A force `F=(3xhati+4hatj)` Newton (where x is in metres) acts on a particle which moves from a position (2m, 3m) to (3m, 0m). Then the work done is

Options

A. 7.5 J
B. `-12J`
C. `-4.5J`
D. `+4.5J`

Answer

To solve the problem of calculating the work done by the force \( F = (3x \hat{i} + 4 \hat{j}) \) Newton as a particle moves from the position \( (2 \, \text{m}, 3 \, \text{m}) \) to \( (3 \, \text{m}, 0 \, \text{m}) \), we can follow these steps:

### Step 1: Understand the Work Done Formula
The work done \( W \) by a force along a path can be calculated using the formula:
\[ W = \int \mathbf{F} \cdot d\mathbf{s} \]
where \( \mathbf{F} \) is the force vector and \( d\mathbf{s} \) is the differential displacement vector.

### Step 2: Define the Displacement Vector
The displacement vector \( d\mathbf{s} \) can be expressed in terms of its components:
\[ d\mathbf{s} = dx \hat{i} + dy \hat{j} \]

### Step 3: Substitute the Force and Displacement
The force vector is given as:
\[ \mathbf{F} = 3x \hat{i} + 4 \hat{j} \]
Now, we can calculate the dot product \( \mathbf{F} \cdot d\mathbf{s} \):
\[ \mathbf{F} \cdot d\mathbf{s} = (3x \hat{i} + 4 \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = 3x \, dx + 4 \, dy \]

### Step 4: Set Up the Integral for Work Done
The total work done can be calculated by integrating the expression obtained:
\[ W = \int (3x \, dx + 4 \, dy) \]

### Step 5: Determine the Path of Integration
The particle moves from \( (2, 3) \) to \( (3, 0) \). We can break this into two segments:
1. Move in the x-direction from \( (2, 3) \) to \( (3, 3) \) (where \( y \) remains constant).
2. Move in the y-direction from \( (3, 3) \) to \( (3, 0) \) (where \( x \) remains constant).

### Step 6: Calculate Work Done in the x-direction
For the first segment, where \( y = 3 \) (constant):
\[ W_1 = \int_{2}^{3} (3x \, dx + 4 \cdot 0) = \int_{2}^{3} 3x \, dx \]
Calculating this integral:
\[ W_1 = 3 \left[ \frac{x^2}{2} \right]_{2}^{3} = 3 \left( \frac{3^2}{2} - \frac{2^2}{2} \right) = 3 \left( \frac{9}{2} - \frac{4}{2} \right) = 3 \left( \frac{5}{2} \right) = \frac{15}{2} \, \text{J} \]

### Step 7: Calculate Work Done in the y-direction
For the second segment, where \( x = 3 \) (constant):
\[ W_2 = \int_{3}^{0} (3 \cdot 3 \cdot 0 + 4 \, dy) = \int_{3}^{0} 4 \, dy \]
Calculating this integral:
\[ W_2 = 4 \left[ y \right]_{3}^{0} = 4 (0 - 3) = -12 \, \text{J} \]

### Step 8: Total Work Done
Now, we can find the total work done by adding the work done in both segments:
\[ W = W_1 + W_2 = \frac{15}{2} - 12 = \frac{15}{2} - \frac{24}{2} = \frac{-9}{2} \, \text{J} = -4.5 \, \text{J} \]

### Final Answer
The total work done by the force is:
\[ \boxed{-4.5 \, \text{J}} \]