[tex]dane:\\h = 5 m\\E_{p} = 1 \ J\\g = 10\frac{N}{kg}\\szukane:\\m = ?\\\\E_{p} = mgh \ \ /:gh\\\\m = \frac{E_{p}}{gh}\\\\m =\frac{1 \ J}{10\frac{N}{kg}\cdot 5 \ m} =\frac{ 1 \ N\cdot m}{10\frac{N}{kg}\cdot5 \ m} \\\\m = \frac{1}{50} \ kg = 0,02 \ kg[/tex]
Odp. Można podniieść masę m = 0,02 kg (2 dag).
