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Očuvanje mehaničke
energije-pokus
NATAŠA KUDLEK
POTKONJAK
Ponovimo pojmove:
 Energija
 ZOE
 Mehanička energija
Pokus: padanje loptice mase
m sa visine h
 Pribor: Loptica, metar, vaga, zaporna ura
 Tijek rada:
1.Izmjerimo masu loptice
2. Odredimo visinu s koje loptica slobodno pada (npr. na zidu
obilježimo kredom visinu s koje želimo bacati lopticu)
3.Pomoču zaporne ure mjerimo vrijeme padanja loptice.
4. Rezultate mjerenja zapisujemo u tablicu
4.Postupak mjerenja ponovimo nekoliko puta( barem tri mjerenja s iste visine)
5.Izračunamo brzinu kojom je kuglica pala na tlo
6.Odredimo gravitacijsku potencijalnu energiju na početku i kinetičku energiju
na kraju padanja
7.Kao konačni rezultat koristimo srednju vrijednost mjerenja
Obrada vježbe sadržava
 1.Opis pokusa
 1.Rezultati mjerenja
 3.Zaključak
Primjer padanja loptice s neke
visine h:
2
2
2
2
2
2
1
1
mv
mgh
mv
mgh 


2
2
)
(
2
1
2
2
2
1
mv
mv
h
h
mg 


h1
h2
v2
v1
1
2
W = Ek
Em1 Em2
Em1 = Em2
Zadatak 1 Kuglica mase 400 g izbačena je s visine 20 m brzinom
40 m s-1. Koliku kinetičku energiju ima tijelo pri udaru o tlo?
Rješenje:
m = 400 g
Ek = ?
Em1 = Em2 ,
2
0
2
k
mv
mgh E
 
Ek = 400J
= 0,400 kg
h = 20 m
v0 = 40 m s-1
2
0
2
k
v
E m gh
 
 
 
 











 m
20
s
m
10
2
)
s
m
(40
kg
400
,
0 2
-
2
-1
Zadatak2: Loptica mase 300 g pada s visine 4 m. Kolike su
gravitacijska potencijalna energija i kinetička energija loptice:
a) kada se nalazi 2 m iznad tla?
b) Neposredno prije udara o tlo?
Zanemarite otpor zraka.
Rješenje:
m = 300 g
h1 = 4 m
= 0,300 kg
h2 = 2m
a)
Egp2 = mgh2 = 0,300 kg10 m s-22 m Egp2 = 6 J
Em = Egp1
Ek2 = Em – Egp2
Ek2 = 6J
b) Egp3 = 0 J
Em = 12 J
= mgh1= 0,300 kg10 m s-24 m
Ek2 = 12 J – 6 J
Ek3 = Em
Ek3 = 12 J
https://phet.colorado.edu/sims/html/energy-
skate-park/latest/energy-skate-
park_en.html
* U simulaciji na adresi https://javalab.org/en/mechanical_energy_en/
izračunati ukupnu mehaničku energiju kada su kolica u najvišoj i najnižoj
točki staze. Kakva se pretvorba energije tu zbiva? Vrijedi li zakon očuvanja
mehaničke energije? Koliki je iznos sile trenja? Zatim kliknuti na Spring i
ponovo odgovoriti na sva pitanja.
* U simulaciji na adresi https://javalab.org/en/energy_conversion_en/
predvidi na kojoj će se visini zaustaviti topovsko tane. Zatim 10 puta povećaj
masu i opet predvidi i provjeri. Objasni. Vrati masu na 1 kg i 2 puta povećaj
početnu brzinu. Predvidi, provjeri i objasni. Spremi tri slike zaslona i potakni
razred da i oni predvide što će se dogoditi. Ako ne znaju, objasni im.
Literatura:
 Jakov Labor: Fizika 1,udžbenik fizike za 1.razred
gimnazije, Alfa
 https://phet.colorado.edu/sims/html/energy-skate-
park/latest/energy-skate-park_en.html
 Alfa portal- fizika

