) = – mg (h 2 – h 1 ) = – mg 2 + mg 1 = mg 1 – mgh 2
Iyada oo lagu saleynayo dib u eegistii hore, waxaa muuqata in xiriir ka dhexeeya shaqada uu sameeyay ciidan ilaaliya shayga iyo isbeddellada ku yimaada tamarta dhaqdhaqaaqa shayga iyo kuwa suurtagalka ah. Haddii xoogga ilaaliya uu sameeyo shaqo togan, tamarta dhaqdhaqaaqa ayaa kordheysa, halka tamarta suurtagalka ah ay hoos u dhacdo. Haddii xoogga ilaaliya uu sameeyo shaqo taban, tamarta dhaqdhaqaaqa ayaa hoos u dhacda, halka tamarta suurtagalka ah ay korodho.
W c = W c
EK 2 – EK 1 = EP 1 – EP 2
EP 1 + EK 1 = EP 2 + EK 2
EM 1 = EM 2
Macluumaadka:
EM 1 = tamar farsamo oo bilow ah, EM 2 = tamar farsamo oo kama dambays ah, EP 1 = tamar suurtagal ah oo bilow ah, EP 2 = tamar suurtagal ah oo kama dambays ah, EK 1 = tamar firfircoon oo bilow ah, EK 2 = tamar firfircoon oo kama dambays ah
Tusaalaha dhibaatooyinka
1. Baloog ayaa la sii daayaa iyada oo aan lahayn xawaare bilow ah oo ku yaal dusha sare ee diyaarad siman (A). Balooggu wuxuu u simbiriirixanayaa salka u janjeera (E).
Haddii AB = BC = CD = DE, markaa saamiga xawaaraha baloogyada C, D iyo E waa…
Dood
Waa la ogyahay:
AB + BC + CD + DE = 1
1/4 + 1/4 + 1/4 + 1/4 = 1
Su'aal: Isbarbardhigga xawaaraha baloogyada C, D iyo E.
Jawaab:
Sharciga ilaalinta tamarta farsamada ayaa sheegaya in tamarta farsamada bilowga ah = tamarta farsamada ugu dambeysa.
Tamar farsamo oo bilow ah = tamar awood u leh cufisjiidadka
Tamarta farsamada ee kama dambaysta ah = tamarta dhaqdhaqaaqa
Marka ugu sarreysa, balooggu wuu nasanayaa, sidaas darteed tamartiisa dhaqdhaqaaqa waa eber, tamarteeda awoodda cufisjiidadkana waa ugu badnaan. Marka ay ka dhaqaaqdo xagga sare ilaa xagga hoose ee u janjeerta, tamarteeda awoodda cufisjiidadka ayaa hoos u dhacda oo loo beddelaa tamar dhaqdhaqaaqa. Marka ay gaarto xagga hoose ee u janjeerta, tamarteeda dhaqdhaqaaqa ayaa ugu badnaan ah, tamarteeda awoodda cufisjiidadkana waa eber.
Tamarta awoodda cufisjiidadka ee barta A = mgh = mg (4/4) = 4/4 mg
Tamarta dhaqdhaqaaqa ee barta A = 1/2 mv 2 = 1/2 m (0 2 ) = 0
Tamarta awoodda cufisjiidadka ee barta B = mgh = mg (3/4) = 3/4 mg
Tamarta dhaqdhaqaaqa ee barta B = 1/2 mv 2
Tamarta awoodda cufisjiidadka ee barta C = mgh = mg (2/4) = 2/4 mg
Tamarta dhaqdhaqaaqa ee barta C = 1/2 mv 2
Tamarta awoodda cufisjiidadka ee barta D = mgh = mg (1/4) = 1/4 mg
Tamarta dhaqdhaqaaqa ee barta D = 1/2 mv 2
Tamarta awoodda cufisjiidadka ee barta E = mgh = mg (0) = 0
Tamarta dhaqdhaqaaqa ee barta E = 1/2 mv 2
Tamar farsamo oo ku taal barta A = Tamarta awoodda cufisjiidadka + tamarta dhaqdhaqaaqa = 4/4 mg + 0 = 4/4 mg = mg
Tamar farsamo oo ku taal barta B = Tamarta awoodda cufisjiidadka + tamarta dhaqdhaqaaqa = 3/4 mg + 1/2 mv 2
Tamar farsamo oo ku taal barta C = Tamarta awoodda cufisjiidadka + tamarta dhaqdhaqaaqa = 2/4 mg + 1/2 mv 2
Tamar farsamo oo ku taal barta D = Tamarta awoodda cufisjiidadka + tamarta dhaqdhaqaaqa = 1/4 mg + 1/2 mv 2
