#51, 52
Net force on the box is given by sum of all forces on the box
As per Newton's II law we can say
[tex]F_{net} = ma[/tex]
it is product of mass and acceleration of the object
[tex]F_{net} = 2.5kg * 2m/s^2[/tex]
given that
mass = 2.5 kg
acceleration = 2 m/s^2
[tex]F_{net} = 5 kg m/s^2[/tex]
here we know that
[tex]1 kg m/s^2 = 1 N[/tex]
so we can say that
[tex]F_{net} = 5 N[/tex]
#53
Now we can say there are two forces acting on the box
1. applied force which is given as 15 N in forward direction
2. friction force on the box in opposite direction.
[tex]F_a - F_f = F_{net}[/tex]
[tex]15 - F_f = 5[/tex]
[tex]F_f = 15 - 5 = 10 N[/tex]
so friction force on this box will be 10 N