A 7.5 kg block rests on a floor with µs = 0.55 and µk = 0.42…

Step 1. The limit. $f_{s,\max} = 0.55 \times 73.5\ \text{N} = 40.4\ \text{N}$.

Step 2. (a) $25\ \text{N}$ is below the limit: friction is $25\ \text{N}$, and $a = 0$.

Step 3. (b) $40\ \text{N}$ is still just below $40.4\ \text{N}$: friction is $40\ \text{N}$, and $a = 0$.

Step 4. (c) $55\ \text{N}$ is above: the block slides. $f_k = 0.42 \times 73.5\ \text{N} = 30.9\ \text{N}$, and $a = \dfrac{55\ \text{N} - 30.9\ \text{N}}{7.5\ \text{kg}} = 3.22\ \text{m/s}^2$.