Force acting on a pasrticle is `vecF = (alpha y hati +beta xy hatj)`. Find the work done by this force, when pasrticle is moved alolng the line `2x = 3y` from origin to the point `(3,2)` {take quantities in `SI` units and `alpha = 1, beta = 1`}

Explanation

To find the work done by the force \(\vec{F} = (\alpha y \hat{i} + \beta xy \hat{j})\) as the particle moves along the line \(2x = 3y\) from the origin to the point \((3, 2)\), we will follow these steps:

### Step 1: Define the Force
Given:
\[ \alpha = 1, \quad \beta = 1 \]
Thus, the force can be expressed as:
\[ \vec{F} = (y \hat{i} + xy \hat{j}) \]

### Step 2: Parameterize the Path
The line \(2x = 3y\) can be rearranged to express \(y\) in terms of \(x\):
\[ y = \frac{2}{3}x \]
We will move from the origin \((0, 0)\) to the point \((3, 2)\).

### Step 3: Express \(dy\) in terms of \(dx\)
Differentiating \(y = \frac{2}{3}x\):
\[ dy = \frac{2}{3}dx \]

### Step 4: Write the Displacement Vector
The small displacement vector \(d\vec{r}\) is given by:
\[ d\vec{r} = dx \hat{i} + dy \hat{j} = dx \hat{i} + \frac{2}{3}dx \hat{j} = \left(dx \hat{i} + \frac{2}{3}dx \hat{j}\right) \]

### Step 5: Calculate the Work Done
The work done \(dW\) by the force along the path is given by the dot product:
\[ dW = \vec{F} \cdot d\vec{r} \]
Substituting \(\vec{F}\) and \(d\vec{r}\):
\[ dW = (y \hat{i} + xy \hat{j}) \cdot \left(dx \hat{i} + \frac{2}{3}dx \hat{j}\right) \]
Calculating the dot product:
\[ dW = y \, dx + xy \left(\frac{2}{3}dx\right) = y \, dx + \frac{2}{3}xy \, dx \]
Factoring out \(dx\):
\[ dW = \left(y + \frac{2}{3}xy\right) dx \]

### Step 6: Substitute \(y\) in terms of \(x\)
Substituting \(y = \frac{2}{3}x\):
\[ dW = \left(\frac{2}{3}x + \frac{2}{3}x \cdot \frac{2}{3}x\right) dx = \left(\frac{2}{3}x + \frac{4}{9}x^2\right) dx \]

### Step 7: Integrate to Find Total Work Done
Now, we integrate from \(x = 0\) to \(x = 3\):
\[ W = \int_{0}^{3} \left(\frac{2}{3}x + \frac{4}{9}x^2\right) dx \]
Calculating the integral:
\[ W = \int_{0}^{3} \frac{2}{3}x \, dx + \int_{0}^{3} \frac{4}{9}x^2 \, dx \]
Calculating each part:
1. \(\int \frac{2}{3}x \, dx = \frac{2}{3} \cdot \frac{x^2}{2} = \frac{1}{3}x^2\)
2. \(\int \frac{4}{9}x^2 \, dx = \frac{4}{9} \cdot \frac{x^3}{3} = \frac{4}{27}x^3\)

Evaluating from \(0\) to \(3\):
\[ W = \left[\frac{1}{3}(3^2) + \frac{4}{27}(3^3)\right] - \left[\frac{1}{3}(0^2) + \frac{4}{27}(0^3)\right] \]
Calculating:
\[ W = \left[\frac{1}{3}(9) + \frac{4}{27}(27)\right] = 3 + 4 = 7 \, \text{Joules} \]

### Final Answer
The work done by the force is \(W = 7 \, \text{J}\).

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