A particle moves along `y=sqrt(1-x^ (2))` between the points (0, -1) m and (0, 1) m under the influence of a force `vec(F)=(y^(2)hat(i)+x^(2)hat(j))N`. Then,

Options

A. the particle is moving along a semi-ellipse
B. the particle is moving along a semicircle
C. work done on the particle by `vec(F)` by `3/4`J
D. work done on the particle by `vec(F)` is `4/3`J

Explanation

To solve the problem step by step, we need to calculate the work done by the force \(\vec{F} = (y^2 \hat{i} + x^2 \hat{j})\) as the particle moves along the path defined by \(y = \sqrt{1 - x^2}\) from the point (0, -1) to (0, 1).

### Step 1: Understand the Path of the Particle
The equation \(y = \sqrt{1 - x^2}\) describes the upper half of a circle with radius 1 centered at the origin. The particle moves from (0, -1) to (0, 1), which means it traces out a semicircular path.

### Step 2: Set Up the Work Done Integral
The work done \(W\) by a force along a path can be calculated using the line integral:
\[ W = \int_C \vec{F} \cdot d\vec{s} \]
where \(d\vec{s} = dx \hat{i} + dy \hat{j}\).

### Step 3: Express the Force in Terms of \(x\)
Since the path is defined by \(y = \sqrt{1 - x^2}\), we can express \(F\) in terms of \(x\):
\[ \vec{F} = (y^2 \hat{i} + x^2 \hat{j}) = ((\sqrt{1 - x^2})^2 \hat{i} + x^2 \hat{j}) = (1 - x^2) \hat{i} + x^2 \hat{j} \]

### Step 4: Calculate \(dy\)
To find \(dy\) in terms of \(dx\), we differentiate \(y\):
\[ y = \sqrt{1 - x^2} \implies dy = \frac{-x}{\sqrt{1 - x^2}} dx \]

### Step 5: Substitute \(dy\) into the Work Integral
Now we substitute \(dy\) into the work integral:
\[ W = \int_{-1}^{1} \left[(1 - x^2) \hat{i} + x^2 \hat{j}\right] \cdot \left[dx \hat{i} + dy \hat{j}\right] \]
This gives us:
\[ W = \int_{-1}^{1} \left[(1 - x^2) dx + x^2 dy\right] \]

### Step 6: Substitute \(dy\) into the Integral
Substituting \(dy\):
\[ W = \int_{-1}^{1} \left[(1 - x^2) dx + x^2 \left(\frac{-x}{\sqrt{1 - x^2}} dx\right)\right] \]
This simplifies to:
\[ W = \int_{-1}^{1} \left[(1 - x^2) - \frac{x^3}{\sqrt{1 - x^2}}\right] dx \]

### Step 7: Evaluate the Integral
Now we can evaluate the integral:
1. The first part:
\[ \int_{-1}^{1} (1 - x^2) dx = \left[x - \frac{x^3}{3}\right]_{-1}^{1} = \left[1 - \frac{1}{3}\right] - \left[-1 + \frac{1}{3}\right] = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \]
2. The second part:
\[ \int_{-1}^{1} -\frac{x^3}{\sqrt{1 - x^2}} dx = 0 \quad (\text{since it's an odd function over a symmetric interval}) \]

### Step 8: Combine Results
Thus, the total work done is:
\[ W = \frac{4}{3} + 0 = \frac{4}{3} \text{ Joules} \]

### Final Answer
The work done on the particle by the force \(\vec{F}\) as it moves along the semicircular path is:
\[ \boxed{\frac{4}{3} \text{ Joules}} \]