[tex]dane:\\p = 400 \ Pa = 400 \ \frac{N}{m^{2}}\\S = 1 \ dm^{2} = 0,01 \ m^{2}\\s = 50 \ cm = 0,5 \ m\\szukane:\\W = ?\\\\p = \frac{F}{S} \ \ /\cdot S\\\\F = p\cdot S=400\frac{N}{m^{2}}\cdot0,01 \ m^{2} = 4 \ N\\\\W = F\cdot s\\\\W = 4 \ N\cdot0,5 \ m\\\\W = 2 \ J[/tex]