Tamar farsamo oo ku taal barta E = Tamarta awoodda cufisjiidadka + tamarta dhaqdhaqaaqa = 0 + 1/2 mv 2 = 1/2 mv 2
Xawaaraha baloogga ee C:
Tamarta farsamada ee barta C = tamar farsamo oo bilow ah (tamar farsamo oo joogto ah)
2/4 mg + 1/2 mv 2 = 4/4 mg
1/2 mv 2 = 4/4 mg – 2/4 mg
1/2 mv 2 = 2/4 mg
1/2 mv 2 = 1/2 mg
mv 2 = mg
v 2 = g
v = √g
Xawaaraha xannibaadda ee D:
Tamarta farsamada ee barta D = tamar farsamo oo bilow ah (tamar farsamo oo joogto ah)
1/4 mg + 1/2 mv 2 = 4/4 mg
1/2 mv 2 = 4/4 mg – 1/4 mg
1/2 mv 2 = 3/4 mg
1/2 v 2 = 3/4 g
v 2 = 2 (3/4) g
v 2 = (6/4) g
v 2 = (3/2) g
v 2 = 1,5g
v = √1,5g
Xawaaraha baloogga ee E:
Tamarta farsamada ee barta E = tamar farsamo oo bilow ah (tamar farsamo oo joogto ah)
1/2 mv 2 = mg
1/2 v 2 = g
v 2 = 2g
v = √2g
Isbarbardhigga xawaaraha baloogyada C, D iyo E:
√ g : √1,5g : √2g
√ 1: √1,5: √2 (ku dhufo 2)
√ 2: √ 3: √ 4
√ 2: √ 3: 2
2. Fiiri sawirka soo socda!
Laba shay ayaa ka soo degaya waddo laga bilaabo barta A. Cufka shayga koowaad waa m 1 = 5 kg shayga labaadna waa m 2 = 15 kg. Haddii dardargelinta cufisjiidadka awgeed ay tahay g = 10 m s -2 , markaa saamiga tamarta dhaqdhaqaaqa Ek 1 : Ek 2 barta B waa…
A. 1: 2
B. 1: 3
C. 1: 9
D. 2: 1
E. 3: 1
Dood
Waa la ogyahay in:
Cufka shayga 1 (m1 ) = 5 kg
Cufka shayga 2 (m2 ) = 15 kg
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s 2
Masaafada u dhaxaysa dhibcaha A iyo B (h) = 40 m – 30 m = 10 m
Su'aal: Isbarbardhigga tamarta dhaqdhaqaaqa Ek 1 : Ek 2 barta B
Jawaab:
Tamarta awoodda cufisjiidadka ee shay A waxaa laga cabbiraa barta B:
EP 1 = m 1 gh = (5 kg)(10 m/s 2 )(10 m) = 500 Joules
EP 2 = m 2 gh = (15 kg)(10 m/s 2 )(10 m) = 1500 Joules
Sharciga ilaalinta tamarta farsamada ayaa sheegaya in tamarta farsamada bilowga ah = tamarta farsamada ugu dambeysa.
Tamar farsamo oo bilow ah = tamar awood u leh cufisjiidadka
Tamarta farsamada ee kama dambaysta ah = tamarta dhaqdhaqaaqa
EP 1 = EK 1
500 Joules = EK 1
EP 2 = EK 2
1500 Joules = EK 2
EX 1 : EX 2
500: 1500
5: 15
1: 3
Jawaabta saxda ah waa B.
3. Fiiri sawirka!
Marmar leh cuf m ayaa hoos u soo dhacaya oo leh wareeg siman oo aan lahayn xawaare bilow ah oo ka imanaya barta A. Xawaaraha marmarka ee barta B waa…
A. 10 ms-1
B. 2 √10 ms -1
C. √ 10 daqiiqo -1
D. 2 √5 ms -1
E. √ 5 mitir -1
Dood
Waa la ogyahay in:
Farqiga dhererka u dhexeeya dhibcaha A iyo B (h) = 4 m – 2 m = 2 mitir
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s 2
Cufka marmarka (m) = m
Su'aal: Xawaaraha marmarka ee barta B
Jawaab:
Tamarta farsamada bilowga ah = tamar farsamo barta A = tamar awood cufisjiidadka = mgh = (m)(10)(2) = 20m
Tamarta farsamada ee kama dambaysta ah = tamar farsamo oo ku taal barta B = tamar firfircoon = 1/2 mv 2
Sharciga ilaalinta tamarta farsamada:
Tamarta farsamada bilowga ah = tamar farsamo oo kama dambays ah
20 m = 1/2 mv 2
20 = 1/2 v 2
2 (20) = v 2
40 = v 2
v = √40
v = √(4)(10)
v = 2√10 m/s
Jawaabta saxda ah waa B.
