1. Use work-energy theorem

The change in kinetic energy is equal to the work done by the force:

$\Delta K = W = \int \vec F \cdot d\vec r$

Given,

$\vec F = 4x\,\hat i + 3y^2\,\hat j$

and

$d\vec r = dx\,\hat i + dy\,\hat j$

So,

$\vec F \cdot d\vec r = 4x\,dx + 3y^2\,dy$

Hence,

$W = \int (4x\,dx + 3y^2\,dy)$

2. Check whether force is conservative

Let

$F_x = 4x, \qquad F_y = 3y^2$

Then,

$\frac{\partial F_x}{\partial y} = 0, \qquad \frac{\partial F_y}{\partial x} = 0$

Since these are equal, the force is conservative. Therefore, work done depends only on initial and final points.

3. Find potential-like function / integrate directly

Since,

$dW = 4x\,dx + 3y^2\,dy$

Integrate termwise from $(1,2)$ to $(2,3)$:

$W = \int_{x=1}^{2} 4x\,dx + \int_{y=2}^{3} 3y^2\,dy$

Now,

$\int_{1}^{2} 4x\,dx = 2x^2\Big|_{1}^{2} = 2(4)-2(1)=8-2=6$

and

$\int_{2}^{3} 3y^2\,dy = y^3\Big|_{2}^{3} = 27-8=19$

Therefore,

$W = 6+19=25\text{ J}$

4. Apply work-energy theorem

$\Delta K = W = 25\text{ J}$

So the kinetic energy increases by

$\boxed{25\text{ J}}$

5. Option check

• A: $50.0\,$J ❌
• B: $12.5\,$J ❌
• C: $25.0\,$J ✅
• D: $0\,$J ❌

Therefore, the correct option is C.