Question

Problema fizica clasa a 9 a

Answer 1

$\displaystyle\it\\descompunem~forta~F~pe~doua~directii,~F_t~paralela\\cu~planul~si~F_n~perpendiculara~pe~el\\pe~axa~Ox~avem:ma=F_t+G_t,~(cele~doua~forte~imprima\\o~acceleratie~corpului).\\F_t=Fcos\alpha,~iar~G_t=mg\sin\alpha.\\Fcos\alpha+mgsin\alpha = ma \implies a= \frac{Fcos\alpha+mgsin\alpha}{m} =\\\frac{Fcos\alpha}{m}+gsin\alpha=\frac{F}{m}\cos\alpha+gsin\alpha \approxeq \boxed{\it10,2m/s^2}~.$

$\displaystyle\it\\Pe~axa~Oy~nu~avem~miscare,~atunci:N+F_n+G_n=0 \implies \\N=G_n-F_n=mgcos\alpha-Fsin\alpha \boxed{\it \approxeq 28,3N}.$