Answer
$1.77\space s$
Work Step by Step
Please see the image first.
We can apply equation $F=ma$ to $m_{1},m_{2}$ separately to find its acceleration as follows.
$m_{2}=\gt\space \rightarrow F=ma$ ; Let's plug known values into this equation.
$T=326\times10^{-3}\space kg\times a-(1)$
$m_{1}=\gt \space \downarrow F=ma$ ; Let's plug known values into this equation.
$14.9\times10^{-3}g-T= 14.9\times10^{-3}\space kg\times a-(2)$
$(1)=\gt(2)$
$14.9\times10^{-3}\times 9.8\space kgm/s^{2}-326\times10^{-3}a\space kg=14.9\times10^{-3}a\space kg$
$146.02\space m/s^{2}=340.9\space a$
$0.43\space m/s^{2}=a$
We can apply the equation $S=ut+\frac{1}{2}at^{2}$ to $m_{2}$
$\rightarrow S=ut+\frac{1}{2}at^{2}$
Let's plug known values into this equation.
$67.2\times10^{-2}\space m=0+\frac{1}{2}(0.43\space m/s^{2})t^{2}$
$t=1.77\space s$
It takes 1.77 seconds $m_{2}$ to reach the end of the track.
