The acceleration $a$ of the block is constant, determined from the velocity-time information:

$$a = \frac{v_s - 0}{t_s - 0} = \frac{v_s}{t_s}$$

Apply Newton's second law. In the vertical direction, the normal force $F_N$ balances the gravitational force $mg$:

$$F_N = mg$$

The kinetic friction force is $f_k = \mu_k F_N = \mu_k mg$.

In the horizontal direction, the net force is the applied force minus friction:

$$\Sigma F_x = F - f_k = ma$$

Substituting the expressions for $f_k$ and $a$:

$$F - \mu_k mg = m \left(\frac{v_s}{t_s}\right)$$

Solving for the coefficient of kinetic friction $\mu_k$:

$$\mu_k mg = F - \frac{mv_s}{t_s}$$ $$\mu_k = \frac{1}{mg} \left( F - \frac{mv_s}{t_s} \right) = \frac{F}{mg} - \frac{v_s}{gt_s}$$