14 Tusaalooyin Su'aalo ku saabsan Dhaqdhaqaaqa Qaybaha
1. Baloogga A oo miisaankiisu yahay 5 kg ayaa la dhigayaa meel siman oo siman, baloogga B oo miisaankiisu yahay 3 kg ayaa lagu dhejiyaa xarig waxaana lagu xiraa baloogga A iyada oo loo marayo jiid, haddii g = 10 m/s 2 go'aami dardargelinta baloogga!
A. 3,50 m/s2
B. 3,75 m/s 2
Q. 4,00 m/s 2
D. 5,00 m/s 2
E. 5,25 m/s 2
Dood
Waa la ogyahay in:
Dusha siman ee siman.
Cufka baloogga A (m A ) = 5 kg
Cufka baloogga B (m B ) = 3 kg
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s 2
Miisaanka baloogga B (w B ) = m B g = (3)(10) = 30 Newton
Su'aal: Dardargelinta baloogga (a)
Jawaab:
Dusha sare ee siman waa siman tahay sidaa darteed ma jirto xoog is jiidjiid ah oo xannibaya dhaqdhaqaaqa baloogga A. Xoogga dhaqaajiya nidaamka baloogga waa miisaanka baloogga B.
ΣF = ma
w B = (m A + m B ) a
30 = (5 + 3) a
30 = 8 a
a = 30/8
a = 3,75 m/s 2
Jawaabta saxda ah waa B.
2. Iyada oo ku saleysan sawirka, waxaa la ogyahay in:
(1) eber dardargelinta shayga
(2) walxaha ku socda xariiq toosan xawaare joogto ah
(3) walxaha ku jira xaalad nasasho
(4) shay ayaa dhaqaaqi doona haddii miisaanka shaygu uu ka yar yahay xoogga jiidista
Weedha saxda ah waa….
A. (1) iyo (2) oo keliya
B. (1) iyo (3) oo keliya
C. (1) iyo (4)
D. (1), (2) iyo (3) oo keliya
E. (1), (2), (3) iyo (4)
Dood
(1) eber dardargelinta shayga
Dardargelinta shay waa eber haddii xoogga ka dhasha uu la mid yahay eber. Xoogga ka dhasha:
ΣF = ma 
dardargelinta (a) = 0
∑ F = 0
F 1 + F 2 – F 3 = 12 + 24 – 36 = 36 – 36 = 0 N
(2) walxaha ku socda xariiq toosan xawaare joogto ah
Xawaaraha eber ee shay wuxuu macnaheedu noqon karaa in shaygu nasanayo ama shaygu uu ku socdo xariiq toosan xawaare joogto ah (shaygu wuxuu ku socdaa xawaare joogto ah).
(3) walxaha ku jira xaalad nasasho
Xoogga eber ee ka dhasha wuxuu macnaheedu noqon karaa in shaygu uu nasanayo.
(4) shay ayaa dhaqaaqi doona haddii miisaanka shaygu uu ka yar yahay xoogga jiidista
Miisaanka walaxdu wuxuu u shaqeeyaa si toosan, halka xoogga jiidista uu u shaqeeyo si siman. Maadaama walaxdu ay si siman u socoto, xoogagga jiifa oo keliya ayaa saameeya walaxda.
Jawaabta saxda ah waa D.
3. Fiiri sawirka dhinaca ku yaal!
Haddii isku-dhafka is-jiidjiidka ee u dhexeeya baloogga A iyo jaantusku yahay 0,1 iyo dardargelinta cufisjiidadka awgeed ay tahay 10 m/s-2 ka dibna xoogga loo baahan yahay in lagu dabaqo A si nidaamku ugu dhaqaaqo bidix iyadoo la dardargelinayo
n 2 ms-2 waa ….
