fisiks

Ho sisinyeha ha sefofane se sekametseng ka mokhoa o batlang o le mothating o itseng ka matla a khohlano – ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa ho sisinyeha

1. Ntho ea boima = 2 kg, ho potlaka ka lebaka la matla a khoheli = 9.8 m/s², coefficient ea khohlano e sa fetoheng = 0.2, coefficient ea kinetic friction = 0.1. Na ntho e phomotse kapa e potlakile? Haeba ntho e potlakile, fumana (a) matla a letlooa (b) boholo le tataiso ea lebokose ho potlaka!

Ho sisinyeha ha sefofane se sekametseng ka matla a khohlano - ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa motsamao 1

tharollo

Ho sisinyeha ha sefofane se sekametseng ka matla a khohlano - ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa motsamao 2

Tse tsejoang:

Boima (m) = 2 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s²

Khokahano ea khohlano e sa fetoheng (μ_s) = 0.2

Khokahano ea kinetic ea khohlano (μ_k) = 0.1

Boima (w) = mg = (2)(9.8) = 19.6 Newton

Karolo e rapameng ea boima ba 'mele (w_x) = sebe sa 30^o = (19.6)(0.5) = 9.8 Newton

Karolo e otlolohileng ea boima ba th (w_y) = w cos 30^o = (19.6)(0.5√3) = 9.8√3 Li-Newton

Matla a tloaelehileng (N) = w_y = 9.8√3 Newton

Matla a khohlano e sa fetoheng (f)_s) = (0.2)(9.8√3) = 1.96√3 Li-Newton = 3.39 Li-Newton

Matla a moferefere oa kinetic (f)_k) = (0.1)(9.8√3) = 0.98√3 Li-Newton = 1.69 Li-Newton

Tharollo:

Ntho e phomotse haeba w_x < f_s, ntho e ntse e theohela tlase haeba w_x > f_s.

w_x = 9.8 Newton le f_s= 3.39 Li-Newton.

(a) matla a letlooa

∑F = w_x- f_k = 9.8 – 1.69 = 8.11 Newton

(b) boholo le tataiso ea ho potlaka

∑F = ma

8.11 = (2) a

= 4.05

Boholo ba ho potlaka = 4.05 m/s²le tataiso ea ho potlaka = ho ea tlase.

2. Boima ba ntho = 4 kg, ho potlaka ka lebaka la matla a khoheli = 9,8 m/s². Koefficient ea kinetic friction = 0.2 le coefficient ea static friction = 0.4. Boholo ba matla F = 40 Newtons. Ntho e phomotse kapa e thella fatše? Haeba ntho e thella tlase, fumana (a) matla a net (b) boholo le tataiso ea ho potlaka!

Ho sisinyeha ha sefofane se sekametseng ka matla a khohlano - ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa motsamao 3

tharollo

Ho sisinyeha ha sefofane se sekametseng ka matla a khohlano - ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa motsamao 4

Tse tsejoang:

Boima (m) = 4 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s²

Coefficient ea khohlano e sa fetoheng (μ_s) = 0.4

Coefficient ea moferefere oa kinetic (μ_k) = 0.2

Boima (w) = mg = (4)(9.8) = 39.2 Li-Newton

Karolo e rapameng ea boima (w_x) = sebe sa 30^o = (39.2)(0.5) = 19.6 Newton

Karolo e otlolohileng ea boima (w_y) = w cos 30^o = (392)(0..5√3) = 19.6√3 Li-Newton

Matla a tloaelehileng (N) = w_y = 19.6√3 Li-Newton = 33.95 Li-Newton

matla a khohlano a sa fetoheng (f)_s) = μ_sN= (0,4)(33.95) = 13.58 Newton

Matla a ho qabana ha kinetic (f)_k) = μ_kN= (0.2)(33.95) = 6.79 Newton

F = 40 Newton

Tharollo:

Ntho e thella fatshe haeba F < w_x +f_sNtho e thellela hodimo haeba F > w_x +f_s.

F = 40 Newton, w_x = 19.6 Newton le f_s= 13.58 Li-Newton.

F e kholo ho feta w_x +f_skahoo ntho eo e thella hodimo.

(a) Matla a letlooa

∑F = F – w_x - f_k = 40 – 19.6 – 6.79 = 13.61 Li-Newton

(b) Boholo le tataiso ea ho potlaka

∑F = ma

6.4 = (4) a

= 1.6

Boholo ba lebelo ke 1.6 m/s²'me tataiso ea ho potlaka e ea holimo.

[ID ea sephutheloana sa wpdm='481′]

Bala haholoanyane

Ho sisinyeha ha sefofane se sekametseng ntle le matla a khohlano – ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa ho sisinyeha

1. Lebokose boima = 2 kg, ho potlaka ka lebaka la matla a khoheli = 9.8 m/s²Fumana (a) matla a letlooa a potlakisang lebokose ho ya tlase (b) boholo ba lebokose ho potlaka.

Ho sisinyeha ha sefofane se sekametseng ntle le matla a khohlano - ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa ho sisinyeha 1

tharollo

Ho sisinyeha ha sefofane se sekametseng ntle le matla a khohlano - ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa ho sisinyeha 2

Tse tsejoang:

Boima (m) = 2 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s²

Weight (w) = mg = (2)(9.8) = 19.6 Li-Newton

w_x = w sebe 30 = (19.6)(0.5) = 9.8 Newton

w_y = w cos 30 = (19.6)(0.5√3) = 9.8√3 Li-Newton

Tharollo:

(A) The letlooa bakeng sace e potlakisang lebokose

Sepakapaka se sekametseng se boreleli, kahoo ha ho na matla a khohlano. Matla feela a sebetsang nthong ke w_x.

∑F = w_x

∑F = 9.8 Newton

(B) boholo ba ho potlaka

∑F = ma

9.8 = (2) a

a = 9.8/2

a = 4.9 m/s²

Boholo ba lebelo ke 4.9 m/s², tsela eo lebelo le tsamayang ka yona e theohela tlase.

2. Sefofane se sekametseng e boreleli kahoo ha ho na matla a khohlanoBoima ba ntho ke 3 kg, lebelo ka lebaka la matla a khoheli ke 9.8 m/s²Fumana boholo ba matla F haeba (a) ntho e phomotse (b) ntho e theohela tlase ka lebelo le sa fetoheng la 2 m/s² (c) ntho e ntse e nyolohela hodimo ka lebelo le sa fetoheng la 2 m/s².

