Isibonelo semibuzo ye-Particle Dynamics

Izibonelo eziyi-14 zemibuzo ye-Particle Dynamics

1. Ibhulokhi A enesisindo esingu-5 kg ​​ibekwa endaweni ebushelelezi eyisicaba, ibhulokhi B enesisindo esingu-3 kg ilenga ngentambo futhi ixhunywe kubhulokhi A nge-pulley, uma u-g = 10 m/s ² unquma ukusheshisa kwebhulokhi!

A. 3,50 m/s²

B. 3,75 m/s ²

C. 4,00 m/s ²

D. 5,00 m/s ²

E. 5,25 m/s ²

Ingxoxo

Kuyaziwa ukuthi:

Ubuso obubushelelezi obuyisicaba.

Isisindo sebhulokhi A (m_A) = 5 kg

Isisindo sebhulokhi B (m_B) = 3 kg

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s ²

Isisindo sebhulokhi B (w)_B) = m_B g = (3)(10) = 30 Newton

Kubuziwe: Ukusheshisa kwebhulokhi (a)

Impendulo:

Ubuso obuyisicaba bubushelelezi ngakho-ke awekho amandla okungqubuzana avimbela ukunyakaza kwebhulokhi A. Amandla ashukumisa uhlelo lwebhulokhi isisindo sebhulokhi 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

Impendulo efanele ingu-B.

2. Ngokusekelwe esithombeni, kwaziwa ukuthi:

(1) ukusheshisa okungenalutho kwento

(2) izinto ezihamba emgqeni oqondile ngesivinini esingaguquki

(3) izinto ezisesimweni sokuphumula

(4) into izohamba uma isisindo sento singaphansi kwamandla okudonsa

Isitatimende esifanele sithi….

A. (1) kanye no-(2) kuphela

B. (1) kanye (3) kuphela

C. (1) kanye no-(4)

D. (1), (2) kanye no-(3) kuphela

E. (1), (2), (3) kanye no (4)

Ingxoxo

(1) ukusheshisa okungenalutho kwento

Ukusheshisa kwento kungu-zero uma amandla aphumayo elingana no-zero. Amandla aphumayo:

∑F = ma ukusheshisa (a) = 0

∑ F = 0

F ₁ + F ₂ – F ₃ = 12 + 24 – 36 = 36 – 36 = 0 N

(2) izinto ezihamba emgqeni oqondile ngesivinini esingaguquki

Ukusheshisa okungelutho kwento kungasho ukuthi into iphumule noma into ihamba ngomugqa oqondile ngesivinini esingaguquki (into ihamba ngesivinini esingaguquki).

(3) izinto ezisesimweni sokuphumula

Amandla angu-zero aphumelayo angasho ukuthi into iphumule.

(4) into izohamba uma isisindo sento singaphansi kwamandla okudonsa

Isisindo sento sisebenza siqonde phezulu, kuyilapho amandla okudonsa esebenza ngokuvundlile. Ngenxa yokuthi into ihamba ngokuvundlile, amandla avundlile kuphela athinta into.

Impendulo efanele ngu-D.

3. Bheka isithombe ohlangothini!

Uma i-coefficient of kinetic friction phakathi kwebhulokhi A kanye nethebula ingu-0,1 futhi ukusheshisa ngenxa yamandla adonsela phansi kungu-10 m/s^-2 bese kuba amandla okumele asetshenziswe ku-A ukuze uhlelo luhambe ngakwesobunxele ngokusheshisan 2 ms^-2 kuyinto ....

A. 70 N

B. 90 N

C. 150 N

D. 250 N

E. 330 N

Ingxoxo

Kuyaziwa ukuthi:

Isisindo sebhulokhi A (m_A) = 30 kg

Isisindo sebhulokhi A (w)_A) = (30 kg)(10 m/s²) = 300 kg m/s² noma amaNewton angu-300

Isisindo sebhulokhi B (m_B) = 20 kg

Isisindo sebhulokhi B (w)_B) = (20 kg)(10 m/s²) = 200 kg m/s2 noma amaNewton angu-200

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s ²

I-coefficient yokungqubuzana kwe-kinetic (= = 0,1

Ukusheshisa kwesistimu (a) = 2 m/s² (isiqondiso sokusheshisa ngakwesobunxele)

Amandla okungqubuzana kwe-kinetic (f)_k) = N= w_A = (0,1)(300) = 30 Newton

Umbuzo: Ubukhulu bamandla F buyini?

