A particle is desplaced from a position `(2hat I + 4hat-3hatk)` metre to another position `(12hati+14hatj+7hatk)` metre under the action of force `(hati+2hatj-hatk)` metre under rthe action of force `(hati+2hatj-hatk)` N. Work done by the force is

Options

A. `5 J`
B. `10 J`
C. `15 J`
D. `20J`

Answer

To solve the problem of finding the work done by the force on the particle, we will follow these steps:

### Step 1: Identify the initial and final positions of the particle.
- The initial position \( \mathbf{r_1} \) is given as:
\[ \mathbf{r_1} = 2\hat{i} + 4\hat{j} - 3\hat{k} \text{ meters} \]
- The final position \( \mathbf{r_2} \) is given as:
\[ \mathbf{r_2} = 12\hat{i} + 14\hat{j} + 7\hat{k} \text{ meters} \]

### Step 2: Calculate the displacement vector \( \mathbf{d} \).
- The displacement \( \mathbf{d} \) is calculated as:
\[ \mathbf{d} = \mathbf{r_2} - \mathbf{r_1} \]
- Performing the subtraction:
\[ \mathbf{d} = (12\hat{i} + 14\hat{j} + 7\hat{k}) - (2\hat{i} + 4\hat{j} - 3\hat{k}) \]
\[ = (12 - 2)\hat{i} + (14 - 4)\hat{j} + (7 - (-3))\hat{k} \]
\[ = 10\hat{i} + 10\hat{j} + 10\hat{k} \]

### Step 3: Identify the force vector \( \mathbf{F} \).
- The force \( \mathbf{F} \) is given as:
\[ \mathbf{F} = \hat{i} + 2\hat{j} - \hat{k} \text{ N} \]

### Step 4: Calculate the work done \( W \) by the force.
- The work done \( W \) is given by the dot product of the force and the displacement:
\[ W = \mathbf{F} \cdot \mathbf{d} \]
- Substituting the values of \( \mathbf{F} \) and \( \mathbf{d} \):
\[ W = (\hat{i} + 2\hat{j} - \hat{k}) \cdot (10\hat{i} + 10\hat{j} + 10\hat{k}) \]
- Calculating the dot product:
\[ W = (1 \cdot 10) + (2 \cdot 10) + (-1 \cdot 10) \]
\[ = 10 + 20 - 10 \]
\[ = 20 \text{ joules} \]

### Final Answer:
The work done by the force is \( 20 \text{ joules} \).

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