A 3.6 kg block lies on a smooth table. A light string joins it, over a smooth pulley at the edge…

Step 1. Acceleration. The driving force is $mg = 1.4\ \text{kg} \times 9.8\ \text{m/s}^2 = 13.72\ \text{N}$, moving $5.0\ \text{kg}$: $a = \dfrac{13.72\ \text{N}}{5.0\ \text{kg}} = 2.744\ \text{m/s}^2$.

Step 2. Tension. $T = Ma = 3.6\ \text{kg} \times 2.744\ \text{m/s}^2 = 9.88\ \text{N}$.

Step 3. Speed. $v = at = 2.744\ \text{m/s}^2 \times 2.5\ \text{s} = 6.86\ \text{m/s}$.