Câu hỏi 31

Ta có: \({W_1} = 3{W_2} \Rightarrow \dfrac{1}{2}k{A_1}^2 = 3.\dfrac{1}{2}k{A_2}^2 \Rightarrow {A_1}^2 = 3{A_2}^2 \Rightarrow {A_1} = {A_2}\sqrt 3 \)

Lại có: \(\begin{array}{l}\left\{ \begin{array}{l}{W_{13}} = 5{W_0}\\{W_{23}} = {W_0}\end{array} \right. \Rightarrow {W_{13}} = 5{W_{23}} \Rightarrow \dfrac{1}{2}k{A_{13}}^2 = 5.\dfrac{1}{2}k{A_{23}}^2\\ \Rightarrow {A_{13}}^2 = 5{A_{23}}^2 \Rightarrow {A_{13}} = {A_{23}}\sqrt 5 \end{array}\)

Ta có giản đồ vecto:

Đặt \({A_{23}} = OB = a \Rightarrow {A_{13}} = OC = a\sqrt 5 \)

Xét \(\Delta OBC:BC = \sqrt {O{C^2} - O{B^2}}  = \sqrt {{{\left( {a\sqrt 5 } \right)}^2} - {a^2}}  = 2a\)

\(\begin{array}{l} \Rightarrow {A_1} + {A_2} = BC = 2a \Rightarrow {A_2}\left( {\sqrt 3  + 1} \right) = a\sqrt 5 \\ \Rightarrow {A_2} = \dfrac{{a\sqrt 5 }}{{\sqrt 3  + 1}} \Rightarrow {A_1} = {A_2}\sqrt 3  = \dfrac{{a\sqrt {15} }}{{\sqrt 3  + 1}}\end{array}\)

Ta có: \(\overrightarrow A  = \overrightarrow {{A_1}}  + \overrightarrow {{A_2}}  + \overrightarrow {{A_3}}  = \overrightarrow {{A_1}}  + \overrightarrow {{A_{23}}} \,\,\left( {\overrightarrow {{A_1}}  \bot \overrightarrow {{A_{23}}} } \right)\)

\( \Rightarrow A = \sqrt {{A_1}^2 + {A_{23}}^2}  = \sqrt {{{\left( {\dfrac{{a\sqrt {15} }}{{\sqrt 3  + 1}}} \right)}^2} + {a^2}}  = 1,7348a \approx 1,73a\)

Lại có: \(\left\{ {\begin{array}{*{20}{l}}{W = \frac{1}{2}k{A^2}} \\ {{W_{23}} = \frac{1}{2}k{A_{23}}^2} \end{array}} \right.\)

\( \Rightarrow \frac{W}{{{W_{23}}}} = {\left( {\frac{A}{{{A_{23}}}}} \right)^2} = {\left( {\frac{{1,73a}}{a}} \right)^2} \approx 3 \Rightarrow W = 3{W_{23}} = 3{W_0}\)

Chọn A.