An object has a displacement from position vector `vecr_(1) = (2 hat i+3hatj)m` to `vecr_(2) = (4hat i + 6 hat j)m` under a force `vec F = (3x^(2) hati + 2y hatj)N`, then work done by the force is

Options

A. 24 J
B. 33 J
C. 83 J
D. 45 J

Explanation

To solve the problem of calculating the work done by the force as an object moves from position vector \(\vec{r}_1 = (2 \hat{i} + 3 \hat{j}) \, m\) to \(\vec{r}_2 = (4 \hat{i} + 6 \hat{j}) \, m\) under the force \(\vec{F} = (3x^2 \hat{i} + 2y \hat{j}) \, N\), we can follow these steps:

### Step 1: Identify the Displacement
The displacement vector \(\vec{d}\) can be calculated as:
\[ \vec{d} = \vec{r}_2 - \vec{r}_1 = (4 \hat{i} + 6 \hat{j}) - (2 \hat{i} + 3 \hat{j}) = (4 - 2) \hat{i} + (6 - 3) \hat{j} = 2 \hat{i} + 3 \hat{j} \, m \]

### Step 2: Set Up the Work Done Integral
The work done \(W\) by the force \(\vec{F}\) as the object moves from \(\vec{r}_1\) to \(\vec{r}_2\) can be expressed as:
\[ W = \int_{\vec{r}_1}^{\vec{r}_2} \vec{F} \cdot d\vec{r} \]
Where \(d\vec{r}\) is the differential displacement vector.

### Step 3: Parameterize the Path
We can parameterize the path of motion. Since the motion is linear, we can express \(x\) and \(y\) in terms of a parameter \(t\):
- Let \(x = 2 + 2t\) (where \(t\) varies from 0 to 1)
- Let \(y = 3 + 3t\) (where \(t\) varies from 0 to 1)

Then, the differential displacement \(d\vec{r}\) can be expressed as:
\[ d\vec{r} = \frac{dx}{dt} dt \hat{i} + \frac{dy}{dt} dt \hat{j} = (2 dt) \hat{i} + (3 dt) \hat{j} \]

### Step 4: Substitute into the Work Integral
Now substituting \(x\) and \(y\) into the force function:
\[ \vec{F} = (3(2 + 2t)^2 \hat{i} + 2(3 + 3t) \hat{j}) = (3(4 + 8t + 4t^2) \hat{i} + (6 + 6t) \hat{j}) = (12 + 24t + 12t^2) \hat{i} + (6 + 6t) \hat{j} \]

### Step 5: Calculate the Dot Product
Now calculate \(\vec{F} \cdot d\vec{r}\):
\[ \vec{F} \cdot d\vec{r} = (12 + 24t + 12t^2)(2 dt) + (6 + 6t)(3 dt) \]
\[ = (24 + 48t + 24t^2 + 18 + 18t) dt = (42 + 66t + 24t^2) dt \]

### Step 6: Integrate
Now we integrate from \(t = 0\) to \(t = 1\):
\[ W = \int_0^1 (42 + 66t + 24t^2) dt \]
Calculating the integral:
\[ = \left[ 42t + 33t^2 + 8t^3 \right]_0^1 = (42 + 33 + 8) - (0) = 83 \, J \]

### Final Answer
The work done by the force is \(W = 83 \, J\).

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