a 15.5 kg box moving on flat ground slows down and comes to a complete stop due to a frictional force. the…

Explanation:

Step1: Identify the frictional - force formula

The frictional force $F_f=\mu_kF_N$, where $\mu_k$ is the coefficient of kinetic friction and $F_N$ is the normal force. On flat ground, $F_N = mg$ (where $m$ is the mass and $g = 9.8\ m/s^2$ is the acceleration due to gravity). So $F_f=\mu_kmg$.

Step2: Apply Newton's second - law

Newton's second - law is $F = ma$. The only horizontal force acting on the box is the frictional force, and it acts in the opposite direction of motion. So $-F_f=ma$. Substituting $F_f=\mu_kmg$ into $-F_f = ma$, we get $-\mu_kmg=ma$.

Step3: Solve for acceleration

We can cancel out the mass $m$ from both sides of the equation $-\mu_kmg=ma$. Then $a=-\mu_kg$. Substitute $\mu_k = 0.200$ and $g = 9.8\ m/s^2$ into the formula: $a=- 0.200\times9.8\ m/s^2=-1.96\ m/s^2$.

Answer:

$-1.96$