Misalai 16 na Tambayoyin Aiki da Makamashi
Ka'idar Makamashin Aiki-Kinetic
1. Ƙarfin da ke ci gaba da aiki na 60 N yana aiki na tsawon daƙiƙa 12 akan abu mai nauyin kilogiram 10.
yana da saurin farko na 6 m/sec a daidai alkiblar da ƙarfin yake.
(1) Aikin da aka yi akan abin shine joules 30.240
(2) Ƙarfin motsi na ƙarshe na abu shine joules 30.240.
(3) Ƙarfin da ake samarwa shine watts 2.520
(4) Ƙaruwar kuzarin motsi na abu shine joules 180.
Bayanin da ya dace shine…
A. 1, 2 da 3
B. 1 da 3
C. 1 da 4
D. 4 kawai
E. duk daidai ne
Tattaunawa
An san cewa:
Ƙarfi (F) = 60 N
Tazarar lokaci (t) = daƙiƙa 12
Nauyin abu (m) = 10 kg
Saurin farko (v)o) = 6 m/s
An tambaya: Wace magana ce daidai?
Amsa:
Haɓaka abu:
ΣF = ma
60 = 10 a
a = 60 / 10 = 6 m/s2
Saurin ƙarshe:
vt = vo + a
vt = 6 + (6)(12)
vt = 6 + 72
vt = 78m/s
Nisa da aka yi a cikin daƙiƙa 12:
s = vo t + 1/2 a2
s = (6)(12) + 1/2 (6)(12)2
s = 72 + (3)(144)
s = 72 + 432
s = mita 504
(1) Aikin da aka yi
W = Fs = (60)(504) = Joules 30.240
(2) Ƙarfin motsi na ƙarshe na abu
EK = 1/2 mvt2 = 1/2 (10)(78)2 = (5)(6084) = Joules 30.420
(3) Wutar lantarki da aka samar
P = W / t = 30.240 / 12 = Joules 2.520/daƙiƙa
(4) Ƙara kuzarin motsi
ΔEK = 1/2 mvt2 – 1/2 mvo2 = 1/2 m (vt2 - vo2) = 1/2 (10)(78)2 - 62) = 5 (6084 –36) = 5 (6048)
ΔEK = Joules 30.240
Amsar da ta dace ita ce B.
2. An gabatar da bayanai kan canjin saurin abu da ke tafiya a layi madaidaiciya kamar haka. Mafi girman aikin ana yin shi ne ta hanyar lambar abu….
A. 1
B. 2
C. 3
D. 4
E. 5
Tattaunawa
Ka'idar makamashin aiki-kinetic:
Jimlar aiki = canjin kuzarin motsi
WTotal = ½ m (v)t2 - vo2)
Manyan kasuwancin sune:
W1 = ½ (8)(4)2 - 22) = (4)(16 – 4) = (4)(12) = Joules 48
W2 = ½ (8)(5)2 -32) = (4)(25 – 9) = (4)(16) = Joules 64
W3 = ½ (10)(6)2 - 52) = (5)(36 – 25) = (5)(11) = Joules 55
W4 = ½ (10)(4)2 - 02) = (5)(16 – 0) = (5)(16) = Joules 80
W5 = ½ (20)(3)2 - 32) = (10)(9 – 9) = (10)(0) = Joules 0
Amsar da ta dace ita ce D.
3. Odi yana tukin mota mai nauyin kilogiram 4000 a kan hanya madaidaiciya a gudun mita 25/s. Ganin cunkoson ababen hawa daga nesa, sai ya yi birki, yana rage saurin motar a hankali zuwa mita 15/s. Aikin da ƙarfin birki ke yi shine….
A. 200 kJ
B. 300 kJ
C. 400 kJ
D. 700 kJ
E. 800 kJ
Tattaunawa
An sani :
Nauyin mota (m) = 4000 kg
Saurin farko na motar (vo) = 25 m/s
Saurin ƙarshe na motar (vt) = 15 m/s
An tambaya : Aiki (W) akan motar?
