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