1. **State the problem:** We have a bag of sand weighing 100 lbs initially, lifted to a height of 35 ft. The sand leaks uniformly so that the bag is empty at 35 ft. We want to find the work done lifting the bag. 2. **Relevant formula:** Work done lifting a variable weight is given by $$W = \int_a^b F(x) \, dx$$ where $F(x)$ is the force (weight) at height $x$. 3. **Understanding the problem:** The weight decreases uniformly from 100 lbs at $x=0$ to 0 lbs at $x=35$ ft. 4. **Express the weight as a function of height:** Since weight decreases linearly, $$F(x) = 100 - \frac{100}{35}x = 100 - \frac{20}{7}x$$ 5. **Calculate the work:** $$W = \int_0^{35} \left(100 - \frac{20}{7}x\right) dx$$ 6. **Integrate:** $$W = \left[100x - \frac{20}{7} \frac{x^2}{2}\right]_0^{35} = \left[100x - \frac{10}{7}x^2\right]_0^{35}$$ 7. **Evaluate at bounds:** $$W = 100(35) - \frac{10}{7}(35)^2 - (0) = 3500 - \frac{10}{7} \times 1225$$ 8. **Simplify:** $$\frac{10}{7} \times 1225 = 10 \times 175 = 1750$$ 9. **Final work done:** $$W = 3500 - 1750 = 1750$$ **Answer:** The work done lifting the bag is $1750$ ft-lbs.