A force of `(4x^(2)+3x)N` acts on a particle which displaces it from x=2m to x=3m. The work done by the force is

Options

A. 32.8 J
B. 3.28 J
C. 0.328J
D. zero

Answer

To find the work done by the force \( F = (4x^2 + 3x) \, \text{N} \) when a particle is displaced from \( x = 2 \, \text{m} \) to \( x = 3 \, \text{m} \), we will use the formula for work done, which is given by the integral of force over displacement:

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

### Step-by-Step Solution:

1. **Identify the Force Function**:
The force acting on the particle is given as:
\[ F(x) = 4x^2 + 3x \]

2. **Set the Limits of Integration**:
The particle is displaced from \( x_1 = 2 \, \text{m} \) to \( x_2 = 3 \, \text{m} \).

3. **Write the Integral for Work Done**:
The work done by the force as the particle moves from \( x = 2 \) to \( x = 3 \) is:
\[ W = \int_{2}^{3} (4x^2 + 3x) \, dx \]

4. **Calculate the Integral**:
We will compute the integral:
\[ W = \int (4x^2 + 3x) \, dx \]
The integral can be solved term by term:
\[ \int 4x^2 \, dx = \frac{4x^3}{3} + C \]
\[ \int 3x \, dx = \frac{3x^2}{2} + C \]
Therefore, the integral becomes:
\[ W = \left[ \frac{4x^3}{3} + \frac{3x^2}{2} \right]_{2}^{3} \]

5. **Evaluate the Integral at the Limits**:
Now we will evaluate this expression from \( x = 2 \) to \( x = 3 \):
\[ W = \left( \frac{4(3)^3}{3} + \frac{3(3)^2}{2} \right) - \left( \frac{4(2)^3}{3} + \frac{3(2)^2}{2} \right) \]

Calculating the first part:
\[ = \frac{4 \cdot 27}{3} + \frac{3 \cdot 9}{2} = 36 + 13.5 = 49.5 \]

Now calculating the second part:
\[ = \frac{4 \cdot 8}{3} + \frac{3 \cdot 4}{2} = \frac{32}{3} + 6 = \frac{32}{3} + \frac{18}{3} = \frac{50}{3} \]

6. **Combine the Results**:
Now, we can combine the results:
\[ W = 49.5 - \frac{50}{3} \]
Converting \( 49.5 \) to a fraction:
\[ 49.5 = \frac{99}{2} \]
Finding a common denominator (which is 6):
\[ W = \frac{99}{2} - \frac{100}{6} \]
Converting \( \frac{99}{2} \) to sixths:
\[ \frac{99}{2} = \frac{297}{6} \]
Thus,
\[ W = \frac{297}{6} - \frac{100}{6} = \frac{197}{6} \approx 32.83 \, \text{J} \]

### Final Answer:
The work done by the force is approximately:
\[ W \approx 32.83 \, \text{J} \]