Ho sisinyeha ha sefofane se sekametseng ntle le matla a khohlano - ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa ho sisinyeha 3

tharollo

Ho sisinyeha ha sefofane se sekametseng ntle le matla a khohlano - ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa ho sisinyeha 4

Tse tsejoang:

Boima (m) = 3 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s²

Boima (w) = mg = (3)(9.8) = 29.4 Li-Newton

w_x = w sebe 30 = (29.4)(0.5) = 14.7 Newton

w_y= w cos 30 = (29.4)(0.5√3) = 14.7√3 Li-Newton

Tharollo:

(a) Boholo ba matla F haeba ntho e phomotse

Molao oa pele oa Newton ea motsamao e bolela hore haeba ntho e phomotse, matla a letlooa a sebetsang nthong ke lefela.

∑F=0

F – w_x = 0

F = w_x

F = 14.7 Newton

(b) Boholo ba matla F haeba ntho e theohela tlase ka lebelo le sa fetoheng la 2 m/s²

∑F = ma

w_x – F = ma

14.7 – F = (3)(2)

14.7 – F = 6

F = 14.7–6

F = 8.7 Newton

∑F = ma

F – w_x = ma

F – 14.7 = (3)(2)

F – 14.7 = 6

F = 14.7 + 6

F = 20.7 Newton

[ID ea sephutheloana sa wpdm='479′]

Bala haholoanyane

Motsamao oa 'mele e 'meli e nang le lebelo le tšoanang holim'a bokaholimo bo bataletseng bo bataletseng ka matla a khohlano - mathata le litharollo

1. Boholo ea lebokose la 1 ke 2 kg, boima ba lebokose la 2 ke 4 kg, ho potlaka ha matla a khoheli ke 10 m/s², boholo ba matla F ke 40 Newtons. Koefficient ea khohlano ea kinetic pakeng tsa lebokose la 1 le fatše ke 0.2 'me koefficient ea khohlano ea kinetic pakeng tsa lebokose la 2 le fatše ke 0.3. Fumana (a) Boholo le tataiso ea lebokose ho potlaka (b) Boholo ba matla a sebelisoang ke lebokose la 1 lebokoseng la 2 (F)₁₂) le boholo ba matla a sebelisoang ke lebokose la 2 lebokoseng la 1 (F₂₁).

Motsamao oa 'mele e 'meli e nang le lebelo le tšoanang holim'a sebaka se bataletseng se bataletseng ka matla a khohlano - mathata le litharollo 1

tharollo

Motsamao oa 'mele e 'meli e nang le lebelo le tšoanang holim'a sebaka se bataletseng se bataletseng ka matla a khohlano - mathata le litharollo 2

Tse tsejoang:

Boima ba lebokose 1 (m₁) = 2 kg

Boima ba lebokose 2 (m₂) = 4 kg

Ho potlaka ha matla a khoheli (g) = 10 m/s²,

Matla F = 40 Newton,

Coefficient of khohlano ea kinetic pakeng tsa lebokose la 1 le mokatong (μ_k1) = 0.2

Koefficient ea khohlano ea kinetic pakeng tsa lebokose la 2 le fatše (μ_k2) = 0.3

The boima ba 'mele ea lebokose la 1 (w₁) = m₁ g = (2)(10) = 20 Newton

Boima ba lebokose 2 (w₂) = m₂ g = (4)(10) = 40 Newton

The matla a tloaelehileng e hlahisitsoeng lebokoseng la 1 (N₁) = w₁ = 20 Newton

Matla a tloaelehileng a sebelisoang lebokoseng la 2 (N₂) = w₂ = 40 Newton

Matla a moferefere oa kinetic o sebelisitsoeng lebokoseng la 1 (f_k1) = (μ_k1)(N₁) = (0.2)(20) = 4 Newton

Matla a moferefere oa kinetic o sebelisitsoeng lebokoseng la 2 (f_k2) = (μ_k1)(N₂) = (0.3)(40) = 12 Newton

Tharollo:

(a) Boholo le tataiso ea lebelo la lebokose

ΣF = ma

F - f_k1 - f_k2= (m₁ +m₂) a

40 – 4 – 12 = (2 + 4) a

24 = 6 a

a = 24/6

a = 4 m/s²

Tsela ea ho potlakisa = tsela ea matla a letlooa = tsela e eang ka ho le letona.

(b) Boholo ba matla a sebelisoang ke lebokose la 1 lebokoseng la 2 (F)₁₂) le boholo ba matla a sebelisoang ke lebokose la 2 lebokoseng la 1 (F₂₁).

Bala boholo ba F₁₂ :

ΣF = ma

F₁₂ - f_k2= (m₂) a

F₁₂ – 12 = (4)(4)

F₁₂ - 12 = 16

F₁₂ = 16 + 12

F₁₂ = 28 Newton

F₁₂ le F₂₁ke matla a ketso le karabelo a sebetsang linthong tse fapaneng.₁₂ le F₂₁ e na le boholo bo tšoanang le lehlakore le fapaneng.

F₁₂ = 28 Newtons = F₂₁ = 28 Li-Newton.

2. Boima ba lebokose la 1 ke 2 kg, boima ba lebokose la 2 ke 4 kg, ho potlaka ha matla a khoheli ke 10 m/s², matla a F ke 40 N. Koefficient ya kgohlano ya kinetic pakeng tsa lebokose la 1 le mokatong ke 0.2 mme koefficient ya kgohlano ya kinetic pakeng tsa lebokose la 2 le mokatong ke 0.3. Fumana (a) Boholo le tataiso ya ho potlakisa (b) Kgatelelo thapong e hokahanyang mabokose. Hlokomoloha boima ba thapo.

Motsamao oa 'mele e 'meli e nang le lebelo le tšoanang holim'a sebaka se bataletseng se bataletseng ka matla a khohlano - mathata le litharollo 3

Tse tsejoang:

Boima ba lebokose 1 (m₁) = 2 kg

Boima ba lebokose 2 (m₂) = 4 kg

Ho potlaka ha matla a khoheli (g) = 10 m/s²,

Matla F = 40 Newton,

Koefficient ea khohlano ea kinetic pakeng tsa lebokose la 1 le fatše ke 0.2 (μ_k1) = 0.2

Koefficient ea khohlano ea kinetic pakeng tsa lebokose la 2 le fatše ke 0.2 (μ_k2) = 0.3

Boima ba lebokose 1 (w₁) = m₁ g = (2)(10) = 20 Newton

Boima ba lebokose 2 (w₂) = m₂ g = (4)(10) = 40 Newton

Matla a tloaelehileng a sebelisoang lebokoseng la 1 (N₁) = w₁ = 20 Newton

Matla a tloaelehileng a sebelisoang lebokoseng la 2 (N₂) = w₂ = 40 Newton

Matla a moferefere oa kinetic o sebelisitsoeng lebokoseng la 1 (f_k1) = (μ_k1)(N₁) = (0.2)(20) = 4 Newton

Matla a moferefere oa kinetic o sebelisitsoeng lebokoseng la 2 (f_k2) = (μ_k1)(N₂) = (0.3)(40) = 12 Newton

Tharollo:

(a) boholo le tataiso ea ho potlaka

ΣF = ma

F - f_k1 - f_k2= (m₁ +m₂) a

40 – 4 – 12 = (2 + 4) a

24 = 6 a

a = 24/6

a = 4 m/s²

Boholo ba lebelo ke 4 m/s², tataiso ea ho potlakisa = tataiso ea matla a letlooa = tataiso e ka ho le letona.

