Friction Physics Example 4

Solution

1. 1
   Constant speed means the net horizontal force is zero, so the kinetic friction force is $29.4 \text{ N}$.
2. 2
   On a level floor, the normal force is $$N = mg = 12 \times 9.8 = 117.6 \text{ N}$$
3. 3
   Use $$f_k = \mu_k N$$, so $$\mu_k = \frac{f_k}{N} = \frac{29.4}{117.6} = 0.25$$