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Zoja S.
fyzika - sš

vypocet

Pomože mi s týmto niekto nejako neviem pohnúť.

Oceľová guľôčka hmotnosti 10 g, je zavesená na ten-
kom vlákne dĺžky 50 cm na pevnom závese. Po vy-
chýlení guľôčky do výšky 6,7 cm nad rovnovážnu po-
lohu a uvoľnení sa začne kývať.
Určte:
1. Celkovú mechanickú energiu sústavy guľôčka-Zem.
2. Veľkosť rýchlosti guľôčky pri prechode rovnováž-
nou polohou.

1 answer
Viliam H.
Viliam H. Contact
Celkova energia je konstanta v tejto sustave a je to Ec=Ep+Ek, potencialna a kineticka. V krajnom bode bude Ep maximalne a Ek nulove pretoze rychlost tu je 0 a vychylka maximalna, naopak ked prechadza bodom 0, resp povodnym rovnovaznym bodom tak ma naopak maximalnu Ek a nulovu Ep, pretoze rychlost tu je maximalna (za rovnovaznou polohou uz teleso spomaluje) a vychylku 0. Takze vo vysledku staci dat Ek=Ep, kde Ek pocitat pre maximalnu vychylku a Ep pri maximalnej rychlosti.
Ek=m*g*h a Ep=1/2*m*v^2.
Celkova energia sa rovna Ep, ktoru vyratas pre maximalnu vychylku, ked je Ek=0 a rychlost zo vztahu Ek=Ep kde si osamostanis v.
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Comments:
Zoja S.
Mohol by si mi prosím , ukázať konkrétny výpočet , lebo počítam ako počítam a nevychádza mi to . Veľmi pekne ďakujem
Viliam H.
Ep(max)=m*g*h=0.01*9.81*0.067=0.006687J Toto je aj celkova energia. Ep(max)=Ek(max) m*g*h=1/2*m*v^2 v=sqrt(2*g*h)=1.15 m/s
Zoja S.
Ďakujem , pomohlo , lebo mňa stále miatla tá dĺžka vlákna 50cm. Ešte raz vďaka
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