(b) Kgatelelo ka hara thapo

Matla a sebetsang lebokoseng la 1 ka lehlakoreng le otlolohileng ke khatello ea 1 (T₁) ka ho le letona le matla a khohlano ea kinetic 1 (f_k1) ka letsohong le letšehali. Sebelisa molao oa bobeli oa Newton:

ΣF = ma

T₁ - f_k1= m₁ a

T₁ - 4 = (2)(4)

T₁ - 4 = 8

T₁= 8 + 4 = 12 Newton

Matla a sebetsang lebokoseng la 2 ka lehlakoreng le otlolohileng ke khatello ea 2 (T₂) ka letsohong le letšehali le matla a khohlano ea kinetic 2 (f_k2) ka ho le letona. Etsa kopo Molao oa bobeli oa Newton :

ΣF = ma

F – T₂ - f_k2= m₂ a

40 – T₂ – 12 = (4)(4)

28 – T₂ = 16

T₂= 28 – 16 = 12 Newton

Khatello e thapong e hokahanyang mabokose = T₁= T₂ = T = 12 Newton.

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Bala haholoanyane

Ho sisinyeha holim'a bokaholimo bo otlolohileng ntle le matla a khohlano - ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa ho sisinyeha

1. Boima ba ntho ya 1 ke 2 kg, boima ba ntho ya 2 ke 4 kg, ho potlaka ha matla a khoheli ke 10 m/s², boholo ba matla F ke 12 Newtons. Fumana boholo le tataiso ea ho potlaka ha lintho.

Ho sisinyeha holim'a bokaholimo bo otlolohileng ntle le matla a khohlano – ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa ho sisinyeha 1

Tse tsejoang:

m₁ = 2 kg, m₂ = 4 kg, g = 10 m/s², F = 12 Newton

o ne a batla : a

Tharollo:

ΣF = ma

F = (m₁ +m₂) a

12 = (2 + 4) a

12 = 6 a

a = 12/6

a = 2 m/s²

Boholo ba lebelo ke 2 m/s², tataiso ea ho potlakisa = tataiso ea matla a letlooa = tataiso e ka ho le letona.

2. Boholo ea ntho ea 1 ke 2 kg, boima ba ntho ea 2 ke 4 kg, ho potlaka ha matla a khoheli ke 10 m/s², boholo ba matla F ke 24 N. Fumana boholo le tataiso ea ho potlaka.

Ho sisinyeha holim'a bokaholimo bo otlolohileng ntle le matla a khohlano – ts'ebeliso ea mathata le litharollo tsa molao oa Newton oa ho sisinyeha 2

Tse tsejoang:

m₁ = 2 kg, m₂ = 4 kg, g = 10 m/s², F = 24 Newton

Oa Batla: ho potlakisa (a)

Tharollo:

ΣF = ma

F = (m₁ +m₂) a

24 = (2 + 4) a

24 = 6 a

a = 24/6

a = 4 m/s²

Tsela eo ho potlakiswang ka yona = tsela eo matla a letlowa a tsamayang ka yona = tsela e yang ka ho le letona.

[ID ea sephutheloana sa wpdm='474′]

Bala haholoanyane

Matla a khohlano e sa fetoheng le ea kinetic - mathata le litharollo

Mathata a rarollotsoeng melaong ea Newton ea motsamao - Matla a khohlano e sa fetoheng le ea kinetic

1. Ntho e lutse mokatong o bataletseng. Koefficient ea khohlano e sa fetoheng ke 0.4 'me ho potlaka ha matla a khoheli ke 9.8 m/s²Fumana (a) Matla a maholo a khohlano e sa fetoheng (b) Matla a bonyane a F

Matla a khohlano e sa fetoheng le e tsamaeang ka kinetic - mathata le litharollo 1

tharollo

Matla a khohlano e sa fetoheng le e tsamaeang ka kinetic - mathata le litharollo 2

Tse tsejoang:

Boholo (m) = 1 kg

Palo ea khohlano e sa fetoheng (μ_s) = 0.4

Ho potlaka ha matla a khoheli (g) = 9.8 m/s²

Weight (w) = mg = (1 kg)(10 m/s²) = 10 kg m/s² = 10 Newton

Matla a tloaelehileng (N) = w = 10 Newton

Ho batloa:

(A) Matla a phahameng ka ho fetisisa a khohlano e sa fetoheng (b) The matla a bonyane a F

Tharollo:

(A) Matla a phahameng ka ho fetisisa a khohlano e sa fetoheng

f_s = μ_sN

f_s = (0.4)(9.8 N) = 3.92 Newton

(b) The matla a bonyane a F

Haeba matla a F a sebediswa nthong empa ntho e sa sisinyehe, kahoo ho tlameha ho be le matla a kgohlano e sa fetoheng a sebediswang ke fatshe nthong eo. Haeba ntho e tla qala ho tsamaya, matla a kgohlano e sa fetoheng a feta, ho tlameha ho be le matla a kgohlano ya kinetic. Ntho e qala ho tsamaya haeba F e le kgolo ho feta matla a hodimo a kgohlano e sa fetoheng.

Kahoo matla a bonyane a F = matla a maholo a khohlano e sa fetoheng = 3.92 Li-Newton.

2. Lebokose la 1 kg le hulwa hodima bokahodimo bo bataletseng ke matla a F, kahoo lebokose le tsamaya ka lebelo le sa fetoheng. Haeba coefficient ya kinetic friction e le 0.1, fumana boholo ba matla a F! (g = 9.8 m/s²)

Matla a khohlano e sa fetoheng le e tsamaeang ka kinetic - mathata le litharollo 3

Tse tsejoang:

Koefficient ea kinetic friction (μ_k= 0.1

Boima ba lebokose (m) = 1 kg

Ho potlaka ha matla a khoheli (g) = 9.8 m/s²

Boima (w) = mg = (1 kg)(9.8 m/s²) = 9.8 kg m/s² = 9.8 Newton

Matla a tloaelehileng (N) = w = 9.8 Newtons

o ne a batla : F

Tharollo:

Molao oa pele oa Newton e bolela hore haeba ho se na matla a letlooa a sebetsang nthong, ntho e 'ngoe le e 'ngoe e tsoela pele boemong ba eona ba ho phomola, kapa lebelo le sa fetoheng moleng o otlolohileng.

