fisiks

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

tharollo

Tse tsejoang:

Boima (m) = 2 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s ²

Koefficient ea khohlano e sa fetoheng ( μs ₎ = 0.2

Koefficient ea kinetic friction ( μk ₎ = 0.1

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

Karolo e rapameng ea boima (w _x ) = w sin 30 ^o = (19.6)(0.5) = 9.8 Newtons

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 Newtons

Matla a khohlano e sa fetoheng (fs ₎ = (0.2)(9.8√3) = 1.96√3 Newton = 3.39 Newton

Matla a khohlano ea kinetic (f _k ) = (0.1)(9.8√3) = 0.98√3 Newton = 1.69 Newton

Tharollo:

Ntho e phomotse haeba w _x < _fs , ntho e ntse e theohela tlase haeba w _x > _fs.

w _x = 9.8 Li-Newton le f _s = 3.39 Li-Newton.

(a) matla a letlooa

∑ F = w _x – f _k = 9.8 – 1.69 = 8.11 Li-Newton

(b) boholo le tataiso ea ho potlaka

∑ F = ma

8.11 = (2) a

= 4.05

Boholo ba ho potlaka = 4.05 m/s ² mme tataiso ya ho potlaka = ho theohela tlase.

[irp]

2. Boima ba ntho = 4 kg, ho potlaka ka lebaka la matla a khoheli = 9,8 m/s ² . 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 tlase? Haeba ntho e thella tlase, fumana (a) matla a net (b) boholo le tataiso ea potlakiso!

tharollo

Tse tsejoang:

Boima (m) = 4 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s ²

Koefficient ea khohlano e sa fetoheng ( μs ₎ = 0.4

Koefficient ea ho ferekana ha kinetic ( μk ₎ = 0.2

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

Karolo e rapameng ea boima (w _x ) = w sin 30 ^o = (39.2)(0.5) = 19.6 Newtons

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 Newtons = 33.95 Newtons

matla a khohlano a sa fetoheng (fs ₎ = μ _s N = (0,4)(33.95) = 13.58 Newtons

Matla a ho qabana ha kinetic (f _k ) = μ _k N = (0.2)(33.95) = 6.79 Newtons

F = 40 Newton

Tharollo:

Ntho e thella tlase haeba F < w _x + fs _. Ntho e thella holimo haeba F > w _x + _fs.

F = 40 Newtons, w _x = 19.6 Newtons le f _s = 13.58 Newtons.

F e kholo ho feta w _x + f _s kahoo ntho e thellela holimo.

(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 ho potlaka ke 1.6 m/s ² mme tataiso ya ho potlaka e hodimo.

[irp]

[ID ea sephutheloana sa wpdm='481′]

1. Boima le boima
2. Matla a tloaelehileng
3. Molao oa bobeli oa Newton oa ho sisinyeha
4. Matla a khohlano
5. Ho sisinyeha holim'a bokaholimo bo otlolohileng ntle le matla a khohlano
6. Motsamao oa 'mele e 'meli e nang le lebelo le tšoanang holim'a bokaholimo bo bataletseng bo bataletseng ka matla a khohlano
7. Ho sisinyeha ha sefofane se sekametseng ntle le matla a khohlano
8. Motsamao sefofaneng se sekametseng ka thata ka matla a khohlano
9. Motsamao ka lifting
10. Motsamao oa 'mele o hokahantsoe ka lithapo le li-pulleys
11. 'Mele e 'meli e nang le boholo bo tšoanang ba ho potlaka
12. Ho potoloha mothapo o bataletseng - matla a motsamao o chitja
13. Ho potoloha mothapo o kobehileng - matla a motsamao o chitja
14. Motsamao o tšoanang ka selikalikoe se otlolohileng
15. Matla a bohareng a motsamao o chitja o tšoanang

Bala haholoanyane

1. Boima ba lebokose = 2 kg, ho potlaka ka lebaka la matla a khoheli = 9.8 m/s ^{2 . Fumana (a) matla a letlooa a potlakisang lebokose ho ya tlase (b) boholo ba} potlakalo ya lebokose.

tharollo

Tse tsejoang:

Boima (m) = 2 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s ²

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

w _x = w sin 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) Letlooa bakeng sa ce le 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/s ²

Boholo ba ho potlaka ke 4.9 m/s ² , tataiso ea ho potlaka e theohela tlase.

