10 Dynamics Chinhu chakabatanidzwa netambo pamusoro pepulley Muchina weAtwood - Matambudziko neMhinduro
1. Bhokisi A rine huremu hwe5 kg, rakaiswa panzvimbo yakatsetseka yakatwasuka. Bhokisi B rine huremu hwe3 kg rakarembera pamucheto wetambo yakabatana nebhokisi A pamusoro pepulley. Kukurumidza kunokonzerwa negiravhiti ndeye 10 m/s 2. Chii chinonzi kukurumidza kwemabhokisi ese ari maviri?
Zvinozivikanwa:
Huremu hwebhuroko A (m)A) = 5 kg
Huremu hwebhuroko B (m B ) = 3 kg
Kukurumidza kunokonzerwa negiravhiti (g) = 10 m/s 2
Kurema kwebhuroko B (w B ) = m B g = (3)(10) = 30 Newtons
Zvinodiwa: Kukurumidzisa mabhuroko ese ari maviri (a)
Solution:
Nzvimbo yakatambanuka yakatsetseka saka hapana simba rekukweshana. Simba rinokurumidzisa mabhuroko ese ari maviri huremu hwebhuroko 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
2.
Zvichibva pamufananidzo uri pamusoro apa, 
(1) kukurumidza kwechinhu = 0
(2) chinhu chinofamba nekumhanya kusingachinji
(3) chinhu chiri pazororo
(4) chinhu chinofamba kana huremu hwechinhu chacho huri hudiki pane simba rinodhonza chinhu chacho.
Solution:
(1) Kukurumidza kwechinhu = 0.
Simba reNet:
∑ F = ma –> kumhanyisa (a) = 0
∑ F = 0
F 1 + F 2 – F 3 = 12 + 24 – 36 = 36 – 36 = 0 N
(2) Chinhu chinofamba nespeed isingachinji
Kusakurumidza zvinoreva chinhu chiri kuzorora kana kufamba nespeed isingachinji.
(3) chinhu chiri pazororo
Hapana simba remagetsi rinoreva chinhu chiri pakuzorora.
(4) chinhu chinofamba kana huremu hwechinhu chacho huri hudiki pane simba rinodhonza chinhu chacho.
Kurema kunoshanda padivi rakatwasuka, ukuwo simba rekudhonza richishanda padivi rakatwasuka.
Chinhu chinofamba nenzira yakatwasuka saka masimba akatwasuka chete ndiwo anoshanda pachinhu chacho
3. Kana coefficient yekinetic friction pakati peblock A netafura iri 0.1. Kukurumidza kunokonzerwa negravity kuri 10 m/s2, saka chii chinonzi simba rinoshanda pablock A kuitira kuti system ifambe ichienda kuruboshwe mu 2 m/s 2.
Zvinozivikanwa:
Huremu hwebhuroko A (m)A) = 30 kg
huremu hwebhuroko A (w A ) = (30 kg)(10 m/s 2 ) = 300 kg m/s 2 kana 300 Newtons
Huremu hwebhuroko B (m B ) = 20 kg
huremu hwebhuroko B (w B ) = (20 kg)(10 m/s 2 ) = 200 kg m/s 2 kana 200 Newtons
Kukurumidza kunokonzerwa negiravhiti (g) = 10 m/s 2
Kukweshana kwekinetic ( μk ) = 0.1
Kukurumidza kwesystem (a) = 2 m/s 2 (kuenda kuruboshwe)
Simba rekukweshana kwekinetic (f k ) = μ k N = μ k w A = (0.1)(300) = 30 Newtons
Zvinodiwa: Hukuru hwesimba F
Solution:
Mutemo wechipiri waNewton:
ΣF = ma
Chinhu A chinoenda kuruboshwe:
F – f k – w B = (m A + m B ) a
F – 30 – 200 = (30 + 20)(2)
F – 230 = (50)(2)
F – 230 = 100
F = 230 + 100
F = 330 Newton
4. Zvinhu zviviri, A = 2 kg uye B = 6 kg, zvakabatanidzwa pamucheto wetambo pamusoro pepulley, sezvakaratidzwa mumufananidzo uri pazasi. Kana kukurumidza nekuda kwesimba regiravhiti kuri 10 ms -2 saka chii chinonzi kukurumidza kwechinhu B.