Kahoo haeba ntho eo e tsamaea ho lebelo le sa fetoheng, ha hoa lokela hore ho be le matla a marang-rang (ΣF = 0)Matla F a sebediswa nthong ka tsela e nepahetseng e le hore matla a kgatello ya kinetic a sebediswa nthong ka lehlakoreng le letshehadi.

∑F=0

F – f_k = 0

F = f_k

Matla a moferefere oa kinetic:

f_k = μ_k N = (0.1)(9.8 N) = 0.98 Li-Newton

ntho e tsamaea ka lebelo le sa fetoheng, F = f_k = 0.98 Newton

3. Ntho e thellela tlase sefofane se sekametseng ka lebelo le sa fetoheng. Fumana coefficient ea kinetic friction (μ_k). g = 9.8 m/s²

Matla a khohlano e sa fetoheng le e tsamaeang ka kinetic - mathata le litharollo 4

tharollo

Matla a khohlano e sa fetoheng le e tsamaeang ka kinetic - mathata le litharollo 5

w = boima, w_x = karolo e bataletseng ea boima, lintlha tse pela moepa, w_y = karolo e otlolohileng ea boima, e otlolohileng ho ea ka sefofane se sekametseng, N = matla a tloaelehileng, f_k = matla a moferefere wa kinetic.

Tse tsejoang:

Boima (m) = 1 kg

Ho potlaka ha matla a khoheli (g) = 9.8 m/s²

boima (w) = mg = (1 kg)(9.8 m/s²) = 9.8 kg m/s² = 9.8 Newton

w_x = sebe sa 30^o = (9.8 N)(0.5) = 4.9 Newton

w_y = w cos 30^o = (9.8 N)(0.5)√3 = 4.9√3 Newton

Matla a tloaelehileng (N) = w_y = 4.9√3 Newton

Ho batloa: coefficient ea khohlano ea kinetic (μ_k)

Tharollo:

Ntho e thellisa fatshe sefofaneng se sekametseng ka lebelo le sa fetoheng e le hore matla a letlowa = 0.

∑F=0

w_x - f_k = 0

w_x = f_k

w_x = μ_k N

5 = μ_k (5√3)

μ_k = 5 / 5√3

μ_k = 1 /√3

μ_k = 0.58

[ID ea sephutheloana sa wpdm='472′]

Bala haholoanyane

Molao oa bobeli oa Newton oa motsamao - mathata le litharollo

Mathata a rarollotsoeng melaong ea Newton ea motsamao – molao oa bobeli oa Newton oa motsamao

1. Ntho e boima ba 1 kg e potlakisang lebelo le tsitsitseng la 5 m/s². Lekanya matla a letlooa a hlokahalang ho potlakisa ntho.

Tse tsejoang:

Boima (m) = 1 kg

matsapa (a) = 5 m/s²

o ne a batla : matla a letlooa (∑F)

Tharollo:

Re sebelisa molao oa bobeli oa Newton ho fumana matla a marang-rang.

∑F = ma

∑F = (1 kg)(5 m/s²) = 5 kg m/s² = 5 Newton

2. Boholo ea ntho = 1 kg, matla a letlooa ∑F = 2 Newtons. Fumana boholo le tataiso ea ho potlaka ha ntho….

Molao oa bobeli oa Newton oa motsamao – mathata le litharollo 1

Tse tsejoang:

Boima (m) = 1 kg

Matla a letlooa (∑F) = 2 Newton

o ne a batla : Boholo le tataiso ea ho potlaka (a)

Tharollo:

a = ∑F / m

a = 2/1

a = 2 m/s²

Tsela ea ho potlakisa = tsela ea matla a letlooa (∑F)

3. Boima ba ntho = 2 kg, F₁ = 5 Newton, F₂ = Li-Newton tse 3. Boholo le tataiso ea ho potlaka ke…

Molao oa bobeli oa Newton oa motsamao – mathata le litharollo 2

Tse tsejoang:

Boima (m) = 2 kg

F₁ = 5 Newton

F₂ = 3 Newton

Ho batloa: Boholo le tataiso ea ho potlaka (a)

Tharollo:

matla a letlooa:

∑F = F₁ - F₂ = 5 – 3 = 2 Newton

Boholo ba ho potlaka ha lebelo:

a = ∑F / m

a = 2/2

a = 1 m/s²

Tataiso ea ho potlakisa = tataiso ea matla a letlooa = tataiso ea F₁

4. Boima ba ntho = 2 kg, F₁ = 10 Newton, F₂ = Li-Newton tse 1. Boholo le tataiso ea ho potlaka ke…

Molao oa bobeli oa Newton oa motsamao – mathata le litharollo 3

Tse tsejoang:

Molao oa bobeli oa Newton oa motsamao – mathata le litharollo 4

Boima (m) = 2 kg

F₂ = 1 Newton

F₁ = 10 Newton

F_1x =F₁ Cos 60^o = (10)(0.5) = 5 Newton

o ne a batla : Boholo le tataiso ea ho potlaka (a)

Tharollo:

Matla a marang-rang:

∑F = F_1x - F₂ = 5 – 1 = 4 Newton

Boholo ba ho potlaka ha lebelo:

a = ∑F / m

a = 4/2

a = 2 m/s²

Tataiso ea ho potlakisa = tataiso ea matla a letlooa = tataiso ea F_1x

5. F₁ = 10 Newton, F₂= 1 Newton, m₁ = 1 kg, m₂ = 2 kg. Boholo le tataiso ea ho potlaka ke…

Molao oa bobeli oa Newton oa motsamao – mathata le litharollo 5

Tse tsejoang:

Boima ba 1 (m₁) = 1 kg

Boima ba 2 (m₂) = 2 kg

F₁ = 10 Newton

F₂ = 1 Newton

o ne a batla : Boholo le tataiso ea ho potlaka (a)

Tharollo:

Matla a marang-rang:

∑F = F₁ - F₂ = 10 – 1 = 9 Newton

Boholo ba ho potlaka ha lebelo:

a = ∑F / (m₁ +m₂)

a = 9 / (1 + 2)

a = 9/3

a = 3 m/s²

Tsela ea ho potlakisa = tsela ea matla a letlooa = tsela ea F₁

Setene sa boima ba 40 kg se potlakisang ka matla a 200 N. Ho potlakisa setene ke 3 m/s²Fumana boholo ba matla a khohlano a bonoang ke boloko.

