A 3.0-kg and a 1.0-kg box rest side-by-side on a smooth, level floor. A horizontal force of 32 N is applied to the 1.0-kg box pushing it against the 3.0-kg box, and, as a result, both boxes slide along the floor. How hard do the two boxes push against each other

Considering both boxes as one body, it would have a total mass of 4.0 kg. By Newton's second law, the 32 N force applies an acceleration a such that

∑ F = 32 N = (4.0 kg) a   →   a = 8.0 m/s²

and both boxes share this acceleration. (There is no friction, so the given force is the only one involved in the direction of the boxes' motion.)

Now consider just the smaller box. It is feeling the effect of the 32 N push in one direction and, as it comes into contact with the larger box, a normal force that points in the opposite direction. Let n be the magnitude of this normal force; this is what you want to find. By Newton's second law,

∑ F = 32 N - n = (1.0 kg) (8.0 m/s²)

n = 32 N - 8.0 N

n = 24 N

Just to make sure that this is consistent: by Newton's third law, the larger box feels the same force but pointing in the opposite direction. On the smaller box, n opposes the pushing force, so points backward. So from the larger box's perspective, n acts on it in the forward direction. This is the only force acting on the larger box, so Newton's second law gives

∑ F = 24 N = (3.0 kg) (8.0 m/s²)

50 inches should be the resultant vector

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