Đáp án:
\(\begin{array}{l}
{A_{ms}} = - 4m(J)\\
{F_{ms}} = \frac{2}{3}m(N)\\
\mu = 0,08333
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{A_P} = P.\sin \alpha .AB = 10m.\frac{{3,6}}{6}.6 = 36m(J)\\
{W_d}' - {W_d} = {A_{ms}} + {A_P}\\
\frac{1}{2}m{v^2} - \frac{1}{2}mv_0^2 = {A_{ms}} + {A_P}\\
{A_{ms}} = \frac{1}{2}m{.8^2} - 36m = - 4m(J)\\
{F_{ms}} = \frac{{{A_{ms}}}}{{AB.\cos 180}} = \frac{{ - 4m}}{{6.\cos 180}} = \frac{2}{3}m(N)\\
{F_{ms}} = \mu N = \mu P\cos \alpha \\
\mu = \frac{{{F_{ms}}}}{{P\sqrt {1 - \sin {\alpha ^2}} }} = \frac{{\frac{2}{3}m}}{{10m\sqrt {1 - {{\frac{{3,6}}{6}}^2}} }} = 0,08333
\end{array}\)