A. 70 N
B. 90 N
C. 150 N
D. 250 N
E. 330 N
Dood
Waa la ogyahay in:
Cufka baloogga A (m A ) = 30 kg
Miisaanka baloogga A (w A ) = (30 kg)(10 m/s 2 ) = 300 kg m/s 2 ama 300 Newton
Cufka baloogga B (m B ) = 20 kg
Miisaanka baloogga B (w B ) = (20 kg)(10 m/s 2 ) = 200 kg m/s2 ama 200 Newtons
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s 2
Isku-darka is-jiidjiidka dhaqdhaqaaqa (
= 0,1
Dardargelinta nidaamka (a) = 2 m/s 2 (jihada dardargelinta ee bidixda)
Xoogga is-khilaafka kinetic-ga (f)k) = N= wA = (0,1)(300) = 30 Newton
Su'aal: Waa maxay baaxadda xoogga F?
Jawaab:
Sharciga Labaad ee Newton:
Σ F = ma
Nidaamku wuxuu u dhaqaaqaa dhanka bidix
F – f k – w B = (m A + m B ) a
F – 30 – 200 = (30 + 20)(2)
F – 230 = (50)(2)
F – 230 = 100
F = 230 + 100
F = 330 Newton
Jawaabta saxda ah waa E.
4. Laba shay A iyo B, oo miisaankoodu yahay 2 kg iyo 6 kg, ayaa lagu xidhaa xadhig iyada oo loo marayo jiid siman sida sawirka ka muuqata. Shayga koowaad ee B ayaa la qabtaa ka dibna la sii daayaa. Haddii g = 10 ms -2 markaa dardargelinta shayga B waa...
A. 8,0 ms-2
B. 7,5 ms -2
C. 6,0 ms -2
D. 5,0 ms -2
E. 4,0 ms -2
Dood
Waa la ogyahay in:
m A = 2 kg, m B = 6 kg, g = 10 m/s 2
w A = (m A )(g) = (2)(10) = 20 N
w B = (m B )(g) = (6)(10) = 60 N
Su'aal: dardargelinta shayga B mise dardargelinta nidaamka?
Jawaab:
w B > w Sidaa darteed A shay B hoos ayuu u socdaa, shay A kor ayuu u socdaa (nidaamku wuxuu u socdaa dhanka saacadda).
Σ F = ma
w B – w A = (m A + m B ) a
60 – 20 = (2 + 6) a
40 = (8) a
a = 5 m/s 2
Jawaabta saxda ah waa D.
5. Shay cuf leh waxaa isku xira xarig dhex mara boolal siman sida sawirka ka muuqata. Haddii m 1 = 1 kg, m 2 = 2 kg, iyo g = 10 ms -2 , markaa xiisadda T waa ….
A. 10,2 N 
B. 13,3 N
C. 15,5 N
D. 18,3 N
E. 20,0 N
Dood
Waa la ogyahay in:
m 1 = 1 kg, m 2 = 2 kg, g = 10 m/s 2
w 1 = m 1 g = (1 kg)(10 m/s 2 ) = 10 kg m/s 2 ama 10 Newtons
w 2 = m 2 g = (2 kg)(10 m/s 2 ) = 20 kg m/s 2 ama 20 Newtons
Su'aal: Waa maxay xoogga xiisadda ee xarigga (T)?
Jawaab:
Dardargelinta nidaamka
w 2 > w 1 sidaas darteed nidaamku wuxuu u socdaa dhanka saacadda (m 2 hoos ayuu u socdaa, m 1 kor ayuu u socdaa).
Sharciga Labaad ee Newton:
Σ F = ma
w 2 – w 1 = (m 1 + m 2 ) a
20 – 10 = (1 + 2) a
10 = (3) a
a = 3,3 m/s 2
Dardargelinta nidaamka waa 3,3 m/s 2.
Xiisadda xarigga?
m 2 ayaa hoos u socda
w 2 – T 2 = m 2 a
20 – T 2 = (2)(3,33)
20 – T 2 = 6,66
T 2 = 20 – 6,66
T 2 = 13,3 Newton
m 1 ayaa kor u kacaya
T 1 – w 1 = m 1 a
T 1 – 10 = (1)(3,3)
T 1 – 10 = 3,33
T 1 = 10 + 3,33
T 1 = 13,3 Newton
Xiisadda xarigga (T) = 13,3 Newtons.
Jawaabta saxda ah waa B.