[irp]

2. Sepakapaka se sekametseng se boreleli kahoo ha ho na matla a khohlano . Boima ba ntho ke 3 kg, ho potlaka ka lebaka la matla a khoheli ke 9.8 m/s ² . Fumana boholo ba matla F haeba (a) ntho e phomotse (b) ntho e theohela tlase ka ho potlaka ho sa feleng ha 2 m/s ² (c) ntho e nyolohela holimo ka ho potlaka ho sa feleng ha 2 m/s ².

tharollo

Tse tsejoang:

Boima (m) = 3 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s ²

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

w _x = w sin 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 oa motsamao o re haeba ntho e phomotse, matla a letlooa a sebetsa nthong eo 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/s ²

∑ F = ma

w _x – F = ma

14.7 – F = (3)(2)

14.7 – F = 6

F = 14.7–6

F = 8.7 Newton

(c) Boholo ba matla F haeba ntho e ntse e nyolohela hodimo ka lebelo le sa fetoheng la 2 m/s ²

∑ F = ma

F – w _x = ma

F – 14.7 = (3)(2)

F – 14.7 = 6

F = 14.7 + 6

F = 20.7 Newton

[irp]

[ID ea sephutheloana sa wpdm='479′]

1. Boima le boima
2. Matla a tloaelehileng
3. Molao oa bobeli oa Newton oa ho sisinyeha
4. Matla a khohlano
5. Ho sisinyeha holim'a bokaholimo bo otlolohileng ntle le matla a khohlano
6. Motsamao oa 'mele e 'meli e nang le lebelo le tšoanang holim'a bokaholimo bo bataletseng bo bataletseng ka matla a khohlano
7. Ho sisinyeha ha sefofane se sekametseng ntle le matla a khohlano
8. Motsamao sefofaneng se sekametseng ka thata ka matla a khohlano
9. Motsamao ka lifting
10. Motsamao oa 'mele o hokahantsoe ka lithapo le li-pulleys
11. 'Mele e 'meli e nang le boholo bo tšoanang ba ho potlaka
12. Ho potoloha mothapo o bataletseng - matla a motsamao o chitja
13. Ho potoloha mothapo o kobehileng - matla a motsamao o chitja
14. Motsamao o tšoanang ka selikalikoe se otlolohileng
15. Matla a bohareng a motsamao o chitja o tšoanang

Bala haholoanyane

1. 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/s ² , boholo ba matla F ke 40 Newtons. Coefficient ea khohlano ea kinetic pakeng tsa lebokose la 1 le fatše ke 0.2 'me coefficient ea khohlano ea kinetic pakeng tsa lebokose la 2 le fatše ke 0.3. Fumana (a) Boholo le tataiso ea ho potlaka ha lebokose (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 (F ₂₁ ).

tharollo

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/s ² ,

Matla F = 40 Newton,

Koefficient ea khohlano ea kinetic pakeng tsa lebokose la 1 le fatše ( μ _k1 ) = 0.2

Koefficient ea khohlano ea kinetic pakeng tsa lebokose la 2 le fatše ( μ _k2 ) = 0.3

Boima ba lebokose 1 (w ₁ ) = m ₁ g = (2)(10) = 20 Newtons

Boima ba lebokose 2 (w ₂ ) = m ₂ g = (4)(10) = 40 Newtons

Matla a tloaelehileng a sebelisoang lebokoseng la 1 (N ₁ ) = w ₁ = 20 Newtons

Matla a tloaelehileng a sebelisoang lebokoseng la 2 (N ₂ ) = w ₂ = 40 Newtons

Matla a khohlano ea kinetic e sebelisitsoeng lebokoseng la 1 (f _k1 ) = ( μ _k1 )(N ₁ ) = (0.2)(20) = 4 Newtons

Matla a khohlano ea kinetic e sebelisitsoeng lebokoseng la 2 (f _k2 ) = ( μ _k1 )(N ₂ ) = (0.3)(40) = 12 Newtons

Tharollo:

(a) Boholo le tataiso ea lebelo la lebokose

ΣF = ma

F – f _k1 – f _k2 = (m ₁ + m ₂ ) a

40 – 4 – 12 = (2 + 4) a

24 = 6 a

a = 24/6

a = 4 m/s ²

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 (F ₂₁ ).

Bala boholo ba F ₁₂ :

ΣF = ma

F ₁₂ – f _k2 = (m ₂ ) a

F ₁₂ – 12 = (4)(4)

F ₁₂ – 12 = 16

F ₁₂ = 16 + 12

F ₁₂ = 28 Newton

F ₁₂ le F ₂₁ ke matla a ketso le karabelo a sebetsang linthong tse fapaneng. F ₁₂ le F ₂₁ li na le boholo bo tšoanang le tataiso e fapaneng.

F ₁₂ = 28 Li-Newton = F ₂₁ = 28 Li-Newton.

[irp]

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/s ² , 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 potlaka (b) Kgatelelo thapong e hokahanyang mabokose. Hlokomoloha boima ba thapo.

