A body displaced from `(1,1)` to `(2,2)` along the line `y=x` under the force `vecF=xhat(j)+yhat(i)` then find the work done by this force for above displacement. (all units are in S.I.)

Options

A. `2` unit
B. `3` unit
C. `4` unit
D. `5` unit

Explanation

To find the work done by the force \(\vec{F} = x \hat{j} + y \hat{i}\) as a body is displaced from the point \((1, 1)\) to \((2, 2)\) along the line \(y = x\), we can follow these steps:

### Step 1: Define the displacement vector
The displacement vector \(d\vec{s}\) along the path can be expressed in terms of \(dx\) and \(dy\). Since the body moves along the line \(y = x\), we have:
\[ dy = dx \]
Thus, the displacement vector can be written as:
\[ d\vec{s} = dx \hat{i} + dy \hat{j} = dx \hat{i} + dx \hat{j} = dx (\hat{i} + \hat{j}) \]

### Step 2: Express the force vector
The force vector is given as:
\[ \vec{F} = x \hat{j} + y \hat{i} \]
Since we are moving along the line \(y = x\), we can substitute \(y\) with \(x\):
\[ \vec{F} = x \hat{j} + x \hat{i} = x (\hat{i} + \hat{j}) \]

### Step 3: Calculate the work done
The work done \(W\) by the force during the displacement is given by the dot product of the force and the displacement vector integrated over the path:
\[ W = \int \vec{F} \cdot d\vec{s} \]
Substituting the expressions for \(\vec{F}\) and \(d\vec{s}\):
\[ W = \int (x (\hat{i} + \hat{j})) \cdot (dx (\hat{i} + \hat{j})) \]
Calculating the dot product:
\[ \vec{F} \cdot d\vec{s} = x \hat{i} \cdot dx \hat{i} + x \hat{j} \cdot dx \hat{j} = x dx + x dx = 2x dx \]
Thus, the work done becomes:
\[ W = \int_{1}^{2} 2x \, dx \]

### Step 4: Perform the integration
Now, we integrate:
\[ W = 2 \int_{1}^{2} x \, dx = 2 \left[ \frac{x^2}{2} \right]_{1}^{2} = 2 \left[ \frac{2^2}{2} - \frac{1^2}{2} \right] = 2 \left[ \frac{4}{2} - \frac{1}{2} \right] = 2 \left[ 2 - 0.5 \right] = 2 \times 1.5 = 3 \]

### Final Answer
The work done by the force for the displacement from \((1, 1)\) to \((2, 2)\) is:
\[ \boxed{3 \, \text{Joules}} \]