6. Fiiri sawirka dhinaca ku yaal! Cufnaanta baloogyada waa m 1 = 6 kg iyo m 2 = 4 kg, cufnaanta baloogyadana waa la iska indho tiraa. Haddii dusha sare ee diyaaraddu siman tahay oo g = 10 m s −2 , markaa dardargelinta nidaamku waa….
A. 0,5 ms-2
B. 2,0 ms −2
C. 2,5 ms −2
D. 4,0 ms −2
E. 5,0 ms −2
Dood
Waa la ogyahay in:
m 1 = 6 kg, m 2 = 4 kg, g = 10 m/s 2
w 1 = m 1 g = (6 kg)(10 m/s 2 ) = 60 kg m/s 2 ama 60 Newtons
w 2 = m 2 g = (4 kg)(10 m/s 2 ) = 40 kg m/s 2 ama 40 Newtons
Waxaa la weydiiyay: dardargelinta nidaamka (a)?
Jawaab:
m 1 wuxuu ku yaal meel siman oo aan lahayn is jiidjiid si nidaamka uu u wado xoogga cufisjiidadka ee baloogga 2.
Ku dabaq sharciga labaad ee Newton:
∑ F = ma
w 2 = (m 1 + m 2 ) a
40 N = (6 kg + 4 kg) a
40 N = (10 kg) a
a = 40 N / 10 kg
a = 4 m/s 2
Jawaabta saxda ah waa D.
7. Laba baloog, mid walba oo culeyskiisu yahay 2 kg, waxaa isku xira xarig iyo baloog sida ku cad sawirka. Dusha sare iyo balooggu waa siman yihiin. Haddii baloogga B lagu jiido xoog jiif ah oo ah 40 N, dardargelinta baloogga waa… (g = 10 m/s 2 )
A. 5 m/s2
B. 7,5 m/s 2
Q. 10 m/s 2
D. 12,5 m/s 2
E. 15 m/s 2
Dood:
Waa la ogyahay in:
m A = m B = 2 kg, g = 10 m/s 2 , F = 40 N
w A = mg = (2)(10) = 20 N
Su'aal: dardargelinta baloogga (a)?
Jawaab:
Dusha sare ee baloogga waa siman tahay, sidaa darteed awoodaha saameeya dhaqdhaqaaqa baloogga waa xoog F oo keliya iyo miisaanka baloogga A.
Ku dabaq sharciga labaad ee Newton:
∑ F = ma
F – w A = (m A + m B ) a
40 – 20 = (2 + 2) a
20 = (4) a
a = 20/4
a = 5 m/s 2
Jawaabta saxda ah waa A.
8. Sawirka soo socda, baloogga A wuxuu leeyahay cufnaan dhan 2 kg iyo baloogga B = 1 kg. Baloogga B wuxuu marka hore taagan yahay ka dibna hoos ayuu u socdaa ilaa uu taabto dabaqa. Haddii g = 10 ms -2 , qiimaha xiisadda xarigga T waa...
A. 20,0 Newton
B. 10,0 Newton
C. 6,7 Newton
D. 3,3 Newton
E. 1,7 Newtons
Dood
Waa la ogyahay in :
Cufka baloogga A (m A ) = 2 kg
Cufka baloogga B (m B ) = 1 kg
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s 2
Miisaanka baloogga B (w B ) = m B g = (1)(10) = 10 Newton
Su'aal : Qiimaha xiisadda xarigga (T)
Jawaab :
Su'aasha ku jirta ma jirto macluumaad ku saabsan is-jiidjiid, markaa iska dhaaf is-jiidjiid.
