Đáp án:
$a=0,4m/{{s}^{2}}$
Giải thích các bước giải:
${{m}_{1}};{{a}_{1}}=2m/{{s}^{2}};{{m}_{2}};{{a}_{2}}=6m/{{s}^{2}};m=4{{m}_{1}}+3{{m}_{2}}$
lực truyền cho vật:
$\begin{align}
& F=m{}_{1}.{{a}_{1}}={{m}_{2}}.{{a}_{2}}=m.a \\
& \Rightarrow \dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{6}{2}=3\Rightarrow {{m}_{1}}=3{{m}_{2}}(1) \\
\end{align}$
ta có: $\begin{align}
& F=m{}_{1}.{{a}_{1}}={{m}_{2}}.{{a}_{2}}=m.a \\
& \Rightarrow \frac{{{m}_{2}}}{m}=\dfrac{a}{{{a}_{2}}}\Leftrightarrow \dfrac{{{m}_{2}}}{4.{{m}_{1}}+3.{{m}_{2}}}=\dfrac{a}{{{a}_{2}}} \\
& \Leftrightarrow \dfrac{{{m}_{2}}}{4.3.{{m}_{2}}+3.{{m}_{2}}}=\dfrac{a}{6} \\
& \Rightarrow a=\dfrac{6}{15}=0,4m/{{s}^{2}} \\
\end{align}$