1. **Problem Statement:** Calculate the force components acting on a 101 kg block on a 15º inclined plane, given the total force $F = 508.94$ N. 2. **Relevant Formulas:** - Gravitational force: $F_g = mg$ - Force component parallel to incline: $F_{\parallel} = mg \sin(\theta)$ - Force component perpendicular to incline: $F_{\perp} = mg \cos(\theta)$ 3. **Given Values:** - Mass $m = 101$ kg - Angle $\theta = 15^\circ$ - Acceleration due to gravity $g = 9.8$ m/s$^2$ 4. **Calculate Gravitational Force:** $$F_g = 101 \times 9.8 = 989.8\, \text{N}$$ 5. **Calculate Parallel Component:** $$F_{\parallel} = 989.8 \times \sin(15^\circ) = 989.8 \times 0.2588 = 255.9\, \text{N}$$ 6. **Calculate Perpendicular Component:** $$F_{\perp} = 989.8 \times \cos(15^\circ) = 989.8 \times 0.9659 = 955.5\, \text{N}$$ 7. **Interpretation:** - The force down the slope due to gravity is approximately 255.9 N. - The force perpendicular to the slope is approximately 955.5 N. - The given force $F = 508.94$ N could represent an applied force or friction opposing motion. **Final answers:** - $F_{\parallel} \approx 255.9$ N - $F_{\perp} \approx 955.5$ N