A. 15 N Molao oa bobeli oa Newton oa motsamao – mathata le litharollo 7

B. 40 N

C. 43 N

D. 80 N

Tse tsejoang:

Boima (m) = 40 kg

Matla (F) = 200 N

Ho potlaka (a) = 3 m/s²

Oa Batla: Matla a khohlano (F_g)

Tharollo:

Tekanyo ea Molao oa bobeli oa Newton oa ho sisinyeha

∑F = ma

∑F = matla a letlooa, m = boima, a = ho potlaka

Tsela ea matla F e lebile ka ho le letona, tsela ea matla a khohlano e lebile ka ho le letšehali (tsela ea matla a khohlano e fapane le tsela ea motsamao oa ntho).

Khetha ho le letona e le ntle le ho le letšehali e le mpe.

∑F = ma

F – F_g = ma

200 – F_g = (40)(3)

200 – F_g = 120

F_g= 200 - 120

F_g= 80 Newton

Karabo e nepahetseng ke D.

7. Boloko A e nang le boima ba digrama tse 100 e behe ka hodima boloko B e nang le boima ba digrama tse 300, ebe boloko B e sutumelletswa ka matla a 5 N hodimo ka ho otloloha. matla a tloaelehileng e sebediswang ke boloko B hodima boloko A.

A. 1 N Molao oa bobeli oa Newton oa motsamao – mathata le litharollo 2

B. 1.25 N

C. 2 N

D. 3 N

Tse tsejoang:

Matla (F) = 5 Newton

Boima ba boloko A (m_A) = digrama tse 100 = 0.1 kg

Boima ba boloko B (m_B) = digrama tse 300 = 0.3 kg

Ho potlaka ha matla a khoheli (g) = 10 m/s²

Weight ea boloko A (w_A) = (0.1 kg)(10 m/s²) = 1 kg m/s² = 1 Newton

Boima ba boloko B (w_B) = (0.3 kg)(10 m/s²) = 3 kg m/s² = 3 Newton

Ho batloa: Matla a tloaelehileng a sebelisoang ke boloko B ho thibela A

Tharollo:

Molao oa bobeli oa Newton oa motsamao – mathata le litharollo 3 Ho na le matla a 'maloa a sebetsang holim'a li-block ka bobeli, joalo ka ha ho bontšitsoe setšoantšong.

F = matla a sutumelletsang (ketso hodima boloko B)

w_A = boima ba boloko A (ketso holim'a boloko A)

w_B = boima ba boloko B (ketso holim'a boloko B)

N_A = matla a tloaelehileng a sebelisoang ke boloko B holim'a boloko A (Etsa joalo ka boloko A)

N_A' = matla a tloaelehileng a sebelisoang ke boloko A holim'a boloko B (Etsa ketso holim'a boloko B)

Sebelisa molao oa bobeli oa Newton oa motsamao holim'a li-block ka bobeli:

∑F = ma

F – w_A - w_B +N_A - N_A' = (m_A +m_B) a

N_A le N_A' ke matla a ketso-karabelo a nang le boholo bo tšoanang empa a fapane ka lehlakoreng leo a tlosoang ho equation.

F – w_A - w_B = (m_A +m_B) a

5 – 1 – 3 = (0.1 + 0.3) a

5 – 4 = (0.4) a

1 = (0.4) a

a = 1/0.4

a = 2.5 m/s²

Sebelisa molao oa bobeli oa Newton oa ho sisinyeha holim'a boloko A:

∑F = ma

N_A - w_A = m_A a

N_A – 1 = (0.1)(2.5)

N_A - 1 = 0.25

N_A = 1 + 0.25

N_A = 1.25 Newton

Karabo e nepahetseng ke B.

8. Ntho e boima ba 4 N e tshehetswang ke thapo le pulley. Matla a 2 N a sebetsa hodima boloko le pheletsong e nngwe ya thapo e hulwang ke matla a 9 N. Fumana hore na matla a letlowa a sebetsa jwang nthong ya X.

A. 3 N ho ea holimo Molao oa bobeli oa Newton oa motsamao – mathata le litharollo 4

B. 4 N ho ea tlase

C. 9 N ho ea holimo

D. 9 N ho ea tlase

Tse tsejoang:

Boima ba X (w)_X) = 4 Newton

Matla a ho hula (F)_x) = 2 Newton

Matla a kgatello (F)_T) = 9 Newton

Oa Batla: Matla a netweke a sebetsa nthong ea X

Tharollo:

Matla a holimo a otlolohileng a sebetsang nthong

Matla a kgatello a na le boholo bo lekanang dikarolong tsohle tsa thapo. Kahoo matla a kgatello ke 9 N.

Matla a theohelang a otlolohileng a sebetsang hodima ntho

Ho na le matla a mabeli a sebetsang nthong ea X 'me matla ka bobeli a shebile tlase ka ho otloloha, karolo e otlolohileng ea boima ba w_x le karolo e rapameng ea matla F_x.

Ketso ea matla a netweke nthong eo

F_T - w_X - F_x= 9 – 4 – 2 = 9 – 6 = 3

Matla a letlooa a sebetsang nthong X ke li-Newton tse 3, a phahame ka holimo.

Karabo e nepahetseng ke A.

9. Ntho e phomotseng qalong hodima bokahodimo bo boreleli bo bataletseng. Matla a 16 N a sebetsa hodima ntho eo kahoo ntho e potlakisa ka lebelo la 2 m/s²Haeba ntho e le 'ngoe e phomotse holim'a sebaka se bataletseng se bataletseng hoo matla a khohlano a sebetsang holim'a ntho e le 2 N, joale fumana hore na ntho e potlakile hakae haeba matla a tšoanang a 16 N a sebetsa holim'a ntho eo.

A. 1.75 m/s²

B. 1.50 m/s²

C. 1.00 m/s²

D. 0.88 m/s²

Tse tsejoang:

Matla (F) = 16 Newtons = 16 kg m/s²

Ho potlaka (a) = 2 m/s²

Matla a khohlano (F)_chelete) = 2 Newton = 2 kg m/s²

Ho batloa: Ho potlaka ha ntho?

Tharollo:

Boreleli bo bataletseng bo otlolohileng (ha ho na matla a khohlano):

Molao oa bobeli oa Newton oa motsamao – mathata le litharollo 5 ∑F = ma

F = ma

16 = (m) 2

m = 16 / 2

m = 8 kg

Boima ba ntho eo ke lik'hilograma tse 8.

Bokaholimo bo bataletseng bo sa tsitsang (ho na le matla a khohlano):

Molao oa bobeli oa Newton oa motsamao – mathata le litharollo 6 ∑F = ma

F – F_chelete = ma

16 – 2 = 8 a

14 = 8 a

a = 14/8

a = 1.75 m/s²

Ho potlaka ha ntho ke 1.75 m/s².

