Đáp án:
\(\begin{array}{l}
1.\\
a.\\
h = 15,625m\\
v = 17,678m/s\\
b.\\
h = 7,8125m\\
v = 21,65m/s\\
c.\\
v = 12,5m/s\\
h = 13,4375m\\
2.\\
a.{h_{\max }} = 9,8m\\
b.h = 4,9m\\
c.{v_{\max }} = 14m/s\\
d.{F_c} = 31,4N\\
\end{array}\)
Giải thích các bước giải:
Bài 1:
a.
Bảo toàn cơ năng:
\(\begin{array}{l}
{W_{d\max }} = {W_{t\max }} \Rightarrow \dfrac{1}{2}mv_{\max }^2 = mg{h_{\max }}\\
\Rightarrow \dfrac{1}{2}{.25^2} = 10{h_{\max }} \Rightarrow {h_{\max }} = 31,25m
\end{array}\)
Ta có:
\(\begin{array}{l}
{W_d} = {W_t} = \dfrac{W}{2} = \dfrac{{{W_{t\max }}}}{2} \Rightarrow mgh = \dfrac{{mg{h_{max}}}}{2}\\
\Rightarrow h = \dfrac{{{h_{\max }}}}{2} = \dfrac{{31,25}}{2} = 15,625m\\
{W_t} = {W_d} \Rightarrow mgh = \dfrac{1}{2}m{v^2}\\
\Rightarrow 10.15,625 = \frac{1}{2}{v^2} \Rightarrow v = 17,678m/s
\end{array}\)
b.
Ta có:
\(\begin{array}{l}
{W_d} = 3{W_t} \Rightarrow {W_t} = \dfrac{W}{4} = \dfrac{{{W_{t\max }}}}{4}\\
\Rightarrow mgh = \dfrac{{mg{h_{\max }}}}{4} \Rightarrow h = \dfrac{{{h_{\max }}}}{4} = \dfrac{{31,25}}{4} = 7,8125m\\
{W_d} = 3{W_t} \Rightarrow \dfrac{1}{2}m{v^2} = 3mgh\\
\Rightarrow \dfrac{1}{2}.{v^2} = 3.10.7,8125 \Rightarrow v = 21,65m/s
\end{array}\)
c.
Ta có:
\(\begin{array}{l}
{W_t} = 3{W_d} \Rightarrow {W_d} = \dfrac{W}{4} = \dfrac{{{W_{d\max }}}}{4}\\
\Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{1}{4}.\dfrac{1}{2}mv_{\max }^2 \Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{1}{2}m{(\dfrac{{{v_{\max }}}}{2})^2}\\
\Rightarrow v = \dfrac{{{v_{\max }}}}{2} = \dfrac{{25}}{2} = 12,5m/s\\
{W_t} = 3{W_d} \Rightarrow mgh = 3.\dfrac{1}{2}m{v^2} \Rightarrow 10.h = \dfrac{3}{2}.12,{5^2}\\
\Rightarrow h = 13,4375m
\end{array}\)
Bài 2:
a.
Bảo toàn cơ năng:
\(\begin{array}{l}
W = {W_{t\max }} \Rightarrow mgh + \dfrac{1}{2}m{v^2} = mg{h_{\max }}\\
\Rightarrow 10.8 + \dfrac{1}{2}{.6^2} = 10{h_{\max }} \Rightarrow {h_{\max }} = 9,8m
\end{array}\)
b.
Ta có:
\(\begin{array}{l}
{W_d} = {W_t} = \dfrac{W}{2} = \dfrac{{{W_{t\max }}}}{2} \Rightarrow mgh = \dfrac{{mg{h_{max}}}}{2}\\
\Rightarrow h = \dfrac{{{h_{\max }}}}{2} = \dfrac{{9,8}}{2} = 4,9m
\end{array}\)
c.
Bảo toàn cơ năng:
\(\begin{array}{l}
{W_{t\max }} = {W_{d\max }} \Rightarrow mg{h_{\max }} = \dfrac{1}{2}mv_{\max }^2\\
\Rightarrow 10.9,8 = \dfrac{1}{2}v_{\max }^2 \Rightarrow {v_{\max }} = 14m/s
\end{array}\)
d
Áp dụng định lý động năng:
\(\begin{array}{l}
{W_d}' - {W_{d\max }} = {A_c} + {A_P}\\
\Rightarrow \dfrac{1}{2}mv{'^2} - \dfrac{1}{2}mv_{\max }^2 = {F_c}.s.\cos 180 + P.s\\
\Rightarrow \dfrac{1}{2}.0,{2.7^2} - \dfrac{1}{2}.0,{2.14^2} = - {F_c}.0,5 + 0,2.10.0,5\\
\Rightarrow {F_c} = 31,4N
\end{array}\)
