Ithiyori yamandla okusebenza-kinetic ithi umsebenzi ophelele, noma umzamo owenziwa amandla e-net entweni, ulingana noshintsho emandleni e-kinetic ento. Uma amandla e-net enza umsebenzi omuhle (amandla e-net asendleleni efanayo nokufuduka), khona-ke amandla e-kinetic ento ayanda. Ngokuphambene nalokho, uma amandla e-net enza umsebenzi ongemuhle (amandla e-net asendleleni ephambene nokufuduka), khona-ke amandla e-kinetic ento ayancipha.
W inani = EK 2 – EK 1 = ½ mv 2 2 – ½ mv 1 2
Uma amandla okugcina kuphela esebenza entweni, njengasendabeni yento ewa ngokukhululekile , khona-ke amandla aphelele alingana namandla okugcina. Isitatimende se-work-kinetic energy theorem singaguqulwa sibe umsebenzi ophelele, noma umsebenzi owenziwe amandla okugcina ulingana noshintsho lwamandla e-kinetic. Uma amandla okugcina enza umsebenzi omuhle (amandla okugcina asendleleni efanayo nokususwa), khona-ke amandla e-kinetic ento ayanda. Ngokuphambene nalokho, uma amandla okugcina enza umsebenzi omubi (amandla okugcina asendleleni ephambene nokususwa), khona-ke amandla e-kinetic ento ayancipha.
W c = EK 2 – EK 1 = ½ mv 2 2 – ½ mv 1 2
Umsebenzi owenziwa amandla okulondoloza entweni ulingana noshintsho olubi emandleni angase abe khona ento. Uma amandla okulondoloza enza umsebenzi omuhle, amandla angase abe khona ayancipha. Ngokuphambene nalokho, uma amandla okulondoloza enza umsebenzi omubi, amandla angase abe khona ayanda.
W c = – (EP 2 – EP 1 ) = – mg (h 2 – h 1 ) = – mgh 2 + mgh 1 = mgh 1 – mgh 2
Ngokusekelwe ekubuyekezweni kwangaphambilini, kubonakala sengathi kunobudlelwano phakathi komsebenzi owenziwe amandla alondolozayo entweni kanye nezinguquko emandleni e-kinetic kanye namandla angase abe khona entweni. Uma amandla alondolozayo enza umsebenzi omuhle, amandla e-kinetic ayanda, kuyilapho amandla angase abe khona ehla. Uma amandla alondolozayo enza umsebenzi omubi, amandla e-kinetic ayancipha, kuyilapho amandla angase abe khona enyuka.
W c = W c
EK 2 – EK 1 = EP 1 – EP 2
I-EP 1 + EK 1 = I-EP 2 + EK 2
I-EM 1 = I-EM 2
Imininingwane:
I-EM 1 = amandla okuqala omshini, i-EM 2 = amandla okugcina omshini, i-EP 1 = amandla okuqala angase abe khona, i-EP 2 = amandla okugcina angase abe khona, i-EK 1 = amandla okuqala omshini, i-EK 2 = amandla okugcina omshini
Isibonelo sezinkinga
1. Ibhulokhi ikhishwa ngaphandle kwesivinini sokuqala phezulu kwendiza ebushelelezi ethambekele (ku-A). Ibhulokhi ishelela phansi kwe-inclination (ku-E).
Uma u-AB = BC = CD = DE, khona-ke isilinganiso sejubane lamabhulokhi ku-C, D kanye no-E singu…
Ingxoxo
Kuyaziwa:
AB + BC + CD + DE = 1
1/4 + 1/4 + 1/4 + 1/4 = 1
Umbuzo: Ukuqhathaniswa kwesivinini samabhulokhi ku-C, D kanye no-E.
Impendulo:
Umthetho wokulondolozwa kwamandla omshini uthi amandla okuqala omshini = amandla okugcina omshini.
Amandla okuqala omshini = amandla adonsela phansi
Amandla okugcina omshini = amandla e-kinetic
Phezulu, ibhloko liphumule, ngakho-ke amandla alo e-kinetic ayi-zero kanti amandla alo e-gravitational potential asezingeni eliphezulu. Njengoba lisuka phezulu liye phansi kokwehla, amandla alo e-gravitational potential ayancipha futhi aguqulwa abe amandla e-kinetic. Uma lifika phansi kokwehla, amandla alo e-kinetic asezingeni eliphezulu kanti amandla alo e-gravitational potential asezingeni eliphezulu.