Zvinozivikanwa:
Huremu hwechinhu A (m)A) = 2 kg, mB = 6 kg, g = 10 m/s2
huremu hwechinhu A (w A ) = (m A )(g) = (2)(10) = 20 N
huremu hwechinhu B (w B ) = (m B )(g) = (6)(10) = 60 N
Zvinodiwa: Kukurumidzisa kwechinhu b (kukurumidzisa kwesystem).
Solution:
w B > w A kuitira kuti chinhu B chifambe pasi, chinhu A chifambe kumusoro
ΣF = ma
w B – w A = (m A + m B ) a
60 – 20 = (2 + 6) a
40 = (8) a
a = 5 m/s 2
5. Zvinhu zviviri zvakabatana netambo pamusoro pepulley yakatsetseka, sezvakaratidzwa mumufananidzo uri pazasi. Kana m 1 = 1 kg, m 2 = 2 kg, uye kukurumidza nekuda kwegiravhiti kuri 10 ms -2 , saka chii chinonzi simba rekumanikidza T.
Zvinozivikanwa:
Huremu hwechinhu 1 (m1) = 1 kg
Huremu hwechinhu 2 (m2 ) = 2 kg
Kukurumidza kunokonzerwa negiravhiti (g) = 10 m/s 2
huremu hwechinhu 1 (w 1 ) = m 1 g = (1 kg)(10 m/s 2 ) = 10 kg m/s 2 kana 10 Newtons
huremu hwechinhu 2 (w 2 ) = m 2 g = (2 kg)(10 m/s 2 ) = 20 kg m/s 2 kana 20 Newtons
Zvinodiwa: Simba rekumanikidza (T)?
Solution:
w 2 > w 1 saka m 2 inodzika pasi , m 1 inokwira kumusoro.
Mutemo wechipiri waNewton wekufamba :
ΣF = ma
w 2 – w 1 = (m 1 + m 2 ) a
20 – 10 = (1 + 2) a
10 = (3) a
a = 3.3 m/s 2
Kukurumidza kwesystem = 3.3 m/s 2.
m2 inodzika pasi:
w 2 – T 2 = m 2 a
20 – T 2 = (2)(3.33)
20 – T 2 = 6.66
T 2 = 20 – 6.66
T 2 = 13.3 Newton
m1 inokwira kumusoro:
T 1 – w 1 = m 1 a
T 1 – 10 = (1)(3.3)
T 1 – 10 = 3.33
T 1 = 10 + 3.33
T 1 = 13.3 Newton
Simba rekumanikidza (T) = 13.3 maNewton.
6. Huremu hwe m 1 = 6 kg uye huremu hwe m 2 = 4 kg. Nzvimbo yakatambanuka yakatsetseka. Kumhanya kunokonzerwa negiravhiti ndeye 10 m/s 2 . Chii chinonzi kukurumidza kwesystem?
Zvinozivikanwa:
Huremu hwe m1 = 6 kg
huremu hwe m2 = 4 kg
kukurumidza kunokonzerwa negiravhiti (g) = 10 m/s 2
huremu hwe w 1 = m 1 g = (6 kg)(10 m/s 2 ) = 60 kg m/s 2 kana 60 Newtons
huremu hwe w 2 = m 2 g = (4 kg)(10 m/s 2 ) = 40 kg m/s 2 kana 40 Newtons
Zvinodiwa: Kukurumidza kwesystem (a)
Solution:
m 1 panhandare yakatsetseka yakatwasuka isina kukweshana kuitira kuti sisitimu ikurumidze nehuremu hwebhuroko 2.