Impendulo:

Umthetho Wesibili KaNewton:

Σ F = ma

Uhlelo luya ngakwesobunxele

F – f_k - w_B = (m_A +m_B) futhi

F – 30 – 200 = (30 + 20)(2)

F – 230 = (50)(2)

F – 230 = 100

F = 230 + 100

F = 330 Newton

Impendulo efanele ngu-E.

4. Izinto ezimbili u-A no-B, ngokulandelana ezinesisindo esingu-2 kg kanye no-6 kg, ziboshwe ngentambo nge-pulley ebushelelezi njengoba kuboniswe esithombeni. Into yokuqala u-B iyabanjwa bese idedelwa. Uma u-g = 10 ms^-2 khona-ke ukusheshisa kwento B…

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

Ingxoxo

Kuyaziwa ukuthi:

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

Umbuzo: ukusheshisa kwento B noma ukusheshisa kwesistimu?

Impendulo:

w_B > w_A Ngakho-ke into B iya phansi, into A iya phezulu (uhlelo luhamba ngokwewashi).

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

Impendulo efanele ngu-D.

5. Into enobunzima ixhunywe ngentambo edlula epulangweni elibushelelezi njengoba kuboniswe esithombeni. Uma u-m₁ = 1 kg, m₂ = 2 kg, kanye no-g = 10 ms^-2, khona-ke i-voltage T ingu ....

A. 10,2 N

B. 13,3 N

C. 15,5 N

D. 18,3 N

E. 20,0 N

Ingxoxo

Kuyaziwa ukuthi:

m ₁ = 1 kg, m ₂ = 2 kg, g = 10 m/s ²

w₁ = m₁ g = (1 kg)(10 m/s²) = 10 kg m/s² noma amaNewton angu-10

w₂ = m₂ g = (2 kg)(10 m/s²) = 20 kg m/s² noma amaNewton angu-20

Umbuzo: Iyini amandla okucindezela entambo (T)?

Impendulo:

Ukusheshisa uhlelo

w ₂ > w ₁ ngakho-ke uhlelo luhamba ngokwewashi (m ₂ uya phansi, m ₁ uya phezulu).

Umthetho Wesibili KaNewton:

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

Ukusheshisa kwesistimu kungu-3,3 m/s².

Ukucindezeleka kwentambo?

m2 _ihlela phansi

w ₂ – T ₂ = m ₂ a

20 – T ₂ = (2)(3,33)

20 – T ₂ = 6,66

Isigaba ₂ = 20 – 6,66

I-T ₂ = 13,3 Newton

m ₁ iya phezulu

I-T ₁ – w ₁ = m ₁ a

I-T ₁ – 10 = (1)(3,3)

Isigaba ₁ – 10 = 3,33

I-T ₁ = 10 + 3,33

I-T ₁ = 13,3 Newton

Ukuqina kwentambo (T) = 13,3 amaNewton.

Impendulo efanele ingu-B.

6. Bheka isithombe ohlangothini! Isisindo sebhulokhi ngayinye singu-m.₁ = 6 kg kanye m₂ = 4 kg futhi isisindo se-pulley asinakwa. Uma ubuso bendiza bubushelelezi futhi g = 10 ms^-2, khona-ke ukusheshisa kohlelo….

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

Ingxoxo

Kuyaziwa ukuthi:

m₁ = 6 kg, m₂ = 4 kg, g = 10 m/s²

w₁ = m₁ g = (6 kg)(10 m/s²) = 60 kg m/s² noma amaNewton angu-60

w₂ = m₂ g = (4 kg)(10 m/s²) = 40 kg m/s² noma amaNewton angu-40

Kubuziwe: ukusheshisa kwesistimu (a)?

Impendulo:

m₁ isendaweni ebushelelezi, eyisicaba engenakho ukungqubuzana ukuze uhlelo luqhutshwe amandla adonsela phansi ebhulokhi 2.

Sebenzisa umthetho wesibili kaNewton:

∑ 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

Impendulo efanele ngu-D.

7. Amabhulokhi amabili, ngalinye linesisindo esingu-2 kg, axhunywe ngentambo kanye ne-pulley njengoba kuboniswe esithombeni. Ubuso kanye ne-pulley kubushelelezi. Uma ibhulokhi B lidonswa ngamandla avundlile angu-40 N, ukusheshisa kwebhulokhi kungu… (g = 10 m/s²)

A. 5 m/s²

B. 7,5 m/s ²

C. 10 m/s ²

D. 12,5 m/s ²

E. 15 m/s ²

Ingxoxo:

Kuyaziwa ukuthi:

m_A = m_B = 2 kg, g = 10 m/s², F = 40 N

w_A = mg = (2)(10) = 20 N

Umbuzo: ukusheshisa kwebhulokhi (a)?