Dardargelinta nidaamka (a)
Marka hore, xisaabi dardargelinta nidaamka adoo isticmaalaya sharciga labaad ee Newton. Baloogga B waa la hakiyey si uu u jiro xoog cufisjiidadka baloogga B oo u dhaqaajinaya baloogga B hoos. Baloogga B iyo baloogga A waxaa ku xiran xarig si baloogga B uu u jiido baloogga A ilaa ay labaduba wada socdaan. Farqiga ayaa ah in baloogga B uu hoos u socdo oo baloogga A uu u dhaqaaqo dhanka midig. Waxaa jira hal xoog oo keliya oo barbar socda dhaqdhaqaaqa labada baloog, waana cufisjiidadka baloogga B (wB). Cufisjiidadka baloogga A waa mid toosan oo u jeeda dhaqdhaqaaqa baloogga A sidaa darteed laguma xisaabtamo xallinta dhibaatada. Xoogga xiisadda ee xariggu wuxuu leeyahay isla cabbirka xarigga iyo jihooyinka ka soo horjeeda si ay isu baabi'iyaan.
∑ F = ma
w B = (m A + m B ) a
10 = (2 + 1) a
10 = 3 a
a = 10/3
Xiisadda xarigga (T)
Xiisadda xarigga waxaa lagu xisaabiyaa iyadoo si gaar ah loo tixgelinayo baloog kasta.
Xiisadda xarigga ee baloogga A
∑ F = ma
T = m A a = (2)(10/3) = 20/3 = 6,7 Newtons
Xiisadda xarigga ee baloogga B
∑ F = ma
w B – T = m B a
10 – T = (1)(10/3)
10 – T = 3,3
T = 10 – 3,3 = 6,7 Newton
Xiisadda xarigga (T) = 6,7 Newton
Jawaabta saxda ah waa C.
9. Sawirka soo socda, baloogga A wuxuu leeyahay cufnaan dhan 2 kg iyo baloogga B = 1 kg. Haddii xoogga isjiidjiid ee u dhexeeya shayga A iyo diyaaradda uu yahay 2,5 Newton, halka xoogga isjiidjiid ee u dhexeeya xarigga iyo jiidaha la iska indho tiro, markaa dardargelinta labada shay waa...
A. 20,0 ms-2
B. 10,0 ms -2
C. 6,7 ms -2
D. 3,3 ms -2
E. 2,5 ms -2
Dood
Waa la ogyahay in :
Cufka baloogga A (m A ) = 2 kg
Cufka baloogga B (m B ) = 1 kg
Xoogga is-jiidjiid ee u dhexeeya baloogga A iyo dusha sare ee siman (f ges A ) = 2,5 Newton
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s 2
Miisaanka baloogga B (w B ) = m B g = (1)(10) = 10 Newton
Su'aal : Dardargelinta labada shay (a)
Jawaab :
Dardargelinta labada shay waxaa lagu xisaabiyaa qaacidada sharciga labaad ee Newton.
∑ F = ma
w B – f ges = (m A + m B ) a
10 – 2,5 = (2 + 1) a
7,5 = 3 a
a = 7,5 / 3 = 2,5 m/s 2
Jawaabta saxda ah waa E.
10. Fiiri sawirka! Baloogga A oo miisaankiisu yahay 30 kg oo ku dul yaal sagxad siman ayaa ku xiran baloogga B oo miisaankiisu yahay 10 kg iyada oo loo marayo barkin. Baloogga B ayaa marka hore la qabtaa ka dibna la sii daayaa si uu hoos ugu dhaqaaqo. Dardargelinta nidaamku waa… (g = 10 ms -2 )
A. 2,5 ms-2
B. 10 ms -2
C. 12 ms -2
D. 15 ms -2
E. 18 ms -2
Dood
Waa la ogyahay in :
Cufka baloogga A (m A ) = 30 kg
Cufka baloogga B (m B ) = 10 kg
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s 2
Miisaanka baloogga B (w B ) = m B g = (10)(10) = 100 Newton
Su'aal : Dardargelinta nidaamka (a)
Jawaab :
∑ F = ma
w B = (m A + m B ) a
100 = (30 + 10) a
100 = 40 a
a = 100/40
a = 2,5 m/s 2
Jawaabta saxda ah waa A.
11. Baloogyada A iyo B, mid walba oo leh cufnaan dhan 8 kg iyo 12 kg, ayaa la dul saaray miis sida ku cad sawirka. Isku-dhafka is-jiidjiidka u dhexeeya baloogga A iyo jadwalka waa 0,3. Baloogga C, oo leh cufnaan dhan 4 kg, ayaa markaa lagu dul dhejiyay baloogga A. Kee baa sax ah weedhaha soo socda?
