The work done by a force `vecF =(-6x^(2)hati) N` in displacing a particle from x = 4mtox = -2mis

Options

A. `-240 J`
B. `360 J`
C. `420 J`
D. `144 J`

Answer

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

### Step 1: Understand the Work Done Formula
The work done \(W\) by a force \(\vec{F}\) when a particle moves through a displacement \(d\vec{x}\) is given by:
\[ dW = \vec{F} \cdot d\vec{x} \]
In this case, since the force is in the \(x\) direction, we can express it as:
\[ dW = F \, dx \]

### Step 2: Substitute the Force
Given the force \(\vec{F} = -6x^2 \hat{i}\), we can write:
\[ dW = -6x^2 \, dx \]

### Step 3: Set Up the Integral for Work Done
To find the total work done as the particle moves from \(x = 4 \, \text{m}\) to \(x = -2 \, \text{m}\), we need to integrate:
\[ W = \int_{4}^{-2} -6x^2 \, dx \]

### Step 4: Perform the Integration
We can factor out the constant:
\[ W = -6 \int_{4}^{-2} x^2 \, dx \]
Now, we calculate the integral:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Thus,
\[ W = -6 \left[ \frac{x^3}{3} \right]_{4}^{-2} \]

### Step 5: Evaluate the Integral at the Limits
Now we substitute the limits into the integrated function:
\[ W = -6 \left( \frac{(-2)^3}{3} - \frac{(4)^3}{3} \right) \]
Calculating the cubes:
\[ (-2)^3 = -8 \quad \text{and} \quad (4)^3 = 64 \]
So we have:
\[ W = -6 \left( \frac{-8}{3} - \frac{64}{3} \right) \]
This simplifies to:
\[ W = -6 \left( \frac{-8 - 64}{3} \right) = -6 \left( \frac{-72}{3} \right) \]

### Step 6: Simplify the Expression
Calculating the fraction:
\[ W = -6 \left( -24 \right) = 144 \, \text{J} \]

### Final Answer
Thus, the work done by the force in displacing the particle from \(x = 4 \, \text{m}\) to \(x = -2 \, \text{m}\) is:
\[ W = 144 \, \text{J} \]