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/s ² ,

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 (w ₁ ) = m ₁ g = (2)(10) = 20 Newtons

Boima ba lebokose 2 (w ₂ ) = m ₂ g = (4)(10) = 40 Newtons

Matla a tloaelehileng a sebelisoang lebokoseng la 1 (N ₁ ) = w ₁ = 20 Newtons

Matla a tloaelehileng a sebelisoang lebokoseng la 2 (N ₂ ) = w ₂ = 40 Newtons

Matla a khohlano ea kinetic e sebelisitsoeng lebokoseng la 1 (f _k1 ) = ( μ _k1 )(N ₁ ) = (0.2)(20) = 4 Newtons

Matla a khohlano ea kinetic e sebelisitsoeng lebokoseng la 2 (f _k2 ) = ( μ _k1 )(N ₂ ) = (0.3)(40) = 12 Newtons

Tharollo:

(a) boholo le tataiso ea ho potlaka

ΣF = ma

F – f _k1 – f _k2 = (m ₁ + m ₂ ) a

40 – 4 – 12 = (2 + 4) a

24 = 6 a

a = 24/6

a = 4 m/s ²

Boholo ba ho potlaka ke 4 m/s ² , tataiso ea ho potlaka = tataiso ea matla a letlooa = ka ho le letona.

(b) Kgatelelo ka hara thapo

Matla a sebetsang lebokoseng la 1 ka lehlakoreng le otlolohileng ke kgatello ya 1 (T ₁ ) e lebileng ka ho le letona le matla a kgatello ya kinetic ya 1 (f _k1 ) e lebileng ka ho le letshehadi. Sebedisa molao wa bobedi wa Newton:

ΣF = ma

T ₁ – f _k1 = m ₁ a

T ₁ – 4 = (2)(4)

T ₁ – 4 = 8

T ₁ = 8 + 4 = 12 Newton

Matla a sebetsang lebokoseng la 2 ka lehlakoreng le otlolohileng ke kgatello ya 2 (T ₂ ) ho ya ka letsohong le letshehadi le matla a kgatello ya kinetic ya 2 (f _k2 ) ho ya ka letsohong le letona. Sebedisa molao wa bobedi wa Newton :

ΣF = ma

F – T ₂ – f _k2 = m ₂ a

40 – T ₂ – 12 = (4)(4)

28 – T ₂ = 16

T ₂ = 28 – 16 = 12 Newton

Kgatelelo e thapong e hokahanyang mabokose = T ₁ = T ₂ = T = 12 Newtons.

[irp]

[ID ea sephutheloana sa wpdm='493′]

1. Boima le boima
2. Matla a tloaelehileng
3. Molao oa bobeli oa Newton oa ho sisinyeha
4. Matla a khohlano
5. Motsamao holim'a bokaholimo bo otlolohileng ntle le matla a khohlano
6. Motsamao oa 'mele e 'meli ka lebelo le tšoanang holim'a sebaka se bataletseng se bataletseng ka matla a khohlano
7. Motsamao sefofaneng se sekametseng ntle le matla a khohlano
8. Motsamao sefofaneng se sekametseng ka thata ka matla a khohlano
9. Motsamao ka lifting
10. Motsamao oa 'mele o hokahaneng ka lithapo le li-pulley
11. 'Mele e 'meli e nang le boholo bo tšoanang ba ho potlaka
12. Ho potoloha mothapo o bataletseng - matla a motsamao o chitja
13. Ho potoloha mothapo o kobehileng - matla a motsamao o chitja
14. Motsamao o tšoanang ka selikalikoe se otlolohileng
15. Matla a bohareng a motsamao o chitja o tšoanang

Bala haholoanyane

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

Tse tsejoang:

m ₁ = 2 kg, m ₂ = 4 kg, g = 10 m/s ² , F = 12 Newtons

Ho batloa : a

Tharollo:

Σ F = ma

F = (m ₁ + m ₂ ) a

12 = (2 + 4) a

12 = 6 a

a = 12/6

a = 2 m/s ²

Boholo ba ho potlaka ke 2 m/s ² , tataiso ea ho potlaka = tataiso ea matla a letlooa = ka ho le letona.

[irp]

2. Boima ba ntho ya 1 ke 2 kg, boima ba ntho ya 2 ke 4 kg, ho potlaka ha matla a kgoheli ke 10 m/s ² , boholo ba matla F ke 24 N. Fumana boholo le tataiso ya ho potlaka.