Jawab :
Ka'idar makamashin aiki-kinetic:
WTotal = ½ m (v)t2 - vo2) = ½(4000)(15)2-252) = (2000)(225-625) = (2000)(-400) = -800.000 Joules = -800 kilo Joules.
Amsar da ta dace ita ce E.
4. Ana jefa ƙwallon da nauyinta ya kai kilogiram 0,1 a kwance a gudun mita 6/s daga rufin gini mai tsawon mita 5. Idan saurin da aka samu sakamakon nauyi a wannan wurin ya kai mita 10/s2, to kuzarin motsi na ƙwallon a tsayin mita 2 shine….
A. 6,8 J
B. 4,8 J
C. 3,8 J
D. 3 J
E. 2 J
Tattaunawa
An sani :
Mass (m) = 0,1 kg
Canjin tsayi (h) = 5 m – 2 m = mita 3
Saurin gudu saboda nauyi (g) = 10 m/s2
An tambaya : kuzarin motsi (EK) a tsayin mita 2?
Jawab :
Hanyar ƙwallon kamar yadda aka nuna a hoton da ke ƙasa.
Motsin parabolic kamar yadda aka nuna a hoton da ke sama haɗuwa ce ta motsi biyu da za a iya bincika daban-daban, inda motsi a cikin alkiblar kwance yake cikin siffar motsi na layi ɗaya da motsi a tsaye a cikin nau'in motsi na faɗuwa kyauta.
Ana magance wannan matsala ta amfani da dokar kiyaye makamashin inji, don haka muna la'akari da motsi ne kawai a tsaye a cikin nau'in motsi na faɗuwa kyauta.
Kowane abu a cikin faɗuwar 'free fall' ba shi da saurin farko (v)o = 0) don haka ƙarfin motsi na farko na abu sifili ne (EK = ½ mv2 = 0). A gefe guda kuma, abin yana fara motsawa daga tsayin mita 5 sama da ƙasa don haka abin yana da ƙarfin ƙarfin nauyi. Don haka ƙarfin injiniya na farko na ƙwallon shine ƙarfin ƙarfin nauyi. Lokacin da ƙwallon ya fara motsawa ƙasa, ƙarfin ƙarfin nauyi na ƙwallon yana canzawa zuwa kuzarin motsi. Ƙarfin ƙarfin nauyi wanda ke canzawa zuwa makamashin motsi lokacin da ƙwallon ya motsa daga tsayin mita 5 sama da ƙasa zuwa tsayin mita 2 sama da ƙasa shine: EP = mgh = (0,1)(10)(3) = Joules 3. Lokacin da ƙwallon ta kasance a tsayin mita 2 sama da ƙasa, ƙarfin motsi na ƙwallon shine Joules 3.
Amsar da ta dace ita ce D.
5. Mota mai nauyin tan 1 ta fara daga hutu. Bayan ɗan lokaci, saurinta zai kai mita 5/s.-1Yawan aikin da injin motar ke yi shine…
A. Joule 1000
B. 2.500 Joule
C. 5.000 Joule
D. 12.500 Joule
E. 25.000 Joule
Tattaunawa
An sani :
Nauyi (m) = tan 1 = kilogiram 1000
Saurin farko (v)o) = 0 (motar tana motsawa daga yanayin da ba a tsayawa ba)
Saurin ƙarshe (v)t) = 5 m/s
An tambaya : Menene aikin (W) da injin motar ke yi?
Jawab :
Ka'idar makamashin aiki-kinetic:
WTotal = ½ m (v)t2 - vo2)
Aikin da injin motar ke yi shine:
WTotal = ½ (1000)(5)2 - 02) = (500)(25 – 0) = (500)(25) = Joules 12.500
Amsar da ta dace ita ce D.
6. Ana jefa ƙwallon da nauyinta ya kai gram 500 a tsaye daga ƙasa da saurin farko na 10 m/s.-2Idan g = 10 ms-2, to aikin da nauyin ƙwallon ya yi idan ya kai matsakaicin tsayinsa shine...
