$\color{pink}{\text{$\textit{Bạn tham khảo:}$}}$

$\text{Tóm tắt: A = 10 (cm) = 0,1 (m)}$

              $\text{$W_{đmax}$ = 200 (mJ) = 200.$10^{-3}$ (J)}$

              $\text{$v_{max}$ = 20$\pi$ (cm/s) = 20$\pi$.$10^{-2}$ (m/s)}$

$\text{Giải: }$

$\text{$v_{max}$ = $\omega$ . A => $\omega$ = 2$\pi$ (rad)}$

$\text{a) $W_{đmax}$ = $\frac{1}{2}$ . m .$v_{max}^{2}$}$ $\\$ $\text{<=> 200.$10^{-3}$ = $\frac{1}{2}$ . m . $(20\pi)^{2}$}$ $\\$ $\text{<=> m $\approx$ 1,01 (kg)}$ $\\$

$\text{b) W = $W_{đmax}$ = $W_{tmax}$ = 200.$10^{-3}$ (J)}$

$\text{c) K = $\omega^{2}$ . m = $(2\pi)^{2}$ . 1,01 $\approx$ 40 (N/m)}$ $\\$ $\text{Ta có: $x^{2}$ + $\frac{v^{2}}{\omega^{2}}$ = $A^{2}$}$ $\\$ $\text{<=> $x^{2}$ + $\frac{(10\pi)^{2}}{(2\pi)^{2}}$ = $10^{2}$}$ $\\$ $\text{<=> $x^{2}$ = 75 (cm) = 7,5.$10^{-3}$ (m)}$ $\\$ $\text{=> $W_t$ = $\frac{1}{2}$ . K .$x^{2}$ = $\frac{1}{2}$ . 40 . (7,5.$10^{-3}$) = 0,15 (J)}$

$\color{pink}{\text{$\textit{Chúc bạn học tốt}$}}$

\(a)\\W_{d_{max}}=200(mJ)=\dfrac{1}{2}m.v_{max}^2=\dfrac{1}{2}m.(20\pi)^2\\\Rightarrow m=0,1(kg)=100(g)\\b)\\W=W_{d_{max}}=200(mJ)=0,2(J)\\c)\\W_t=W-W_d=200-\dfrac{1}{2}.0,1.(10\pi)^2=150(mJ)\)

Cách 2:

\(v_{max}=20\pi=A\omega =10\omega\\\Rightarrow \omega =2\pi(rad/s)\\v=10\pi=\dfrac{v_{max}}{2}\\\Rightarrow x=\dfrac{A\sqrt{3}}{2}=5\sqrt{3}(cm)\\\Rightarrow W_t=\dfrac{1}{2}m\omega^2x^2=\dfrac{1}{2}.0,1.(2\pi)^2.(5\sqrt{3})^2=150(mJ)\)