A particle experiences a variable force `vecF = (4xhati + 3y^2hatj)` in a horizontal x-y plane. Assume distance in meters and force is newton. If the particle moves from point (1, 2) to point (2,3) in the r-y plane, then Kinetic Energy changes by :

Options

A. 50.0 J
B. 12. 5 J
C. 25.0 J
D. 0 J

Answer

To find the change in kinetic energy of the particle moving under the influence of a variable force \(\vec{F} = (4x \hat{i} + 3y^2 \hat{j})\), we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Identify the initial and final positions of the particle**:
- Initial position: \( (x_1, y_1) = (1, 2) \)
- Final position: \( (x_2, y_2) = (2, 3) \)

2. **Write the expression for work done by the force**:
- The work done \(W\) by a variable force is given by:
\[ W = \int \vec{F} \cdot d\vec{s} \]
- Here, \(d\vec{s} = (dx \hat{i} + dy \hat{j})\).

3. **Express the force in terms of its components**:
- The force can be expressed as:
\[ \vec{F} = 4x \hat{i} + 3y^2 \hat{j} \]

4. **Calculate the work done**:
- The work done can be calculated as:
\[ W = \int_{(1,2)}^{(2,3)} (4x \, dx + 3y^2 \, dy) \]
- We can split this into two integrals:
\[ W = \int_{1}^{2} 4x \, dx + \int_{2}^{3} 3y^2 \, dy \]

5. **Evaluate the first integral**:
- For the first integral:
\[ \int 4x \, dx = 2x^2 \]
- Evaluating from \(x = 1\) to \(x = 2\):
\[ W_1 = \left[ 2x^2 \right]_{1}^{2} = 2(2^2) - 2(1^2) = 8 - 2 = 6 \]

6. **Evaluate the second integral**:
- For the second integral:
\[ \int 3y^2 \, dy = y^3 \]
- Evaluating from \(y = 2\) to \(y = 3\):
\[ W_2 = \left[ y^3 \right]_{2}^{3} = 3^3 - 2^3 = 27 - 8 = 19 \]

7. **Calculate the total work done**:
- The total work done \(W\) is:
\[ W = W_1 + W_2 = 6 + 19 = 25 \]

8. **Apply the work-energy theorem**:
- According to the work-energy theorem:
\[ W = \Delta K = K_f - K_i \]
- Assuming the particle starts from rest, \(K_i = 0\):
\[ \Delta K = K_f - 0 = 25 \text{ J} \]

### Final Answer:
The change in kinetic energy of the particle is \(25 \text{ J}\).

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