1. Understand the force variation

The motion happens in two parts:

• From $0$ to $15\,\text{m}$, the person applies a constant force of $200\,\text{N}$.
• From $15$ to $30\,\text{m}$, the force decreases linearly from $200\,\text{N}$ to $100\,\text{N}$.

We need total work done by the person:

$W = \int F\,dx$

For a force-vs-distance graph, work equals the area under the graph.

2. Work done in the first 15 m

Since force is constant,

$W_1 = F \cdot s = 200 \times 15 = 3000\,\text{J}$

3. Work done in the next 15 m

Here force decreases linearly from $200\,\text{N}$ to $100\,\text{N}$.

So the average force over this interval is

$F_{\text{avg}} = \frac{200 + 100}{2} = 150\,\text{N}$

Hence work done in this part is

$W_2 = F_{\text{avg}} \cdot s = 150 \times 15 = 2250\,\text{J}$

4. Total work done

$W = W_1 + W_2 = 3000 + 2250 = 5250\,\text{J}$

5. Check options

• A: $5690\,\text{J}$
• B: $5250\,\text{J}$
• C: $2780\,\text{J}$
• D: $3280\,\text{J}$

So the correct option is:

$\boxed{\text{B: } 5250\,\text{J}}$