1. **Problem statement:** We have two blocks: $M_1=1.5$ kg on top of $M_2=5$ kg. A horizontal force $F=10$ N acts on $M_2$. The surface under $M_2$ is frictionless. The coefficients of friction between $M_1$ and $M_2$ are static friction $\mu_s=0.5$ and kinetic friction $\mu_k=0.25$. We want to analyze the motion and forces, including friction, acceleration, and tension. 2. **Identify forces and assumptions:** - Since the surface under $M_2$ is frictionless, no friction opposes $M_2$'s motion. - Friction between $M_1$ and $M_2$ can prevent $M_1$ from slipping. - The maximum static friction force is $f_s^{max} = \mu_s N$, where $N$ is the normal force between $M_1$ and $M_2$. - Normal force $N = M_1 g = 1.5 \times 9.8 = 14.7$ N. - So, $f_s^{max} = 0.5 \times 14.7 = 7.35$ N. 3. **Calculate acceleration if blocks move together:** - Total mass $M = M_1 + M_2 = 6.5$ kg. - Acceleration $a = \frac{F}{M} = \frac{10}{6.5} \approx 1.54$ m/s$^2$. 4. **Check friction force needed to accelerate $M_1$ at $a$:** - Force needed on $M_1$ to accelerate: $F_{needed} = M_1 a = 1.5 \times 1.54 = 2.31$ N. - Since $2.31$ N $< f_s^{max} = 7.35$ N, static friction can hold $M_1$ without slipping. 5. **Friction force acts on $M_1$ to accelerate it, and reaction friction acts on $M_2$:** - Friction force $f = 2.31$ N (to the right on $M_1$). - On $M_2$, friction acts left with $2.31$ N. 6. **Calculate acceleration of $M_2$ alone:** - Net force on $M_2$ is $F - f = 10 - 2.31 = 7.69$ N. - Acceleration $a_2 = \frac{7.69}{5} = 1.54$ m/s$^2$ (same as total, confirming blocks move together). 7. **Summary:** - Both blocks accelerate together at approximately $1.54$ m/s$^2$. - Static friction force between blocks is $2.31$ N, less than maximum static friction. - No slipping occurs. **Final answer:** $$a = 1.54\ \text{m/s}^2, \quad f = 2.31\ \text{N (static friction force)}$$