Đáp án:
\(\begin{array}{l}
a.\\
{v_0} = 20m/s\\
b.\\
h = 10m\\
v = 10\sqrt 2 m/s\\
c.\\
v' = 10\sqrt 3 m/s\\
d.\\
h'' = 18,75m
\end{array}\)
Giải thích các bước giải:
a.
Bảo toàn cơ năng:
\(\begin{array}{l}
{W_{t\max }} = {W_{d\max }}\\
\Rightarrow mg{h_{\max }} = \dfrac{1}{2}mv_{\max }^2 = \dfrac{1}{2}mv_0^2\\
\Rightarrow g{h_{\max }} = \dfrac{1}{2}v_0^2\\
\Rightarrow {v_0} = \sqrt {2g{h_{\max }}} = \sqrt {2.10.20} = 20m/s
\end{array}\)
b.
Ta có:
\(\begin{array}{l}
{W_d} = {W_t} \Rightarrow {W_t} = \dfrac{1}{2}W = \dfrac{{{W_{t\max }}}}{2}\\
\Rightarrow mgh = \dfrac{{mg{h_{\max }}}}{2}\\
\Rightarrow h = \dfrac{{{h_{\max }}}}{2} = \dfrac{{20}}{2} = 10m\\
{W_d} = {W_t}\\
\Rightarrow \dfrac{1}{2}m{v^2} = mgh\\
\Rightarrow v = \sqrt {2.gh} = \sqrt {2.10.10} = 10\sqrt 2 m/s
\end{array}\)
c.
Bảo toàn cơ năng:
\(\begin{array}{l}
{W_{t\max }} = W' = {W_t}' + {W_d}'\\
\Rightarrow mg{h_{\max }} = mgh' + \dfrac{1}{2}mv{'^2}\\
\Rightarrow 10.20 = 10.5 + \dfrac{1}{2}v{'^2}\\
\Rightarrow v' = 10\sqrt 3 m/s
\end{array}\)
d.
Bảo toàn cơ năng:
\(\begin{array}{l}
{W_{t\max }} = W'' = {W_t}'' + {W_d}''\\
\Rightarrow mg{h_{\max }} = mgh'' + \dfrac{1}{2}mv'{'^2}\\
\Rightarrow 10.20 = 10.h'' + \dfrac{1}{2}{5^2}\\
\Rightarrow h'' = 18,75m
\end{array}\)