Tse tsejoang:

m ₁ = 2 kg, m ₂ = 4 kg, g = 10 m/s ² , F = 24 Newtons

Ho batloa: ho potlakisa (a)

Tharollo:

Σ F = ma

F = (m ₁ + m ₂ ) a

24 = (2 + 4) a

24 = 6 a

a = 24/6

a = 4 m/s ²

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

[irp]

[ID ea sephutheloana sa wpdm='474′]

1. Boima le boima
2. Matla a tloaelehileng
3. Molao oa bobeli oa Newton oa ho sisinyeha
4. Matla a khohlano
5. Ho sisinyeha holim'a bokaholimo bo otlolohileng ntle le matla a khohlano
6. Motsamao oa 'mele e 'meli e nang le lebelo le tšoanang holim'a bokaholimo bo bataletseng bo bataletseng ka matla a khohlano
7. Ho sisinyeha ha sefofane se sekametseng ntle le matla a khohlano
8. Motsamao sefofaneng se sekametseng ka thata ka matla a khohlano
9. Motsamao ka lifting
10. Motsamao oa 'mele o hokahantsoe ka lithapo le li-pulleys
11. 'Mele e 'meli e nang le boholo bo tšoanang ba ho potlaka
12. Ho potoloha mothapo o bataletseng - matla a motsamao o chitja
13. Ho potoloha mothapo o kobehileng - matla a motsamao o chitja
14. Motsamao o tšoanang ka selikalikoe se otlolohileng
15. Matla a bohareng a motsamao o chitja o tšoanang

Bala haholoanyane

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/s ^2. Fumana (a) Matla a maholo a khohlano e sa fetoheng (b) Matla a bonyane a F

tharollo

Tse tsejoang:

Boima (m) = 1 kg

Koefficient ea khohlano e sa fetoheng (μ _s ) = 0.4

Ho potlaka ha matla a khoheli (g) = 9.8 m/s ²

Boima (w) = mg = (1 kg)(10 m/s ² ) = 10 kg m/s ² = 10 Li-Newton

Matla a tloaelehileng (N) = w = 10 Newtons

Ho batloa:

(a) Matla a maholo a khohlano e sa fetoheng (b) Matla a bonyane a F

Tharollo:

(a) Matla a maholohali a khohlano e sa fetoheng

fs = μs N_​​

f _s = (0.4)(9.8 N) = 3.92 Newton

(b) 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 Newtons.

[irp]

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/s ² )

Tse tsejoang:

_{Koefficient ea kinetic} friction ( μk ) = 0.1

Boima ba lebokose (m) = 1 kg

Ho potlaka ha matla a khoheli (g) = 9.8 m/s ²

Boima (w) = mg = (1 kg)(9.8 m/s ² ) = 9.8 kg m/s ² = 9.8 Newtons

Matla a tloaelehileng (N) = w = 9.8 Newtons

Ho batloa : F

Tharollo:

Molao oa pele oa Newton o re 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 e tsamaya ka lebelo le sa fetoheng , ha hoa lokela hore ho be le matla a letlooa ( Σ F = 0) . Matla F a sebelisoa nthong ka tsela e nepahetseng e le hore matla a kinetic friction a sebelisoe nthong ka lehlakoreng le letšehali.

∑ 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 Newtons

ntho e tsamaya ka lebelo le sa fetoheng, F = f _k = 0.98 Newtons

[irp]

3. Ntho e thella fatshe sefofaneng se sekametseng ka lebelo le sa fetoheng. Fumana coefficient ya kinetic friction ( _μk ) . g = 9.8 m/s ²

tharollo

w = boima, w _x = karolo e rapameng ea boima, lintlha tse pela moepa, w _y = karolo e otlolohileng ea boima, e otlolohileng ho sefofane se sekametseng, N = matla a tloaelehileng, f _k = matla a khohlano ea kinetic.

Tse tsejoang:

Boima (m) = 1 kg

Ho potlaka ha matla a khoheli (g) = 9.8 m/s ²

boima (w) = mg = (1 kg)(9.8 m/s ² ) = 9.8 kg m/s ² = 9.8 Li-Newton

w _x = w sin 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 Li-Newton

Matla a tloaelehileng (N) = w _y = 4.9 √ 3 Newtons

Ho batloa : coefficient ea kinetic friction ( μk ₎

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

[irp]

[ID ea sephutheloana sa wpdm='472′]

1. Boima le boima
2. Matla a tloaelehileng
3. Molao oa bobeli oa Newton oa ho sisinyeha
4. Matla a khohlano
5. Motsamao holim'a bokaholimo bo otlolohileng ntle le matla a khohlano
6. Motsamao oa 'mele e 'meli ka lebelo le tšoanang holim'a sebaka se bataletseng se bataletseng ka matla a khohlano
7. Motsamao sefofaneng se sekametseng ntle le matla a khohlano
8. Motsamao sefofaneng se sekametseng ka thata ka matla a khohlano
9. Motsamao ka lifting
10. Motsamao oa 'mele o hokahaneng ka lithapo le li-pulley
11. 'Mele e 'meli e nang le boholo bo tšoanang ba ho potlaka
12. Ho potoloha mothapo o bataletseng - matla a motsamao o chitja
13. Ho potoloha mothapo o kobehileng - matla a motsamao o chitja
14. Motsamao o tšoanang ka selikalikoe se otlolohileng
15. Matla a bohareng a motsamao o chitja o tšoanang

Bala haholoanyane

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

1. Ntho e boima ba 1 kg e potlakisitsoeng ka lebelo le sa fetoheng la 5 m/s ² . Hakanya matla a letlooa a hlokahalang ho potlakisa ntho eo.