A. 2,5 J
B. 5,0 J
C. 25 J
D. 50 J
E. 500 J
Tattaunawa
Yi watsi da juriyar iska.
An sani :
Nauyin ƙwallon (m) = gram 500 = 0,5 kg
Saurin farko (v)o) = 10 m/s2
Saurin ƙarshe (v)t) = 0. A matsakaicin tsayi, abin yana nan na ɗan lokaci kafin ya juya alkibla.
Saurin gudu saboda nauyi (g) = 10 m/s2
An tambaya : Aikin (W) an yi shi ne ta hanyar nauyin ƙwallon
Jawab :
Ka'idar makamashin aiki-kinetic :
Ka'idar makamashin aiki-kinetic ta bayyana cewa jimlar aikin ko ƙoƙarin da ƙarfin da aka samu akan wani abu yayi daidai da canjin kuzarin motsi na abu. Tsarin ka'idar makamashin aiki-kinetic shine:
WTotal = ΔEK = EKt – EKo
WTotal = ½ mvt2 – ½ mvo2 = ½ m (v)t2 - vo2)
Bayani:
EKt = makamashin motsi na ƙarshe, EKo = kuzarin motsi na farko, m = nauyin abu, vt = saurin ƙarshe na abu, vo = saurin farko na abu.
Jimillar ƙoƙari :
Aikin da nauyin ƙwallon ke yi tun daga lokacin da aka jefa ƙwallon har zuwa lokacin da ƙwallon ta kai matsakaicin tsayinta shine:
WTotal = ½ m (v)t2 - vo2) = ½ (0,5)(0)2 - 102)
WTotal = (0,25)(-100) = -25 Joule
Alamar mara kyau tana nuna cewa alkiblar da ƙwallon ke motsawa ta saba da alkiblar nauyin ƙwallon. Alkiblar da ƙwallon ke motsawa tana sama, yayin da alkiblar nauyin ƙwallon take ƙasa. Alkiblar da ƙwallon ke motsawa tana aiki mara kyau akan ƙwallon, ma'ana aikin da nauyi ke yi yana ƙara ƙarfin ƙarfin nauyi na ƙwallon kuma yana rage ƙarfin kuzarinsa. Alkiblar da ƙwallon ke motsawa tana aiki mai kyau idan alkiblar nauyin ƙwallon yayi daidai da alkiblar da ƙwallon ke motsawa (ƙwallon yana motsawa ƙasa).
Amsar da ta dace ita ce C.
7. Ana jefar da ƙwallon da nauyinta ya kai kilogiram 1 ba tare da saurin farko ba daga saman gini ta taga A a saman bene zuwa taga B a ƙasan bene tare da bambancin tsayi na mita 2,5 (g = 10 ms)-2) Nawa ne ƙoƙari ake buƙata don motsa ƙwallon daga taga A zuwa taga B?
A. Joule 5
B. 15 Joule
C. 20 Joule
D. 25 Joule
E. 50 Joule
Tattaunawa
An sani :
Nauyin ƙwallon (m) = 1 kg
Saurin farko (v)o) = 0 m/s
Tsawo (h) = mita 2,5
Saurin gudu saboda nauyi (g) = 10 m/s2
An tambaya : Jimillar aiki yayin motsi na ƙwallon
Jawab :
Saurin ƙarshe na ƙwallon (v)t)
Kwallo tana faɗuwa ba tare da wani saurin farko ba, don haka ana ɗaukar motsinta a matsayin faɗuwa kyauta. Da farko, ƙididdige saurin ƙarshe na ƙwallon ta amfani da dabarar faɗuwa kyauta. Hanzarta saboda nauyi (g) shine 10 m/s.2, canjin tsayin ƙwallon (h) = mita 2,5 kuma an tambayi gudun ƙarshe (v)t(don haka, yi amfani da dabarar) vt2 = 2 gh
vt2 = 2 gh = 2(10)(2,5) = 2(25)
vt = √2(25) = 5√2
Don haka saurin ƙarshe na ƙwallon shine 5√2 m/s
Jimlar aiki = canjin kuzarin motsi
WTotal = ΔEK = ½ m (v)t2 - vo2) = ½ (1){(5√2)2 - 02}
WTotal = ½ (25)(2) = Joule 25
Amsar da ta dace ita ce D.
