[tex]dane:\\m = 43 \ kg\\s = 10 \ m\\g = 10\frac{N}{kg}\\\alpha = 60^{o}\\sin60^{o} = 0,5\\szukane:\\W = ?\\\\Rozwiazanie\\\\a)\\W = F\cdot s\\\\F = m\cdot g = 43 \ kg \cdot10\frac{N}{kg} = 430 \ N\\\\W = 430 \ N\cdot10 \ m\\\\W = 4 \ 300 \ J = 4,3 \ kJ\\\\(1 kJ = 1000 \ J)\\\\b)\\W = F\cdot s\cdot cos \alpha\\\\W = 430 \ N\cdot10 \ m\cdot0,5\\\\W = 2 \ 150 \ J = 2,15 \ kJ[/tex]