Đáp án:

\(\begin{array}{l}
a.W = 0,05J\\
b.{v_{\max }} = 1m/{s^2}\\
c.v' = \frac{{\sqrt 3 }}{2}m/s\\
d.\Delta l = 5cm
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
W = {W_{t\max }} = \frac{1}{2}k\Delta l_{\max }^2 = \frac{1}{2}.10.0,{1^2} = 0,05J\\
b.\\
{W_{d\max }} = W\\
\frac{1}{2}m{v_{\max }}^2 = 0,05\\
\frac{1}{2}.0,1{v_{\max }}^2 = 0,05\\
 \Rightarrow {v_{\max }} = 1m/{s^2}\\
c.\\
W = W' = \frac{1}{2}mv{'^2} + \frac{1}{2}k\Delta l{'^2}\\
0,05 = \frac{1}{2}.0,1.v{'^2} + \frac{1}{2}.10.0,{05^2}\\
 \Rightarrow v' = \frac{{\sqrt 3 }}{2}m/s\\
d.\\
W = {W_t} + {W_d} = {W_t} + 3{W_t} = 4{W_t} = 2k\Delta l{^2}\\
W = W''\\
0,05 = 2.10.\Delta l{^2}\\
\Delta l = 0,05m = 5cm
\end{array}\)