8. Abu mai nauyin kilogiram 2 yana fara motsawa da sauri na kilomita 72/h.-1Bayan motsi mita 400, saurin abu zai zama kilomita 144/h-1 kuma (g = 10 ms-2) Jimillar aikin da abin ya yi a wancan lokacin shine…
A. 20 J
B. 60 J
C. 1.200 J
D. 2.000 J
E. 2.400 J
Tattaunawa
An sani :
Nauyin abu (m) = 2 kg
Saurin farko (v)o) = 72 km/h = 20 m/s
Saurin ƙarshe (v)t) = 144 km/h = 40 m/s
Nisa da aka yi tafiya (s) = mita 400
Saurin gudu saboda nauyi (g) = 10 m/s2
An tambaya : Jimillar ƙoƙari
Jawab :
Jimlar aiki = canjin kuzarin motsi
WTotal = ΔEK = ½ m (v)t2 - vo2) = ½ (2)(40)2 - 202}
WTotal = ½ (2)(1600 – 400) = Joules 1200
Amsar da ta dace ita ce C.
9. Abu mai nauyin kilogiram 2 yana motsawa akan wani wuri mai santsi a gudun mita 2/s.-1Idan an yi joules 21 na aiki akan wani abu, saurin abu zai canza zuwa...
A. 1 ms-1
B. 2 ms-1
C. 3 ms-1
D. 5 ms-1
E. 17 ms-1
Tattaunawa
An san cewa:
Mass (m) = 2 kg
Saurin farko (v)o) = 2 m/s
Aiki (W) = Joules 21
An tambaya: Saurin ƙarshe (v)t)
Amsa:
Ka'idar makamashin aiki-kinetic ta bayyana cewa jimlar aikin da aka yi akan wani abu daidai yake da canjin kuzarin motsi na abu. A lissafi:
Jimilla = ΔEK
Jimilla = 1/2 mvt2 -1/2 mvo2
Jimilla = 1/2 m (v)t2 - vo2)
21 = 1/2 (2) (vt2 - 22)
21 = (vt2 - 22)
21 = kut2 - 4
vt2 = 21 + 4 = 25
vt = √25
vt = 5m/s
Amsar da ta dace ita ce D.
10. Ana jefa ƙwallon da nauyinta ya kai gram 500 a tsaye daga ƙasa da saurin farko na 10 ms.-2Idan g = 10 ms-2, to aikin da nauyin ƙwallon ya yi idan ya kai matsakaicin tsayinsa shine...
A. 2,5 J
B. 5,0 J
C. 25 J
D. 50 J
E. 500 J
Tattaunawa
Yi watsi da juriyar iska.
An sani :
Nauyin ƙwallon (m) = gram 500 = 0,5 kg
Saurin farko (v)o) = 10 m/s2
Saurin ƙarshe (v)t) = 0. A matsakaicin tsayi, abin yana nan na ɗan lokaci kafin ya juya alkibla.
Saurin gudu saboda nauyi (g) = 10 m/s2
An tambaya : Aikin (W) an yi shi ne ta hanyar nauyin ƙwallon
Jawab :
Ka'idar makamashin aiki-kinetic :
Ka'idar makamashin aiki-kinetic ta bayyana cewa jimlar aikin ko kasuwanci Tsarin ka'idar makamashin aiki-kinetic don ƙarfin da aka samu akan wani abu iri ɗaya ne da canjin kuzarin motsi na abu.
