Answer:
[tex]\mu_k=0.27[/tex]
Explanation:
According to the free body diagram, in this case, we have:
[tex]\sum F_x:-F_k=ma\\\sum F_y:N=mg[/tex]
Recall that the force of friction is given by:
[tex]F_k=\mu_k N[/tex]
Replacing and solving for the coefficient of kinetic friction:
[tex]-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}[/tex]
We have an uniformly accelerated motion. Thus, the acceleration is defined as:
[tex]a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}[/tex]
Finally, we calculate [tex]\mu_k[/tex]:
[tex]\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27[/tex]
