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To solve the problem, we need to determine the work done on the body by the time-dependent force \( F(t) \) over the 20-second period shown in the graph. Work is calculated as:

\[
W = \int F(t) \cdot v(t) \, dt
\]

However, the direct approach here is to calculate the work as the area under the force-time graph. This is because work done by a force is equivalent to the area under the force vs. time graph when multiplied by velocity.

### Step 1: Break the time intervals
The graph can be divided into different segments based on the changes in force:
- From 0 to 5 seconds, the force is \( +20 \, \text{N} \)
- From 5 to 10 seconds, the force is decreasing linearly to \( -20 \, \text{N} \)
- From 10 to 15 seconds, the force is \( -20 \, \text{N} \)
- From 15 to 20 seconds, the force increases linearly back to \( 0 \, \text{N} \)

### Step 2: Calculate the areas under each segment
The work done is the area under the curve:
1. **0 to 5 seconds**: Rectangle with height \( 20 \, \text{N} \) and base \( 5 \, \text{s} \):
   \[
   A_1 = 20 \times 5 = 100 \, \text{J}
   \]

2. **5 to 10 seconds**: Triangle with height \( 20 \, \text{N} \) and base \( 5 \, \text{s} \) (negative work):
   \[
   A_2 = \frac{1}{2} \times 20 \times 5 = 50 \, \text{J} \, (\text{negative, so } -50 \, \text{J})
   \]

3. **10 to 15 seconds**: Rectangle with height \( -20 \, \text{N} \) and base \( 5 \, \text{s} \) (negative work):
   \[
   A_3 = -20 \times 5 = -100 \, \text{J}
   \]

4. **15 to 20 seconds**: Triangle with height \( 20 \, \text{N} \) and base \( 5 \, \text{s} \):
   \[
   A_4 = \frac{1}{2} \times 20 \times 5 = 50 \, \text{J}
   \]

### Step 3: Total work done
Now sum the areas:
\[
W_{\text{total}} = 100 + (-50) + (-100) + 50 = 0 \, \text{J}
\]

### Conclusion
The total work done on the body in 20 seconds is \( 0 \, \text{J} \). Thus, the correct answer does not match any of the options provided. There might be an issue with the options, but based on the calculation, the total work done is zero.

Do you want further details on the steps or explanations? Here are some related questions to explore:

1. How do we calculate work from force-time graphs in general?
2. What is the relationship between force, displacement, and work?
3. How would changing the mass or initial velocity affect the total work done?
4. What happens if the force remains constant instead of varying with time?
5. How would you calculate the change in kinetic energy from the work done?

**Tip:** Always consider both positive and negative areas when interpreting force-time graphs to calculate work.

A body of mass 2 kg starts with an initial velocity 5 m/s. If the body is acted upon by a time-dependent force (F) as shown in the figure, then what is the work done on the body in 20 seconds?

This problem involves calculating the work done on a 2 kg mass over 20 seconds, given a time-dependent force as illustrated. Using the Work-Energy Theorem and graphing techniques, this detailed solution explains how to interpret the force-time graph and compute the total work done. Suitable for Grades 11-12 students.

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