A particle moves along the X-axis x=0 to x=5 m under the influence of a force given by `F=10-2x+3x^(2)`. Work done in the process is

Options

A. 70 units
B. 270 units
C. 35 units
D. 150 units

Explanation

To find the work done by the force \( F = 10 - 2x + 3x^2 \) as a particle moves from \( x = 0 \) to \( x = 5 \) meters, we will follow these steps:

### Step 1: Understand the Work Done Formula
The work done \( W \) by a force when moving an object from position \( x_1 \) to \( x_2 \) is given by the integral of the force over that distance:
\[ W = \int_{x_1}^{x_2} F \, dx \]

### Step 2: Set Up the Integral
Given the force \( F = 10 - 2x + 3x^2 \), we need to integrate this force from \( x = 0 \) to \( x = 5 \):
\[ W = \int_{0}^{5} (10 - 2x + 3x^2) \, dx \]

### Step 3: Perform the Integration
Now, we will integrate the function:
\[ W = \int (10) \, dx - \int (2x) \, dx + \int (3x^2) \, dx \]
Calculating each integral:
- The integral of \( 10 \) is \( 10x \).
- The integral of \( 2x \) is \( x^2 \).
- The integral of \( 3x^2 \) is \( x^3 \).

So, combining these results, we have:
\[ W = 10x - x^2 + x^3 \]

### Step 4: Evaluate the Integral from 0 to 5
Now we will evaluate this expression from \( x = 0 \) to \( x = 5 \):
\[ W = \left[ 10(5) - (5)^2 + (5)^3 \right] - \left[ 10(0) - (0)^2 + (0)^3 \right] \]
Calculating the first part:
\[ W = 10(5) - 25 + 125 \]
\[ W = 50 - 25 + 125 = 150 \]

### Step 5: Conclusion
Thus, the work done by the force as the particle moves from \( x = 0 \) to \( x = 5 \) meters is:
\[ W = 150 \text{ units} \]