Umthetho Wesibili KaNewton ungomunye wemibono eyisisekelo ku-physics elawula ubudlelwano phakathi kwamandla, ubuningi, kanye nokusheshisa. Lo mthetho uthi ukusheshisa kwento kuhambisana namandla asebenzayo kuyo futhi kuhambisana nobukhulu bayo. Ngokwezibalo, uMthetho Wesibili KaNewton uchazwa kanje:
\[ F = ma \]
Di mana:
– \( F \) amandla aphelele asebenza entweni (kumaNewton, N).
– \( m \) isisindo sento (ngamakhilogremu, kg).
– \( a \) ukusheshisa kwento (ngamamitha ngomzuzwana oyisikwele, \( m/s^2 \)).
Kulesi sihloko, sizoxoxa ngezibonelo ezithile zoMthetho Wesibili kaNewton ukuze siqonde ukusebenza kwawo ezimweni ezahlukene.
Isibonelo Umbuzo 1: Ukuphoqelela Imoto Esheshisayo
Umbuzo:
Imoto enesisindo esingu-1000 kg ishesha ukusuka ekuphumuleni iye esivinini esingama-20 m/s ngemizuzwana emi-5. Bala amandla adingekayo ukuze kufezwe lokhu kusheshisa.
Isixazululo:
Okokuqala, sidinga ukubala ukusheshisa kwemoto. Ukusheshisa (\( a \)) kungabalwa kusetshenziswa ifomula:
\[ a = \frac{\Delta v}{\Delta t} \]
Di mana:
– \(\Delta v\) ushintsho lwejubane.
– \(\Delta t\) ushintsho lwesikhathi.
Faka amanani aziwayo esikhundleni sawo:
\[ a = \frac{20 \, \text{m/s} – 0 \, \text{m/s}}{5 \, \text{seconds}} \]
\[ a = \frac{20 \, \text{m/s}}{5 \, \text{seconds}} \]
\[ a = 4 \, \text{m/s}^2 \]
Manje, singabala amandla adingekayo sisebenzisa uMthetho Wesibili kaNewton:
\[ F = ma \]
\[ F = (1000 \, \text{kg})(4 \, \text{m/s}^2) \]
\[ F = 4000 \, \umbhalo{N} \]
Ngakho-ke, amandla adingekayo ukusheshisa imoto angama-4000 N.
Isibonelo Umbuzo 2: Amandla Okungqubuzana Ebhokisini
Umbuzo:
Ibhokisi elinesisindo esingama-50 kg licindezelwa endaweni elukhuni ngamandla angama-300 N. Uma amandla okungqubuzana phakathi kwebhokisi nendawo engama-100 N, bala ukusheshisa kwebhokisi.
Isixazululo:
Okokuqala, sibala amandla e-net asebenza ebhokisini. Amandla e-net (\( F_{\text{net}} \)) amandla aphelele asebenza entweni ngemva kokucabangela wonke amandla asebenza kuyo, okuhlanganisa nokungqubuzana.
\[ F_{\text{net}} = F_{\text{push}} – F_{\text{friction}} \]
\[ F_{\text{net}} = 300 \, \text{N} – 100 \, \text{N} \]
\[ F_{\text{net}} = 200 \, \text{N} \]
Manje, singabala ukusheshisa kwebhokisi sisebenzisa uMthetho Wesibili kaNewton:
\[ F_{\text{net}} = ma \]
\[ 200 \, \umbhalo{N} = (50 \, \umbhalo{kg})a \]
\[ a = \frac{200 \, \text{N}}{50 \, \text{kg}} \]
\[ a = 4 \, \text{m/s}^2 \]
Ngakho-ke, ukusheshisa kwebhokisi kungu-4 m/s².
Isibonelo Umbuzo 3: Ukubala Amandla Adingekayo Ukuphakamisa Umthwalo
Umbuzo:
I-crane iphakamisa umthwalo onesisindo esingama-200 kg phezulu ngesivinini esingu-1,5 m/s². Bala amandla adingekayo yi-crane ukuphakamisa umthwalo.
