pocisk o masie 5g leci z predkoscia 800m/s i przebija drzewo grubosci 20 cm. Ile wynosza sily oporu dzialajace na pocisk, jezeli hego predkosc zmniejszyla sie do 400m/s.

m = 5g = 0,005kg v₁ = 800[latex] frac{m}{s} [/latex] v₂ = 400[latex] frac{m}{s} [/latex] d = 20cm = 0,2m W = [latex] E_{k} [/latex] W = F * d [latex]E_{k}= E_{k1} - E_{k2} = frac{[(m(v_{1})^{2} - (v_{2})^{2})]}{2} [/latex] [latex]F* s =frac{[(m(v_{1})^{2} - (v_{2})^{2})] }{2} [/latex] [latex]F* s =frac{[(m(v_{1})^{2} - (v_{2})^{2})] }{2} /:s[/latex] [latex]F=frac{[(m(v_{1})^{2} - (v_{2})^{2})] }{2s}[/latex] [latex]F=frac{[(0,005kg* (800 frac{m}{s})^{2} - (400 frac{m}{s})^{2})] }{2 * 0,2m}[/latex] [latex]F=frac{[0,005kg* (640000 frac{m^2}{s^2} - 160000 frac{m^2}{s^2})] }{2 * 0,2m}[/latex] [latex]F=frac{[0,005kg* 480000frac{m^2}{s^2}] }{0,4m}[/latex] [latex]F=6000N[/latex] Odp: Siły oporu wynoszą 6000N.