xin hay nhất ạ !

Đáp án:

 0,36J

Giải thích các bước giải:

\(\begin{align}
  & d_{max}^{2}={{A}^{2}}+{{(A\sqrt{3})}^{2}}-2A.A\sqrt{3}\text{cos }\!\!\Delta\!\!\text{ }\varphi  \\ 
 & \to \text{cos }\!\!\Delta\!\!\text{ }\varphi =\dfrac{\sqrt{3}}{2}\to \left| \text{ }\!\!\Delta\!\!\text{ }\varphi  \right|=\dfrac{\pi }{6}. \\ 
\end{align}\)
\(\begin{align}
  & \frac{{{\text{W}}_{dmax}}}{{{\text{W}}_{N}}}=\frac{{{A}^{2}}}{{{(A\sqrt{3})}^{2}}}=\frac{1}{3}\Leftrightarrow \frac{0,16}{{{\text{W}}_{N}}}=\frac{1}{3} \\ 
 & \to {{\text{W}}_{N}}=0,48(J). \\ 
\end{align}\)
\({{\text{(}{{\text{W}}_{dM}})}_{\text{max}}}\Rightarrow {{x}_{N}}=\dfrac{{{A}_{N}}}{2}=\dfrac{A\sqrt{3}}{2}\)

\(\begin{align}
  & {{\text{W}}_{tN}}=\dfrac{1}{2}k{{\left( \dfrac{A\sqrt{3}}{2} \right)}^{2}}=\dfrac{1}{2}k{{A}^{2}}(\frac{3}{4}) \\ 
 & ={{\text{W}}_{M}}.\frac{3}{4}=0,12(J) \\ 
\end{align}\)

động năng của N: 
\({{W}_{N~}}=\text{ }{{W}_{N}}~\text{ }{{W}_{tN}}~=\text{ 0,48-0,12=0,36 }\left( J \right)\)