To calculate the work done by the two forces over the 20-meter distance, we'll break the problem into two parts:

1. First part (0 to 10 meters): The woman applies a constant force of 100 N over the first 10 meters.

   The work done in this region is calculated by: $W_1 = F_1 \cdot d_1 = 100 \, \text{N} \cdot 10 \, \text{m} = 1000 \, \text{J}$

2. Second part (10 to 20 meters): In this section, the force decreases linearly from 100 N to 50 N. The work done by a linearly varying force can be calculated using the average force over the distance.

   The average force is: $F_{\text{avg}} = \frac{F_{\text{initial}} + F_{\text{final}}}{2} = \frac{100 \, \text{N} + 50 \, \text{N}}{2} = 75 \, \text{N}$

   The distance is 10 meters, so the work done in this section is: $W_2 = F_{\text{avg}} \cdot d_2 = 75 \, \text{N} \cdot 10 \, \text{m} = 750 \, \text{J}$

3. Total work done: $W_{\text{total}} = W_1 + W_2 = 1000 \, \text{J} + 750 \, \text{J} = 1750 \, \text{J}$

Thus, the total work done by the forces over the 20 meters is 1750 J.

The correct answer is: (a) 1750 J

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Do you need further details or have any questions? Here are some related problems to explore:

1. What is the work done if the force applied over the entire 20 m distance was constant at 100 N?
2. How does the work-energy theorem apply to this scenario?
3. If the surface were frictionless, how would the work done change?
4. What is the total energy expended if the woman's force reduced non-linearly?
5. How would the solution change if the force applied was proportional to the distance?

Tip: When forces vary over distance, using the average force is a quick way to calculate work in uniformly varying force situations.