Question

A force of 21.0 n is required to start a 2.8 kg box moving across a horizontal concrete floor. (a) what is the coefficient of static friction between the box and the floor? (b) if the 21.0 n force continues, the box accelerates at 0.50 m/s2. what is the coefficient of kinetic friction?

Answer 1

(a)
$F_f = u_sN$
Assuming the box is on a horizontal surface, the Normal Force would be equal to the Gravitational force which is 2.8 * 9.8 = 27.44N. You are already given the force, so we are ready to solve:
$21 = u_s(27.44)$
$u_s=0.77$
Since a coefficient of friction doesn't have units, your answer is 0.77.

(b)
We can use F = ma to get the Force, and then the same Normal force to get the kinetic coefficient of friction.

$F = ma = 2.8(0.5) = 1.4N$

We can use that to formulate the kinetic coefficient of friction:
$1.4 = u_k(27.44)$
$u_k = 0.051$

And those are your answers. Hope this helps. I'm not a college student, but I'm taking AP Physics, so I have an idea.