Here as we can see there we be net horizontal force acting on it
[tex]F_x = F_1cos20 + F_2cos20[/tex]
also we know that
[tex]F_1 = F_2 = 80 N[/tex]
now we will have
[tex]F_x = 2(80cos20)[/tex]
[tex]F_x = 150.35 N[/tex]
now for net force of box we know
[tex]F_x - F_f = ma[/tex]
[tex]150.35 - 100 = ma[/tex]
[tex]20a = 50.35[/tex]
[tex]a = 2.52 m/s^2[/tex]
so acceleration of the box will be 2.52 m/s/s
