Hướng dẫn
\(\omega = \sqrt {\frac{k}{{M + m}}} = \sqrt {\frac{{200}}{{0,9 + 0,1}}} = 10\sqrt 2 rad/s\)
\(x = \frac{{mg}}{k} = \frac{{0,1.10}}{{200}} = 0,005m = 0,5cm\)
\({F_{qt\max }} \le mg \Rightarrow m{\omega ^2}A \le mg \Rightarrow A \le \frac{g}{{{\omega ^2}}} = \frac{{10}}{{{{\left( {10\sqrt 2 } \right)}^2}}} = 0,05m = 5cm\)
\(v = \omega \sqrt {{A^2} - {x^2}} = 10\sqrt 2 .\sqrt {{5^2} - {{0,5}^2}} = 15\sqrt {22} cm/s\)
\({v_m} = \frac{{\left( {M + m} \right)v}}{m} = \frac{{\left( {0,9 + 0,1} \right).15\sqrt {22} }}{{0,1}} = 150\sqrt {22} cm/s = 1,5\sqrt {22} m/s\)
\({h_{\max }} = \frac{{v_m^2}}{{2g}} = \frac{{{{\left( {1,5\sqrt {22} } \right)}^2}}}{{2.10}} = 2,475m\). Chọn B
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