A particle moves from point `(hati+hatj)`m to a point `(6hati-2hatj)`m under the action of force `(10hati-2hatj)`N. The work done by the force is

Options

A. 8 J
B. 56 J
C. 64 J
D. Zero

Answer

To find the work done by the force on the particle as it moves from one point to another, we can use the formula for work done:

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

where \( \mathbf{F} \) is the force vector and \( \mathbf{d} \) is the displacement vector.

### Step 1: Identify the initial and final positions
The initial position vector \( \mathbf{r_1} \) is given as:
\[ \mathbf{r_1} = \hat{i} + \hat{j} \]

The final position vector \( \mathbf{r_2} \) is given as:
\[ \mathbf{r_2} = 6\hat{i} - 2\hat{j} \]

### Step 2: Calculate the displacement vector \( \mathbf{d} \)
The displacement vector \( \mathbf{d} \) is calculated as:
\[ \mathbf{d} = \mathbf{r_2} - \mathbf{r_1} \]
Substituting the values:
\[ \mathbf{d} = (6\hat{i} - 2\hat{j}) - (\hat{i} + \hat{j}) \]
\[ \mathbf{d} = (6 - 1)\hat{i} + (-2 - 1)\hat{j} \]
\[ \mathbf{d} = 5\hat{i} - 3\hat{j} \]

### Step 3: Identify the force vector \( \mathbf{F} \)
The force vector \( \mathbf{F} \) is given as:
\[ \mathbf{F} = 10\hat{i} - 2\hat{j} \]

### Step 4: Calculate the work done \( W \)
Now we calculate the work done using the dot product:
\[ W = \mathbf{F} \cdot \mathbf{d} \]
\[ W = (10\hat{i} - 2\hat{j}) \cdot (5\hat{i} - 3\hat{j}) \]

Calculating the dot product:
\[ W = (10 \cdot 5) + (-2 \cdot -3) \]
\[ W = 50 + 6 \]
\[ W = 56 \, \text{J} \]

### Conclusion
The work done by the force is:
\[ W = 56 \, \text{J} \]

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