Tse tsejoang:

Boima (m) = 1 kg

Ho potlaka (a) = 5 m/s ²

Ho batloa : 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/s ² ) = 5 kg m/s ² = 5 Newtons

[irp]

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

Tse tsejoang:

Boima (m) = 1 kg

Matla a letlooa (∑F) = 2 Newton

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

Tharollo:

a = ∑F / m

a = 2/1

a = 2 m/s ²

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

[irp]

3. Boima ba ntho = 2 kg, F ₁ = 5 Newtons, F ₂ = 3 Newton. Boholo le tataiso ea ho potlaka ke…

Tse tsejoang:

Boima (m) = 2 kg

F ₁ = 5 Newton

F ₂ = 3 Newton

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

Tharollo:

matla a letlooa:

∑ F = F ₁ – F ₂ = 5 – 3 = 2 Newton

Boholo ba ho potlaka ha lebelo:

a = ∑F / m

a = 2/2

a = 1 m/s ²

Tsela ea ho potlakisa = tsela ea matla a letlooa = tsela ea F ₁

[irp]

4. Boima ba ntho = 2 kg, F ₁ = 10 Newtons, F ₂ = 1 Newton. Boholo le tataiso ea ho potlaka ke…

Tse tsejoang:

Boima (m) = 2 kg

F ₂ = 1 Newton

F ₁ = 10 Newton

F _1x = F ₁ cos 60 ^o = (10)(0.5) = 5 Newtons

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

Tharollo:

Matla a marang-rang:

∑ F = F _1x – F ₂ = 5 – 1 = 4 Newton

Boholo ba ho potlaka ha lebelo:

a = ∑F / m

a = 4/2

a = 2 m/s ²

Tsela ea ho potlakisa = tsela ea matla a letlooa = tsela ea F _1x

[irp]

5. F ₁ = 10 Newtons, F ₂ = 1 Newton, m ₁ = 1 kg, m ₂ = 2 kg. Boholo le tataiso ya ho potlaka ke…

Tse tsejoang:

Boima ba 1 (m1 ₎ = 1 kg

Boima ba 2 (m2 ₎ = 2 kg

F ₁ = 10 Newton

F ₂ = 1 Newton

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

Tharollo:

Matla a marang-rang:

∑ F = F ₁ – F ₂ = 10 – 1 = 9 Newton

Boholo ba ho potlaka ha lebelo:

a = ∑F / (m ₁ + m ₂ )

a = 9 / (1 + 2)

a = 9/3

a = 3 m/s ²

Tsela ea ho potlakisa = tsela ea matla a letlooa = tsela ea F ₁

[irp]

6.

Sekotwana sa boima ba 40-kg se potlakisang ka matla a 200 N. Ho potlakisa sekotwana ke 3 m/ s ^2. Fumana boholo ba matla a khohlano a fumanwang ke sekotwana.

A. 15 N

B. 40 N

C. 43 N

D. 80 N

Tse tsejoang:

Boima (m) = 40 kg

Matla (F) = 200 N

Ho potlaka (a) = 3 m/s ²

Ho batloa: Matla a khohlano (F _g )

Tharollo:

Tekanyo ea molao oa bobeli oa Newton oa motsamao

∑ 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. Fumana matla a tlwaelehileng a sebediswang ke boloko B ho boloko A.

A. 1 N

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/s ²

Boima ba boloko A (w _A ) = (0.1 kg)(10 m/s ² ) = 1 kg m/s ² = 1 Newton

Boima ba boloko B (w _B ) = (0.3 kg)(10 m/s ² ) = 3 kg m/s ² = 3 Newtons

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

Tharollo:

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 (sebetsa hodima 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 karabelo ya ketso a nang le boholo bo tshwanang empa a fapane ka lehlakoreng kahoo a tloswa 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/s ²

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)

NA – ₁ = 0.25

NA = 1 + _0.25

NA = 1.25 _Newtons

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

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 Newtons

Ho batloa: Matla a marang-rang a sebetsa nthong

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 theohela tlase ka ho otloloha, karolo e bataletseng ea boima ba w _x le karolo e bataletseng 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 potlaka ka lebelo la 2 m/s ² . Haeba ntho e le nngwe e phomotseng hodima bokahodimo bo bataletseng bo bataletseng kahoo matla a kgohlano a sebetsa hodima ntho eo ke 2 N, jwale fumana hore na matla a tshwanang a 16 N a sebetsa hodima ntho eo.

