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Updated Apr 2, 2026

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Osnove kinematike in dinamike

Kinematika in dinamika točke sta osnova fizike, ki jo potrebuješ... Show more

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# Kinematika in dinamika točke

Uvodni pregled

Kinematika in dinamika točke sta osnova mehanike. Kinematika opisuje kako
se telesa gibljejo

Osnove kinematike in dinamike

Če se ti zdi fizika zapletena, te lahko pomirimo – kinematika je preprosto opisovanje gibanja, dinamika pa razlaganje vzrokov za to gibanje. Pri kinematiki se ukvarjaš s potjo, hitrostjo in pospeškom, pri dinamiki pa s silami.

Masna točka je tvoj najboljši prijatelj pri reševanju nalog. To je poenostavitev, kjer zanemariš velikost telesa in se osredotočiš samo na njegovo maso ter gibanje.

Najpomembnejše vektorske količine so lega (kje se nahaja točka), hitrost (kako hitro in kam se giblje) in pospešek (kako se spreminja hitrost). Pomni: pospešek nastane zaradi sile – to je ključna povezava med kinematiko in dinamiko.

Nasvet: Vektorske količine imajo smer IN velikost – to je razlika med hitrostjo (vektor) in brzino (skalar).

Kinematika - gibanje brez sil

Pri enakomernem premočrtnem gibanju se premikаš s konstantno hitrostjo. Formula je enostavna: s = v·t. Če pa imaš konstanten pospešek, uporabiš tri glavne enačbe.

Enakomerno pospešeno gibanje obvladaš s tremi formulami: v(t) = v₀ + at za hitrost, s(t) = v₀t + ½at² za pot in v² = v₀² + 2as (Torricellijeva enačba), ko čas ni pomemben.

Te enačbe delujejo za gibanje v eni dimenziji. Pri dvodimenzionalnem gibanju (kot je poševni met) razdeli gibanje na x in y komponento ter obravnavaj vsako posebej.

Pozor: V eni smeri je lahko gibanje enakomerno, v drugi pa pospešeno – tipičen primer je poševni met!

Newtonovi zakoni - temelj dinamike

Prvi Newtonov zakon (vztrajnost) pravi, da telo ohranja svoje gibalno stanje, dokler nanj ne deluje zunanja sila. Če je F_rez = 0, potem je v = konstanta.

Drugi Newtonov zakon je srce dinamike: F_rez = m·a. To je vektorska enačba, zato jo vedno razstavi na komponente! Za vsako smer posebej: ΣF_x = m·a_x in ΣF_y = m·a_y.

Tretji Newtonov zakon opisuje akcijo in reakcijo: F₁→₂ = -F₂→₁. Pozor na pogosto napako – sili akcije in reakcije delujeta na različni telesi, zato se nikoli ne seštejeta v rezultanto!

Ključno: Drugi zakon je osnova za reševanje nalog. Vedno narisi diagram sil in koordinatni sistem!

Vrste sil in pristop k nalogam

Glavne sile, ki jih srečaš: sila teže $F_g = mg$ , oporna sila $F_N, pravokotno na podlago$ , trenje $F_tr = μ·F_N, nasprotno gibanju$ in sila vrvice $F_v, vzdolž vrvice$ .

Sistematičen pristop k reševanju: nariši skico, izoliraj vsako telo posebej in nariši diagram sil (najpomembnejši korak!), postavi koordinatni sistem, zapiši 2. Newtonov zakon in ga razstavi na komponente.

Pri problemih s klancem postavi eno os vzporedno s klancem, drugo pravokotno nanj. To občutno poenostavi računanje, ker razbije silo teže na uporabne komponente.

Praktičen nasvet: Pri klancu je x-os vzporedna s klancem, y-os pravokotna nanj – tako se izogneš zapletenim izračunom!

Praktičen primer: klada na klancu

Klada z maso 5 kg na klancu 30° s koeficientom trenja 0,2. Zunanja sila 50 N deluje navzgor po klancu. Kako velik je pospešek?

Sile na klado: teža (mg navzdol), oporna sila $F_N pravokotno na klanec$ , zunanja sila (F vzporedno navzgor), trenje $F_tr vzporedno navzdol$ .

Razstavitev sil: Sila teže ima komponenti mg·sin α (vzporedno s klancem navzdol) in mg·cos α (pravokotno na klanec). Oporna sila je F_N = mg·cos α = 42,4 N.

Po x-osi (vzporedno s klancem): F - mg·sin α - μ_k·F_N = ma. Vstaviš vrednosti: 50 - 24,5 - 8,48 = 5a, torej a = 3,4 m/s².

Sistematičnost: Vedno razstavi sile na komponente glede na izbran koordinatni sistem!

Atwoodov stroj - povezani telesi

Dve telesi (2 kg in 3 kg) povezani z vrvico preko škripca. Ker je vrvica neraztegljiva, imata telesi enak pospešek. Ker je sila v idealni vrvici povsod enaka, uporabiš isto F_v za obe telesi.

Za lažje telo (m₁): F_v - m₁g = m₁a. Za težje telo (m₂): m₂g - F_v = m₂a. Seštej enačbi, da se F_v izniči: $m₂ - m₁$ g = $m₁ + m₂$ a.