4. Kubbad ayaa ku dul simbiriirixanaysa waddo simbiriirixan sida ka muuqata sawirka hoose.
Haddii xawaaraha kubadda ee barta A uu yahay 6 ms -1 , barta B waa √ 92 ms -1 , iyo g = 10 ms -2 , markaa dhererka barta B ee laga bilaabo salka wadada waa...
A. 0,5 mitir
B. 0,5 √ 2 m
C. 1 mitir
D. √ 2 mitir
E. 2,8 mitir
Dood
Waa la ogyahay in:
Xawaaraha kubadda ee barta A (v A ) = 6 ms -1
Xawaaraha kubbadda ee barta B (v B ) = √ 92 ms -1
Dardargelinta cufisjiidadka (g) = 10 ms -2
Dhererka A (h A ) = 5,6 mitir
Dhererka B (h B ) = h
Su'aal: Dhererka barta B laga bilaabo salka wadada
Jawaab:
Jidku waa simbiriirixan yahay, sidaa darteed ma jiro wax is jiidjiid ah. Sidaa darteed, xoogga kaliya ee ku shaqeeya shayga waa cufisjiidadka. Cufisjiidadka waa xoog muxaafid ah. Shaqada ay qabato ciidan muxaafid ah kuma xirna qaabka wadada laakiin waxay ku xiran tahay isbeddelka booska.
Markaa waxaan xisaabin karnaa dhererka barta B laga bilaabo salka wadada annagoo adeegsanayna qaacidada sharciga ilaalinta tamarta farsamada.
Tamar farsamo oo bilow ah = tamar awood u leh cufisjiidadka
Marka ay ku taal barta A, kubbadu weli ma dhaqaaqdo sidaas darteed xawaarihiisu waa eber. Xawaaraha kubbadu waa eber sidaa darteed tamarta dhaqdhaqaaqa kubbadu waa eber. Tamarta dhaqdhaqaaqa: EK = 1/2 mv 2 = 1/2 m (0) = 0.
Laakiin kubbadu waxay joogtaa meel dhererkeedu yahay 5,6 mitir marka laga eego xagga hoose, sidaas darteed kubbadu waxay leedahay tamar cufisjiidadka. Tamar cufisjiidadka: EP = mgh = m (10)(5,6) = 56 m
Tamar farsamo oo bilow ah = Tamarta awoodda cufisjiidadka + Tamarta dhaqdhaqaaqa = 56 m + 0 = 56 m
Tamarta farsamada ee kama dambaysta ah = Tamarta awoodda cufisjiidadka + Tamarta Kinetic
Marka la gaaro barta B, dhererka kubbadu waa h. Tamarta awoodda cufisjiidadka: EP = mgh = m (10) h = 10 mh
Dhibcaha B aad bay uga hooseeyaan dhibicda A sidaa darteed kubaddu wali waxay ku socotaa xawaare gaar ah. Kubaddu wali waxay ku socotaa xawaare gaar ah sidaa darteed kubaddu waxay leedahay tamar dhaqdhaqaaqa. Tamarta dhaqdhaqaaqa: EK = 1/2 mv 2 = 1/2 m ( √ 92 ) 2 = 1/2 m ( 92 ) = 46 m
Tamarta farsamada ee kama dambaysta ah = Tamarta awoodda cufisjiidadka + Tamarta dhaqdhaqaaqa = 10 mh + 46 m = m (10 saacadood + 46)
Sharciga ilaalinta tamarta farsamada:
Tamarta farsamada bilowga ah = Tamarta farsamada kama dambaysta ah
56 m = m (10 saacadood + 46)
56 = 10 saacadood + 46
56 – 46 = 10 saacadood
10 = 10 saacadood
h = 10/10 = 1 mitir
Jawaabta saxda ah waa C.
Isha su'aasha:
Su'aalaha Imtixaanka Qaranka ee Fiisigiska ee Dugsiga Sare/Dugsiga Sare ee Xirfadda
