I found this on archive. I just post for 3 seconds.
sorry...
[tex]
Hmm--I prefer to keep the static friction coefficients as \mu _{S,1} and \mu _{S,2} until the end, where <br />
\left\{ \begin{gathered}<br />
\mu _{S,1} = 0.60 \hfill \\<br />
\mu _{S,2} = 0.20 \hfill \\ <br />
\end{gathered} \right\} <br />
and F_{net,1} is the net force for the top block, F_{net,2} is the net force for the bottom block. Here's my solution:<br />
\left\{ \begin{gathered}<br />
F_{net,1} = a \cdot 4.0kg = F_T - 39N\mu _{S,1} \hfill \\<br />
F_{net,2} = a \cdot 3.0kg = 39N\mu _{S,1} - 69N\mu _{S,2} \hfill \\ <br />
\end{gathered} \right\} \Rightarrow <br />
a = \frac{{F_T - 39N\mu _{S,1} }}{{4.0kg}} \Rightarrow <br />
<br />
3.0kg\frac{{F_T - 39N\mu _{S,1} }}{{4.0kg}} = 39N\mu _{S,1} - 69N\mu _{S,2} \Rightarrow <br />
\frac{{3.0}}{{4.0}}\left( {F_T - 23N} \right) = 9.6N \Rightarrow F_T = 36N <br />
\therefore a = \frac{{36N - 23N}}{{4.0kg}} = 3.3\frac{m}{{s^2 }} <br />
<br />
*Because v_0 = 0 \frac{m}{s} , and I like to set x_0 = 0 m ,<br />
\therefore v = \left( {3.3\frac{m}{{s^2 }}} \right)t \Rightarrow x = 5.0m = \left( {3.3\frac{m}{{s^2 }}} \right)\frac{{t^2 }}{2} \Rightarrow t = 1.7s [/tex]