Impendulo:

Ubuso bebhulokhi bubushelelezi, ngakho-ke amandla athinta ukunyakaza kwebhulokhi ngamandla F kuphela kanye nesisindo sebhulokhi A.

Sebenzisa umthetho wesibili kaNewton:

∑ 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

Impendulo efanele ngu-A.

8. Kusukela esithombeni esilandelayo, ibhulokhi A inesisindo esingu-2 kg kanti ibhulokhi B = 1 kg. Ibhulokhi B liqala limile bese liyehla lize lithinte phansi. Uma g = 10 ms^-2, inani lokucindezeleka kwentambo u-T liyi…

A. 20,0 AmaNewton

B. 10,0 AmaNewton

C. 6,7 AmaNewton

D. 3,3 AmaNewton

E. 1,7 AmaNewton

Ingxoxo

Kuyaziwa ukuthi :

Isisindo sebhulokhi A (m_A) = 2 kg

Isisindo sebhulokhi B (m_B) = 1 kg

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s ²

Isisindo sebhulokhi B (w)_B) = m_B g = (1)(10) = 10 Newton

Kubuziwe : Inani lokucindezeleka kwentambo (T)

Impendulo :

Embuzweni akukho lwazi mayelana nokungqubuzana, ngakho-ke vele ungayinaki ingqubuzana.

Ukusheshisa uhlelo (a)

Ukubala kokuqala ukusheshisa uhlelo Kusetshenziswa umthetho wesibili kaNewton, ibhlogo B lilenga ukuze isisindo sebhloko B sihambe ibhloko B siye phansi. Amabhlogo B kanye nebhloko A axhunywe ngentambo ukuze ibhloko B lidonse ibhloko A kuze kube yilapho zombili zihamba ndawonye. Umehluko ukuthi ibhloko B liyehla bese ibhloko A liye kwesokudla. Kukhona amandla elilodwa kuphela ahambisana nokunyakaza kwamabhlogo womabili, okungukuthi isisindo sebhloko B (wB). Isisindo sebhloko A siqonde ngqo esiqondisweni sokunyakaza. Ukuhamba kwebhulokhi A akucatshangelwa ekuxazululeni inkinga. Amandla okucindezela entanjeni anobukhulu obufanayo ngobude bentambo futhi ahlukene ngakho ayahlukana.

∑ F = ma

w _B = (m _A + m _B ) a

10 = (2 + 1) a

10 = 3 a

a = 10/3

Ukuqina kwentambo (T)

Ukucindezeleka entanjeni kubalwa ngokucabangela ibhulokhi ngayinye ngokwehlukana.

Ukucindezeleka entanjeni ebhulokini A

∑ F = ma

T = m _A a = (2)(10/3) = 20/3 = 6,7 AmaNewton

Ukucindezeleka entanjeni ebhulokini 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

Ukuqina kwentambo (T) = 6,7 Newton

Impendulo efanele ngu-C.

9. Kusukela esithombeni esilandelayo, ibhulokhi A inesisindo esingu-2 kg kanti ibhulokhi B = 1 kg. Uma amandla okungqubuzana phakathi kwento A nendiza engu-2,5 Newton, kuyilapho amandla okungqubuzana entambo ene-pulley enganakwa, khona-ke ukusheshisa kwezinto zombili ku…

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

Ingxoxo

Kuyaziwa ukuthi :

Isisindo sebhulokhi A (m_A) = 2 kg

Isisindo sebhulokhi B (m_B) = 1 kg

Amandla okungqubuzana phakathi kwebhulokhi A kanye nobuso obuyisicaba (f_{i-ges A}) = 2,5 Newton

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s ²

Isisindo sebhulokhi B (w)_B) = m_B g = (1)(10) = 10 Newton

Kubuziwe Ukusheshisa kwezinto zombili (a)

Impendulo :

Ukusheshisa kwezinto zombili kubalwa kusetshenziswa ifomula yomthetho wesibili kaNewton.

∑ F = ma

w_B - f_ges = (m_A +m_B) futhi

10 – 2,5 = (2 + 1) a

7,5 = 3 a

a = 7,5 / 3 = 2,5 m/s ²

Impendulo efanele ngu-E.

