A uniform force of `(3 hat(i)+hat(j))` newton acts on a particle of mass `2 kg`. Hence the particle is displaced from position `(2 hat(i)+hat(j))` meter to position `(4 hat(i)+ 3hat(j)-hat(k))` meter. The work done by the force on the particle is `:`

Options

A. ` 15 J`
B. `9 J`
C. `6 J`
D. `13 J`

Explanation

To find the work done by a uniform force on a particle, we can use the formula:

\[ W = \mathbf{F} \cdot \mathbf{S} \]

where:
- \( W \) is the work done,
- \( \mathbf{F} \) is the force vector,
- \( \mathbf{S} \) is the displacement vector.

### Step 1: Identify the Force Vector
The force vector given in the problem is:

\[ \mathbf{F} = 3\hat{i} + \hat{j} \, \text{N} \]

### Step 2: Identify the Initial and Final Position Vectors
The initial position vector is:

\[ \mathbf{r_i} = 2\hat{i} + \hat{j} \, \text{m} \]

The final position vector is:

\[ \mathbf{r_f} = 4\hat{i} + 3\hat{j} - \hat{k} \, \text{m} \]

### Step 3: Calculate the Displacement Vector
The displacement vector \( \mathbf{S} \) can be calculated as:

\[ \mathbf{S} = \mathbf{r_f} - \mathbf{r_i} \]

Calculating this gives:

\[ \mathbf{S} = (4\hat{i} + 3\hat{j} - \hat{k}) - (2\hat{i} + \hat{j}) \]
\[ = (4 - 2)\hat{i} + (3 - 1)\hat{j} + (0 - 0)\hat{k} \]
\[ = 2\hat{i} + 2\hat{j} - \hat{k} \]

### Step 4: Calculate the Work Done
Now we can calculate the work done using the dot product:

\[ W = \mathbf{F} \cdot \mathbf{S} \]
\[ = (3\hat{i} + \hat{j}) \cdot (2\hat{i} + 2\hat{j} - \hat{k}) \]

Calculating the dot product:

\[ W = (3 \cdot 2) + (1 \cdot 2) + (0 \cdot -1) \]
\[ = 6 + 2 + 0 \]
\[ = 8 \, \text{J} \]

### Final Answer
Thus, the work done by the force on the particle is:

\[ W = 8 \, \text{J} \]