A particle moves along x - axis under the action of a position dependent force `F=(5x^(2)-2x)N`. Work done by force on the particle when it moves from origin to x = 3 m is

Options

A. 45 J
B. 36 J
C. 32 J
D. 42 J

Answer

To find the work done by the force \( F = (5x^2 - 2x) \, \text{N} \) as the particle moves from the origin \( x = 0 \) to \( x = 3 \, \text{m} \), we will use the concept of work done by a variable force, which is given by the integral of the force with respect to displacement.

### Step-by-Step Solution:

1. **Identify the Work Done Formula**:
The work done \( W \) by a force \( F \) when moving from position \( x_1 \) to \( x_2 \) is given by:
\[ W = \int_{x_1}^{x_2} F \, dx \]

2. **Set the Limits of Integration**:
In this case, the particle moves from the origin \( x_1 = 0 \) to \( x_2 = 3 \, \text{m} \).

3. **Substitute the Force Function**:
The force is given as \( F = 5x^2 - 2x \). Thus, we can write:
\[ W = \int_{0}^{3} (5x^2 - 2x) \, dx \]

4. **Integrate the Force Function**:
We will integrate each term separately:
\[ W = \int_{0}^{3} 5x^2 \, dx - \int_{0}^{3} 2x \, dx \]
- The integral of \( 5x^2 \) is:
\[ \int 5x^2 \, dx = \frac{5}{3} x^3 \]
- The integral of \( 2x \) is:
\[ \int 2x \, dx = x^2 \]

5. **Evaluate the Integrals**:
Now we evaluate the definite integrals:
\[ W = \left[ \frac{5}{3} x^3 \right]_{0}^{3} - \left[ x^2 \right]_{0}^{3} \]

- For \( \frac{5}{3} x^3 \):
\[ = \frac{5}{3} (3^3) - \frac{5}{3} (0^3) = \frac{5}{3} \times 27 = 45 \]
- For \( x^2 \):
\[ = (3^2) - (0^2) = 9 - 0 = 9 \]

6. **Combine the Results**:
Now we can combine the results from the two integrals:
\[ W = 45 - 9 = 36 \, \text{J} \]

### Final Answer:
The work done by the force on the particle when it moves from the origin to \( x = 3 \, \text{m} \) is \( 36 \, \text{J} \).

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