[tex]dane:\\V = 0,2 \ m^{3}\\d_{d} = 800\frac{kg}{m^{3}}\\d_{w} = 1000\frac{kg}{m^{3}}\\h = 40 \ cm = 0,4 \ m\\t = 2 \ s\\g = 10\frac{N}{kg}\\szukane:\\P = ?\\\\Rozwiazanie\\\\P=\frac{W}{t}\\\\d = \frac{m}{V} \ \ \rightarrow \ \ m = d\cdot V\\\\m = 800\frac{kg}{m^{3}}\cdot0,2 \ m^{3}}=160 \ kg[/tex]
Na deskę działa siła grawitacji ↓
[tex]F_{g} = m\cdot g = 160 \ kg\cdot10\frac{N}{kg} = 1600 \ N[/tex]
oraz siła wyporu ↑
[tex]F_{w} = d_{w}\cdot g\cdot V = 1000\frac{kg}{m^{3}}\cdot10\frac{N}{kg}\cdot0,2 \ m^{3} = 2000 \ N[/tex]
Wypadkowa tych sił:
[tex]F = F_{w}-F_{g} = 2000 \ J - 1600 \ J = 400 \ J[/tex]
[tex]W = F\cdot s\\\\W = 400 \ N\cdot0,4 \ m = 160 \ J\\\\\\P = \frac{W}{t}\\\\P = \frac{160 \ J}{2 \ s}\\\\\underline{P = 80 \ W}[/tex]