WTotal = ΔEK = EKt – EKo
WTotal = ½ mvt2 – ½ mvo2 = ½ m (v)t2 - vo2)
Bayani:
EKt = makamashin motsi na ƙarshe, EKo = kuzarin motsi na farko, m = nauyin abu, vt = gudun ƙarshen abu, vo = saurin farko na abu.
Jimillar ƙoƙari :
Aikin da nauyin ƙwallon ke yi tun daga lokacin da aka jefa ƙwallon har zuwa lokacin da ƙwallon ta kai matsakaicin tsayinta shine:
WTotal = ½ m (v)t2 - vo2) = ½ (0,5)(0)2 - 102)
WTotal = (0,25)(-100) = -25 Joule
Alamar mara kyau tana nuna cewa alkiblar da ƙwallon ke bi ta saba da alkiblar nauyin ƙwallon. Alkiblar da ƙwallon ke bi tana sama, yayin da alkiblar nauyin ƙwallon take ƙasa. Nauyi Yin aiki mara kyau akan ƙwallon yana nufin cewa aikin da ƙarfin nauyi ke yi yana sa ƙarfin ƙarfin nauyi na ƙwallon ya ƙaru kuma kuzarin motsi na ƙwallon ya ragu. Ƙarfin nauyi yana yin aiki mai kyau idan alkiblar nauyi na ƙwallon iri ɗaya ce da alkiblar motsawar ƙwallon (ƙwallon yana motsawa ƙasa).
Amsar da ta dace ita ce C.
11. Ana jefar da ƙwallon da nauyinta ya kai kilogiram 1 ba tare da saurin farko ba daga saman gini ta taga A a saman bene zuwa taga B a ƙasan bene tare da bambancin tsayi na mita 2,5 (g = 10 ms)-2) Nawa ne ƙoƙari ake buƙata don motsa ƙwallon daga taga A zuwa taga B?
A. Joule 5
B. 15 Joule
C. 20 Joule
D. 25 Joule
E. 50 Joule
Tattaunawa
An sani :
Nauyin ƙwallon (m) = 1 kg
Saurin farko (v)o) = 0 m/s
Tsawo (h) = mita 2,5
Saurin gudu saboda nauyi (g) = 10 m/s2
An tambaya : Jimillar aiki yayin motsi na ƙwallon
Jawab :
Saurin ƙarshe na ƙwallon (v)t)
Kwallo ta faɗi ba tare da wani saurin farko ba, don haka ana ɗaukar motsin ƙwallon a matsayin faɗuwa kyauta. Da farko, ƙididdige saurin ƙarshe na ƙwallon ta amfani da dabarar motsi na faɗuwa kyautaAn san cewa saurin gudu saboda nauyi (g) = 10 m/s2, canjin tsayin ƙwallon (h) = mita 2,5 kuma an tambayi gudun ƙarshe (v)t(don haka, yi amfani da dabarar) vt2 = 2 gh
vt2 = 2 gh = 2(10)(2,5) = 2(25)
vt = √2(25) = 5√2
Don haka saurin ƙarshe na ƙwallon shine 5√2 m/s
Jimlar aiki = canjin kuzarin motsi
WTotal = ΔEK = ½ m (v)t2 - vo2) = ½ (1){(5√2)2 - 02}
WTotal = ½ (25)(2) = Joule 25
Amsar da ta dace ita ce D.
12. Abu mai nauyin kilogiram 2 yana fara motsawa da sauri na kilomita 72/h.-1Bayan motsi mita 400, saurin abu zai zama kilomita 144/h-1 kuma (g = 10 ms-2) Jimillar aikin da abin ya yi a wancan lokacin shine…
A. 20 J
B. 60 J
C. 1.200 J
D. 2.000 J
E. 2.400 J
Tattaunawa
An sani :
Nauyin abu (m) = 2 kg
Saurin farko (v)o) = 72 km/h = 20 m/s
Saurin ƙarshe (v)t) = 144 km/h = 40 m/s
Nisa da aka yi tafiya (s) = mita 400
Saurin gudu saboda nauyi (g) = 10 m/s2
An tambaya : Jimillar ƙoƙari
Jawab :
Jimlar aiki = canjin kuzarin motsi
WTotal = ΔEK = ½ m (v)t2 - vo2) = ½ (2)(40)2 - 202}
WTotal = ½ (2)(1600 – 400) = Joules 1200
Amsar da ta dace ita ce C.
