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mrs_skeptik [129]
3 years ago
12

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?
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
4 0

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

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