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 follow these steps:

### Step 1: Understand the relationship between work and force
The work done \( W \) by a variable force can be calculated using the integral of the force over the displacement. Mathematically, this is expressed as:
\[ W = \int F \, dx \]

### Step 2: Substitute the given force into the equation
Given the force \( F = x + x^3 \), we can substitute this into the work equation:
\[ W = \int (x + x^3) \, dx \]

### Step 3: Set the limits of integration
The limits of integration are from \( x = 0 \) to \( x = 2 \):
\[ W = \int_{0}^{2} (x + x^3) \, dx \]

### Step 4: Break the integral into two parts
We can split the integral into two separate integrals:
\[ W = \int_{0}^{2} x \, dx + \int_{0}^{2} x^3 \, dx \]

### Step 5: Calculate the first integral
The first integral \( \int x \, dx \) can be computed as follows:
\[ \int x \, dx = \frac{x^2}{2} \Big|_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = \frac{4}{2} - 0 = 2 \]

### Step 6: Calculate the second integral
The second integral \( \int x^3 \, dx \) can be computed as:
\[ \int x^3 \, dx = \frac{x^4}{4} \Big|_{0}^{2} = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} - 0 = 4 \]

### Step 7: Combine the results
Now, we can combine the results of the two integrals to find the total work done:
\[ W = 2 + 4 = 6 \, \text{Joules} \]

### 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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