A. 1.75 m/s ²

B. 1.50 m/s ²

C. 1.00 m/s ²

D. 0.88 m/s ²

Tse tsejoang:

Matla (F) = 16 Newtons = 16 kg m/s ²

Ho potlaka (a) = 2 m/s ²

Matla a khohlano (F _fric ) = 2 Li-Newton = 2 kg m/s ²

Ho Batloa: Ho potlaka ha ntho?

Tharollo:

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

∑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):

∑F = ma

F – F _fric = ma

16 – 2 = 8 a

14 = 8 a

a = 14/8

a = 1.75 m/s ²

Ho potlaka ha ntho ke 1.75 m/s ².

Karabo e nepahetseng ke A.

10. Tom le Andrew ba sutumelletsa ntho fatše le boreleli. Tom o sutumelletsa ntho ka matla a 5.70 N. Haeba boima ba ntho e le 2.00 kg mme lebelo le bonoang ke ntho e le 2.00 ms ^-2 , joale fumana boholo le tataiso ea ketso ea matla ke 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 sebetsoang ke Andrew (F ₁ ) = 5.70 Newton

Boima ba ntho (m) = 2.00 kg

Ho potlaka (a) = 2.00 m/s ²

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

Tharollo:

Sebelisa molao oa bobeli oa Newton oa motsamao:

∑ F = ma

F ₁ + F ₂ = ma

5.70 + F ₂ = (2)(2)

5.70 + F ₂ = 4

F ₂ = 4 – 5.70

F ₂ = – 1.7 Newton

Letšoao la ho tlosa le bontšitse hore (F ₂ ) le fapane le ketso ea Andrew ea ho sutumelletsa matla (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?

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.

[irp]

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

Ho batloa: Ho potlakisa 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/s ²

[irp]

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) Libuka tse pampiring ha li oe ha pampiri e huloa kapele

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

(4) Li- ars 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

[ID ea sephutheloana sa wpdm='470′]

1. Boima le boima
2. Matla a tloaelehileng
3. Molao oa bobeli oa Newton oa ho sisinyeha
4. Matla a khohlano
5. Ho sisinyeha holim'a bokaholimo bo otlolohileng ntle le matla a khohlano
6. Motsamao oa 'mele e 'meli e nang le lebelo le tšoanang holim'a bokaholimo bo bataletseng bo bataletseng ka matla a khohlano
7. Ho sisinyeha ha sefofane se sekametseng ntle le matla a khohlano
8. Motsamao sefofaneng se sekametseng ka thata ka matla a khohlano
9. Motsamao ka lifting
10. Motsamao oa 'mele o hokahantsoe ka lithapo le li-pulleys
11. 'Mele e 'meli e nang le boholo bo tšoanang ba ho potlaka
12. Ho potoloha mothapo o bataletseng - matla a motsamao o chitja
13. Ho potoloha mothapo o kobehileng - matla a motsamao o chitja
14. Motsamao o tšoanang ka selikalikoe se otlolohileng
15. Matla a bohareng a motsamao o chitja o tšoanang

Bala haholoanyane

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 ke 1 kg. Ho potlaka ha matla a khoheli ke 9.8 m/s ² . Fumana matla a tloaelehileng a sebelisoang nthong eo tafoleng.

Tse tsejoang:

Boima (m) = 1 kg

Ho potlaka ha matla a khoheli (g) = 9.8 m/s ²

Boima (w) = mg = (1 kg)(9.8 m/s ² ) = 9.8 kg m/s ² = 9.8 Newtons

Ho batloa: matla a tloaelehileng (N)

Tharollo:

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 ka ho otloloha 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 phomotseng 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 ema ka ho sa feleng.

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.

[irp]

2. Lintho tse peli tse lutseng tafoleng. Boima ba ntho 1 (m ₁ ) = 1 kg, boima ba ntho 2 (m ₂ ) = 2 kg, ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s ^2. Fumana boholo le tataiso ea matla a tloaelehileng a sebelisoang ke m ₂ ho m ₁ le matla a tloaelehileng a sebelisoang ke tafole ho m ₂.

tharollo

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/s ²

Boima ba ntho 1 (w ₁ ) = m ₁ g = (1)(9.8 m/s ² ) = 9.8 kg m/s ² = 9.8 Newtons

Boima ba ntho 2 (w ₂ ) = m ₂ g = (2)(9.8 m/s ² ) = 19.6 kg m/s ² = 19.6 Newtons

Ho batloa: N ₁ le N ₂

Tharollo:

(a) Matla a tloaelehileng a sebelisoang ke m ₂ ho m ₁ (N ₁ )

N ₁ = w ₁ = 9.8 Newton

Tsela ea N1 _e ea holimo.

