1. The problem statement, all variables and given/known data Two boxes stay beside each other touching on a flat ground. The ground is frictionless. ________ _______ |~~~~~||~~~~~| |~1.5KG`||~1.0KG| |~BoxA~||~BoxB~| |_______||______| A) If 5.0N of force is applied to BoxA what is the acceleration of BoxB B) During this, what is their contact force? M1 = 1.5KG M2 = 1.0KG Fapp = 5.0N 3. The attempt at a solution a) so what i did was add the boxes together so total mass is 2.5KG then the Fnet will equal applied force since no other forces are relevant so by second law Fnet = ma 5.0 = 2.5a a = 2.0 m/s^2 the second box will accelerate at 2.0 m/s^2. b) since only applied force of 5N is on BoxA the third law states that for every action there is an equal and opposite reaction so if the boxA is pushed by 5N that means boxB will back on boxA with 5N therefore the contact force = applied force = 5.0 N did i do it right?