A. Xiisadda xariggu way ka weyn tahay sidii hore, dardargelintu way ka yar tahay sidii hore
B. Xiisadda xarigga iyo dardargelinta nidaamku isma beddelaan.
C. Xiisadda xariggu way ka yar tahay sidii hore, dardargelintu way ka weyn tahay sidii hore.
D. Xiisadda xariggu way ka weyn tahay sidii hore, dardargelintu way joogto tahay.
E. Xiisadda xariggu way isbeddeshaa halka dardargelintu ay ka yar tahay sidii hore.
Dood
Waa la ogyahay in:
Cufka baloogga A (m A ) = 8 kg
Cufka baloogga B (m B ) = 12 kg
Cufka baloogga C (m C ) = 4 kg
Isugeynta is-jiidjiidka u dhexeeya baloogga A iyo jadwalka ( μ k ) = 0,3
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s 2
Miisaanka baloogga A (w A ) = m A g = (8 kg)(10 m/s 2 ) = 80 kg m/s 2
Miisaanka baloogga B (w B ) = m B g = (12 kg)(10 m/s 2 ) = 120 kg m/s 2
Xoogga is-jiidjiidka ee u dhexeeya baloogga A iyo jadwalka (f k ) = μ k N A = μ k w A = (0,3)(80) = 24 N
Jawaab:
Dardargelinta nidaamka:
Σ F = ma
w B – f k = (m A + m B ) a
120 – 24 = ( 8 + 12 ) a
96 = ( 20 ) a
a = 96/20
a = 4,8 m/s 2
Xiisadda xarigga:
Ka fiirso mid ka mid ah walxaha, tusaale ahaan B.
Σ F = ma
w B – T = (m B ) a
120 – T = ( 12 ) 4,8
120 – T = 57,6
T = 120 – 57,6
T = 62,4 Newtons
Baloogga C oo miisaankiisu yahay 4 kg ayaa markaa lagu dul dhejiyaa baloogga A dushiisa.
Dardargelinta nidaamka:
Σ F = ma
w B – f k = (m A + m B + m C ) a
120 – 24 = ( 8 + 12 + 4 ) a
96 = ( 24 ) a
a = 96/24
a = 4 m/s 2
Xiisadda xarigga:
Ka fiirso mid ka mid ah walxaha, tusaale ahaan B.
Σ F = ma
w B – T = (m B ) a
120 – T = ( 12 ) 4
120 – T = 48
T = 120 – 48
T = 72 Newtons
Jawaabta saxda ah waa A.
12. Sawirka soo socda, baloogga A wuxuu leeyahay cufnaan dhan 2 kg iyo baloogga B = 1 kg. Baloogga B wuxuu marka hore taagan yahay ka dibna hoos ayuu u socdaa ilaa uu taabto dabaqa. Haddii g = 10 ms -2 , qiimaha xiisadda xarigga T waa...
A. 20,0 Newton
B. 10,0 Newton
C. 6,7 Newton
D. 3,3 Newton
E. 1,7 Newtons
Dood
Waa la garanayaa :
Cufka baloogga A (m)A) = 2 kg
Cufka baloogga B (m)B) = 1 kg
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s2
Miisaanka baloogga B (w)B) = mB g = (1)(10) = 10 Newton
La weydiiyay Qiimaha xiisadda xarigga (T)
Jawab :
Su'aasha ku jirta ma jirto macluumaad ku saabsan is-jiidjiid, markaa iska dhaaf is-jiidjiid.
