berapakah percepatan benda diatas setelah diberi gaya f3= 20N, m= 80kg ,f2= 20N, f1=500N?

GAYA

F1 = 500 N
F2 = 20 N
F3 = 20 N
m = 80 Kg
a = ?

Jawab:
a = F/m

Karena gaya searah maka F (gaya) di total
a = (500 + 20 + 20)/80
a = 540/80
a = 6,72 m/s²