Amandla adonsela phansi endaweni A = mgh = mg (4/4) = 4/4 mg
Amandla e-kinetic endaweni A = 1/2 mv 2 = 1/2 m (0 2 ) = 0
Amandla adonsela phansi endaweni B = mgh = mg (3/4) = 3/4 mg
Amandla e-kinetic endaweni B = 1/2 mv 2
Amandla adonsela phansi endaweni C = mgh = mg (2/4) = 2/4 mg
Amandla e-kinetic endaweni C = 1/2 mv 2
Amandla adonsela phansi endaweni D = mgh = mg (1/4) = 1/4 mg
Amandla e-kinetic endaweni D = 1/2 mv 2
Amandla adonsela phansi endaweni ethi E = mgh = mg (0) = 0
Amandla e-kinetic endaweni E = 1/2 mv 2
Amandla omshini endaweni A = Amandla adonsela phansi + amandla e-kinetic = 4/4 mg + 0 = 4/4 mg = mg
Amandla omshini endaweni B = Amandla adonsela phansi + amandla e-kinetic = 3/4 mg + 1/2 mv 2
Amandla omshini endaweni C = Amandla adonsela phansi + amandla e-kinetic = 2/4 mg + 1/2 mv 2
Amandla omshini endaweni D = Amandla adonsela phansi + amandla e-kinetic = 1/4 mg + 1/2 mv 2
Amandla omshini endaweni E = Amandla adonsela phansi + amandla e-kinetic = 0 + 1/2 mv 2 = 1/2 mv 2
Isivinini sebhulokhi ku-C:
Amandla omshini endaweni C = amandla okuqala omshini (amandla omshini ahlala njalo)
2/4 mg + 1/2 mv 2 = 4/4 mg
1/2 mv 2 = 4/4 mg – 2/4 mg
1/2 mv 2 = 2/4 mg
1/2 mv 2 = 1/2 mg
i- mv 2 = mg
v 2 = g
v = √g
Isivinini sebhulokhi ku-D:
Amandla omshini endaweni D = amandla okuqala omshini (amandla omshini ahlala njalo)
1/4 mg + 1/2 mv 2 = 4/4 mg
1/2 mv 2 = 4/4 mg – 1/4 mg
1/2 mv 2 = 3/4 mg
1/2 v 2 = 3/4 g
v 2 = 2 (3/4) g
v 2 = (6/4) g
v 2 = (3/2) g
v 2 = 1,5g
v = √1,5g
Isivinini sebhulokhi ku-E:
Amandla omshini endaweni E = amandla okuqala omshini (amandla omshini ahlala njalo)
1/2 mv 2 = mg
1/2 v 2 = g
v 2 = 2g
v = √2g
Ukuqhathanisa isivinini samabhulokhi ku-C, D kanye no-E:
√ g : √1,5g : √2g
√ 1 : √1,5 : √2 (phinda ngo-2)
√ 2: √ 3: √ 4
√ 2: √ 3: 2
2. Bheka isithombe esilandelayo!
Izinto ezimbili zehla ngendlela esuka endaweni A. Isisindo sento yokuqala singu-m 1 = 5 kg kanti into yesibili ingu-m 2 = 15 kg. Uma ukusheshisa okubangelwa amandla adonsela phansi kungu-g = 10 ms s -2 , khona-ke isilinganiso samandla e-kinetic Ek 1 : Ek 2 endaweni B singu…
A. 1: 2
B. 1: 3
C. 1: 9
D. 2: 1
Isahluko 3: 1
Ingxoxo
Kuyaziwa ukuthi:
Isisindo sento 1 (m 1 ) = 5 kg
Isisindo sento 2 (m 2 ) = 15 kg
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Ibanga phakathi kwamaphuzu A no-B (h) = 40 m – 30 m = 10 m
Umbuzo: Ukuqhathaniswa kwamandla e-kinetic Ek 1 : Ek 2 endaweni B
Impendulo:
Amandla adonsela phansi ento ku-A alinganiswa kusukela endaweni B:
I-EP 1 = m 1 gh = (5 kg)(10 m/s 2 )(10 m) = 500 ama-Joules
I-EP 2 = m 2 gh = (15 kg)(10 m/s 2 )(10 m) = 1500 ama-Joules
Umthetho wokulondolozwa kwamandla omshini uthi amandla okuqala omshini = amandla okugcina omshini.
Amandla okuqala omshini = amandla adonsela phansi
Amandla okugcina omshini = amandla e-kinetic
I-EP 1 = EK 1
AmaJoules angu-500 = EK 1
I-EP 2 = EK 2
AmaJoules angu-1500 = EK 2
EX 1 : EX 2
500: 1500
5: 15
1: 3
Impendulo efanele ingu-B.