Shandisa mutemo wechipiri waNewton:
∑ 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
7. Mabhuroko maviri, bhuroko rimwe nerimwe rine huremu hwe2 kg, rakabatanidzwa netambo pamusoro pepulley, sezvakaratidzwa mumufananidzo uri pazasi. Nzvimbo yakatambanuka nepulley zvakatsetseka. Kana bhuroko B rikadhonzwa nesimba rakatambanuka re40 Newtons, saka chii chinonzi kukurumidza kwebhuroko. Kukurumidza kunokonzerwa negiravhiti 10 m/s 2.
Zvinozivikanwa:
huremu hwebhuroko A (mA) = huremu hwebhuroko B (mB) = 2 kg
Kukurumidza kunokonzerwa negiravhiti (g) = 10 m/s 2
Simba reF = 40 N
huremu hwechinhu A (w A ) = mg = (2)(10) = 20 N
Zvinodiwa: Kukurumidza kwesystem (a)?
Solution:
Shandisa mutemo wechipiri waNewton:
∑ 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
8. Huremu hwebhuroko A = 2 kg uye huremu hwebhuroko B = 1 kg. Bhuroko B pakutanga rakazorora, rozokurumidza kudzika kusvika rasvika pasi. Kumhanya kunokonzerwa negiravhiti i10 m/s 2. Simba rekumanikidzana rinokura sei?
Zvinozivikanwa:
Huremu hwebhuroko A (m)A) = 2 kg
Huremu hwebhuroko B (m B ) = 1 kg
Kukurumidza kunokonzerwa negiravhiti (g) = 10 m/s 2
Kurema kwebhuroko B (w B ) = m B g = (1)(10) = 10 Newtons
Zvinodiwa: Hukuru hwesimba rekumanikidzwa (T)
Solution:
Usafuratira simba rekukweshana.
Kukurumidza kwesystem (a)
∑ F = ma
w B = (m A + m B ) a
10 = (2 + 1) a
10 = 3 a
a = 10/3
Simba rekumanikidza (T)
Simba rekumanikidzwa pabhuroko A:
∑ F = ma
T = m A a = (2)(10/3) = 20/3 = 6.7 Newtons
Simba rekumanikidzana pabhuroko 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
Simba rekumanikidza (T) = 6.7 Newtons
9. Huremu hwebhuroko A = 2 kg uye huremu hwebhuroko B = 1 kg. Simba rekukweshana riri pakati pechinhu A nenzvimbo yakatwasuka = 2.5 Newtons. Usafuratire kukweshana papulley netambo. Chii chinonzi kukurumidza kwemabhuroko ese ari maviri?
Zvinozivikanwa:
Huremu hwebhuroko A (m)A) = 2 kg
Huremu hwebhuroko B (m B ) = 1 kg
Simba rekukweshana pakati pekiyi a nendege yakatwasuka (f kA ) = 2.5 Newtons
Kukurumidza kunokonzerwa negiravhiti (g) = 10 m/s 2
Kurema kwebhuroko (w B ) = m B g = (1)(10) = 10 Newtons
Zvinodiwa: Kukurumidzisa mabhuroko ese ari maviri (a)
Solution:
Shandisa mutemo wechipiri waNewton:
∑ F = ma
w B – f k = (m A + m B ) a
10 – 2.5 = (2 + 1) a
7.5 = 3 a
a = 7.5 / 3
a = 2.5 m/s 2
10. Huremu hwebhuroko a = 30 kg, mira panzvimbo yakatwasuka yakabatana nebhuroko B nehuremu hwe10 kg pamusoro pepulley. Chii chinonzi kukurumidza kwesystem? Kukurumidza kunokonzerwa negiravhiti ndeye 10 ms -2.
Zvinozivikanwa:
Huremu hwebhuroko A (m)A) = 30 kg
Huremu hwebhuroko B (m B ) = 10 kg
Kukurumidza kunokonzerwa negiravhiti (g) = 10 m/s 2
huremu hwebhuroko B (w B ) = m B g = (10)(10) = 100 Newtons
Zvinodiwa: Kukurumidza kwesystem (a)
Solution:
∑ 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