(b) Matla a tloaelehileng a sebelisoang ke tafole ho m ₂ (N ₂ )

N ₂ = w ₁ + w ₂ = 9.8 Newton + 19.6 Newton = 29.4 Newton

Tsela ea N2 _e ea holimo.

[irp]

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

tharollo

Tse tsejoang:

Boima ba ntho (m) = 2 kg

Ho potlaka ka lebaka la matla a khoheli (g) = 9.8 m/s ²

Boima (w) = mg = (2 kg)(9.8 m/s ² ) = 19.6 kg m/s ² = 19.6 Newtons

Matla F (F) = 10 Newton

Ho batloa : 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

[irp]

4. Ntho e phomotseng tafoleng. Boima ba ntho ke 1 kg, ho potlaka ka lebaka la matla a khoheli ke 9,8 m/s ² , matla F ₁ ke 10 N mme matla F ₂ ke 20 N. Fumana boholo le tataiso ea matla a tloaelehileng a sebelisoang ke tafole nthong eo. g = 9.8 m/s ²

tharollo

Tse tsejoang:

Boima (m) = 1 kg

Ho potlaka ha matla a khoheli (g) = 9.8 m/s ²

Boima (w) = mg = (1 kg)(9.8 m/s ² ) = 9.8 kg m/s ² = 9.8 Newtons

F ₁ = 10 Newton

F ₂ = 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 – F ₂ – w + F ₁ = 0

N = F ₂ + w – F ₁

N = 20 Newtons + 9.8 Newtons – 10 Newtons

N = 19.8 Newton

[irp]

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

Tharollo:

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/s ²

boima (w) = mg = (2 kg)(9.8 m/s ² ) = 19.6 kg m/s ² = 19.6 Li-Newton

w _x = w sin 60 ^o = (19.6 N)(0.5) √ 3 = 9.8 √ 3 Li-Newton

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

Ho batloa: matla a tloaelehileng (N)

Tharollo:

∑ F = 0

N – w _y = 0

N = w _y

N = 9.8 Newton

[irp]

[ID ea sephutheloana sa wpdm='467′]

1. Boima le boima
2. Matla a tloaelehileng
3. Molao oa bobeli oa Newton oa ho sisinyeha
4. Matla a khohlano
5. Ho sisinyeha holim'a bokaholimo bo otlolohileng ntle le matla a khohlano
6. Motsamao oa 'mele e 'meli e nang le lebelo le tšoanang holim'a bokaholimo bo bataletseng bo bataletseng ka matla a khohlano
7. Ho sisinyeha ha sefofane se sekametseng ntle le matla a khohlano
8. Motsamao sefofaneng se sekametseng ka thata ka matla a khohlano
9. Motsamao ka lifting
10. Motsamao oa 'mele o hokahantsoe ka lithapo le li-pulleys
11. 'Mele e 'meli e nang le boholo bo tšoanang ba ho potlaka
12. Ho potoloha mothapo o bataletseng - matla a motsamao o chitja
13. Ho potoloha mothapo o kobehileng - matla a motsamao o chitja
14. Motsamao o tšoanang ka selikalikoe se otlolohileng
15. Matla a bohareng a motsamao o chitja o tšoanang

Bala haholoanyane

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/s ²

Tse tsejoang:

Boima (m) = 1 kg

Ho potlaka ho bakoang ke matla a khoheli holim'a Lefatše (g) = 9.8 m/s ²

Ho batloa: 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/s ² )

w = boima (Yuniti ea SI ea w ke kg m/s ² kapa Newton)

Boima ba 'mele:

w = (1 kg)(9.8 m/s ² ) = 9.8 kg m/s ² = 9.8 Li-Newton

[irp]

2.

(a) Thala matla a khoheli (boima) a sebetsang nthong ha ntho e phomotse tafoleng, joalo ka ha ho bontšitsoe setšoantšong sa (a).

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

tharollo

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

[irp]

3. Boima ba lebokose ke 1 kg mme lebelo ka lebaka la matla a khoheli ke 9.8 m/s ^2. Fumana (a) boima (b) karolo e rapameng le karolo e otlolohileng ea boima.

tharollo

Boima : w = mg = (1 kg)(9.8 m/s ² ) = 9.8 kg m/s ² = 9.8 Newtons

Karolo e rapameng ea boima:

w _x = w sin 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

[irp]

[ID ea sephutheloana sa wpdm='458′]

