A particle moves from a point `(-2hati+5hatj)` to `(4hatj+3hatk)` when a force of `(4hati+3hatj)N` is applied. How much work has been done by the force ?

Options

A. 8 J
B. 11 J
C. 5 J
D. 2 J

Answer

To find the work done by the force when a particle moves from one point to another, we can use the formula:

\[ W = \vec{F} \cdot \vec{d} \]

where \( W \) is the work done, \( \vec{F} \) is the force vector, and \( \vec{d} \) is the displacement vector.

### Step-by-Step Solution:

**Step 1: Identify the initial and final positions of the particle.**

The initial position \( \vec{r_1} \) is given as:
\[ \vec{r_1} = -2\hat{i} + 5\hat{j} \]

The final position \( \vec{r_2} \) is given as:
\[ \vec{r_2} = 4\hat{j} + 3\hat{k} \]

**Step 2: Calculate the displacement vector \( \vec{d} \).**

The displacement \( \vec{d} \) can be calculated as:
\[ \vec{d} = \vec{r_2} - \vec{r_1} \]

Substituting the values:
\[ \vec{d} = (4\hat{j} + 3\hat{k}) - (-2\hat{i} + 5\hat{j}) \]
\[ = 4\hat{j} + 3\hat{k} + 2\hat{i} - 5\hat{j} \]
\[ = 2\hat{i} - \hat{j} + 3\hat{k} \]

**Step 3: Identify the force vector \( \vec{F} \).**

The force vector \( \vec{F} \) is given as:
\[ \vec{F} = 4\hat{i} + 3\hat{j} \]

**Step 4: Calculate the work done \( W \) using the dot product.**

The work done is given by:
\[ W = \vec{F} \cdot \vec{d} \]

Calculating the dot product:
\[ W = (4\hat{i} + 3\hat{j}) \cdot (2\hat{i} - \hat{j} + 3\hat{k}) \]
\[ = (4 \cdot 2) + (3 \cdot -1) + (0 \cdot 3) \]
\[ = 8 - 3 + 0 \]
\[ = 5 \text{ joules} \]

### Final Answer:
The work done by the force is \( 5 \) joules.

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