Dardargelinta nidaamka (a)
Tirada koowaad dardargelinta nidaamka iyadoo la adeegsanayo qaacidada Sharciga labaad ee Newton. Baloogga B waa la sudhay si uu u jiro xoog cufisjiidadka baloogga B ah oo baloogga B hoos u dhaqaajinaya. Baloogga B iyo baloogga A waxaa ku xiran xarig si baloogga B uu u jiido baloogga A ilaa labaduba ay wada socdaan. Farqiga ayaa ah in baloogga B uu hoos u socdo baloogga A-na uu u dhaqaaqo dhanka midig. Waxaa jira hal xoog oo keliya oo barbar socda dhaqdhaqaaqa labada baloog, waana xoogga cufisjiidadka baloogga B (wB). Cufisjiidadka Baloogga A wuxuu si toosan ugu toosan yahay jihada dhaqdhaqaaqa baloogga A sidaa darteed looma tixgelinayo xallinta dhibaatada. Xoogga xiisadda ee xariggu wuxuu leeyahay cabbir isku mid ah dhererka xarigga iyo jihooyinka ka soo horjeeda sidaas darteed way is burinayaan.
∑F = ma
wB = (mA +mB) iyo
10 = (2 + 1) a
10 = 3 a
a = 10/3
Xiisadda xarigga (T)
Xiisadda xarigga waxaa lagu xisaabiyaa iyadoo si gaar ah loo tixgelinayo baloog kasta.
Xiisadda xarigga ee baloogga A
∑F = ma
T = mA a = (2)(10/3) = 20/3 = 6,7 Newtons
Xiisadda xarigga ee baloogga B
∑F = ma
wB – T = mB a
10 – T = (1)(10/3)
10 – T = 3,3
T = 10 – 3,3 = 6,7 Newton
Xiisadda xarigga (T) = 6,7 Newton
Jawaabta saxda ah waa C.
13. Sawirka soo socda, baloogga A wuxuu leeyahay miisaan dhan 2 kg iyo baloogga B = 1 kg. Marka xoogga is jiidka inta u dhaxaysa shayga A iyo diyaaradda Newton ee 2,5, halka xoogga is jiidka ee xarigga iyo jiidka la iska indho tiro, markaa dardargelinta labada shay waa...
A. 20,0 ms-2
B. 10,0 ms-2
C. 6,7 ms-2
D. 3,3 ms-2
E. 2,5 ms-2
Dood
Waa la garanayaa :
Cufka baloogga A (m)A) = 2 kg
Cufka baloogga B (m)B) = 1 kg
Xoogga is-jiidjiid ee u dhexeeya baloogga A iyo dusha sare ee siman (f)ges A) = 2,5 Newton
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s2
Miisaanka baloogga B (w)B) = mB g = (1)(10) = 10 Newton
La weydiiyay : Dardargelinta labada shay (a)
Jawab :
Dardargelinta labada shay waxaa lagu xisaabiyaa qaacidada sharciga labaad ee Newton.
∑F = ma
wB - fges = (mA +mB) iyo
10 – 2,5 = (2 + 1) a
7,5 = 3 a
a = 7,5/3 = 2,5 m/s2
Jawaabta saxda ah waa E.
14. 
Fiiri sawirka! Baloogga A oo miisaankiisu yahay 30 kg oo ku dul yaal sagxad siman ayaa ku xiran baloogga B oo miisaankiisu yahay 10 kg iyada oo loo marayo barkin. Baloogga B ayaa marka hore la qabtaa ka dibna la sii daayaa si uu hoos ugu dhaqaaqo. Dardargelinta nidaamku waa… (g = 10 m/s-2)
A. 2,5 ms-2
B. 10 ms-2
C. 12 ms-2
D. 15 ms-2
Q. 18 ms-2
Dood
Waa la garanayaa :
Cufka baloogga A (m)A) = 30 kg
Cufka baloogga B (m)B) = 10 kg
Dardargelinta cufisjiidadka awgeed (g) = 10 m/s2
Miisaanka baloogga B (w)B) = mB g = (10)(10) = 100 Newton
La weydiiyay : Dardargelinta nidaamka (a)
Jawab :
∑F = ma
wB = (mA +mB) iyo
100 = (30 + 10) a
100 = 40 a
a = 100/40
a = 2,5 m/s2
Jawaabta saxda ah waa A.
Isha su'aasha:
Su'aalaha Imtixaanka Qaranka ee Fiisigiska ee Dugsiga Sare/Dugsiga Sare ee Xirfadda