Karabo e nepahetseng ke A.

10. Tom le Andrew ba sutumelletsa ntho fatše e boreleli. Tom o sutumelletsa ntho ka matla a 5.70 N. Haeba boima ba ntho e le 2.00 kg mme lebelo le fihlellehang ke ntho e le 2.00 ms^-2, ebe o fumana boholo le tataiso ea ketso ea matla ea Tom.

A. 1.70 N mme tataiso ya yona e fapane le kgatello e entsweng ke Andre.w

B. 1.70 N le tataiso ea eona li tšoana le matla a entsoeng ke Andrew

C. 2.30 N mme tataiso ya yona e fapane le kgatello e entsweng ke Andrew.

D. 2.30 N le tataiso ea eona li tšoana le matla a entsoeng ke Andrew.

Tse tsejoang:

Matla a sutumelletsang a etswang ke Andrew (F)₁) = 5.70 Newton

Boima ba ntho (m) = 2.00 kg

Ho potlaka (a) = 2.00 m/s²

Ho batloa: Boholo le tataiso ea matla a entsoeng ke Tom (F)₂)?

Tharollo:

Sebelisa molao oa bobeli oa Newton oa motsamao:

∑F = ma

F₁ +F₂ = ma

5.70 + F₂ = (2)(2)

5.70 + F₂ = 4

F₂ = 4 - 5.70

F₂ = – 1.7 Newton

Letšoao la ho tlosa le ne le bontša hore (F)₂) e fapane le ketso ea ho sutumelletsa ka matla ea Andrew (F)₁).

Karabo e nepahetseng ke A.

11. Haeba boima ba boloko bo tshwana, ke setshwantsho sefe se bontshang lebelo le lenyenyane ka ho fetisisa?

Molao oa pele oa Newton le molao oa bobeli oa Newton 2

tharollo

Matla a nete A:

ΣF = 4 N + 2 N – 3 N = 6 N – 3 N = 3 Newtons, ka ho le letshehadi

Matla a letlooa B:

ΣF = 2 N + 3 N – 4 N = 5 N – 4 N = 1 Newtons, ka ho le letona

Matla a nete C:

ΣF = 4 N + 3 N – 2 N = 7 N – 2 N = 5 Newtons, ka ho le letona

Matla a nete D:

ΣF = 3 N + 4 N + 2 N = 9 Newtons, ka ho le letona

Tekanyo ea molao oa bobeli oa Newton:

ΣF = ma

a = ΣF / m

a = ho potlaka, ΣF = matla a letlooa, m = boima

Ho latela foromo e ka hodimo, ho potlakisa (a) ho lekana ka ho toba le matla a letlowa (ΣF) mme ho lekana ka tsela e fapaneng le boima (m). Haeba boima ba ntho bo tshwana, ha matla a hlahang a le maholo, ho potlakisa ho le hoholo kapa ha matla a hlahang a le manyane, ho potlakisa ho le honyenyane.
Ho latela palo e ka holimo, matla a manyane ka ho fetisisa a letlooa ke 1 Newton kahoo ho potlakisa le hona ke ho honyenyane ka ho fetisisa.

Karabo e nepahetseng ke B.

12. Matla a mang a sebetsa nthong e boima ba 20 kg, joalo ka ha ho bontšitsoe setšoantšong se ka tlase.

Fumana lebelo la ntho.

Tse tsejoang:

Boima ba ntho (m) = 20 kg

Matla a marang-rang (ΣF) = 25 N + 30 N – 15 N = 40 N

Oa Batla: Ho potlaka ha ntho

Tharollo:

Ho potlakisa ha ntho ho balwa ho sebediswa equation ya molao wa bobedi wa Newton:

ΣF = ma

a = ΣF / m = 40 N / 20 kg = 2 N/kg = 2 m/s²

13. Ke polelo efe e ka tlase e hlalosang molao oa boraro oa Newton?

(1) Bapalami ba ile ba sutumelletsa pele ha bese e roba brake ka tshohanyetso

(2) Blibuka pampiring ha li oe ha pampiri e huloa kapele

(3) Ha o bapala skateboarding ha leoto le sutumelletsa fatshe morao, skateboard e tla thellela pele

(4) Olikepe li sutumelletsoa morao, likepe li ea pele

Tharollo:

(1) Molao oa pele oa Newton

(2) Molao oa pele oa Newton

(3) Molao oa boraro oa Newton

(4) Molao oa boraro oa Newton

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Bala haholoanyane

Matla a tloaelehileng - mathata le litharollo

Mathata a rarollotsoeng melaong ea Newton ea motsamao – Matla a tloaelehileng

1. Ntho e phomotseng tafoleng, e bontšitsoeng setšoantšong se ka tlase. Boima ba ntho eo ke 1 kg. Ho potlaka ha matla a khoheli ke 9.8 m/s²Fumana matla a tloaelehileng a sebelisoang nthong e haufi le tafole.

Matla-a tloaelehileng-–-mathata-le-tharollo-1-1

Tse tsejoang:

Boima (m) = 1 kg

Ho potlaka ha matla a khoheli (g) = 9.8 m/s²

Boima (w) = mg = (1 kg)(9.8 m/s²) = 9.8 kg m/s² = 9.8 Newton

Oa Batla: matla a tloaelehileng (N)

Tharollo:

Matla a tloaelehileng - mathata le litharollo 2

Ntho e phomotse tafoleng, kahoo matla a letlooa holim'a ntho ke lefela (molao oa pele kapa oa bobeli oa Newton). Boima ba ntho bo sebetsa bo otlolohile ho ea tlase, ho ea bohareng ba Lefatše. Ho tlameha ho be le matla a mang holim'a ntho ho leka-lekanya matla a khoheli. Ntho e lutseng tafoleng, e le hore tafole e sebelise matla ana a holimo. Matla a sebelisoang ke tafole hangata a bitsoa matla a tloaelehileng (N). Tloaelehileng e bolela ho otloloha.

Khetha tataiso e yang hodimo e le tataiso e ntle ya y. Matla a sebele nthong ke:

∑Fy = 0

N – w = 0

N = w

N = mg

N = 9.8 Newton

Matla a tloaelehileng nthong, a sebelisoang ke tafole ke 9.8 N ho ea holimo.

2. Lintho tse peli tse phomotseng tafoleng. Boholo ea ntho ea 1 (m₁) = 1 kg, boima ba ntho 2 (m₂) = 2 kg, ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s²Fumana boholo le tataiso ea matla a tloaelehileng a sebelisoang ke m₂ holim'a m₁ le matla a tloaelehileng a sebelisoang ke tafole e ho m₂.

