Đáp án:
\(\begin{array}{l}
a.\\
N = 5\sqrt 3 N\\
{v_B} = 2,875m/s\\
b.\\
a = - 0,4m/{s^2}\\
{k_2} = 0,04
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
\vec P + {{\vec F}_{ms}} + \vec N = m\vec a\\
N = P\cos 30 = mg\cos 30 = 1.10.\cos 30 = 5\sqrt 3 N\\
P\sin 30 - {F_{ms}} = ma\\
a = \frac{{P\sin 30 - {k_1}N}}{m} = \frac{{1.10.\sin 30 - 0,1.5\sqrt 3 }}{1} = 4,134m/{s^2}\\
a = \frac{{{v_B}^2 - v_A^2}}{{2AB}}\\
4,134 = \frac{{v_B^2}}{{2.1}}\\
{v_B} = 2,875m/s\\
b.\\
a' = \frac{{v_C^2 - v_B^2}}{{2BC}} = \frac{{ - 2,{{875}^2}}}{{2.10,35}} = - 0,4m/{s^2}\\
\vec P + {{\vec F}_{ms}}' + \vec N = m\vec a'\\
- {F_{ms}} = ma'\\
- {k_2}mg = ma'\\
{k_2} = \frac{{a'}}{{ - g}} = \frac{{ - 0,4}}{{ - 10}} = 0,04
\end{array}\)