1. **State the problem:** A 5 kg block is pushed 3 m across a horizontal floor by a 25 N force applied at an angle of 30° above the horizontal. The coefficient of kinetic friction between the block and the floor is 0.2. We need to find the net work done on the block. 2. **Relevant formulas and concepts:** - Work done by a force: $$W = Fd\cos(\theta)$$ where $F$ is the force magnitude, $d$ is displacement, and $\theta$ is the angle between force and displacement. - Frictional force: $$f_k = \mu_k N$$ where $\mu_k$ is the coefficient of kinetic friction and $N$ is the normal force. - Normal force on a horizontal surface with an angled force: $$N = mg - F\sin(\theta)$$ because the vertical component of the applied force reduces the normal force. - Net work done is the sum of work done by all forces (applied force and friction). 3. **Calculate the normal force:** $$N = mg - F\sin(30^\circ) = 5 \times 9.8 - 25 \times 0.5 = 49 - 12.5 = 36.5\,N$$ 4. **Calculate frictional force:** $$f_k = \mu_k N = 0.2 \times 36.5 = 7.3\,N$$ 5. **Calculate work done by the applied force:** The horizontal component of the applied force is $$F_x = F \cos(30^\circ) = 25 \times \frac{\sqrt{3}}{2} \approx 21.65\,N$$ Work done by applied force: $$W_{applied} = F_x \times d = 21.65 \times 3 = 64.95\,J$$ 6. **Calculate work done by friction:** Friction force acts opposite to displacement, so angle is 180°: $$W_{friction} = f_k \times d \times \cos(180^\circ) = 7.3 \times 3 \times (-1) = -21.9\,J$$ 7. **Calculate net work done:** $$W_{net} = W_{applied} + W_{friction} = 64.95 - 21.9 = 43.05\,J$$ **Final answer:** The net work done on the block is approximately **43.05 joules**.