1. **Problem statement:** A bag of sand weighing 100 lbs is lifted to a height of 35 ft. The sand leaks out uniformly so that the bag is empty at 35 ft. We need to find the work done in lifting the bag. 2. **Formula and concepts:** Work done by a variable force is given by the integral $$W = \int F(x) \, dx$$ where $F(x)$ is the force at height $x$. 3. **Force function:** The weight decreases uniformly from 100 lbs at $x=0$ to 0 lbs at $x=35$. The weight at height $x$ is a linear function: $$F(x) = 100 - \frac{100}{35}x = 100 - \frac{20}{7}x$$ 4. **Calculate work:** Work done is $$W = \int_0^{35} F(x) \, dx = \int_0^{35} \left(100 - \frac{20}{7}x\right) dx$$ 5. **Integrate:** $$W = \left[100x - \frac{20}{7} \frac{x^2}{2}\right]_0^{35} = \left(100 \times 35 - \frac{20}{7} \times \frac{35^2}{2}\right) - 0$$ 6. **Simplify:** $$100 \times 35 = 3500$$ $$\frac{20}{7} \times \frac{35^2}{2} = \frac{20}{7} \times \frac{1225}{2} = \frac{20}{7} \times 612.5 = 20 \times 87.5 = 1750$$ 7. **Final work:** $$W = 3500 - 1750 = 1750$$ **Answer:** The work done in lifting the bag is $1750$ ft-lbs.