10. Bheka isithombe! Ibhulokhi A enesisindo esingama-30 kg elele phansi ebushelelezi ixhunywe kubhulokhi B enesisindo esingama-10 kg nge-pulley. Ibhulokhi B ibanjwa ekuqaleni bese idedelwa ukuze yehle. Ukusheshisa kwesistimu… (g = 10 m/s^-2)

A. 2,5 ms^-2

B. 10 ms ^-2

C. 12 ms ^-2

D. 15 ms ^-2

E. 18 ms ^-2

Ingxoxo

Kuyaziwa ukuthi :

Isisindo sebhulokhi A (m_A) = 30 kg

Isisindo sebhulokhi B (m_B) = 10 kg

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s ²

Isisindo sebhulokhi B (w)_B) = m_B g = (10)(10) = 100 Newton

Kubuziwe : Ukusheshisa uhlelo (a)

Impendulo :

∑ 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

Impendulo efanele ngu-A.

11. Amabhulokhi A no-B, ngalinye linesisindo esingu-8 kg kanye no-12 kg, abekwe etafuleni njengoba kuboniswe esithombeni. I-coefficient of friction phakathi kwebhulokhi A kanye nethebula ingu-0,3. Ibhulokhi C inesisindo esingama-4 kg bese ibekwa phezu kwebhulokhi A. Yisiphi isitatimende esilandelayo esinembile?

A. Ukucindezeleka entanjeni kukhulu kunangaphambili, ukusheshisa kuncane kunangaphambili

B. Ukucindezeleka entanjeni kanye nokusheshisa kohlelo akushintshi.

C. Ukucindezeleka entanjeni kuncane kunangaphambili, ukusheshisa kukhulu kunangaphambili.

D. Ukucindezeleka entanjeni kukhulu kunangaphambili, ukusheshisa kuhlala kungaguquki.

E. Ukucindezeleka entanjeni kuhlala kungaguquguquki kuyilapho ukusheshisa kuncane kunangaphambili.

Ingxoxo

Kuyaziwa ukuthi:

Isisindo sebhulokhi A (m_A) = 8 kg

Isisindo sebhulokhi B (m_B) = 12 kg

Isisindo sebhulokhi C (m_C) = 4 kg

I-coefficient yokungqubuzana phakathi kwebhulokhi A nethebula (μ_k= = 0,3

Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s ²

Isisindo sebhulokhi A (w)_A) = m_A g = (8 kg)(10 m/s²) = 80 kg m/s²

Isisindo sebhulokhi B (w)_B) = m_B g = (12 kg)(10 m/s²) = 120 kg m/s²

Amandla okungqubuzana phakathi kwebhulokhi A netafula (f_k) = μ_k N_A = μ_k w_A = (0,3)(80) = 24 N

Impendulo:

Ukusheshisa uhlelo:

Σ F = ma

w _B – f _k = (m _A + m _B ) a

120 – 24 = (8 + 12) futhi

96 = (20) futhi

a = 96/20

a = 4,8 m/s ²

Ukucindezeleka kwentambo:

Cabanga ngenye yezinto, isibonelo u-B.

Σ F = ma

w_B - T = (m_B) futhi

120 – T = (124,8

120 – T = 57,6

T = 120 – 57,6

T = 62,4 Newton

Ibhulokhi C elinobunzima obungama-4 kg libe selibekwa phezu kwebhulokhi A.

Ukusheshisa uhlelo:

Σ F = ma

w_B - f_k = (m_A +m_B +m_C) futhi

120 – 24 = (8+12+4) futhi

96 = (24) futhi

a = 96/24

a = 4 m/s ²

Ukucindezeleka kwentambo:

Cabanga ngenye yezinto, isibonelo u-B.

Σ F = ma

w_B - T = (m_B) futhi

120 – T = (124

120 – T = 48

T = 120 – 48

T = 72 Newton

Impendulo efanele ngu-A.

12. Kusukela esithombeni esilandelayo, ibhulokhi A inesisindo esingu-2 kg kanti ibhulokhi B = 1 kg. Ibhulokhi B liqala limile bese liyehla lize lithinte phansi. Uma g = 10 ms^-2, inani lokucindezeleka kwentambo u-T liyi…

A. 20,0 AmaNewton
B. 10,0 AmaNewton
C. 6,7 AmaNewton
D. 3,3 AmaNewton
E. 1,7 AmaNewton

Ingxoxo
I-Diketahui :
Isisindo sebhulokhi A (m_A) = 2 kg
Isisindo sebhulokhi B (m_B) = 1 kg
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s²
Isisindo sebhulokhi B (w)_B) = m_B g = (1)(10) = 10 Newton
Kubuziwe : Inani lokucindezeleka kwentambo (T)
Jawab :

Embuzweni akukho lwazi mayelana nokungqubuzana, ngakho-ke vele ungayinaki ingqubuzana.

