A particle moves from position `r_1=(3hati+2hatj-6hatk)` m to position `r_2=(14hati+13hatj+9hatk)` m under the action of a force `(4hati+hatj-3hatk)` N , then the work done is

Options

A. 46 J
B. 56 J
C. 86 J
D. 10 J

Answer

To find the work done by the force on the particle as it moves from position \( \mathbf{r_1} \) to position \( \mathbf{r_2} \), we can follow these steps:

### Step 1: Identify the positions and force
Given:
- Initial position \( \mathbf{r_1} = 3\hat{i} + 2\hat{j} - 6\hat{k} \) m
- Final position \( \mathbf{r_2} = 14\hat{i} + 13\hat{j} + 9\hat{k} \) m
- Force \( \mathbf{F} = 4\hat{i} + \hat{j} - 3\hat{k} \) N

### Step 2: Calculate the displacement \( \Delta \mathbf{r} \)
The displacement \( \Delta \mathbf{r} \) is given by:
\[ \Delta \mathbf{r} = \mathbf{r_2} - \mathbf{r_1} \]
Calculating this:
\[ \Delta \mathbf{r} = (14\hat{i} + 13\hat{j} + 9\hat{k}) - (3\hat{i} + 2\hat{j} - 6\hat{k}) \]
\[ = (14 - 3)\hat{i} + (13 - 2)\hat{j} + (9 + 6)\hat{k} \]
\[ = 11\hat{i} + 11\hat{j} + 15\hat{k} \]

### Step 3: Calculate the work done \( W \)
The work done \( W \) by the force is given by the dot product of the force and the displacement:
\[ W = \mathbf{F} \cdot \Delta \mathbf{r} \]
Calculating the dot product:
\[ W = (4\hat{i} + \hat{j} - 3\hat{k}) \cdot (11\hat{i} + 11\hat{j} + 15\hat{k}) \]
\[ = 4 \cdot 11 + 1 \cdot 11 + (-3) \cdot 15 \]
\[ = 44 + 11 - 45 \]
\[ = 10 \text{ Joules} \]

### Final Answer
The work done by the force is \( W = 10 \) Joules.

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