upuszczona z okna wieżowca kula o masie 200g spada swobodnie na ziemie przez 3 sekundy.Ile wynosiła energia potencjalna kuli, w chwili gdy kula zaczęła spadać?
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m = 200 g= 0,2 kg
t = 3 s
g = 10 m/s²
V₀ = 0 m/s
Ep = ?
Ep = m * g * h
1. obl. wysokość
h = g * t² /2
h = 10 m/s² * ( 3 s )² / 2
h = 45 m
2. obl. Ep
Ep = 0,2 kg * 10 m/s² * 45 m
Ep = 90 J
[ kg * m/s² * m ] = [ N * m ] = [ J ]
Ep = mgh
g = 10N/kg
m = 200g = 0,2kg
h = ?
t = 3s
h = ½gt² = ½*10*3² = 5*9 = 45m
Ep = 0,2*10*45 = 90J