Ka'idar makamashin aiki-kinetic
13. Mota mai nauyin tan 1, tana motsawa daga hutu. Bayan ɗan lokaci, saurinta ya kai mita 5/s.-1Babba kasuwanci abin da injin motar ke yi shi ne…
A. Joule 1000
B. 2.500 Joule
C. 5.000 Joule
D. 12.500 Joule
E. 25.000 Joule
Tattaunawa
An sani :
Nauyi (m) = tan 1 = kilogiram 1000
Saurin farko (v)o) = 0 (motar tana motsawa daga yanayin da ba a tsayawa ba)
Saurin ƙarshe (v)t) = 5 m/s
An tambaya : Menene aikin (W) da injin motar ke yi?
Jawab :
Ka'idar makamashin aiki-kinetic:
WTotal = ½ m (v)t2 - vo2)
Aikin da injin motar ke yi shine:
WTotal = ½ (1000)(5)2 - 02) = (500)(25 – 0) = (500)(25) = Joules 12.500
Amsar da ta dace ita ce D.
Ka'idar Aikin Inji-Makamashi
14. Kalli hoton motsin abubuwa a ƙasa!
Nauyin abu = 1 kg. Idan ma'aunin gogayya tsakanin tubalin da bene shine 0,2, to ƙimar matsar da abu (s) shine… g = 10 m/s2\
A. 10 m
B. 15 m
C. mita 20
D. 25 m
E. 30 m
Tattaunawa:
An sani :
m = 1 kg, g = 10 m/s2ao = 10 m/s, vt = 0
An tambaya :
Menene matsar da abu(s)?
Jawab :
Aiki - ka'idar makamashin inji :
Aikin da ƙarfin da ba na masu ra'ayin mazan jiya ba ke yi = canjin kuzarin injiniya na wani abu.
WNC = EM2 – EM1
WNC = (EP + EK)2 – (EP + EK)1 —- Daidaito ta 1
Ƙarfin da ba na ra'ayin mazan jiya ba wanda ke aiki akan abu shine ƙarfin gogayya na motsi:
WNC = fk s —- Daidaito na 2
Tsarin gogayya na motsa jiki :
Sauya fk a cikin lissafi na 2 da fk a cikin lissafi na 3:
WNC = fk s = 2 ku
Yayin da ake motsi, abu yana fuskantar canji a kuzarin motsi (ƙarfin motsi yana raguwa) yayin da kuzarin ƙarfin nauyi na abu baya canzawa (canjin kuzarin ƙarfin nauyi sifili ne).
WNC = (EK)2 – (EK)1 = ½ mvt2 – ½ mvo2 = ½ m (v)t2 - vo2)
2 s = ½(1)(0)2 - 102)
2 s = (0,5)(100)
2 s = 50
s = 50 / 2 = mita 25
Amsar da ta dace ita ce D.
Dokar kiyaye makamashin inji
15. Mai wasan kankara yana zamewa daga tsayin A kamar yadda aka nuna a hoton da ke ƙasa:
Idan saurin farko na mai tsere kan dusar ƙanƙara = sifili kuma saurin da aka samu sakamakon nauyi shine 10 ms-2, to saurin ɗan wasan a tsayin B shine...
A. √2 ms-1
B. 5√2 ms-1
C. 10√2 ms-1
D. 20√2 ms-1
E. 25√2 ms-1
Tattaunawa
An sani :
Saurin farko (v)o) = 0
Saurin gudu saboda nauyi (g) = 10 m/s2
Canjin tsayi = mita 50 – mita 10 = mita 40
An tambaya : saurin ɗan wasan a tsayin B?
