Đáp án:
\(\begin{array}{l}
21.B\\
{W_t} = {W_d} = \dfrac{W}{2} = \dfrac{{{W_{d\max }}}}{2}\\
\Rightarrow mgh = \dfrac{{\frac{1}{2}mv_0^2}}{2} \Rightarrow 10.h = \dfrac{1}{4}{.10^2}\\
\Rightarrow h = 2,5m\\
{W_t} = 4{W_d} \Rightarrow {W_t} = \dfrac{{4W}}{5} = \dfrac{{4{W_{d\max }}}}{5}\\
\Rightarrow mgh = \dfrac{{4.\frac{1}{2}mv_0^2}}{5} \Rightarrow 10.h = \dfrac{2}{5}{.10^2}\\
\Rightarrow h = 4m\\
22.A\\
W = mgh + \dfrac{1}{2}m{v^2} = 0,02.9,8.1,6 + \dfrac{1}{2}.0,{02.4^2} = 0,4736J\\
W = {W_{t\max }} = mg{h_{\max }} \Rightarrow 0,4736 = 0,02.9,8{h_{\max }}\\
\Rightarrow {h_{\max }} = 2,42m\\
10.C\\
{\rm{W}} = {W_{t\max }} = mg{h_{\max }} = 0,4.10.20 = 80J\\
{{\rm{W}}_t} = mgh = 0,4.10(20 - 12) = 32J\\
{{\rm{W}}_d} = W - {W_t} = 80 - 32 = 48J\\
23.B\\
{W_t} = mgh \Rightarrow h = \dfrac{{{W_t}}}{{mg}} = \dfrac{4}{{9,8.2}} = 0,204m\\
12.B\\
{{\rm{W}}_t} = \dfrac{1}{2}k\Delta {l^2} \Rightarrow k = \dfrac{{2{W_t}}}{{\Delta {l^2}}} = \dfrac{{2.0,18}}{{0,{{03}^2}}} = 400N/m\\
24.C\\
{{\rm{W}}_t} = \dfrac{1}{2}k\Delta {l^2} = \dfrac{1}{2}.{F_{dh}}.\Delta l = \dfrac{1}{2}.3.0,02 = 0,03J\\
25.D\\
{F_{dh}} = \dfrac{1}{2}k\Delta {l^2} = \dfrac{1}{2}k{(l - {l_0})^2} = \dfrac{1}{2}150{(0,14 - 0,1)^2} = 0,12J\\
26.\\
{\rm{W = }}{{\rm{W}}_{t1}} = mg{h_1} \Rightarrow 600 = 3.10.{h_1} \Rightarrow {h_1} = 20m\\
{W_{t2}} = mg{h_2} \Rightarrow - 900 = 3.10.{h_2} \Rightarrow {h_2} = - 30m\\
h = {h_1} + |{h_2}| = 20 + | - 30| = 50m
\end{array}\)
