A particle of mass `0.5 kg` is displaced from position `vec r_(1)(2,3,1)` to`vec r_(2)(4,3,2)` by applying a force of magnitude `30 N` which is acting along `(hati + hatj + hatk)`. The work done by the force is

Options

A. `10 sqrt(3) J`
B. `30 sqrt(3)`
C. 30 J
D. 40 J

Explanation

To find the work done by the force on the particle, we can follow these steps:

### Step 1: Identify the initial and final positions
The initial position vector \( \vec{r_1} \) is given as \( (2, 3, 1) \) and the final position vector \( \vec{r_2} \) is \( (4, 3, 2) \).

### Step 2: Calculate the displacement vector \( \vec{r} \)
The displacement vector \( \vec{r} \) is calculated as:
\[ \vec{r} = \vec{r_2} - \vec{r_1} = (4 - 2) \hat{i} + (3 - 3) \hat{j} + (2 - 1) \hat{k} = 2 \hat{i} + 0 \hat{j} + 1 \hat{k} = 2 \hat{i} + 1 \hat{k} \]

### Step 3: Determine the direction of the force vector
The force has a magnitude of \( 30 \, \text{N} \) and acts along the direction \( (\hat{i} + \hat{j} + \hat{k}) \). We need to find the unit vector in this direction:
\[ \text{Magnitude of the direction vector} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
Thus, the unit vector \( \hat{F} \) in the direction of the force is:
\[ \hat{F} = \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) \]

### Step 4: Calculate the force vector \( \vec{F} \)
The force vector \( \vec{F} \) can be calculated as:
\[ \vec{F} = 30 \hat{F} = 30 \cdot \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k}) = 10\sqrt{3} (\hat{i} + \hat{j} + \hat{k}) \]

### Step 5: Calculate the work done \( W \)
The work done by the force is given by the dot product of the force vector and the displacement vector:
\[ W = \vec{F} \cdot \vec{r} \]
Substituting the values:
\[ \vec{F} = 10\sqrt{3} (\hat{i} + \hat{j} + \hat{k}), \quad \vec{r} = 2 \hat{i} + 0 \hat{j} + 1 \hat{k} \]
Calculating the dot product:
\[ W = (10\sqrt{3} \hat{i} + 10\sqrt{3} \hat{j} + 10\sqrt{3} \hat{k}) \cdot (2 \hat{i} + 0 \hat{j} + 1 \hat{k}) \]
\[ W = 10\sqrt{3} \cdot 2 + 10\sqrt{3} \cdot 0 + 10\sqrt{3} \cdot 1 \]
\[ W = 20\sqrt{3} + 0 + 10\sqrt{3} = 30\sqrt{3} \, \text{J} \]

### Final Answer
The work done by the force is \( 30\sqrt{3} \, \text{J} \).

---