3. Bheka isithombe!
Imabula enobunzima buka-m igoqa phansi endaweni ebushelelezi eyindilinga ngaphandle kwejubane lokuqala kusukela endaweni A. Ijubane lemabula endaweni B lingu…
A. 10 ms-1
B. 2 √10 ms -1
C. √ 10 ms.s -1
D. 2 √5 ms -1
E. √ 5 ms.s -1
Ingxoxo
Kuyaziwa ukuthi:
Umehluko ngokuphakama phakathi kwamaphuzu A no-B (h) = 4 m – 2 m = 2 amamitha
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 m/s 2
Isisindo semabula (m) = m
Umbuzo: Isivinini semabula endaweni B
Impendulo:
Amandla okuqala omshini = amandla omshini endaweni A = amandla adonsela phansi = mgh = (m)(10)(2) = 20m
Amandla okugcina omshini = amandla omshini endaweni B = amandla omshini = 1/2 mv 2
Umthetho wokulondolozwa kwamandla emishini:
Amandla okuqala omshini = amandla okugcina omshini
20 m = 1/2 mv 2
20 = 1/2 v 2
2 (20) = v 2
40 = v 2
v = √40
v = √(4)(10)
v = 2√10 m/s
Impendulo efanele ingu-B.
4. Ibhola liyashelela endaweni eshelelayo njengoba kuboniswe esithombeni esingezansi.
Uma ijubane lebhola endaweni A lingu-6 ms -1 , endaweni B lingu- √ 92 ms -1 , kanye no-g = 10 ms -2 , khona-ke ukuphakama kwephuzu B kusukela phansi kwethrekhi...
A. 0,5 m
B. 0,5 √ 2 m
C. 1 m
D. √ 2 m
E. 2,8 m
Ingxoxo
Kuyaziwa ukuthi:
Isivinini sebhola endaweni A (v A ) = 6 ms -1
Isivinini sebhola endaweni B (v B ) = √ 92 ms -1
Ukusheshisa ngenxa yamandla adonsela phansi (g) = 10 ms -2
Ukuphakama A (h A ) = amamitha angu-5,6
Ukuphakama B (h B ) = h
Umbuzo: Ukuphakama kwephuzu B kusukela phansi kwethrekhi
Impendulo:
Umzila uyashelela, ngakho akukho ukungqubuzana. Ngakho-ke, amandla odwa asebenza entweni amandla adonsela phansi. Amandla adonsela phansi angamandla alondolozayo. Umsebenzi owenziwa amandla alondolozayo awuxhomekile ekubunjweni komzila kodwa kunalokho ekushintsheni kwesikhundla.
Ngakho-ke singabala ukuphakama kwephuzu B kusukela phansi komzila sisebenzisa ifomula yomthetho wokulondolozwa kwamandla omshini.
Amandla okuqala omshini = Amandla adonsela phansi
Uma kufika iphuzu A, ibhola alikahambi okwamanje ngakho ijubane lalo liyi-zero. Ijubane lebhola liyi-zero ngakho amandla e-kinetic ebhola ayi-zero. Amandla e-kinetic: EK = 1/2 mv 2 = 1/2 m (0) = 0.
Kodwa ibhola liphakeme ngamamitha angu- 5,6 ukusuka phansi ngakho ibhola linamandla adonsela phansi. Amandla adonsela phansi: EP = mgh = m (10)(5,6) = 56 m
Amandla okuqala omshini = Amandla adonsela phansi + Amandla e-Kinetic = 56 m + 0 = 56 m
Amandla okugcina omshini = Amandla adonsela phansi + Amandla e-Kinetic
Uma kufikwa endaweni B, ukuphakama kwebhola kungu-h. Amandla adonsela phansi: EP = mgh = m (10) h = 10 mh
Iphuzu B lingaphansi kakhulu kwephuzu A ngakho ibhola lisahamba ngesivinini esithile. Ibhola lisahamba ngesivinini esithile ngakho ibhola linamandla e-kinetic. Amandla e-kinetic: EK = 1/2 mv 2 = 1/2 m ( √ 92 ) 2 = 1/2 m ( 92) = 46 m
Amandla okugcina omshini = Amandla adonsela phansi + Amandla e-Kinetic = 10 mh + 46 m = m (10 h + 46)
Umthetho wokulondolozwa kwamandla emishini:
Amandla okuqala omshini = Amandla okugcina omshini
56 m = m (10 amahora + 46)
56 = 10 amahora + 46
56 – 46 = amahora ayi-10
10 = amahora angu-10
h = 10/10 = 1 imitha
Impendulo efanele ngu-C.
Umthombo wombuzo:
Imibuzo Yefiziksi Yezivivinyo Zikazwelonke Zesikole Samabanga Aphezulu/Isikole Samabanga Aphezulu Sokufundela Umsebenzi
