The work done by a force `vec(F)=(-6x^(3)hat(i))` N in displacing a particle from x = 4m to x = - 2m is

Options

A. `240 J`
B. 360 J
C. 420 J
D. will depend upon the path

Answer

To find the work done by the force \( \vec{F} = -6x^3 \hat{i} \) N in displacing a particle from \( x = 4 \, \text{m} \) to \( x = -2 \, \text{m} \), we will use the concept of work done by a variable force. The work done, \( W \), can be calculated using the integral of the force over the displacement.

### Step-by-Step Solution:

1. **Identify the Force Function**:
The force is given as:
\[ \vec{F} = -6x^3 \hat{i} \]

2. **Set Up the Work Integral**:
The work done by the force when moving from position \( x_1 \) to \( x_2 \) is given by:
\[ W = \int_{x_1}^{x_2} F_x \, dx \]
Here, \( F_x = -6x^3 \) and we need to integrate from \( x_1 = 4 \, \text{m} \) to \( x_2 = -2 \, \text{m} \).

3. **Write the Integral**:
\[ W = \int_{4}^{-2} -6x^3 \, dx \]

4. **Factor Out the Constant**:
Since \(-6\) is a constant, we can factor it out of the integral:
\[ W = -6 \int_{4}^{-2} x^3 \, dx \]

5. **Calculate the Integral**:
The integral of \( x^3 \) is:
\[ \int x^3 \, dx = \frac{x^4}{4} \]
Therefore, we have:
\[ W = -6 \left[ \frac{x^4}{4} \right]_{4}^{-2} \]

6. **Evaluate the Definite Integral**:
Now we evaluate the integral at the limits:
\[ W = -6 \left( \frac{(-2)^4}{4} - \frac{(4)^4}{4} \right) \]
Calculate \( (-2)^4 = 16 \) and \( (4)^4 = 256 \):
\[ W = -6 \left( \frac{16}{4} - \frac{256}{4} \right) \]
Simplifying:
\[ W = -6 \left( 4 - 64 \right) = -6 \left( -60 \right) = 360 \, \text{J} \]

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