1. **Problem statement:** A block weighing 5.0 N is pushed horizontally against a vertical wall with a force of 12 N. The coefficient of static friction is 0.60, and the coefficient of kinetic friction is 0.40. The block is initially at rest. We need to determine (a) if the block will move and (b) the force on the block from the wall in unit-vector notation. 2. **Relevant formulas and concepts:** - Weight $W = mg = 5.0$ N acts downward. - Normal force $N$ is the horizontal force pushing the block against the wall, here $N = 12$ N. - Maximum static friction force $f_s^{max} = \mu_s N$ where $\mu_s = 0.60$. - Kinetic friction force $f_k = \mu_k N$ where $\mu_k = 0.40$. - The block will move if the weight exceeds the maximum static friction force. 3. **Calculate maximum static friction:** $$f_s^{max} = 0.60 \times 12 = 7.2\text{ N}$$ 4. **Compare weight and static friction:** - Weight $W = 5.0$ N downward. - Maximum static friction $7.2$ N upward (friction opposes motion). Since $5.0 < 7.2$, the static friction can hold the block in place, so the block will not move. 5. **Force from the wall on the block:** - The wall exerts a normal force equal and opposite to the applied horizontal force: $\vec{N} = -12\hat{i}$ N (to the left). - The friction force balances the weight to prevent motion, so friction force $\vec{f} = +5.0\hat{j}$ N (upward). 6. **Total force from the wall:** $$\vec{F}_{wall} = \vec{N} + \vec{f} = -12\hat{i} + 5.0\hat{j}$$ **Final answers:** (a) The block will not move. (b) The force on the block from the wall is $\boxed{\vec{F}_{wall} = -12\hat{i} + 5.0\hat{j}}$ N.