Matla a tloaelehileng - mathata le litharollo 3

tharollo

Matla a tloaelehileng - mathata le litharollo 4

Tse tsejoang:

Boima ba ntho 1 (m₁) = 1 kg

Boima ba ntho 2 (m₂) = 2 kg

Ho potlaka ha matla a khoheli (g) = 9.8 m/s²

Weight ea ntho ea 1 (w₁) = m₁ g = (1)(9.8 m/s²) = 9.8 kg m/s² = 9.8 Newton

Boima ba ntho 2 (w₂) = m₂ g = (2)(9.8 m/s²) = 19.6 kg m/s² = 19.6 Newton

Ho batloa: N₁ le N₂

Tharollo:

(a) Matla a tloaelehileng a sebelisoang ke m₂ ho ea ho m₁ (N₁)

N₁ = w₁ = 9.8 Newton

Tataiso ea N₁ e holimo.

(b) Matla a tloaelehileng a sebelisoang ke tafole ho m₂ (N₂)

N₂ = w₁ +w₂ = 9.8 Li-Newton + 19.6 Li-Newton = 29.4 Li-Newton

Tataiso ea N₂ e holimo.

3. Ntho e phomotseng tafoleng. Boima ba ntho ke 2 kg, lebelo ka lebaka la matla a khoheli ke 9.8 m/s²Boholo ba matla F ke 10 Newtons. Fumana boholo le tataiso ea matla a tloaelehileng a sebelisoang ke tafole holim'a ntho.

Matla a tloaelehileng - mathata le litharollo 5

tharollo

Matla a tloaelehileng - mathata le litharollo 6

Tse tsejoang:

Boima ba ntho (m) = 2 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s²

Boima (w) = mg = (2 kg)(9.8 m/s²) = 19.6 kg m/s² = 19.6 Newton

Matla F (F) = 10 Newton

o ne a batla : boholo le tataiso ea matla a tloaelehileng (N)

Tharollo:

tataiso ea matla a tloaelehileng e ea holimo.

Boholo ba matla a tloaelehileng:

∑F=0

N – F – w = 0

N = F + w

N = 10 Newtons + 20 Newtons

N = 30 Newton

4. Ntho e phomotseng tafoleng. Boima ba ntho ke 1 kg, lebelo ka lebaka la matla a khoheli ke 9,8 m/s², qobella F₁ ke 10 N le matla a F₂ ke 20 N. Fumana boholo le tataiso ea matla a tloaelehileng a sebelisoang ke tafole holim'a ntho. g = 9.8 m/s²

Matla a tloaelehileng - mathata le litharollo 7

tharollo

Matla a tloaelehileng - mathata le litharollo 8

Tse tsejoang:

Boima (m) = 1 kg

Ho potlaka ha matla a khoheli (g) = 9.8 m/s²

Boima (w) = mg = (1 kg)(9.8 m/s²) = 9.8 kg m/s² = 9.8 Newton

F₁ = 10 Newton

F₂ = 20 Newton

Ho batloa: boholo le tataiso ea matla a tloaelehileng (N)

Tharollo:

Tsela eo matla a tloaelehileng a lebang ho yona e ya hodimo.

Boholo ba matla a tloaelehileng:

∑F=0

N – F₂ – w + F₁ = 0

N = F₂ + w – F₁

N = 20 Newtons + 9.8 Newtons – 10 Newtons

N = 19.8 Newton

5. Boima ba ntho (m) = 2 kg, ho potlaka ha matla a khoheli (g) = 9.8 m/s², sekhutlo = 30^oFumana boholo le tataiso ea matla a tloaelehileng a sebelisoang nthong eo.

Matla a tloaelehileng - mathata le litharollo 9

Tharollo:

Matla a tloaelehileng - mathata le litharollo 10

w ke boima, w_x ke karolo e rapameng ea boima, w_y ke karolo e otlolohileng ea boima, N ke matla a tloaelehileng.

Tse tsejoang:

boima (m) = 2 kg

ho potlaka ha matla a khoheli (g) = 9.8 m/s²

boima (w) = mg = (2 kg)(9.8 m/s²) = 19.6 kg m/s² = 19.6 Newton

w_x = sebe sa 60^o = (19.6 N)(0.5)√3= 9.8√3 Newton

w_y = w cos 60 = (19.6 N)(0.5) = 9.8 Newton

Oa Batla: matla a tloaelehileng (N)

Tharollo:

∑F=0

N – w_y = 0

N = w_y

N = 9.8 Newton

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Bala haholoanyane

Boima le boima - mathata le litharollo

Mathata a rarollotsoeng melaong ea Newton ea motsamao - Boima, le boima

1. Boima ba boima ba 1 kg holim'a Lefatše ke… g = 9.8 m/s²

Tse tsejoang:

Boima (m) = 1 kg

The ho potlaka ka lebaka la matla a khoheli holim'a Lefatše (g) = 9.8 m/s²

Oa Batla: boima (w)

Tharollo:

w = mg

m = boima (Yuniti ea SI ea boima ke kilogram, kg)

g = ho potlaka ka lebaka la matla a khoheli (Yuniti ea SI ea g ke m/s²)

w = boima (Yuniti ea SI ea w ke kg m/s² kapa Newton)

Boima ba 'mele:

w = (1 kg)(9.8 m/s²) = 9.8 kg m/s² = 9.8 Newton

(a) Thala matla a khoheli (boima) e sebetsang hodima ntho ha ntho e phomotse tafoleng, jwalo ka ha ho bontshitswe setshwantshong sa (a).

(b) Thala matla a khoheli (boima) le likarolo tsa ona tse sebetsang nthong e thellang nthong sefofane se sekametseng, joalo ka ha ho bontšitsoe setšoantšong (b)

Boima le boima - mathata le litharollo 1

tharollo

Boima le boima - mathata le litharollo 2

Tsela eo boima bo tsamaeang ka yona e theohela tlase bohareng ba Lefatshe.

w_x = karolo e rapameng ea boima le w_y = karolo e otlolohileng ea boima

3. Boima ba lebokose ke 1 kg mme lebelo ka lebaka la matla a khoheli ke 9.8 m/s²Fumana (a) boima (b) karolo e rapameng le karolo e otlolohileng ea boima.

