Đáp án:
Câu 10: D
$v=A.\sqrt{\dfrac{4}{3}.\dfrac{k}{m}}$
Câu 11: B
$v=\pm \dfrac{\pi }{\sqrt{2}}(m/s)$
Giải thích các bước giải:
Câu 10: ${{\text{W}}_{d}}=3{{\text{W}}_{t}}$
cơ năng:
${{\text{W}}_{d}}=3{{\text{W}}_{t}}$
mà:
$\begin{align}
& {{A}_{_{{{S}_{1}}{{S}_{2}}}}}=2a; \\
& {{\text{W}}_{d}}=3{{\text{W}}_{t}} \\
& \dfrac{1}{2}m.v_{max}^{2}=\dfrac{4}{3}.\dfrac{1}{2}.m.{{v}^{2}} \\
& \Rightarrow v=\sqrt{\dfrac{3}{4}.v_{max}^{2}}=A.\sqrt{\dfrac{4}{3}.\dfrac{k}{m}} \\
\end{align}$
Câu 11:
$\begin{align}
& x=0,1.cos(10\pi t-\dfrac{\pi }{6}) \\
& {{\text{W}}_{d}}=\dfrac{1}{2}\text{W} \\
\end{align}$
Ta có:
$\begin{align}
& \dfrac{1}{2}.m.{{v}^{2}}=\dfrac{1}{2}.\dfrac{1}{2}.m.v_{max}^{2} \\
& \Leftrightarrow {{v}^{2}}=\dfrac{1}{2}.{{(A.\omega )}^{2}} \\
& \Rightarrow v=\pm \sqrt{\dfrac{1}{2}.{{(0,1.10\pi )}^{2}}}=\pm \dfrac{\pi }{\sqrt{2}}(m/s) \\
\end{align}$