Ukusheshisa uhlelo (a)
Ukubala kokuqala ukusheshisa uhlelo usebenzisa ifomula Umthetho wesibili kaNewton. Ibhulokhi B ilengiswe ukuze kube namandla adonsela phansi ebhulokhi B ahambisa ibhulokhi B phansi. Ibhulokhi B kanye nebhulokhi A zixhunywe ngentambo ukuze ibhulokhi B lidonse ibhulokhi A kuze kube yilapho zombili zihamba ndawonye. Umehluko ukuthi ibhulokhi B liyehla bese ibhulokhi A liye kwesokudla. Kukhona amandla ahambisana nokunyakaza kwamabhulokhi womabili, okungukuthi amandla adonsela phansi ebhulokhi B (wB). Amandla adonsela phansi Ibhulokhi A iqonde ngqo endaweni yokunyakaza kwebhulokhi A ngakho ayicatshangelwa ekuxazululeni inkinga. Amandla okucindezela entanjeni anobukhulu obufanayo ngobude bentambo futhi ahlukene ngakho ayahlukana.
∑F = ma
w_B = (m_A +m_B) futhi
10 = (2 + 1) a
10 = 3 a
a = 10/3

Ukuqina kwentambo (T)
Ukucindezeleka entanjeni kubalwa ngokucabangela ibhulokhi ngayinye ngokwehlukana.
Ukucindezeleka entanjeni ebhulokini A
∑F = ma
T = m_A a = (2)(10/3) = 20/3 = 6,7 amaNewton
Ukucindezeleka entanjeni ebhulokini 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
Ukuqina kwentambo (T) = 6,7 Newton
Impendulo efanele ngu-C.

13. Kusukela esithombeni esilandelayo, ibhulokhi A inesisindo esingu-2 kg kanti ibhulokhi B = 1 kg. Uma amandla okungqubuzana phakathi kwento A kanye nendiza engu-2,5 Newton, kuyilapho amandla okungqubuzana entambo kanye ne-pulley enganakwa, khona-ke ukusheshisa kwezinto ezimbili...

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

Ingxoxo
I-Diketahui :
Isisindo sebhulokhi A (m_A) = 2 kg
Isisindo sebhulokhi B (m_B) = 1 kg
Amandla okungqubuzana phakathi kwebhulokhi A kanye nobuso obuyisicaba (f_{i-ges A}) = 2,5 Newton
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s²
Isisindo sebhulokhi B (w)_B) = m_B g = (1)(10) = 10 Newton
Kubuziwe Ukusheshisa kwezinto zombili (a)
Jawab :
Ukusheshisa kwezinto zombili kubalwa kusetshenziswa ifomula yomthetho wesibili kaNewton.
∑F = ma
w_B - f_ges = (m_A +m_B) futhi
10 – 2,5 = (2 + 1) a
7,5 = 3 a
a = 7,5 / 3 = 2,5 m/s²
Impendulo efanele ngu-E.

14. Bheka isithombe! Ibhulokhi A enesisindo esingama-30 kg elele phansi ebushelelezi ixhunywe kubhulokhi B enesisindo esingama-10 kg nge-pulley. Ibhulokhi B ibanjwa ekuqaleni bese idedelwa ukuze yehle. Ukusheshisa kwesistimu… (g = 10 m/s^-2)
A. 2,5 ms^-2
B. 10 ms^-2
C. 12 ms^-2
D. 15 ms^-2
E. 18 ms^-2
Ingxoxo
I-Diketahui :
Isisindo sebhulokhi A (m_A) = 30 kg
Isisindo sebhulokhi B (m_B) = 10 kg
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s²
Isisindo sebhulokhi B (w)_B) = m_B g = (10)(10) = 100 Newton
Kubuziwe : Ukusheshisa uhlelo (a)
Jawab :
∑F = ma
w_B = (m_A +m_B) futhi
100 = (30 + 10) a
100 = 40 a
a = 100/40
a = 2,5 m/s²
Impendulo efanele ngu-A.

Umthombo wombuzo:

Imibuzo Yefiziksi Yezivivinyo Zikazwelonke Zesikole Samabanga Aphezulu/Isikole Samabanga Aphezulu Sokufundela Umsebenzi