A 2.5 kg block has a force of kinetic friction between it and a level surface of 3.5 N. This block is pulled by an applied force of 14 N. a)Find the coefficient of kinetic friction between block and surface b)What is the unbalanced force as the block is moved? c) What is the acceleration of the block?

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

it has to be b, about 97 percent.