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

Options

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

Answer

To find the work done by the force \( F = (5x^2 - 2x) \, \text{N} \) on the particle as it moves from the origin (x = 0) to x = 3 m, we will follow these steps:

### Step 1: Understand the Work Done Formula
The work done \( W \) by a force when moving an object along a path can be calculated using the integral of the force with respect to displacement. For a force that is position-dependent, the work done is given by:
\[ W = \int_{x_1}^{x_2} F(x) \, dx \]
where \( F(x) \) is the force as a function of position \( x \), and \( x_1 \) and \( x_2 \) are the initial and final positions, respectively.

### Step 2: Set Up the Integral
In this case, the force is given by:
\[ F(x) = 5x^2 - 2x \]
We need to evaluate the work done from \( x = 0 \) to \( x = 3 \):
\[ W = \int_{0}^{3} (5x^2 - 2x) \, dx \]

### Step 3: Calculate the Integral
Now, we will compute the integral:
\[ W = \int_{0}^{3} (5x^2 - 2x) \, dx = \int_{0}^{3} 5x^2 \, dx - \int_{0}^{3} 2x \, dx \]

Calculating each integral separately:
1. For \( \int 5x^2 \, dx \):
\[ \int 5x^2 \, dx = 5 \cdot \frac{x^3}{3} = \frac{5x^3}{3} \]

2. For \( \int 2x \, dx \):
\[ \int 2x \, dx = 2 \cdot \frac{x^2}{2} = x^2 \]

Putting it all together:
\[ W = \left[ \frac{5x^3}{3} - x^2 \right]_{0}^{3} \]

### Step 4: Evaluate the Limits
Now we will evaluate the expression from 0 to 3:
\[ W = \left( \frac{5(3)^3}{3} - (3)^2 \right) - \left( \frac{5(0)^3}{3} - (0)^2 \right) \]
Calculating the upper limit:
\[ W = \left( \frac{5 \cdot 27}{3} - 9 \right) - 0 \]
\[ W = \left( 45 - 9 \right) = 36 \, \text{J} \]

### Final Answer
Thus, the work done by the forces on the particle when it moves from the origin to \( x = 3 \, \text{m} \) is:
\[ \boxed{36 \, \text{J}} \]