1. **Problem statement:** Find the coefficient of static friction $\mu_s$ for a system with a smooth pulley, a rough horizontal plane, and masses 25 kg on the plane and 15 kg hanging. 2. **Given:** - Mass on horizontal plane, $m_1 = 25$ kg - Hanging mass, $m_2 = 15$ kg - The system is about to move, so friction force is at maximum static friction. 3. **Formula and concepts:** - The frictional force $f = \mu_s N$, where $N$ is the normal force. - On the horizontal plane, $N = m_1 g$. - The tension $T$ in the rope balances the forces at the verge of motion. - For the hanging mass, weight $W = m_2 g$ acts downward. 4. **Equilibrium conditions:** - Horizontal mass: $T = f = \mu_s m_1 g$ - Hanging mass: $T = m_2 g$ 5. **Equate tensions:** $$\mu_s m_1 g = m_2 g$$ 6. **Simplify:** $$\mu_s = \frac{m_2}{m_1} = \frac{15}{25} = \frac{3}{5}$$ 7. **Answer:** The coefficient of static friction is $\boxed{\frac{3}{5}}$. This corresponds to option (b).