Question

Va rog ajutați-ma!.........

Answer 1

$\displaystyle m=5~ kg;~ v_o=10~ \frac{m}{s};~AB=d=5~ m;~\alpha =30^ o\\ \\ \mu_1=0,5\rightarrow plan~ orizontal;~ \mu_2=0,58\left(\cong \frac{1}{\sqrt{3} } \right)\rightarrow plan~ inclinat\\ \\ a.~ L_{FfAB}=-F_{fAB}\cdot d\\ \\ F_{fAB}=\mu_1mg\Rightarrow F_{fAB}=0,5 \cdot 5 \cdot 10\Rightarrow F_{fAB}=2,5\cdot10\Rightarrow F_{fAB}=25~ N\\ \\ L_{FfAB}=-25\cdot5\Rightarrow \mathbf{L_{FfAB}=-125~ J}$

$\displaystyle b.~\Delta E_c=L \Rightarrow E_{cB}-E_{cA}=L_{FfAB}\Rightarrow E_{cB}=L_{FfAB}+E_{cA}\Rightarrow\\ \\ \Rightarrow E_{cB}=L_{FfAB}+\frac{mv_0^ 2}{2} \Rightarrow E_{cB}=-125+\frac{5 \cdot 10^2}{2} \Rightarrow \\ \\ \Rightarrow E_{cB}=-125+\frac{5\cdot 100}{2} \Rightarrow E_{cB}=-125+\frac{500}{2} \Rightarrow E_{cB}=-125+250\Rightarrow \\ \\ \Rightarrow \mathbf{E_{cB}=125~ J}$

$\displaystyle c.~\Delta E_c=L_{total}\Rightarrow E_{cC}-E_{cB}=L_G+L_{FfBC}\Rightarrow 0-E_{cB}=L_G+L_{FfBC}\Rightarrow \\ \\ \Rightarrow -E_{cB}=L_G+L_{FfBC} \Rightarrow -E_{cB}=-mgh-F_{fBC}\cdot d \\ \\ sin\alpha =\frac{h}{d} \Rightarrow sin30^ o=\frac{h}{d} \Rightarrow \frac{1}{2} =\frac{h}{d}\Rightarrow d=2h\\ \\ -E_{cB}=-mgh-\mu_2mgcos\alpha 2h\Rightarrow -E_{cB}=-mgh(1+2\mu_2 cos\alpha )\Rightarrow \\ \\ \Rightarrow E_{cB}=mgh(1+2\mu_2cos\alpha) \Rightarrow h=\frac{E_{cB}}{mg(1+2\mu_2cos\alpha) } \Rightarrow$

$\displaystyle \Rightarrow h=\frac{125}{5 \cdot 10\left(1+2 \cdot \frac{1}{\sqrt{3} }\cdot cos30^ o\right) } \Rightarrow h=\frac{125}{50\left(1+\frac{2}{\sqrt{3} }\cdot \frac{\sqrt{3} }{2} \right) } \Rightarrow \\ \\ \Rightarrow h=\frac{125}{50(1+1)} \Rightarrow h=\frac{125}{50 \cdot 2} \Rightarrow h=\frac{125}{100} \Rightarrow \mathbf{h=1,25~ m}$

$d.~E_{mC}=E_c+E_p\Rightarrow E_{mC}=E_p\Rightarrow E_{mC}=mgh \Rightarrow E_{mC}=5 \cdot 10 \cdot 1,25 \Rightarrow \\ \\ \Rightarrow E_{mC}=50 \cdot 1,25 \Rightarrow \mathbf{E_{mC}=62,5~ J}$