A body is displaced from `(0,0)` to `(1 m, 1m)` along the path `x = y` by a force `F=(x^(2) hatj + y hati)N`. The work done by this force will be

Options

A. `(4)/(3) J`
B. `(5)/(6) J`
C. `(3)/(2) J`
D. `(7)/(5) J`

Explanation

To find the work done by the force \( \mathbf{F} = (x^2 \hat{j} + y \hat{i}) \, \text{N} \) as a body moves from the point \( (0,0) \) to \( (1,1) \) along the path \( x = y \), we can follow these steps:

### Step 1: Understand the Work Done Formula
The work done by a force along a path is given by the line integral:
\[ W = \int_C \mathbf{F} \cdot d\mathbf{r} \]
where \( C \) is the path taken, \( \mathbf{F} \) is the force vector, and \( d\mathbf{r} \) is the differential displacement vector along the path.

### Step 2: Parameterize the Path
Since the body moves along the line \( x = y \), we can parameterize the path using a single variable \( x \):
\[ \mathbf{r}(t) = (t, t) \quad \text{for } t \in [0, 1] \]
Thus, \( d\mathbf{r} = (dx, dy) = (dt, dt) \).

### Step 3: Express the Force in Terms of the Parameter
Substituting \( x = t \) and \( y = t \) into the force vector:
\[ \mathbf{F} = (t^2 \hat{j} + t \hat{i}) \, \text{N} \]

### Step 4: Calculate the Dot Product
Now, we need to compute the dot product \( \mathbf{F} \cdot d\mathbf{r} \):
\[ \mathbf{F} \cdot d\mathbf{r} = (t \hat{i} + t^2 \hat{j}) \cdot (dt \hat{i} + dt \hat{j}) = (t \cdot dt + t^2 \cdot dt) = (t + t^2) dt \]

### Step 5: Set Up the Integral for Work Done
Now we can set up the integral for the work done:
\[ W = \int_0^1 (t + t^2) \, dt \]

### Step 6: Evaluate the Integral
We can split the integral:
\[ W = \int_0^1 t \, dt + \int_0^1 t^2 \, dt \]
Calculating each integral separately:
1. \( \int_0^1 t \, dt = \left[ \frac{t^2}{2} \right]_0^1 = \frac{1}{2} \)
2. \( \int_0^1 t^2 \, dt = \left[ \frac{t^3}{3} \right]_0^1 = \frac{1}{3} \)

Thus,
\[ W = \frac{1}{2} + \frac{1}{3} \]

### Step 7: Combine the Results
To combine the fractions, we find a common denominator:
\[ W = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \, \text{Joules} \]

### Final Answer
The work done by the force is:
\[ W = \frac{5}{6} \, \text{Joules} \]

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