A particle moves along the X-axis from x=0 to x=5 m under the influence of a force given by `F=7-2x+3x^(2).` Find the work done in the process.

Options

A. 70
B. 135
C. 270
D. 35

Answer

To find the work done by the force \( F = 7 - 2x + 3x^2 \) as a particle moves from \( x = 0 \) m to \( x = 5 \) m, we will use the work integral for variable forces. The work done \( W \) is given by:

\[ W = \int_{x_1}^{x_2} F \, dx \]

where \( F \) is the force and \( x_1 \) and \( x_2 \) are the initial and final positions, respectively.

### Step 1: Set up the integral

Given \( F = 7 - 2x + 3x^2 \), we can write the work done as:

\[ W = \int_{0}^{5} (7 - 2x + 3x^2) \, dx \]

### Step 2: Break down the integral

We can separate the integral into three parts:

\[ W = \int_{0}^{5} 7 \, dx - \int_{0}^{5} 2x \, dx + \int_{0}^{5} 3x^2 \, dx \]

### Step 3: Calculate each integral

1. **First Integral**:

\[ \int_{0}^{5} 7 \, dx = 7x \Big|_{0}^{5} = 7(5) - 7(0) = 35 \]

2. **Second Integral**:

\[ \int_{0}^{5} 2x \, dx = 2 \cdot \frac{x^2}{2} \Big|_{0}^{5} = x^2 \Big|_{0}^{5} = 5^2 - 0^2 = 25 \]

3. **Third Integral**:

\[ \int_{0}^{5} 3x^2 \, dx = 3 \cdot \frac{x^3}{3} \Big|_{0}^{5} = x^3 \Big|_{0}^{5} = 5^3 - 0^3 = 125 \]

### Step 4: Combine the results

Now, substituting back into the work equation:

\[ W = 35 - 25 + 125 \]

Calculating this gives:

\[ W = 35 - 25 + 125 = 135 \, \text{Joules} \]

### Final Answer

The work done by the force as the particle moves from \( x = 0 \) m to \( x = 5 \) m is:

\[ \boxed{135 \, \text{J}} \]