Pospešek sistema: a = $m₂ - m₁$ g/ $m₁ + m₂$ = 1,96 m/s². Sila v vrvici: F_v = m₁ $a + g$ = 23,52 N.

Ključna ideja: Pri povezanih telesih je pospešek enak, sile v vrvici pa povsod enake!

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Fizika

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Updated Apr 2, 2026

•

9 pages

Osnove kinematike in dinamike

Kinematika in dinamika točke sta osnova fizike, ki jo potrebuješ za razumevanje gibanja. Kinematika se ukvarja s tem, kako se telesa gibljejo, dinamika pa razloži, zakaj se gibljejo.

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Osnove kinematike in dinamike

Masna točka je tvoj najboljši prijatelj pri reševanju nalog. To je poenostavitev, kjer zanemariš velikost telesa in se osredotočiš samo na njegovo maso ter gibanje.

Nasvet: Vektorske količine imajo smer IN velikost – to je razlika med hitrostjo (vektor) in brzino (skalar).

Sign up to see the contentIt's free!

Access to all documents

Improve your grades

Join milions of students

Kinematika - gibanje brez sil

Pri enakomernem premočrtnem gibanju se premikаš s konstantno hitrostjo. Formula je enostavna: s = v·t. Če pa imaš konstanten pospešek, uporabiš tri glavne enačbe.

Enakomerno pospešeno gibanje obvladaš s tremi formulami: v(t) = v₀ + at za hitrost, s(t) = v₀t + ½at² za pot in v² = v₀² + 2as (Torricellijeva enačba), ko čas ni pomemben.

Te enačbe delujejo za gibanje v eni dimenziji. Pri dvodimenzionalnem gibanju (kot je poševni met) razdeli gibanje na x in y komponento ter obravnavaj vsako posebej.

Pozor: V eni smeri je lahko gibanje enakomerno, v drugi pa pospešeno – tipičen primer je poševni met!

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Newtonovi zakoni - temelj dinamike

Prvi Newtonov zakon (vztrajnost) pravi, da telo ohranja svoje gibalno stanje, dokler nanj ne deluje zunanja sila. Če je F_rez = 0, potem je v = konstanta.

Drugi Newtonov zakon je srce dinamike: F_rez = m·a. To je vektorska enačba, zato jo vedno razstavi na komponente! Za vsako smer posebej: ΣF_x = m·a_x in ΣF_y = m·a_y.

Ključno: Drugi zakon je osnova za reševanje nalog. Vedno narisi diagram sil in koordinatni sistem!

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Vrste sil in pristop k nalogam

Glavne sile, ki jih srečaš: sila teže $F_g = mg$ , oporna sila $F_N, pravokotno na podlago$ , trenje $F_tr = μ·F_N, nasprotno gibanju$ in sila vrvice $F_v, vzdolž vrvice$ .

Pri problemih s klancem postavi eno os vzporedno s klancem, drugo pravokotno nanj. To občutno poenostavi računanje, ker razbije silo teže na uporabne komponente.

Praktičen nasvet: Pri klancu je x-os vzporedna s klancem, y-os pravokotna nanj – tako se izogneš zapletenim izračunom!

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Praktičen primer: klada na klancu

Klada z maso 5 kg na klancu 30° s koeficientom trenja 0,2. Zunanja sila 50 N deluje navzgor po klancu. Kako velik je pospešek?

Sile na klado: teža (mg navzdol), oporna sila $F_N pravokotno na klanec$ , zunanja sila (F vzporedno navzgor), trenje $F_tr vzporedno navzdol$ .

Razstavitev sil: Sila teže ima komponenti mg·sin α (vzporedno s klancem navzdol) in mg·cos α (pravokotno na klanec). Oporna sila je F_N = mg·cos α = 42,4 N.

Po x-osi (vzporedno s klancem): F - mg·sin α - μ_k·F_N = ma. Vstaviš vrednosti: 50 - 24,5 - 8,48 = 5a, torej a = 3,4 m/s².

Sistematičnost: Vedno razstavi sile na komponente glede na izbran koordinatni sistem!

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Access to all documents

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Atwoodov stroj - povezani telesi

Dve telesi (2 kg in 3 kg) povezani z vrvico preko škripca. Ker je vrvica neraztegljiva, imata telesi enak pospešek. Ker je sila v idealni vrvici povsod enaka, uporabiš isto F_v za obe telesi.

Za lažje telo (m₁): F_v - m₁g = m₁a. Za težje telo (m₂): m₂g - F_v = m₂a. Seštej enačbi, da se F_v izniči: $m₂ - m₁$ g = $m₁ + m₂$ a.

Pospešek sistema: a = $m₂ - m₁$ g/ $m₁ + m₂$ = 1,96 m/s². Sila v vrvici: F_v = m₁ $a + g$ = 23,52 N.

Ključna ideja: Pri povezanih telesih je pospešek enak, sile v vrvici pa povsod enake!

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Access to all documents

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Join milions of students

Sign up to see the contentIt's free!

Access to all documents

Improve your grades

Join milions of students

Sign up to see the contentIt's free!

Access to all documents

Improve your grades

Join milions of students