Question

Pe un plan orizontal se afla initial in repaus un corp cu masa m=2, F=15, alfa=30, miu=0,2 ,iar d=2. Sa se afle viteza corpului,energia cinetica

Answer 1

$\Delta Ec=L=L_F+L_N+L_G+L_{Ff}\\v_o=0\\\Delta Ec=Ec_f-Ec_o=\frac{mv^2}{2}-\frac{mv_o^2}{2}=\frac{mv^2}{2}\\L_N=0\\L_G=0\\L_F=Fdcos\alpha\\L_{Ff}=-F_f*d=-\mu Nd$$\vec{F}+\vec{F_f}+\vec{G}+\vec{N}=0\\Oy:N+F_y-mg=0=>N=mg-F_y=mg-Fsin\alpha\\L_{Ff}=-\mu Nd=-\mu (mg-Fsin\alpha)d\\[tex]\Delta Ec=L_F+L_N+L_G+L_{Ff}<=>\frac{mv^2}{2}=Fdcos\alpha-\mu (mg-Fsin\alpha)d=>v^2=2(Fdcos\alpha-\mu (mg-Fsin\alpha)d)=\frac{2d}{m}[F(cos\alpha+\mu sin\alpha)-\mu g]=>v=\sqrt{\frac{2d}{m}[F(cos\alpha+\mu sin\alpha)-\mu g]}\\Ec=\frac{mv^2}{2}=\frac{\frac{2d}{m}[F(cos\alpha+\mu sin\alpha)-\mu g]m}{2}=[ F(cos\alpha+\mu sin\alpha)-\mu g]d$[/t$P_m=\frac{L_F}{\Delta t}=\frac{Fdcos\alpha}{\Delta t}\\d=v_m\Delta t\\v_m=\frac{v+v_o}{2}=\frac{0+v}{2}=\frac{v}{2}\\P_m=Fv_mcos\alpha=\frac{Fvcos\alpha}{2}$