In the vertical direction, the block is in equilibrium:
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0
where <em>n</em> is the magnitude of the normal force exerted by the surface on the block, and <em>mg</em> is the block's weight. It follows that <em>n</em> = <em>mg</em> = 24.5 N.
In the horizontal direction, the applied force of 14 N is opposed on by the 3.5 N force due to friction, so that the net force gives the block an acceleration <em>a</em> such that
∑ <em>F</em> (horizontal) = 14 N - 3.5 N = <em>ma</em>
(a) The magnitude of friction <em>f</em> is proportional to the magnitude of the normal force such that <em>f</em> = <em>µn</em>, where <em>µ</em> is the coefficient of kinetic friction. So we have
3.5 N = <em>µ</em> (24.5 N) ==> <em>µ</em> ≈ 0.14
(b) The net force on the block is 14 N - 3.5 N = 10.5 N pointing in the direction of the block's motion.
(c) Solve for <em>a</em> above:
10.5 N = (2.5 kg) <em>a</em> ==> <em>a</em> = 4.2 m/s²