Boima le boima - mathata le litharollo 3 tharollo

Boima: w = mg = (1 kg)(9.8 m/s²) = 9.8 kg m/s² = 9.8 Newton

Karolo e rapameng ea boima:

w_x = sebe sa 30^o = (9,8 N)(0,5) = 4.9 Newton

Karolo e otlolohileng ea boima:

w_y = w cos 30^o = (9.8 N)(0.5√3) = 4.9√3 Li-Newton

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Bala haholoanyane

Ho tsamaea holimo le tlase nakong ea ho oa ka bolokolohi - mathata le litharollo

Mathata a Rarollotsoeng ho Motsamao o Mola - Motsamao o Holimo le o Tlase nakong ea ho oa ho lokolohileng

1. Motho o lahlela bolo holimo moeeng ka lebelo la pele la 20 m/s. Bala hore na e ea holimo hakae. Hlokomoloha ho hanyetsa metsi. Ho potlakisa ka baka la matla a khoheli (g) = 10 m/s².

tharollo

Re sebelisa e 'ngoe ea li-equation tsena tsa kinematic bakeng sa motsamao o potlakileng kamehla, joalokaha ho bontšitsoe ka tlase.

v_t = v_o + ho

s = v_o t + ½ ho²

v_t² = v_o² + Li-axle tse 2

Tse tsejoang:

Re khetha tsela e eang holimo e le e ntle le e theohelang e le e mpe.

Lebelo la pele (v)_o) = 20 m/s (e ntle ho ea holimo)

Ho potlaka ha matla a khoheli (g) = – 10 m/s²(e mpe e theohela tlase).

Lebelo la ho qetela (v)_t) = 0 (lebelo la yona ke lefela bakeng sa motsotsoana sebakeng se phahameng ka ho fetisisa)

Ho batloa: Bolelele bo phahameng ka ho fetisisa (h)

Tharollo:

v_t² = v_o² + 2 gh

0 = (20)²) + 2(-10) lihora

0 = 400 – 20 dihora

400 = lihora tse 20

h = 400 / 20 = 40 / 2 = limithara tse 20

2. Motho o lahlela lejoe holimo ka lebelo la 20 m/s ha a eme moeling oa lefika, e le hore lejoe le ka oela botlaaseng ba lefika limithara tse 100 ka tlase.

(a) Bolo e nka nako e kae ho fihla botlaaseng ba lefika (b) Lebelo la ho qetela pele lejoe le otla mobu. Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s²Hlokomoloha khanyetso ea moea.

Tse tsejoang:

Re khetha tsela e eang holimo e le e ntle le e theohelang e le e mpe.

Holimo (h) = -100 metres (e mpe hobane boemo ba ho qetela bo ka tlase ho boemo ba pele)

Qalong Velocity (v_o) = 20 m/s (e ntle ho ea holimo)

Ho potlaka ha matla a khoheli (g) = -10 m/s²(e mpe ho theoha)

Ho batloa:

(a) Nako moeeng kapa nako e sa fellang (t)

(b) Lebelo la ho qetela (v)_t)

Tharollo:

(a) Nako e sa fellang (t)

Tse tsejoang:

Holimo (h) = -100 metres (e mpe hobane boemo ba ho qetela bo ka tlase ho boemo ba pele)

Lebelo la pele (v)_o) = 20 m/s (e ntle ho ea holimo), Ho potlaka ha matla a khoheli (g) = -10 m/s² (e mpe e theohela tlase).

h = v_o t + ½ gt²

-100 = (20) t + ½ (-10) t²

-100 = 20 t – 5 t²

-5 t² + 20 t + 100 = 0

Re sebelisa foromo ea quadratic:

Motsamao o nyolohang le o theohang mathateng le ditharollong tsa ho wa ka bolokolohi 1

(b) Lebelo la ho qetela

v_t² = v_o² + 2 gh

v_t² = (20²) + 2 (-10)(-100)

v_t² = 400 + 2000

v_t² = 2400

v_t = 49 m/s

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Ho sisinyeha ha ho oa ka bolokolohi - mathata le litharollo

Mathata a Rarollotsoeng ho Motsamao o Molelele – Motsamao o theohelang tlase nakong ea ho oa ha mahala

1. Bolo e lahlelwa fatshe ka lebelo la pele la 10 m/s mme e fihla fatshe ka metsotswana e 2. Fumana lebelo la ho qetela pele bolo e thula fatshe. Ho potlaka ha matla a khoheli (g) = 10 m/s²Hlokomoloha khanyetso ea moea.

Tse tsejoang:

Lebelo la pele (v)_o) = 10 m/s

Nako e fetileng (t) = metsotsoana e 2

Ho potlaka ha matla a khoheli (g) = 10 m/s²

Ho batloa: Lebelo la ho qetela (v_t)

Tharollo:

Ho potlaka 10 m/s²ho bolela keketseho ea lebelo ka 10 m/s ka motsotsoana. Kamora metsotsoana e 3, lebelo = 30 m/s.

Lebelo la ho qetela = 10 m/s + 20 m/s = 30 m/s.

Li-equation tsa Kinematic bakeng sa motsamao o potlakileng kamehla, jwalo ka ha ho bontshitswe ka tlase:

v_t = v_o + ho …………. 1

h = v_o t + ½ ho²…………. 2

v_t² = v_o² + 2 ah ………. 3

v_t = v_o + gt

v_t = 10 + (10)(2)

v_t = 10 + 20 = 30 m/s

Lebelo la ho qetela = v_t = 30 m/s

2. Lejoe le lahleloa ka ho otloloha tlase ho tloha borokhong ka lebelo la pele la 5 m/s mme le fihla metsing ka metsotsoana e 2. Bala bolelele ba borokho.

Tse tsejoang:

Lebelo la pele (v)_o) = 5 m/s

Nako e fetileng (t) = metsotsoana e 2

Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s²

Ho batloa: bophahamo ba borokho (h)

Tharollo:

h = v_o t + ½ gt²

h = (5)(2) + ½ (10)(2)²

h = 10 + (5)(4)

h = 10 + 20

h = limithara tse 30

3. Bolo e lahlelwa fatshe ka lebelo la pele la 10 m/s ho tloha bophahamong ba limithara tse 80. Fumana (a) Nako moyeng (b) Lebelo la ho qetela pele bolo e otla fatshe.

Tse tsejoang:

bophahamo (h) = limithara tse 80

Lebelo la pele (v)_o) = 10 m/s

Ho potlaka ha matla a khoheli (g) = 10 m/s²

Ho batloa:

(a) Nako e sa fellang (t)

(b) Lebelo la ho qetela (v)_t)

Tharollo:

(a) Nako e sa fellang (t)

Lebelo la ho qetela:

v_t² = v_o² + 2 gh

v_t² = (10)² + 2(10)(80) = 100 + 1600 = 1700

v_t= 41 m/s

Nako ea ho arohana (t):

v_t = v_o + gt

41 = 10 + (10)(t)

41 – 10 = 10 t

31 = 10 t

t = 31 / 10 = metsotsoana e 3,1

(b) Lebelo la ho qetela (v)_t₎ ?

v_t = 41 m/s

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