The work done by a force `vecF = (-6xx x^(3)hati ) N` in displacing a particle from x = 4 m to x = - 2 m is

Options

A. 360 J
B. 240 J
C. `-240 J`
D. `-360 J`

Explanation

To find the work done by the force \(\vec{F} = -6x^3 \hat{i}\) in displacing a particle from \(x = 4 \, m\) to \(x = -2 \, m\), we can follow these steps:

### Step 1: Understand the Work Done Formula
The work done \(W\) by a variable force is given by the integral of the force over the displacement:
\[ W = \int_{x_1}^{x_2} \vec{F} \cdot d\vec{x} \]
Here, \(d\vec{x} = dx \hat{i}\) since the motion is along the x-axis.

### Step 2: Set Up the Integral
Given the force \(\vec{F} = -6x^3 \hat{i}\), we can set up the integral for work done as:
\[ W = \int_{4}^{-2} (-6x^3) \, dx \]

### Step 3: Change the Limits of Integration
Since we are integrating from \(x = 4\) to \(x = -2\), we can rewrite the integral:
\[ W = -6 \int_{4}^{-2} x^3 \, dx \]
To make the integration easier, we can switch the limits and change the sign:
\[ W = 6 \int_{-2}^{4} x^3 \, dx \]

### Step 4: Calculate the Integral
Now we compute the integral:
\[ \int x^3 \, dx = \frac{x^4}{4} \]
Thus,
\[ W = 6 \left[ \frac{x^4}{4} \right]_{-2}^{4} \]

### Step 5: Evaluate the Integral at the Limits
Now we evaluate the integral at the limits:
\[ W = 6 \left( \frac{4^4}{4} - \frac{(-2)^4}{4} \right) \]
Calculating \(4^4\) and \((-2)^4\):
\[ 4^4 = 256 \quad \text{and} \quad (-2)^4 = 16 \]
Substituting these values back:
\[ W = 6 \left( \frac{256}{4} - \frac{16}{4} \right) = 6 \left( 64 - 4 \right) = 6 \times 60 \]

### Step 6: Final Calculation
Now calculate:
\[ W = 360 \, \text{Joules} \]

### Conclusion
The work done by the force in displacing the particle from \(x = 4 \, m\) to \(x = -2 \, m\) is:
\[ \boxed{360 \, \text{J}} \]