A person pushes a box on a rough horizontal platform surface. He applies a force of 200 N over a distance of 15 m. Thereafter, he gets progressively tired and his applied force reduces linearly with distance to 100 N. The total distance through which the box has been moved is 30 m. What is the work done by the person during the total movement of the box?

We first recall the definition of mechanical work. The work $$W$$ done by a variable or constant force acting along the line of motion is given by the formula

$$W = \int F\,\mathrm{d}x,$$

where $$F$$ is the instantaneous force and $$x$$ is the displacement. When the force remains constant over a segment, the integral reduces to the simple product of the constant force and the distance covered, $$W = F \times x$$. When the force varies linearly, the integral is numerically equal to the product of the average force and the distance.

According to the statement, the motion of the box can be divided into two successive parts:

(i) A first stretch of $$15\ \text{m}$$ during which the person keeps pushing with a constant force of $$200\ \text{N}$$.

(ii) A second and equal stretch of $$15\ \text{m}$$ during which the force decreases uniformly (linearly) from $$200\ \text{N}$$ down to $$100\ \text{N}$$ because the person is getting tired.

We now compute the work done in each part and then add the two contributions.

Work during the first 15 m

The force is constant, so we apply $$W = F \times x$$ directly:

$$W_1 = 200\ \text{N} \times 15\ \text{m} = 3000\ \text{J}.$$

Work during the next 15 m

Here the force varies linearly from $$200\ \text{N}$$ at the start of the segment to $$100\ \text{N}$$ at the end. For a linear variation, the average (mean) force $$F_{\text{avg}}$$ is simply the arithmetic mean:

$$F_{\text{avg}} = \dfrac{F_{\text{initial}} + F_{\text{final}}}{2} = \dfrac{200\ \text{N} + 100\ \text{N}}{2} = 150\ \text{N}.$$

Using again the constant-force formula with this average value gives the work for that segment:

$$W_2 = F_{\text{avg}} \times x = 150\ \text{N} \times 15\ \text{m} = 2250\ \text{J}.$$

Total work

Now we add the two pieces together:

$$W_{\text{total}} = W_1 + W_2 = 3000\ \text{J} + 2250\ \text{J} = 5250\ \text{J}.$$

Hence, the correct answer is Option D.