A particle is displaced from (1, 2)m to (0,0)m along the path `y= 2x^(3)`. Find the work done by a force `vec(F)= (x^(3)hat(j) + y hat(i))N` acting on the particle, during this displacement is

Explanation

To find the work done by the force \(\vec{F} = (x^3 \hat{j} + y \hat{i})\) during the displacement of the particle from \((1, 2)\) m to \((0, 0)\) m along the path \(y = 2x^3\), we can follow these steps:

### Step 1: Understand the Work Done Formula
The work done by a force along a path can be calculated using the formula:
\[ W = \int \vec{F} \cdot d\vec{r} \]
where \(d\vec{r} = dx \hat{i} + dy \hat{j}\).

### Step 2: Substitute the Force and Displacement
Given the force \(\vec{F} = (x^3 \hat{j} + y \hat{i})\), we can express the dot product:
\[ \vec{F} \cdot d\vec{r} = (y \hat{i} + x^3 \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = y \, dx + x^3 \, dy \]

### Step 3: Express \(dy\) in terms of \(dx\)
Since the path is defined by \(y = 2x^3\), we can differentiate this to find \(dy\):
\[ dy = \frac{d}{dx}(2x^3) = 6x^2 \, dx \]

### Step 4: Substitute \(y\) and \(dy\) into the Work Integral
Now we can substitute \(y\) and \(dy\) into the work integral:
\[ W = \int (2x^3 \, dx + x^3 \cdot 6x^2 \, dx) = \int (2x^3 + 6x^5) \, dx \]

### Step 5: Set the Limits of Integration
The particle moves from \((1, 2)\) to \((0, 0)\). Therefore, the limits for \(x\) will be from \(1\) to \(0\).

### Step 6: Calculate the Integral
Now we can compute the integral:
\[ W = \int_{1}^{0} (2x^3 + 6x^5) \, dx \]
Calculating the integral:
\[ = \left[ \frac{2x^4}{4} + \frac{6x^6}{6} \right]_{1}^{0} = \left[ \frac{x^4}{2} + x^6 \right]_{1}^{0} \]

Evaluating at the limits:
\[ = \left( \frac{0^4}{2} + 0^6 \right) - \left( \frac{1^4}{2} + 1^6 \right) = 0 - \left( \frac{1}{2} + 1 \right) = -\frac{3}{2} \]

### Step 7: Final Result
Thus, the work done by the force during the displacement is:
\[ W = -\frac{3}{2} \text{ Joules} = -1.5 \text{ Joules} \]