Bài 3:
l = OB = OA = 50 cm = 0,5 m
g = 10 $
m\mathrm{/}{s}^{2}
$
Ta có:
OC = OB.cos60 = 0,5.cos60 = 0,25(m)
AC=h = l - OC = 0,5 - 0,25 = 0,25 (m)
Chọn mốc thế năng tại vị trí cân bằng A
Cơ năng tại vị trí B là:
$
{W}_{B}
\begin{array}{l}
{\mathrm{{=}}{W}_{d}\mathrm{{+}}{W}_{t}\mathrm{{=}}{0}\mathrm{{+}}{mgh}}\\
{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\mathrm{{=}}{m}{\mathrm{.}}{\mathrm{10}}{\mathrm{.}}{0}{\mathrm{,}}{\mathrm{25}}\mathrm{{=}}{2}{\mathrm{,}}{5}{m}\hspace{0.33em}{\mathrm{(}}{J}{\mathrm{)}}}
\end{array}
$
Cơ năng tại vị trí A là:
$
\begin{array}{l}
{{W}_{A}\mathrm{{=}}{W}_{d}\mathrm{{+}}{W}_{t}\mathrm{{=}}\frac{1}{2}{mv}^{2}\mathrm{{+}}{0}}\\
{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\mathrm{{=}}\frac{1}{2}{mv}^{2}\hspace{0.33em}{\mathrm{(}}}
\end{array}
$
Bảo toàn cơ năng ta có:
$
{W}_{A}
\begin{array}{l}
{\mathrm{{=}}{W}_{B}\mathrm{\Longleftrightarrow}\frac{1}{2}{mv}^{2}\mathrm{{=}}{2}{\mathrm{,}}{5}{m}}\\
{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\mathrm{\Longleftrightarrow}{v}^{2}\mathrm{{=}}{5}}\\
{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\mathrm{\Longleftrightarrow}{v}\mathrm{{=}}\sqrt{5}{\mathrm{(}}{m}{\mathrm{/}}{s}{\mathrm{)}}}
\end{array}
$
