Đáp án:
\(\begin{array}{l}
5.\\
A = 400J\\
6.\\
{W_d}' = 8{W_d}\\
7.\\
h = 0,1m\\
8.\\
h = 15m\\
9.\\
h = 0,6m
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
5.\\
A = Fs\cos 6 = 100.8.\cos 60 = 400J\\
6.\\
\frac{{{W_d}'}}{{{W_d}}} = \frac{{\frac{1}{2}.\frac{m}{2}.{{(4v)}^2}}}{{\frac{1}{2}m{v^2}}} = 8\\
\Rightarrow {W_d}' = 8{W_d}\\
7.\\
W = {W_d} + {W_t} = {W_t} + {W_t} = 2{W_t} = 2mgh\\
{W_{d\max }} = W\\
\frac{1}{2}mv_{\max }^2 = 2mgh\\
\frac{1}{2}{.2^2} = 2.10.h\\
h = 0,1m\\
8.\\
W = {W_d} + {W_t} = 3{W_t} + {W_t} = 4{W_t} = 4mgh\\
{W_{t\max }} = W\\
mg{h_{\max }} = 4mgh\\
h = \frac{{{h_{\max }}}}{4} = \frac{{60}}{4} = 15m\\
9.\\
W = {W_d} + {W_t} = 2{W_t} + {W_t} = 3{W_t} = 3mgh\\
{W_{d\max }} = W\\
\frac{1}{2}mv_{\max }^2 = 3mgh\\
\frac{1}{2}{.6^2} = 3.10.h\\
h = 0,6m
\end{array}\)