Jawab :
Dokar kiyaye makamashin inji Ya ce makamashin inji na farko = makamashin inji na ƙarshe.
Lokacin da aka fara motsawa daga wuri na A, saurin farko na mai tsere kan dusar ƙanƙara = vo = 0 don haka ƙarfin motsi na farko = EKo = ½ mvo2 = 0. Akasin haka, idan a wurin A, mai tseren kankara yana da ƙarfin ƙarfin nauyi na farko = EPo = mgh, inda h = mita 50. Don haka makamashin injiniya na farko (EM)o) = ƙarfin ƙarfin nauyi na farko (EP)o) A lokacin motsi daga A zuwa B, tsayin mai tseren kankara yana raguwa, don haka kuzarin ƙarfin nauyi yana raguwa. Ƙarfin ƙarfin nauyi ba ya ɓacewa amma ana canza shi zuwa kuzarin motsi. Ƙarfin kuzarin motsi yana nuna ƙaruwar saurin mai tseren kankara. Idan mai tseren kankara ya kai ƙasan karkata, duk ƙarfin ƙarfin nauyi yana canzawa zuwa kuzarin motsi. A ƙasan karkata, ƙarfin ƙarfin nauyi sifili ne, yayin da kuzarin motsi yana kan iyakarsa. Don haka makamashin injiniya na ƙarshe (EM) shinet) = makamashin motsi na ƙarshe (EK)t).
Yaya batun kuzarin ƙarfin nauyi da kuzarin motsi a wurin B? A wurin B, wasu daga cikin kuzarin ƙarfin nauyi suna raguwa kuma suna canzawa zuwa kuzarin motsi. Musamman, kuzarin motsi a wurin B = raguwar kuzarin ƙarfin nauyi idan an ɗauka cewa abu yana cikin faɗuwa kyauta daga tsayin mita 40. Me zai hana a yi amfani da tsawon jirgin sama mai karkata amma tsayin mita 40? Idan ba ku fahimci wannan ba, ya kamata ku sake nazarin kayan. ƙarfin mazan jiya da alaƙarsa da makamashi mai yuwuwa sairta dokar kiyaye makamashin inji.
Gudun mai tsere a tsayin B:
Makamashin injiniya na ƙarshe = makamashin injiniya na farko
Makamashin motsi a wuri na B = makamashin ƙarfin nauyi a tsayin mita 40
EK = EP
½ mvt2 = mgh
½ vt2 = gh
½ vt2 = (10)(50-10)
½ vt2 = (10)(40)
½ vt2 = 400
vt2 = (2)(400) = 800
vt = √800 = √(2)(400) = 20√2 m/s
Amsar da ta dace ita ce D.
16. Abu yana motsawa daga wuri A ba tare da saurin farko ba. Idan babu gogayya yayin motsi, saurin abu a mafi ƙasƙanci shine...
A. 8 ms-1
B. 12 ms-1
C. 20 ms-1
D. 24 ms-1
E. 30 ms-1
Tattaunawa
An sani :
Nauyin abu = m
Saurin farko (v)o) = 0
Tsawo (h) = mita 20
Saurin gudu saboda nauyi (g) = 10 m/s2
An tambaya : saurin ƙarshe (vt)?
Jawab :
Makamashin injina na farko (EM)1) = makamashin ƙarfin nauyi a A (EPA) = mgh = (m)(10)(20) = 200 m
Makamashin injiniya na ƙarshe (EM)2) = kuzarin motsi (EK) = ½ mvt2
Saurin abu a mafi ƙasƙanci ko saurin ƙarshe na abu vt)?
Dokar kiyaye makamashin inji:
EM1 = EM2
200 m = ½ mvt2
200 = ½ vt2
400 = kut2
vt= 20m/s
Amsar da ta dace ita ce C.
Tushen tambaya:
Tambayoyin Nazarin Fizik na Ƙasa ga Makarantar Sakandare ta Babbar Sakandare/Makarantar Sakandare ta Sana'a
