[tex]dane:\\m = 2 \ g\\\Delta E_{p} = 0,4 \ kJ = 400 \ J\\g = 10\frac{m}{s^{2}}\\szukane:\\h = ?\\\\\Delta E_{p} = m\cdot g\cdot h \ \ /:(m\cdot g)\\\\h = \frac{\Delta E_{p}}{m\cdot g}\\\\h = \frac{400 \ N\cdot m}{2 \ kg\cdot10\frac{m}{s^{2}}}\\\\h = 20 \ m[/tex]
Odp. Ciało podniesiono na wysokość 20 m.
