Answer
$ F = 6. 15$ $N$
Work Step by Step
The force exerted on the top link can be calculated by adding the weights of all links to the force required to accelerate all the links at $2.50m/s^{2}$,
$F=(m_{1}+m_{2}+m_{3}+m_{4}+m_{5})g +
(m_{1}+m_{2}+m_{3}+m_{4}+m_{5})a$
$F=(m_{1}+m_{2}+m_{3}+m_{4}+m_{5} ) (g +a)$
$F =[(0.500kg) (9.8m/s^{2}+2.5m/s^{2})]=6.15$ $N$
