1. **State the problem:** A box of mass 4.9 kg is pulled up an inclined plane at constant velocity by a force of 64 N parallel to the slope. The slope is inclined at 20° and the vertical height gained is 5.8 m. We need to find the work done against friction. 2. **Identify known values:** - Mass, $m = 4.9$ kg - Force applied, $F = 64$ N - Angle of incline, $\theta = 20^\circ$ - Vertical height, $h = 5.8$ m 3. **Find the length of the slope (distance moved along the incline):** Using trigonometry, vertical height $h = L \sin \theta$ where $L$ is the length of the slope. $$L = \frac{h}{\sin \theta} = \frac{5.8}{\sin 20^\circ}$$ Calculate $\sin 20^\circ \approx 0.3420$: $$L = \frac{5.8}{0.3420} \approx 16.96 \text{ m}$$ 4. **Since the box moves at constant velocity, net force along the slope is zero:** $$F_{applied} = F_{friction} + F_{gravity, slope}$$ 5. **Calculate the component of gravitational force along the slope:** $$F_{gravity, slope} = mg \sin \theta = 4.9 \times 9.8 \times \sin 20^\circ$$ Calculate: $$F_{gravity, slope} = 4.9 \times 9.8 \times 0.3420 \approx 16.43 \text{ N}$$ 6. **Calculate frictional force:** $$F_{friction} = F_{applied} - F_{gravity, slope} = 64 - 16.43 = 47.57 \text{ N}$$ 7. **Calculate work done against friction:** Work done $W = F_{friction} \times L$ $$W = 47.57 \times 16.96 \approx 807 \text{ J}$$ **Final answer:** The work done against the frictional force is approximately **807 J**.