Isixazululo:
Okokuqala, sidinga ukubala amandla adonsela phansi asebenza emthwalweni. Amandla adonsela phansi (\( F_g \)) angabalwa kusetshenziswa ifomula:
\[ F_g = mg \]
Di mana:
– \( g \) ukusheshisa okubangelwa amandla adonsela phansi (\( 9,8 \, \text{m/s}^2 \)).
Faka amanani aziwayo esikhundleni sawo:
\[ F_g = (200 \, \text{kg})(9,8 \, \text{m/s}^2) \]
\[ F_g = 1960 \, \umbhalo{N} \]
Manje, sibala amandla aphelele (\( F \)) adingwa yi-crane ukuphakamisa umthwalo sicabangela ukusheshisa okwengeziwe:
\[ F = ma + F_g \]
\[ F = (200 \, \text{kg})(1,5 \, \text{m/s}^2) + 1960 \, \text{N} \]
\[ F = 300 \, \umbhalo{N} + 1960 \, \umbhalo{N} \]
\[ F = 2260 \, \umbhalo{N} \]
Ngakho-ke, amandla adingekayo yi-crane ukuphakamisa umthwalo angama-2260 N.
Isibonelo Inkinga 4: Amandla Ohlelo Lwezinto Ezimbili Ezixhunywe Yintambo
Umbuzo:
Izinto ezimbili ezinesisindo esingu-10 kg kanye no-20 kg ngokulandelana zixhunywe ngentambo elula futhi zilengiswe ku-pulley. Bala ukusheshisa kwesistimu kanye nokucindezeleka entanjeni lapho uhlelo lukhishwa ekuphumuleni.
Isixazululo:
Okokuqala, ake sichaze amandla asebenza kuzo zombili izinto. Ake sibize inqwaba \( m_1 = 10 \, \text{kg} \) kanye nenqwaba \( m_2 = 20 \, \text{kg} \). Amandla adonsela phansi asebenza kuzo zombili izinto yilawa:
\[ F_{g1} = m_1 g = (10 \, \text{kg})(9,8 \, \text{m/s}^2) = 98 \, \text{N} \]
\[ F_{g2} = m_2 g = (20 \, \text{kg})(9,8 \, \text{m/s}^2) = 196 \, \text{N} \]
Njengoba uhlelo lukhishwa ekuphumuleni, ukusheshisa kohlelo kungabalwa kusetshenziswa uMthetho Wesibili kaNewton. Ukusheshisa okuphelele (\( a \)) kohlelo yilokhu:
\[ (m_1 + m_2)a = F_{g2} – F_{g1} \]
\[ (10 \, \text{kg} + 20 \, \text{kg})a = 196 \, \text{N} – 98 \, \text{N} \]
\[ 30 \, \text{kg} \cdot a = 98 \, \text{N} \]
\[ a = \frac{98 \, \text{N}}{30 \, \text{kg}} \]
\[ a = 3,27 \, \text{m/s}^2 \]
Manje, sibala ukucindezeleka entanjeni (\( T \)). Ukucindezeleka entanjeni kungabalwa kusetshenziswa uMthetho Wesibili kaNewton kwenye yezinhlobo, isibonelo \( m_1 \):
\[ T – m_1 g = m_1 a \]
\[ T – 98 \, \text{N} = (10 \, \text{kg})(3,27 \, \text{m/s}^2) \]
\[ T – 98 \, \umbhalo{N} = 32,7 \, \umbhalo{N} \]
\[ T = 32,7 \, \umbhalo{N} + 98 \, \umbhalo{N} \]
\[ T = 130,7 \, \umbhalo{N} \]
Ngakho-ke, ukusheshisa kwesistimu kungu-3,27 m/s² kanti ukucindezeleka entanjeni kungu-130,7 N.
Isiphetho
Ngezibonelo ezahlukahlukene zoMthetho Wesibili kaNewton, sifunde ukuthi lesi simiso sisetshenziswa kanjani ekubaleni amandla, ukusheshisa, kanye nokucindezeleka ezimweni ezahlukene. UMthetho Wesibili kaNewton awubalulekile nje kuphela kwi-physics yethiyori kodwa futhi unezindlela eziningi ezisebenzayo empilweni yansuku zonke kanye nobuchwepheshe. Ngokuqonda nokusebenzisa uMthetho Wesibili kaNewton, singaxazulula izinkinga ezahlukene zemishini ngempumelelo nangokunembile.
