Analysis of forces on the box:
The component of weight parallel to the incline is mg sin30° = 50 × $\frac{1}{2}$ = 25 N (directed downwards along the incline).
Newton's second law applied along the incline (with the upward direction as positive):
T − mg sin 30° = ma
30 N − 25 N = 5 kg × a
a = $\frac{5}{5}$ = 1 m s-2