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

Options

A. a)`360 J`
B. b)`85 J`
C. c)`185 J`
D. d)`135 J`

Explanation

To solve the problem, we need to calculate the work done by the force \( F = 7 - 2x + 3x^2 \) as the particle moves from \( x = 0 \) to \( x = 5 \) meters. The work done by a force is given by the integral of the force over the distance moved.

### Step-by-step Solution:

1. **Write the expression for work done**:
The work done \( W \) by the force \( F \) when moving from \( x = a \) to \( x = b \) is given by:
\[ W = \int_{a}^{b} F \, dx \]
In this case, \( a = 0 \) and \( b = 5 \), so:
\[ W = \int_{0}^{5} (7 - 2x + 3x^2) \, dx \]

2. **Integrate the force function**:
We will integrate the function term by term:
\[ W = \int_{0}^{5} 7 \, dx - \int_{0}^{5} 2x \, dx + \int_{0}^{5} 3x^2 \, dx \]

3. **Calculate each integral**:
- For the first term:
\[ \int 7 \, dx = 7x \Big|_{0}^{5} = 7(5) - 7(0) = 35 \]
- For the second term:
\[ \int 2x \, dx = x^2 \Big|_{0}^{5} = 5^2 - 0^2 = 25 \]
- For the third term:
\[ \int 3x^2 \, dx = x^3 \Big|_{0}^{5} = 5^3 - 0^3 = 125 \]

4. **Combine the results**:
Now substituting back into the expression for work:
\[ W = 35 - 25 + 125 \]

5. **Calculate the final work done**:
\[ W = 35 - 25 + 125 = 135 \, \text{Joules} \]

### Final Answer:
The work done by the applied force on the particle is \( 135 \, \text{Joules} \).