1. **State the problem:** A 5 kg block is pushed 3 m across a horizontal floor by a 25 N force at an angle of 30° above the horizontal. The coefficient of kinetic friction is 0.2. We need to find the friction force, normal reaction, net work done, and gravitational force. 2. **Given data:** - Mass, $m = 5$ kg - Displacement, $d = 3$ m - Applied force, $F = 25$ N - Angle of force, $\theta = 30^\circ$ - Coefficient of kinetic friction, $\mu_k = 0.2$ - Acceleration due to gravity, $g = 9.8$ m/s$^2$ 3. **Find the gravitational force:** $$F_g = mg = 5 \times 9.8 = 49 \text{ N}$$ This force acts vertically downward. 4. **Calculate the normal reaction force $N$:** The vertical forces must balance since the block moves horizontally. The vertical component of the applied force is upward: $$F_y = F \sin \theta = 25 \times \sin 30^\circ = 25 \times 0.5 = 12.5 \text{ N}$$ Normal force balances the weight minus this upward component: $$N = F_g - F_y = 49 - 12.5 = 36.5 \text{ N}$$ 5. **Calculate the friction force $f_k$:** Friction force is: $$f_k = \mu_k N = 0.2 \times 36.5 = 7.3 \text{ N}$$ 6. **Calculate the horizontal component of the applied force:** $$F_x = F \cos \theta = 25 \times \cos 30^\circ = 25 \times 0.866 = 21.65 \text{ N}$$ 7. **Calculate the net force in horizontal direction:** $$F_{net} = F_x - f_k = 21.65 - 7.3 = 14.35 \text{ N}$$ 8. **Calculate the net work done $W_{net}$:** Work done by net force over displacement $d$: $$W_{net} = F_{net} \times d = 14.35 \times 3 = 43.05 \text{ J}$$ **Final answers:** - Friction force $f_k = 7.3$ N - Normal reaction $N = 36.5$ N - Net work done $W_{net} = 43.05$ J - Gravitational force $F_g = 49$ N