Consider a force `overset ( r)(F)= -x^(3) hat( i) +y^(2) hat(j) `.The work done by this force in moving a particle from A(1,0) to B(0,1) along a straight line is `:` ( All quantities are in SI units )

Options

A. 1
B. `5/12`
C. `7/12`
D. `3/7`

Explanation

To find the work done by the force \(\overset{\rightarrow}{F} = -x^3 \hat{i} + y^2 \hat{j}\) in moving a particle from point A(1,0) to point B(0,1) along a straight line, we will use the formula for work done by a variable force, which is given by:

\[ W = \int_{A}^{B} \overset{\rightarrow}{F} \cdot d\overset{\rightarrow}{s} \]

### Step 1: Define the displacement vector \(d\overset{\rightarrow}{s}\)

Since we are moving from point A(1,0) to point B(0,1) along a straight line, we can express the displacement vector \(d\overset{\rightarrow}{s}\) in terms of \(dx\) and \(dy\). The path can be parameterized as follows:

Let \(x = 1 - t\) and \(y = t\), where \(t\) varies from 0 to 1. This gives us the following relationships:
- At \(t=0\), \(A(1,0)\)
- At \(t=1\), \(B(0,1)\)

The differential displacement vector is given by:

\[ d\overset{\rightarrow}{s} = \frac{dx}{dt} dt \hat{i} + \frac{dy}{dt} dt \hat{j} = (-1 \hat{i} + 1 \hat{j}) dt \]

### Step 2: Substitute the force and displacement into the work integral

Now we can substitute the force and the displacement into the work integral. The force \(\overset{\rightarrow}{F}\) can be expressed in terms of \(t\):

\[ \overset{\rightarrow}{F} = -x^3 \hat{i} + y^2 \hat{j} = -(1-t)^3 \hat{i} + t^2 \hat{j} \]

Now we can compute the dot product \(\overset{\rightarrow}{F} \cdot d\overset{\rightarrow}{s}\):

\[ \overset{\rightarrow}{F} \cdot d\overset{\rightarrow}{s} = [-(1-t)^3 \hat{i} + t^2 \hat{j}] \cdot [(-1 \hat{i} + 1 \hat{j}) dt] \]

Calculating the dot product:

\[ = [-(1-t)^3 \cdot (-1) + t^2 \cdot 1] dt = (1-t)^3 + t^2 dt \]

### Step 3: Set up the integral for work done

Now we can set up the integral for work done:

\[ W = \int_{0}^{1} \left[(1-t)^3 + t^2\right] dt \]

### Step 4: Evaluate the integral

Now we will evaluate the integral:

\[ W = \int_{0}^{1} (1-t)^3 dt + \int_{0}^{1} t^2 dt \]

Calculating the first integral:

\[ \int (1-t)^3 dt = \frac{(1-t)^4}{4} \Big|_0^1 = \frac{(1-1)^4}{4} - \frac{(1-0)^4}{4} = 0 - \frac{1}{4} = -\frac{1}{4} \]

Calculating the second integral:

\[ \int t^2 dt = \frac{t^3}{3} \Big|_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \]

### Step 5: Combine the results

Now we combine the results from both integrals:

\[ W = -\frac{1}{4} + \frac{1}{3} \]

To combine these fractions, we find a common denominator (12):

\[ W = -\frac{3}{12} + \frac{4}{12} = \frac{1}{12} \]

### Final Answer

Thus, the work done by the force in moving the particle from A to B is:

\[ W = \frac{1}{12} \text{ Joules} \]