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F = 120 N + 2 . 120 N . cos30° ≈ 328 N
La fuerza neta sobre la caja es:
328 N - 300 kg . 9,80 m/s² = - 2612 N
Por lo tanto caja bajará acelerada con:
a = 2612 N / 300 kg = 8,70 m/s²
Saludos Herminio