The total work done by a variable force is represented by the area under the force-displacement graph.

1. Area 1 (from $x=0$ m to $x=3$ m): This segment forms a rectangle. Its length (displacement) is $3$ m and its height (force) is $5$ N. $A_1 = \text{length} \times \text{height} = (3 \text{ m}) \times (5 \text{ N}) = 15 \text{ J}$

2. Area 2 (from $x=3$ m to $x=5$ m): This segment forms a triangle. Its base (displacement) is $(5-3)=2$ m and its height (force) is $5$ N (at $x=3$ m, decreasing to $0$ N at $x=5$ m). $A_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2 \text{ m}) \times (5 \text{ N}) = 5 \text{ J}$

Total work done $W_{\text{total}}$ is the sum of these areas: $W_{\text{total}} = A_1 + A_2 = 15 \text{ J} + 5 \text{ J} = 20 \text{ J}$

Why other options are incorrect:

• Option A ($10$ J): This value does not correspond to the correct area calculation for the given force-displacement graph.
• Option B ($15$ J): This represents only the work done in the first segment ($0$ m to $3$ m, the rectangular area), neglecting the work done in the second segment.
• Option C ($17.5$ J): This would be an incorrect calculation, possibly due to a miscalculation of one of the areas (e.g., if the triangle area was $2.5$ J instead of $5$ J, leading to $15+2.5=17.5$ J).