The work done by an applied variable force `F=x+x^(3)` from x = 0 m to x = 2m, where x is displacement, is

Options

A. 6 J
B. 8 J
C. 10 J
D. 12 J

Answer

To find the work done by the variable force \( F = x + x^3 \) from \( x = 0 \, \text{m} \) to \( x = 2 \, \text{m} \), we will use the concept of integration. The work done \( W \) by a force is given by the integral of the force over the displacement:

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

### Step-by-Step Solution:

1. **Identify the Force Function**:
The given force function is \( F = x + x^3 \).

2. **Set Up the Integral**:
We need to integrate the force from \( x = 0 \) to \( x = 2 \):
\[ W = \int_{0}^{2} (x + x^3) \, dx \]

3. **Split the Integral**:
We can split the integral into two parts:
\[ W = \int_{0}^{2} x \, dx + \int_{0}^{2} x^3 \, dx \]

4. **Calculate Each Integral**:
- For the first integral:
\[ \int x \, dx = \frac{x^2}{2} \]
Evaluating from 0 to 2:
\[ \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} - 0 = 2 \]

- For the second integral:
\[ \int x^3 \, dx = \frac{x^4}{4} \]
Evaluating from 0 to 2:
\[ \left[ \frac{x^4}{4} \right]_{0}^{2} = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} - 0 = 4 \]

5. **Combine the Results**:
Now, we add the results of both integrals:
\[ W = 2 + 4 = 6 \, \text{Joules} \]

### Final Answer:
The work done by the applied variable force from \( x = 0 \, \text{m} \) to \( x = 2 \, \text{m} \) is \( 6 \, \text{J} \).

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