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`

Answer

To solve the problem of calculating the work done by the force \( F = (x^2 \hat{j} + y \hat{i}) \) as a body is displaced from \( (0,0) \) to \( (1,1) \) along the path \( x = y \), we can follow these steps:

### Step 1: Define the Work Done
The work done \( W \) by a force along a path can be calculated using the formula:
\[ W = \int \mathbf{F} \cdot d\mathbf{r} \]
where \( \mathbf{F} \) is the force vector and \( d\mathbf{r} \) is the differential displacement vector.

### Step 2: Express the Force and Displacement Vectors
Given the force:
\[ \mathbf{F} = y \hat{i} + x^2 \hat{j} \]
The differential displacement vector \( d\mathbf{r} \) can be expressed as:
\[ d\mathbf{r} = dx \hat{i} + dy \hat{j} \]

### Step 3: Substitute the Path Equation
Since the path is defined by \( x = y \), we can express \( y \) in terms of \( x \) (or vice versa). Thus, we can set \( y = x \). Therefore, \( dy = dx \).

### Step 4: Rewrite the Work Integral
Substituting \( y = x \) into the force vector:
\[ \mathbf{F} = x \hat{i} + x^2 \hat{j} \]
Now, we can substitute this into the work integral:
\[ W = \int \left( x \hat{i} + x^2 \hat{j} \right) \cdot \left( dx \hat{i} + dy \hat{j} \right) \]
This simplifies to:
\[ W = \int (x \, dx + x^2 \, dy) \]
Since \( dy = dx \), we have:
\[ W = \int (x \, dx + x^2 \, dx) = \int (x + x^2) \, dx \]

### Step 5: Set the Limits of Integration
The limits of integration will be from \( x = 0 \) to \( x = 1 \):
\[ W = \int_0^1 (x + x^2) \, dx \]

### Step 6: Calculate the Integral
Now we can calculate the integral:
\[ W = \int_0^1 x \, dx + \int_0^1 x^2 \, dx \]
Calculating each part:
1. \(\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}\)
2. \(\int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3}\)

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

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

### Conclusion
The total work done by the force as the body is displaced from \( (0,0) \) to \( (1,1) \) along the path \( x = y \) is:
\[ W = \frac{5}{6} \text{ Joules} \]