In the vertical direction, the block is in equilibrium:
∑ F (vertical) = n - mg = 0
where n is the magnitude of the normal force exerted by the surface on the block, and mg is the block's weight. It follows that n = mg = 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 a such that
∑ F (horizontal) = 14 N - 3.5 N = ma
(a) The magnitude of friction f is proportional to the magnitude of the normal force such that f = µn, where µ is the coefficient of kinetic friction. So we have
3.5 N = µ (24.5 N) ==> µ ≈ 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 a above:
10.5 N = (2.5 kg) a ==> a = 4.2 m/s²