1. Boima le boima
2. Matla a tloaelehileng
3. Molao oa bobeli oa Newton oa ho sisinyeha
4. Matla a khohlano
5. Ho sisinyeha holim'a bokaholimo bo otlolohileng ntle le matla a khohlano
6. Motsamao oa 'mele e 'meli e nang le lebelo le tšoanang holim'a bokaholimo bo bataletseng bo bataletseng ka matla a khohlano
7. Ho sisinyeha ha sefofane se sekametseng ntle le matla a khohlano
8. Motsamao sefofaneng se sekametseng ka thata ka matla a khohlano
9. Motsamao ka lifting
10. Motsamao oa 'mele o hokahantsoe ka lithapo le li-pulleys
11. 'Mele e 'meli e nang le boholo bo tšoanang ba ho potlaka
12. Ho potoloha mothapo o bataletseng - matla a motsamao o chitja
13. Ho potoloha mothapo o kobehileng - matla a motsamao o chitja
14. Motsamao o tšoanang ka selikalikoe se otlolohileng
15. Matla a bohareng a motsamao o chitja o tšoanang

Bala haholoanyane

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 potlaka ka lebaka la matla a khoheli (g) = 10 m/s ².

tharollo

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

v _t = v _o + ho

s = vo _ot + ½ ho ²

v _t ² = v _o ² + 2 jwalo ka

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 (le holimo ho ea holimo)

Ho potlaka ha matla a khoheli (g) = – 10 m/s ² (e mpe ho ya tlase).

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

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

Tharollo:

v _t ² = v _o ² + 2 gh

0 = (20 ² ) + 2(-10) h

0 = 400 – 20 dihora

400 = lihora tse 20

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

[irp]

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/s ² . Hlokomoloha ho hanyetsa metsi.

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)

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

Ho potlaka ha matla a khoheli (g) = -10 m/s ² (e mpe ho ea tlase)

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 (vo _o ) = 20 m/s (ntle ho ya hodimo), Ho potlaka ha matla a khoheli (g) = -10 m/s ² (hampe ho ya tlase).

h = vot ₊ ½ gt ²

-100 = (20) t + ½ (-10) t ²

-100 = 20 t – 5 t ²

-5 t ² + 20 t + 100 = 0

Re sebelisa foromo ea quadratic:

(b) Lebelo la ho qetela

v _t ² = v _o ² + 2 gh

v _t ² = (20 ² ) + 2 (-10)(-100)

v _t ² = 400 + 2000

v _t ² = 2400

v _t = 49 m/s

[irp]

[ID ea sephutheloana sa wpdm='515′]

[ID ea sephutheloana sa wpdm='517′]

1. Sebaka le ho falla
2. Lebelo le tloaelehileng le lebelo le tloaelehileng
3. Lebelo le sa fetoheng
4. Ho potlaka kamehla
5. Motsamao oa hoetla o lokolohileng
6. Motsamao oa ho theoha ha ho oa ka bolokolohi
7. Motsamao o yang hodimo le tlase nakong ya ho wa ha mahala

Bala haholoanyane

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/s ² . Hlokomoloha ho hanyetsa metsi.

Tse tsejoang:

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

Nako e fetileng (t) = metsotsoana e 2

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

Ho batloa: Lebelo la ho qetela (v _t )

Tharollo:

Ho potlaka ha 10 m/s ² ho 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 ka ho potlaka ho sa khaotseng , joalo ka ha ho bontšitsoe ka tlase:

v _t = vo _o + ho ………. 1

h = vo _o t + ½ ho ² ………. 2

v _t ² = v _o ² + 2 ah ………. 3

v _t = v _o + g t

v _t = 10 + (10)(2)

v _t = 10 + 20 = 30 m/s

Lebelo la ho qetela = v _t = 30 m/s

[irp]

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 (vo _o ) = 5 m/s

Nako e fetileng (t) = metsotsoana e 2

Ho potlaka ka lebaka la matla a khoheli (g) = 10 m/s ²

Ho batloa: bophahamo ba borokho (h)

Tharollo:

h = vot ₊ ½ gt ²

h = (5)(2) + ½ (10)(2) ²

h = 10 + (5)(4)

h = 10 + 20

h = limithara tse 30

[irp]

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 (vo _o ) = 10 m/s

Ho potlaka ha matla a khoheli (g) = 10 m/s ²

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 _t ² = v _o ² + 2 gh

v _t ² = (10) ² + 2(10)(80) = 100 + 1600 = 1700

v _t = 41 m/s

Nako ea ho arohana (t):

v _t = v _o + g t

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

[irp]

[ID ea sephutheloana sa wpdm='513′]

[ID ea sephutheloana sa wpdm='517′]

1. Sebaka le ho falla
2. Lebelo le tloaelehileng le lebelo le tloaelehileng
3. Lebelo le sa fetoheng
4. Ho potlaka kamehla
5. Motsamao oa hoetla o lokolohileng
6. Motsamao oa ho theoha ha ho oa ka bolokolohi
7. Motsamao o yang hodimo le tlase nakong ya ho wa ha mahala

Bala haholoanyane