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ZOME 31.5.22 (1).ppt

  • 2. Ponovimo pojmove:  Energija  ZOE  Mehanička energija
  • 3. Pokus: padanje loptice mase m sa visine h  Pribor: Loptica, metar, vaga, zaporna ura
  • 4.  Tijek rada: 1.Izmjerimo masu loptice 2. Odredimo visinu s koje loptica slobodno pada (npr. na zidu obilježimo kredom visinu s koje želimo bacati lopticu) 3.Pomoču zaporne ure mjerimo vrijeme padanja loptice. 4. Rezultate mjerenja zapisujemo u tablicu
  • 5. 4.Postupak mjerenja ponovimo nekoliko puta( barem tri mjerenja s iste visine) 5.Izračunamo brzinu kojom je kuglica pala na tlo 6.Odredimo gravitacijsku potencijalnu energiju na početku i kinetičku energiju na kraju padanja 7.Kao konačni rezultat koristimo srednju vrijednost mjerenja
  • 6. Obrada vježbe sadržava  1.Opis pokusa  1.Rezultati mjerenja  3.Zaključak
  • 7. Primjer padanja loptice s neke visine h: 2 2 2 2 2 2 1 1 mv mgh mv mgh    2 2 ) ( 2 1 2 2 2 1 mv mv h h mg    h1 h2 v2 v1 1 2 W = Ek Em1 Em2 Em1 = Em2
  • 8. Zadatak 1 Kuglica mase 400 g izbačena je s visine 20 m brzinom 40 m s-1. Koliku kinetičku energiju ima tijelo pri udaru o tlo? Rješenje: m = 400 g Ek = ? Em1 = Em2 , 2 0 2 k mv mgh E   Ek = 400J = 0,400 kg h = 20 m v0 = 40 m s-1 2 0 2 k v E m gh                     m 20 s m 10 2 ) s m (40 kg 400 , 0 2 - 2 -1
  • 9. Zadatak2: Loptica mase 300 g pada s visine 4 m. Kolike su gravitacijska potencijalna energija i kinetička energija loptice: a) kada se nalazi 2 m iznad tla? b) Neposredno prije udara o tlo? Zanemarite otpor zraka. Rješenje: m = 300 g h1 = 4 m = 0,300 kg h2 = 2m a) Egp2 = mgh2 = 0,300 kg10 m s-22 m Egp2 = 6 J Em = Egp1 Ek2 = Em – Egp2 Ek2 = 6J b) Egp3 = 0 J Em = 12 J = mgh1= 0,300 kg10 m s-24 m Ek2 = 12 J – 6 J Ek3 = Em Ek3 = 12 J
  • 11. * U simulaciji na adresi https://javalab.org/en/mechanical_energy_en/ izračunati ukupnu mehaničku energiju kada su kolica u najvišoj i najnižoj točki staze. Kakva se pretvorba energije tu zbiva? Vrijedi li zakon očuvanja mehaničke energije? Koliki je iznos sile trenja? Zatim kliknuti na Spring i ponovo odgovoriti na sva pitanja. * U simulaciji na adresi https://javalab.org/en/energy_conversion_en/ predvidi na kojoj će se visini zaustaviti topovsko tane. Zatim 10 puta povećaj masu i opet predvidi i provjeri. Objasni. Vrati masu na 1 kg i 2 puta povećaj početnu brzinu. Predvidi, provjeri i objasni. Spremi tri slike zaslona i potakni razred da i oni predvide što će se dogoditi. Ako ne znaju, objasni im.
  • 12. Literatura:  Jakov Labor: Fizika 1,udžbenik fizike za 1.razred gimnazije, Alfa  https://phet.colorado.edu/sims/html/energy-skate- park/latest/energy-skate-park_en.html  Alfa portal- fizika