Answer
$F = 4.92$ $N$
Work Step by Step
The force that link 5 exerts on link 4 can be calculated by adding the weights of link 1 through link 4 to the force required to accelerate the links 1 through 4 together at $2.50m/s^{2}$,
$F_{54}=(m_{1}+m_{2}+m_{3}+m_{4})g + (m_{1}+m_{2}+m_{3}+m_{4})a$
$F_{54}=(m_{1}+m_{2}+m_{3}+m_{4}) (g +a)$
$F_{54}=[(0.400kg) (9.8m/s^{2}+2.5m/s^{2})]=4.92$ $N$
