There would be two forces acting on the box parallel to the floor, with a net force of
∑ <em>F</em> = <em>p</em> - <em>f</em> = <em>m a</em>
where <em>p</em> = magnitude of the push, <em>f</em> = mag. of friction, <em>m</em> = mass of the box, and <em>a</em> = acceleration. To find <em>p</em>, we first need <em>f</em> .
There are also only two forces acting on the box perpendicular to the floor, with net force
∑ <em>F</em> = <em>n</em> - <em>w</em> = 0
where <em>n</em> = mag. of normal force of the floor on the box and <em>w</em> = weight of the box. The net force is 0 because the box is only accelerating parallel to the floor.
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<em>w</em> = <em>m g</em>, where <em>g</em> = 9.8 m/s² is the mag. of the acceleration due to gravity, so we can solve for <em>n</em> :
<em>n</em> = <em>w</em> = <em>m g</em>
<em>n</em> = (22 kg) (9.8 m/s²)
<em>n</em> = 215.6 N
The kinetic friction is proportional to the normal force by a factor of the given coefficient of friction, <em>µ</em> = 0.17, such that
<em>f</em> = <em>µ</em> <em>n</em>
<em>f</em> = 0.17 (215.6 N)
<em>f</em> = 36.652 N
Now solve for the required pushing force:
<em>p</em> - 36.652 N = (22 kg) (1.9 m/s²)
<em>p</em> ≈ 78 N
