1. **State the problem:** We have a particle of mass 6 kg moving up a rough slope inclined at 15° to the horizontal. A force of 30 N pulls the particle up the slope, and the particle moves at constant speed. We need to find the magnitude of the resistance due to friction. 2. **Analyze the forces:** Since the particle moves at constant speed, the net force along the slope is zero. 3. **Set up the force balance along the slope:** Let $F_r$ be the resistance due to friction. The forces along the slope are: - Pulling force up the slope: $30$ N - Component of weight down the slope: $mg \sin 15^\circ$ - Friction force down the slope: $F_r$ Since the particle moves at constant speed, the sum of forces along the slope is zero: $$30 = mg \sin 15^\circ + F_r$$ 4. **Calculate the weight component:** Given $m=6$ kg and $g=9.8$ m/s$^2$, $$mg \sin 15^\circ = 6 \times 9.8 \times \sin 15^\circ$$ Calculate $\sin 15^\circ \approx 0.2588$, $$mg \sin 15^\circ = 6 \times 9.8 \times 0.2588 = 15.21 \text{ N (approx)}$$ 5. **Solve for friction force $F_r$:** $$F_r = 30 - 15.21 = 14.79 \text{ N}$$ **Final answer:** The magnitude of the resistance due to friction is approximately $14.79$ N.