Question

A 6 kg box with initial speed 5 m/s slides across the floor and comes to a stop after 1.9 s. What is the coefficient of kinetic friction?

Answer 1

Answer:

$\mu_k=0.27$

Explanation:

According to the free body diagram, in this case, we have:

$\sum F_x:-F_k=ma\\\sum F_y:N=mg$

Recall that the force of friction is given by:

$F_k=\mu_k N$

Replacing and solving for the coefficient of kinetic friction:

$-\mu_kN=ma\\-\mu_k(mg)=ma\\\mu_k=-\frac{a}{g}$

We have an uniformly accelerated motion. Thus, the acceleration is defined as:

$a=\frac{v_f-v_0}{t}\\a=\frac{0-5\frac{m}{s}}{1.9s}\\a=-2.63\frac{m}{s^2}$

Finally, we calculate $\mu_k$:

$\mu_k=-\frac{-2.63\frac{m}{s^2}}{9.8\frac{m}{s^2}}\\\mu_k=0.27$